# Martingale inequalities

Pavel Zorin-Kranich

(University of Bonn, Winter term 2021/22)

These lecture notes cover a few techniques for proving $L^{p}$ estimates for
martingales and the most basic applications to Itô integration and rough
paths. They have been created for a course that consisted of 13 lectures á 90
minutes. Due to leaving academia, I didn’t have time to polish these notes
much, but you are welcome to, because:

This work is licensed under the Creative Commons Attribution 4.0 International
License. To view a copy of this license, visit

https://creativecommons.org/licenses/by/4.0/

or send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042,
USA.

##### Literature

There are two books whose titles include our topic [Gar73, Lon93]. They can be
useful for some of the basics, but are overall outdated.

For vector-valued martingales, modern references are [Hyt+16, Pis16].

The only book about sharp constants in martingale inequalties is [Osę12].

An extensive treatment of rough paths can be found in [FH20].

One stochastic analysis text that pays attention to inequalities is [Kal21].

###### Contents

  1. 0 Review of martingale basics
  2. 1 Maximal and square functions
    1. 1.1 Predictable square function
  3. 2 Lépingle inequality
    1. 2.1 Martingale convergence
  4. 3 Vector-valued inequalities
    1. 3.1 Weighted and vector-valued Doob inequalities
  5. 4 Rough paths
    1. 4.1 Young integration with Hölder functions
    2. 4.2 Young integration with $V^{r}$ functions
    3. 4.3 Rough paths
      1. 4.3.1 Rough differential equations
      2. 4.3.2 Operations on controlled paths
      3. 4.3.3 Contractive iteration
      4. 4.3.4 Lipschitz dependence on data
  6. 5 Martingale transforms
    1. 5.1 Davis decomposition
    2. 5.2 Vector-valued maximal paraproduct estimate
    3. 5.3 Stopping time construction
  7. 6 Itô integration
    1. 6.1 Quadratic covariation
  8. 7 A sharp inequality for the square function
    1. 7.1 A Bellman function
    2. 7.2 Sharp constant in the Davis inequality for the square function

##  0 Review of martingale basics

Random variables will be usually defined on a filtered probability space
$(\Omega,(\mathcal{F}_{n})_{n\in\mathbb{N}},\mu)$. We denote by
$\mathcal{F}_{\infty}$ the $\sigma$-algebra generated by
$\cup_{n\in\mathbb{N}}\mathcal{F}_{n}$. The following examples are useful to
keep in mind.

######  Example 0.1 (Dyadic filtration).

$\Omega=[0,1]$, $\mu$ Lebesgue measure, $\mathcal{F}_{n}$ is the
$\sigma$-algebra generated by the dyadic intervals of length $2^{-n}$, that
is, intervals of the form $[2^{-n}k,2^{-n}(k+1)]$ with $k\in\mathbb{Z}$.

A higher dimensional version involves dyadic cubes. One can also construct
similar filtrations on more general manifolds, or even metric measure spaces.

######  Example 0.2 (Atomic filtrations).

$\Omega=[0,1]$, $\mu$ Lebesgue measure, $\mathcal{F}_{n}$ is a
$\sigma$-algebra generated by finitely many intervals.

We recall that an _atom_ for a measure $\mu$ on a $\sigma$-algebra
$\mathcal{F}$ is a set $A\in\mathcal{F}$ such that $\mu(A)>0$ and, for every
$A^{\prime}\in\mathcal{F}$ with $A^{\prime}\subseteq A$, we have
$\mu(A^{\prime})\in\\{0,\mu(A)\\}$. In an atomic $\sigma$-algebra, every
measurable set is a finite union of atoms.

One can view martingale analysis as analysis of atomic filtrations, if we do
not allow any constants to depend on the filtration. All results that we are
interested in can be transferred from atomic filtrations to general
filtrations. However, it is technically convenient to always use general
filtrations, since they can appear in applications.

######  Example 0.3.

$\Omega=[0,1]^{\mathbb{N}}$, $\mu$ the product of Lebesgue measures,

$\mathcal{F}_{n}=\\{B\times[0,1]^{\\{n,n+1,\dotsc\\}}\nonscript\>|\allowbreak\nonscript\>\mathopen{}B\subseteq[0,1]^{\\{0,\dotsc,n-1\\}}\text{
Borel}\\}.$ (0.1)

This filtration appears in the analysis of independent random variables, and
gives a good idea of how a general filtration looks like.

An _adapted process_ is a sequence of functions $(f_{n})$ such that, for every
$n\in\mathbb{N}$, the function $f_{n}$ is $\mathcal{F}_{n}$-measurable.

An adapted process $f$ is called _predictable_ if, for every $n>0$, the
function $f_{n}$ is $\mathcal{F}_{n-1}$-measurable.

For nested $\sigma$-algebras $\mathcal{F}^{\prime}\subseteq\mathcal{F}$ on
$\Omega$, the _conditional expectation_ is the orthogonal projection

$\mathbb{E}(\cdot|\mathcal{F}^{\prime}):L^{2}(\Omega,\mathcal{F},\mu)\to
L^{2}(\Omega,\mathcal{F}^{\prime},\mu).$

The conditional expectation has the following properties. Here and later, all
identities and inequalities are meant to hold almost surely, unless mentioned
otherwise.

  1. 1.

$\mathbb{E}(\mathbf{1}|\mathcal{F}^{\prime})=\mathbf{1}$.

  2. 2.

For every $p\in[1,\infty]$, $\mathbb{E}$ extends to a contraction
$L^{p}(\mathcal{F})\to L^{p}(\mathcal{F}^{\prime})$

  3. 3.

$\int\mathbb{E}(f|\mathcal{F}^{\prime})=\int f$ for every $f\in
L^{1}(\mathcal{F})$.

  4. 4.

Positivity: $f\geq 0\implies\mathbb{E}(f|\mathcal{F}^{\prime})\geq 0$.

  5. 5.

Assume that $f\in L^{1}(\mathcal{F})$, $g\in L^{0}(\mathcal{F}^{\prime})$, and
either $fg\in L^{1}(\mathcal{F})$ or $f\geq
0,\mathbb{E}(f|\mathcal{F}^{\prime})g\in L^{1}(\mathcal{F}^{\prime})$. Then

$\mathbb{E}(fg|\mathcal{F}^{\prime})=\mathbb{E}(f|\mathcal{F}^{\prime})g\text{
in }L^{1}(\mathcal{F}^{\prime}).$

Suppose that $(X,\mathcal{F},\mu)$ is a regular measure space and
$\mathcal{F}^{\prime}\subseteq\mathcal{F}$ is a sub-$\sigma$-algebra. Then
there exists an essentially unique measurable map
$(X,\mathcal{F}^{\prime})\to\mathcal{M}(X)$, $y\mapsto\mu_{y}$ such that, for
every $f\in L^{1}(\mathcal{F})$, for $\mu$-a.e. $y\in X$, we have $f\in
L^{1}(X,\mu_{y})$, and $\int
f\mathop{}\\!\mathrm{d}\mu_{y}=\mathbb{E}(f|\mathcal{F}^{\prime})(y)$. This
map is called a _measure disintegration_. A measure disintegration satisfies
$\mu_{x}=\mu_{y}$ for $\mu$-a.e. $y$ and $\mu_{y}$-a.e. $x$.

######  Example 0.4.

If $\mathcal{F}^{\prime}$ is an atomic filtration, then we can choose a
collection of disjoint atoms $\mathcal{A}\subseteq\mathcal{F}^{\prime}$ with
$\mu(\cup_{A\in\mathcal{A}}A)=1$. Then, for each $x\in A\in\mathcal{A}$, we
can set
$\mathop{}\\!\mathrm{d}\mu_{x}:=\mu(A)^{-1}\mathbf{1}_{A}\mathop{}\\!\mathrm{d}\mu$,
and

$\mathbb{E}(f|\mathcal{F}^{\prime})(x)=\mu(A)^{-1}\int_{A}f(x^{\prime})\mathop{}\\!\mathrm{d}\mu(x^{\prime}).$

######  Example 0.5.

Let $\Omega=[0,1]^{2}$ with the Lebesgue measure, $\mathcal{F}$ the Borel
$\sigma$-algebra, and $\mathcal{F}^{\prime}$ the Borel $\sigma$-algebra in the
first variable, like in (7.5). Then, we can choose $\mu_{(x,y)}$ to be the
Lebesgue measure on $\\{x\\}\times[0,1]$, and

$\mathbb{E}(f|\mathcal{F}^{\prime})(x,y)=\int_{0}^{1}f(x,y^{\prime})\mathop{}\\!\mathrm{d}y^{\prime}.$

A _martingale_ is an adapted process with values in $\mathbb{C}$ (later also
in a Banach space) such that, for every $m\leq n$, the function $f_{n}$ is
integrable, and we have

$f_{m}=\mathbb{E}(f_{n}|\mathcal{F}_{m}).$

######  Example 0.6.

Sums of independent random variables

######  Example 0.7 (Dyadic martingale).

Let $f$ be an integrable function on $[0,1]$ with the Lebesgue measure and
$f_{n}=\mathbb{E}(f|\mathcal{F}_{n})$ (the same definition works for any
filtration).

This construction can often be used to transfer results from martingales to a
real analysis setting. The main difference between the analysis of dyadic
martingales and general martingales is that the Calderón–Zygmund decomposition
does not work for general martingales. A substitute that does work for general
martingales is the Gundy decomposition.

The main reason why this construction does not produce all possible
martingales is that it may happen that $\lim_{n\to\infty}\lVert
f_{n}\rVert=\infty$, in which case there might be no function $f$ with
$f_{n}=\mathbb{E}(f|\mathcal{F}_{n})$. A well-known example involves a
doubling betting strategy.

######  Example 0.8.

Any integrable process can be written as the sum of a process with predictable
jumps and a martingale.

A _stopping time_ is a function
$\tau:\Omega\to\bar{\mathbb{N}}=\mathbb{N}\cup\\{\infty\\}$ such that, for
every $n\in\mathbb{N}$, we have $\\{\tau\leq n\\}\in\mathcal{F}_{n}$.

######  Example 0.9.

Any constant function is a stopping time.

######  Example 0.10 (Hitting time).

If $f$ is an adapted process with values in a metric space $X$ and $B\subseteq
X$ is a Borel set, then

$\tau:=\inf\\{t\nonscript\>|\allowbreak\nonscript\>\mathopen{}f_{t}\in B\\}$

is a stopping time, called the _first hitting time_ of $B$.

If $\sigma,\tau$ are stopping times, then $\sigma\wedge\tau$ and
$\sigma\vee\tau$ are also stopping times.

For a stopping time $\tau$ and a process $f$, the _stopped process_ $f^{\tau}$
is defined by

$f^{\tau}_{t}:=f_{\tau\wedge t}.$

If $f$ is a martingale, then the stopped process $f^{\tau}$ is again a
martingale.

If $\tau$ is a stopping time, the corresponding $\sigma$-algebra is defined by

$\mathcal{F}_{\tau}:=\\{A\in\mathcal{F}_{\infty}\nonscript\>|\allowbreak\nonscript\>\mathopen{}(\forall
n)A\cap\\{\tau\leq n\\}\in\mathcal{F}_{n}\\}.$

The function $\tau$ is $\mathcal{F}_{\tau}$-measurable. We abbreviate
$\mathbb{E}_{\tau}f:=\mathbb{E}(f|\mathcal{F}_{\tau})$.

The _optional sampling_ theorem says that, for every discrete time martingale
$f$, bounded stopping time $\tau$, and another stopping time $\sigma$, we have

$f_{\sigma\wedge\tau}=\mathbb{E}_{\sigma}f_{\tau}.$

##  1 Maximal and square functions

You are probably already familiar with the Lebesgue differentialtion theorem.
The best proof of that theorem uses the maximal operator to absorb error
terms. This idea is used in many places in analysis, and the study of
martingales is one of these places.

######  Definition 1.1 (Maximal operator).

For an adapted process $f$ with values in a normed space, we write

$Mf_{n}=(Mf)_{n}:=\sup_{k\leq n}\lvert f_{k}\rvert.$

######  Remark.

Since our martingales are indexed by a countable set, we can use a pointwise
supremum here. In continuous time, we would need a lattice supremum.

A _submartingale_ is an adapted process with values in $\mathbb{R}_{\geq 0}$
such that, for every $m\leq n$, we have

$f_{m}\leq\mathbb{E}_{m}f_{n}.$

######  Example 1.2.

If $(f_{n})$ is a martingale, then $(\lvert f_{n}\rvert)$ is a submartingale.

######  Example 1.3.

If $f$ is any adapted process, then $Mf$ is a submartingale.

If $f$ is a submartingale, then, for any $p\in[1,\infty]$ and $m\leq n$, we
have

$\lVert f_{m}\rVert_{p}\leq\lVert f_{n}\rVert_{p}.$

We define the $L^{p}$ norm of a (sub-)martingale by

$\lVert f\rVert_{p}:=\sup_{n}\lVert f_{n}\rVert_{p}.$

With this definition, the maximal operator is clearly bounded on $L^{\infty}$.
The next result looks a lot like an $L^{1}\to L^{1,\infty}$ bound for $M$, but
is in fact stronger and more convenient to use.

######  Lemma 1.4.

Let $f$ be a submartingale with values in $\mathbb{R}_{\geq 0}$. Then, for
every $\lambda>0$ and $n\in\mathbb{N}$, we have

$\lambda\lvert\\{Mf_{n}>\lambda\\}\rvert\leq\int_{\\{Mf_{n}>\lambda\\}}f_{n}\mathop{}\\!\mathrm{d}\mu.$

###### Proof.

For a fixed $\lambda$, define the stopping time

$\tau:=\inf\\{k\nonscript\>|\allowbreak\nonscript\>\mathopen{}f_{k}>\lambda\\}.$

Then, $\\{Mf_{n}>\lambda\\}=\\{\tau\leq n\\}$. Hence,

$\displaystyle\lambda\lvert\\{Mf_{n}>\lambda\\}\rvert$
$\displaystyle=\lambda\sum_{k\leq n}\lvert\\{\tau=k\\}\rvert$
$\displaystyle\leq\sum_{k\leq
n}\int_{\\{\tau=k\\}}f_{k}\mathop{}\\!\mathrm{d}\mu$ (submartingale property)
$\displaystyle\leq\sum_{k\leq
n}\int_{\\{\tau=k\\}}f_{n}\mathop{}\\!\mathrm{d}\mu$
$\displaystyle=\int_{\\{Mf_{n}>\lambda\\}}f_{n}\mathop{}\\!\mathrm{d}\mu.\qed$

######  Remark ($L^{1,\infty}$ norm).

By Chebychev’s inequality, for any measurable function $g$, we have

$\sup_{\lambda>0}\lambda^{-1}\lvert\\{g>\lambda\\}\rvert\leq\lVert
g\rVert_{1}.$

The converse inequality is false with any constant, as shown by the example
$g(x)=1/x$ on the measure space $\mathbb{R}_{>0}$ with the Lebesgue measure.
The left-hand side of the above inequality is the so-called $L^{1,\infty}$
seminorm of $g$. As suggested by the notation, $L^{1,\infty}$ is part of a
larger family of Lorentz spaces, which is in turn contained in Orlicz spaces,
but we will not talk about such generalities.

######  Theorem 1.5 (Doob’s maximal inequality).

Let $f$ be a submartingale with values in $\mathbb{R}_{\geq 0}$. Then, for
every $p\in(1,\infty]$, we have

$\lVert Mf\rVert_{p}\leq p^{\prime}\lVert f\rVert_{p}.$

Here and later, $p^{\prime}$ denotes the Hölder conjugate:
$1/p+1/p^{\prime}=1$.

###### Proof.

By the layer cake formula and Lemma 1.4, we write

$\displaystyle\lVert Mf_{n}\rVert_{p}^{p}$
$\displaystyle=\int_{0}^{\infty}p\lambda^{p-1}\mu(Mf_{n}>\lambda)\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\leq\int_{0}^{\infty}p\lambda^{p-2}\int_{\\{Mf_{n}>\lambda\\}}f_{n}\mathop{}\\!\mathrm{d}\mu\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=\int_{\Omega}\int_{0}^{Mf_{n}}p\lambda^{p-2}f_{n}\mathop{}\\!\mathrm{d}\lambda\mathop{}\\!\mathrm{d}\mu$
$\displaystyle=\frac{p}{p-1}\int_{\Omega}(Mf_{n})^{p-1}f_{n}\mathop{}\\!\mathrm{d}\mu$
$\displaystyle\leq\frac{p}{p-1}(\int_{\Omega}f_{n}^{p}\mathop{}\\!\mathrm{d}\mu)^{1/p}(\int_{\Omega}(Mf_{n})^{p}\mathop{}\\!\mathrm{d}\mu)^{1-1/p}.$

Since $f_{n}\in L^{p}$ implies $f_{k}\in L^{p}$ for all $k\leq n$ by $L^{p}$
contractivity of conditional expectations, we have $Mf_{n}\in L^{p}$. Hence,
we can cancel suitable powers of $\lVert Mf_{n}\rVert_{p}$ on both sides and
obtain the claim. ∎

There are other ways to deduce Theorem 1.5 from Lemma 1.4, such as real
interpolation with the $L^{\infty}$ estimate. But the above proof does not
rely on the $L^{\infty}$ estimate, only on the $L^{1}$-like estimate in Lemma
1.4.

It is often the case in martingale theory that inequalities near $L^{1}$ are
the most powerful ones. One can justify this by the observation that $L^{1}$
is the minimal assumption required to even define a martingale.

If $f$ is a martingale, then the increments $df_{k}$ are orthogonal in the
Hilbert space $L^{2}(\Omega)$. Indeed, more generally, for any $n$ and $g\in
L^{0}(\mathcal{F}_{n-1})$, we have

$\mathbb{E}(gdf_{n})=\mathbb{E}(\mathbb{E}(gdf_{n}|\mathcal{F}_{n-1}))=\mathbb{E}(g\mathbb{E}(df_{n}|\mathcal{F}_{n-1}))=\mathbb{E}(g\cdot
0)=0.$ (1.1)

######  Definition 1.6 (Square function).

For a martingale $f$, we write

$Sf_{n}:=\bigl{(}\sum_{k\leq n}\lvert df_{k}\rvert^{2}\bigr{)}^{1/2}.$

In continuous time, the analog of $Sf$ is denoted by $[f]$ and called the
_quadratic variation_. We will construct it in (6.12).

Here and later, for notational simplicity, we will consider martingales with
$f_{0}=0$. Since martingale increments $df_{k}$ are orthogonal, we have

$\lVert f_{n}\rVert_{2}=\lVert Sf_{n}\rVert_{2}.$

Next, we will see that $S$ is bounded from $L^{1}$ to $L^{1,\infty}$. The
proof that we present for this fact uses a summation by parts identity of a
kind that also appears e.g. in the Itô formula.

######  Lemma 1.7.

Let $f$ be a martingale, $\lambda>0$, and
$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lvert
f_{n}\rvert>\lambda\\}$. Then,

$\sum_{k}\mathbb{E}(\lvert df_{k}\rvert^{2}\mathbf{1}_{\tau>k})\leq
2\lambda\lVert f\rVert_{1}.$

On the left-hand side of the above estimate, we are taking the expectation of
$S_{\tau-1}$. Note that $\tau-1$ is not a stopping time. Such predictability
emulation will also be important in the BDG inequality below.

###### Proof.

By the monotone convergence theorem, it suffices to consider the case
$f_{N}=f_{N+1}=\dotsb$ for an arbitrarily large $N$. Then, we can replace
$\tau$ by $\tau\wedge N+1$ without changing the inequality. This is necessary
to apply the optional sampling theorem.

We use the identity

$\sum_{k<\tau}\lvert df_{k}\rvert^{2}+\lvert
f_{\tau-1}\rvert^{2}=2f_{\tau}f_{\tau-1}-2\sum_{k\leq\tau}f_{k-1}df_{k}.$

Note that

$\\{k\leq\tau\\}=\Omega\setminus\\{\tau<k\\}=\Omega\setminus\\{\tau\leq
k-1\\}\in\mathcal{F}_{k-1}.$

Therefore, by (1.1), we have

$\mathbb{E}(\mathbf{1}_{k\leq\tau}f_{k-1}df_{k})=0.$

Hence,

$\mathbb{E}\sum_{k<\tau}\lvert df_{k}\rvert^{2}+\mathbb{E}\lvert
f_{\tau-1}\rvert^{2}=2\mathbb{E}(f_{\tau}f_{\tau-1}).$

Therefore,

$\sum_{k}\mathbb{E}(\lvert df_{k}\rvert^{2}\mathbf{1}_{\tau>k})\leq
2\mathbb{E}\lvert f_{\tau}f_{\tau-1}\rvert\leq 2\mathbb{E}\lvert
f_{\tau}\lambda\rvert\leq 2\lambda\lVert f\rVert_{1},$

where the last step follows from the optional sampling theorem, which gives in
particular $f_{\tau}=\mathbb{E}_{\tau}f_{N+1}$. ∎

######  Corollary 1.8.

Let $f$ be a martingale and $\lambda>0$. Then,

$\lvert\\{Sf>\lambda\\}\rvert\leq 3\lambda^{-1}\lVert f\rVert_{1}.$

###### Proof.

Let $\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lvert
f_{n}\rvert>\lambda\\}$. Then

$\lvert\\{Sf>\lambda\\}\rvert\leq\lvert\\{Sf_{\tau-1}>\lambda\\}\rvert+\lvert\\{\tau<\infty\\}\rvert\leq\lambda^{-2}\lVert
Sf_{\tau-1}\rVert_{2}^{2}+\lambda^{-1}\lVert f\rVert_{1}.$

By Lemma 1.7, the first summand is $\leq 2\lambda^{-1}\lVert f\rVert_{1}$. ∎

######  Lemma 1.9 (Davis decomposition).

For every martingale $(f_{n})$, there exists a decomposition
$f=f^{\mathrm{pred}}+f^{\mathrm{bv}}$ as a sum of two martingales such that
$f^{\mathrm{pred}}$ has a predictable bound on jumps:

$\lvert df^{\mathrm{pred}}_{n}\rvert\leq 2Mdf_{n-1},$ (1.2)

and $f^{\mathrm{bv}}$ has bounded variation:

$\mathbb{E}\sum_{n}\lvert df^{\mathrm{bv}}_{n}\rvert\leq 2\mathbb{E}Mdf.$
(1.3)

###### Proof.

Let

$\displaystyle dg_{n}$ $\displaystyle:=\min(1,\frac{Mdf_{n-1}}{\lvert
df_{n}\rvert})df_{n},$ $\displaystyle df^{\mathrm{pred}}_{n}$
$\displaystyle:=dg_{n}-\mathbb{E}_{n-1}(dg_{n}),$ $\displaystyle dh_{n}$
$\displaystyle:=df_{n}-dg_{n}=\max(0,1-\frac{Mdf_{n-1}}{\lvert
df_{n}\rvert})df_{n},$ $\displaystyle df^{\mathrm{bv}}_{n}$
$\displaystyle:=dh_{n}-\mathbb{E}_{n-1}(dh_{n}).$

Then, by definition,

$\lvert dg_{n}\rvert\leq Mdf_{n-1},$

and, by positivity of conditional expectation, also

$\lvert\mathbb{E}_{n-1}(dg_{n})\rvert\leq\mathbb{E}_{n-1}(\lvert
dg_{n}\rvert)\leq\mathbb{E}_{n-1}(Mdf_{n-1})=Mdf_{n-1}.$

This implies (1.2).

On the other hand, we have the telescoping bound

$\lvert dh_{n}\rvert=\max(0,\lvert df_{n}\rvert-Mdf_{n-1})=Mdf_{n}-Mdf_{n-1},$

which implies (1.3). ∎

######  Theorem 1.10 (Davis inequalities).

Let $f$ be a martingale with $f_{0}=0$. Then,

$\mathbb{E}Sf\sim\mathbb{E}Mf.$

Here and later, $A\lesssim B$ means that $A\leq CB$ with an absolute constant
$C$, and $A\sim B$ means that $A\lesssim B$ and $B\lesssim A$.

###### Proof.

It suffices to consider martingales with $f_{N}=f_{N+1}=\dotsb=:f_{\infty}$
for some $N$, as long as we show bounds independent of $N$.

Let $f=f^{\mathrm{pred}}+f^{\mathrm{bv}}$ be a Davis decomposition as in Lemma
1.9. Then, for every $n>0$, we have the predictable bound

$\lvert f^{\mathrm{pred}}_{n}\rvert\leq\lvert
f^{\mathrm{pred}}_{n-1}\rvert+\lvert df^{\mathrm{pred}}_{n}\rvert\leq\lvert
f^{\mathrm{pred}}_{n-1}\rvert+2\lvert Mdf_{n-1}\rvert=:\rho_{n-1}.$

Let $\lambda>0$ and
$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}\rho_{n}>\lambda\\}$.
Then,

$\displaystyle\lvert\\{Sf^{\mathrm{pred}}>\lambda\\}\rvert$
$\displaystyle\leq\lvert\\{Sf^{\mathrm{pred}}>\lambda,\tau=\infty\\}\rvert+\lvert\\{\tau<\infty\\}\rvert$
$\displaystyle\leq\lvert\\{Sf^{\mathrm{pred},(\tau)}>\lambda\\}\rvert+\lvert\\{\tau<\infty\\}\rvert$
$\displaystyle\leq\lambda^{-2}\lVert
Sf^{\mathrm{pred},(\tau)}\rVert_{2}^{2}+\lvert\\{\tau<\infty\\}\rvert$
$\displaystyle=\lambda^{-2}\lVert
f^{\mathrm{pred}}_{\tau}\rVert^{2}+\lvert\\{\tau<\infty\\}\rvert.$

On the set $\\{\tau=\infty\\}$, we use the bound $Mf^{\mathrm{pred}}\leq
M\rho\leq\lambda$, while on the set $\\{\tau<\infty\\}$, we use the bound
$\lvert f^{\mathrm{pred}}_{\tau}\rvert\leq\rho_{\tau-1}\leq\lambda$ (at this
point, predictability is essential). This gives

$\displaystyle\lvert\\{Sf^{\mathrm{pred}}>\lambda\\}\rvert$
$\displaystyle\leq\lambda^{-2}\int_{\\{\tau=\infty\\}}\lvert
f^{\mathrm{pred}}_{\tau}\rvert^{2}+\lambda^{-2}\int_{\\{\tau<\infty\\}}\lvert
f^{\mathrm{pred}}_{\tau}\rvert^{2}+\lvert\\{\tau<\infty\\}\rvert$
$\displaystyle\leq\lambda^{-2}\int_{\\{Mf^{\mathrm{pred}}\leq\lambda\\}}\lvert
Mf^{\mathrm{pred}}\rvert^{2}+\lvert\\{\tau<\infty\\}\rvert+\lvert\\{\tau<\infty\\}\rvert.$

Note that $\\{\tau<\infty\\}=\\{M\rho>\lambda\\}$. Inserting this in the above
inequality and integrating in $\lambda$, we obtain

$\displaystyle\mathbb{E}Sf^{\mathrm{pred}}$
$\displaystyle=\int_{0}^{\infty}\lvert\\{Sf^{\mathrm{pred}}>\lambda\\}\rvert\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\leq\int_{0}^{\infty}\lambda^{-2}\int_{\\{Mf^{\mathrm{pred}}\leq\lambda\\}}\lvert
Mf^{\mathrm{pred}}\rvert^{2}\mathop{}\\!\mathrm{d}\lambda+2\int_{0}^{\infty}\lvert\\{M\rho>\lambda\\}\rvert\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=\mathbb{E}\lvert
Mf^{\mathrm{pred}}\rvert^{2}\int_{Mf^{\mathrm{pred}}}^{\infty}\lambda^{-2}\mathop{}\\!\mathrm{d}\lambda+2\mathbb{E}M\rho$
$\displaystyle=\mathbb{E}Mf^{\mathrm{pred}}+2\mathbb{E}M\rho$
$\displaystyle\leq 3\mathbb{E}Mf^{\mathrm{pred}}+4\mathbb{E}Mdf$
$\displaystyle\leq 3\mathbb{E}Mf+3\mathbb{E}\sum_{n}\lvert
df^{\mathrm{bv}}_{n}\rvert+4\mathbb{E}Mdf$ $\displaystyle\leq
3\mathbb{E}Mf+10\mathbb{E}Mdf.$

This, together with the simple bound

$\mathbb{E}Sf^{\mathrm{bv}}\leq\mathbb{E}\sum_{n}\lvert
df^{\mathrm{bv}}_{n}\rvert\leq 2\mathbb{E}Mdf,$

implies $\mathbb{E}Sf\lesssim\mathbb{E}Mf$.

The proof of the converse inequality is similar and uses

$Sf^{\mathrm{pred}}_{n}=\bigl{(}(Sf^{\mathrm{pred}}_{n-1})^{2}+(df^{\mathrm{pred}}_{n})^{2})^{1/2}\leq\lvert
Sf^{\mathrm{pred}}_{n-1}\rvert+\lvert df^{\mathrm{pred}}_{n}\rvert\leq\lvert
Sf^{\mathrm{pred}}_{n-1}\rvert+2\lvert Mdf_{n-1}\rvert=:\rho_{n-1}.$

∎

######  Lemma 1.11 (Garsia–Neveu).

Let $W,Z$ be positive random variables such that, for every $\lambda>0$, we
have

$\mathbb{E}(\mathbf{1}_{W>\lambda}(W-\lambda))\leq\mathbb{E}(\mathbf{1}_{W>\lambda}Z).$
(1.4)

Then, for every $p\geq 1$, we have

$\lVert W\rVert_{p}\leq p\lVert Z\rVert_{p}.$

###### Proof.

For $p=1$, it suffices to take $\lambda\to 0$, so assume $p>1$. Suppose first
that $W$ is bounded by $\Lambda$ and (1.4) holds for all
$\lambda\in(0,\Lambda)$. We will use the formula

$t^{p}=p(1-p)\int_{0}^{t}(t-\lambda)\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda.$

It yields

$\displaystyle\mathbb{E}W^{p}$
$\displaystyle=p(p-1)\mathbb{E}\int_{0}^{W}(W-\lambda)\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=p(p-1)\int_{0}^{\Lambda}\mathbb{E}((W-\lambda)\mathbf{1}_{W>\lambda})\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\leq
p(p-1)\int_{0}^{\Lambda}\mathbb{E}(Z\mathbf{1}_{W>\lambda})\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=p(p-1)\mathbb{E}\int_{0}^{W}Z\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=p\mathbb{E}ZW^{p-1}$ $\displaystyle\leq
p(\mathbb{E}Z^{p})^{1/p}(\mathbb{E}W^{p})^{1-1/p}.$

Since $W$ is bounded, we can cancel a suitable power of $\mathbb{E}W^{p}$ on
both sides and obtain the claim.

For general $W$, we apply the bounded case to $\min(W,\Lambda)$ and let
$\Lambda\to\infty$. ∎

######  Corollary 1.12.

Let $(A_{t})$ be an increasing predictable process with $A_{0}=0$ and $\xi$ a
positive random variable such that, for everty $t$, we have

$\mathbb{E}_{t}(A_{\infty}-A_{t})\leq\mathbb{E}_{t}(\xi).$

Then, for every $p\geq 1$, we have

$\lVert A_{\infty}\rVert_{p}\leq p\lVert\xi\rVert_{p}.$

###### Proof.

For $\lambda>0$, let
$\tau:=\inf\\{t\nonscript\>|\allowbreak\nonscript\>\mathopen{}A_{t+1}>\lambda\\}$.
Then,

$\mathbb{E}\max(A_{\infty}-\lambda,0)=\mathbb{E}(A_{\infty}-\lambda)\mathbf{1}_{\tau<\infty}\leq\mathbb{E}(A_{\infty}-A_{\tau})\mathbf{1}_{\tau<\infty}\leq\mathbb{E}(\xi\mathbf{1}_{\tau<\infty})\leq\mathbb{E}(\xi\mathbf{1}_{A_{\infty}>\lambda}),$

so we can apply Lemma 1.11. ∎

######  Corollary 1.13 (Burkholder–Davis–Gundy inequalities).

For every $p\in[1,\infty)$ and every martingale $f$, we have

$\lVert Sf\rVert_{p}\sim\lVert Mf\rVert_{p}.$

###### Proof.

For any $t_{0}$ and $B\in\mathcal{F}_{t_{0}}$, we can apply Theorem 1.10 to
the martingale $\mathbf{1}_{B}(f-f^{(t_{0})})_{t\geq t_{0}}$, and obtain

$\mathbb{E}_{t_{0}}(M(f-f^{t_{0}}))\sim\mathbb{E}_{t_{0}}(S(f-f^{t_{0}})).$

Note that

$Mf-Mf_{t_{0}}\leq M(f-f^{t_{0}})\leq 2Mf,\quad Sf-Sf_{t_{0}}\leq
S(f-f^{t_{0}})\leq Sf,$

so the conditions of Corollary 1.12 hold for instance for
$\tilde{A}_{t}=Mf_{t}$ and $\xi=Sf$, with exception of the fact that
$\tilde{A}_{t}$ is not predictable. This can be remedied by taking
$A_{t}=Mf_{t-1}$ and $\xi=Sf+Mdf\leq 3Sf$. Corollary 1.12 now shows that
$\lVert Mf\rVert_{p}\lesssim p\lVert Sf\rVert_{p}$.

The proof of the converse inequality is similar. ∎

###  1.1 Predictable square function

######  Definition 1.14 (Predictable square function).

For a martingale $f$, let

$sf_{n}:=\bigl{(}\sum_{k=1}^{n}\mathbb{E}_{k-1}(\lvert
df_{k}\rvert^{2})\bigr{)}^{1/2}.$

The continuous time version of $sf$ is denoted by $\left\langle
f\right\rangle$ and called the _predictable quadratic variation_. We mainly
discuss $sf$ because the continuous time analog is sometimes used in
stochastic analysis, and in order to present a few techniques.

The next lemma looks similar to $L^{p}$ contractivity of conditional
expectation, but it cannot be proved by using this contractivity for each
invidual summand.

######  Lemma 1.15.

Let $z_{1},z_{2},\dotsc$ be positive random variables. Then, for every
$p\in[1,\infty)$, we have

$\mathbb{E}(\sum_{k=1}^{\infty}\mathbb{E}_{k-1}(z_{k}))^{p}\leq
p^{p}\mathbb{E}(\sum_{k=1}^{\infty}z_{k})^{p}.$

###### Proof.

The hypothesis of Corollary 1.12 holds with equality for $A_{n}=W_{n}$ and
$\xi=Z_{\infty}$, where

$W_{n}:=\sum_{k=1}^{n}\mathbb{E}_{k-1}(z_{k}),\quad
Z_{n}:=\sum_{k=1}^{n}z_{k}.$ (1.5)

∎

######  Corollary 1.16.

Let $p\in[2,\infty)$ and $f$ be a martignale. Then, we have

$\lVert sf\rVert_{p}\leq(p/2)^{1/2}\lVert Sf\rVert_{p}.$

######  Lemma 1.17.

Let $Z,W$ be positive random variables such that, for some $C<\infty$ and all
$\lambda>0$, we have

$\mathbb{E}(Z\wedge\lambda)\leq C\mathbb{E}(W\wedge\lambda).$

Then, for every $p\in(0,1]$, we have

$\mathbb{E}Z^{p}\leq C\mathbb{E}W^{p}.$

###### Proof.

For $p=1$, it suffices to let $\lambda\to\infty$ in the hypothesis.

For $p<1$, we use the formula

$t^{p}=p(1-p)\int_{0}^{\infty}(t\wedge\lambda)\lambda^{p-2}\mathop{}\\!\mathrm{d}\lambda.$

Chaning the order of integration and applying the hypothesis for each
$\lambda$, we obtain the claim. ∎

######  Corollary 1.18.

Let $z_{1},z_{2},\dotsc$ be positive random variables. Then, for every
$p\in(0,1]$, we have

$\mathbb{E}(\sum_{k=1}^{\infty}z_{k})^{p}\leq
2\mathbb{E}(\sum_{k=1}^{\infty}\mathbb{E}_{k-1}(z_{k}))^{p}.$

Thus, we see that, for $p\in(0,2]$ and any martingale $f$, we have

$\lVert Sf\rVert_{p}\lesssim_{p}\lVert sf\rVert_{p}.$

###### Proof.

Define $W,Z$ as in (5.17). Let
$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}W_{n+1}>\lambda\\}$.
Then,

$\displaystyle\mathbb{E}(Z_{\infty}\wedge\lambda)$
$\displaystyle\leq\mathbb{E}(Z_{\tau}+\lambda\mathbf{1}_{\tau<\infty})$
$\displaystyle=\sum_{k}\mathbb{E}(\mathbf{1}_{k\leq\tau}z_{k})+\lambda\lvert\\{W_{\infty}>\lambda\\}\rvert$
$\displaystyle\leq\sum_{k}\mathbb{E}(\mathbf{1}_{k\leq\tau}\mathbb{E}_{k-1}z_{k})+\mathbb{E}(W_{\infty}\wedge\lambda)$
$\displaystyle\leq 2\mathbb{E}(W_{\infty}\wedge\lambda).$

We conclude by Lemma 1.17. ∎

######  Corollary 1.19.

Let $p\in(0,2]$ and $f$ be a martingale. Then

$\lVert Mf\rVert_{p}\leq 5^{1/p}\lVert sf\rVert_{p}.$

###### Proof.

Let
$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}sf_{n+1}>\lambda\\}$.
Then,

$\displaystyle\mathbb{E}((Mf)^{2}\wedge\lambda^{2})$
$\displaystyle\leq\mathbb{E}[(Mf^{\tau})^{2}]+\lambda^{2}\mathbb{E}\mathbf{1}_{\tau<\infty}$
(Doob’s inequality) $\displaystyle\leq
4\mathbb{E}[(f^{\tau})^{2}]+\lambda^{2}\mathbb{E}\mathbf{1}_{sf_{\infty}>\lambda}$
$\displaystyle\leq 5\mathbb{E}((sf_{\infty})^{2}\wedge\lambda^{2}).$

We conclude by Lemma 1.17. ∎

######  Lemma 1.20 (Good-$\lambda$ inequality).

Let $f,g$ be positive random variables, $p\in(0,\infty)$, and suppose that we
are given $\beta>1$ and $\delta,\varepsilon\in\mathbb{R}_{>0}$ with
$\beta^{p}\varepsilon<1$. Assume that, for every $\lambda>0$, we have

$\mu\\{g>\beta\lambda,f\leq\delta\lambda\\}\leq\varepsilon\mu\\{g>\lambda\\}.$

Then,

$\mathbb{E}g^{p}\leq\frac{\delta^{-p}}{\beta^{-p}-\varepsilon}\mathbb{E}f^{p}.$

###### Proof.

The hypothesis implies

$\displaystyle\mu\\{g>\beta\lambda\\}$
$\displaystyle=\mu\\{g>\beta\lambda,f\leq\delta\lambda\\}+\mu\\{g>\beta\lambda,f>\delta\lambda\\}$
$\displaystyle\leq\varepsilon\mu\\{g>\lambda\\}+\mu\\{f>\delta\lambda\\}.$

Using the formula
$t^{p}=p\int_{0}^{t}\lambda^{p-1}\mathop{}\\!\mathrm{d}\lambda$, we obtain

$\displaystyle\mathbb{E}g^{p}$
$\displaystyle=p\mathbb{E}\int_{0}^{g}\lambda^{p-1}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=p\int_{0}^{\infty}\mu\\{g>\lambda\\}\lambda^{p-1}\mathop{}\\!\mathrm{d}\lambda.$

Inserting the above estimate, we obtain

$\mathbb{E}(g/\beta)^{p}\leq\varepsilon\mathbb{E}g^{p}+\mathbb{E}(f/\delta)^{p}.$

Rearranging, we obtain the claim. ∎

######  Lemma 1.21.

Let $z_{1},z_{2},\dotsc$ be positive random variables. Then, for every
$p\in(0,\infty)$, we have

$\mathbb{E}(\sum_{k=1}^{\infty}z_{k})^{p}\lesssim_{p}\mathbb{E}(Mz\vee\sum_{k=1}^{\infty}\mathbb{E}_{k-1}(z_{k}))^{p}.$

Thus, we see that, for $p\in[2,\infty)$ and any martingale $f$, we have

$\lVert Sf\rVert_{p}\lesssim_{p}\lVert sf\vee Mdf\rVert_{p}.$

###### Proof.

We continue using notation (5.17). We will verify the hypothesis of Lemma
1.20. Let $\beta,\delta\in\mathbb{R}_{>0}$ with $\beta>\delta+1$. Consider the
stopping times

$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}Z_{n}>\lambda\\},\tau_{\beta}:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}Z_{n}>\beta\lambda\\},\sigma:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}(W_{n+1}\vee
z_{n})>\delta\lambda\\}$

and the process

$h_{n}=\sum_{k\leq n}\mathbf{1}_{\tau<k\leq\tau_{\beta}\wedge\sigma}z_{k}.$

Then, if $Z>\beta\lambda$ and $W\vee Mz\leq\delta\lambda$, we have
$\sigma=\infty$, so that

$h_{\infty}=\sum_{\tau<k\leq\tau_{\beta}}z_{k}\geq\beta\lambda-\sum_{k<\tau}z_{k}-z_{\tau}\geq\beta\lambda-\lambda-\delta\lambda=(\beta-1-\delta)\lambda.$

Hence,

$\displaystyle\mu\\{Z>\beta\lambda,W\vee Mz\leq\delta\lambda\\}$
$\displaystyle\leq\mu\\{h_{\infty}>(\beta-1-\delta)\lambda\\}$
$\displaystyle\leq\frac{1}{(\beta-1-\delta)\lambda}\mathbb{E}h_{\infty}$
$\displaystyle=\frac{1}{(\beta-1-\delta)\lambda}\sum_{k}\mathbb{E}(\mathbf{1}_{\tau<k\leq\tau_{\beta}\wedge\sigma}z_{k})$
$\displaystyle=\frac{1}{(\beta-1-\delta)\lambda}\sum_{k}\mathbb{E}(\mathbf{1}_{\tau<k\leq\tau_{\beta}\wedge\sigma}\mathbb{E}_{k-1}z_{k})$
$\displaystyle\leq\frac{1}{(\beta-1-\delta)\lambda}\mathbb{E}(\mathbf{1}_{\tau<\infty}\delta\lambda)$
$\displaystyle=\frac{\delta}{\beta-1-\delta}\mu\\{Z>\lambda\\}.$

Now it suffices to take any $\beta>1$ and $\delta$ sufficiently small
depending on $\beta$. ∎

######  Remark.

Most of the material in this section is from [Bur73].

##  2 Lépingle inequality

For $0<r<\infty$ and a sequence of random variables $f=(f_{n})_{n}$, the
$r$-variation of $f$ on the interval $[t^{\prime},t]$ is defined by

$V^{r}f_{t^{\prime},t}:=\sup_{J,t^{\prime}\leq u_{0}<\dotsb<u_{J}\leq
t}\Bigl{(}\sum_{j=1}^{J}\lvert f_{u_{j-1}}-f_{u_{j}}\rvert^{r}\Bigr{)}^{1/r},$
(2.1)

where the supremum is taken over arbitrary increasing sequences. Analogously,
$V^{\infty}f_{t^{\prime},t}:=\sup_{t^{\prime}\leq u^{\prime}<u\leq t}\lvert
f_{u^{\prime}}-f_{u}\rvert$.

We abbreviate $V^{r}f_{t}:=V^{r}f_{0,t}$ and $V^{r}f:=V^{r}f_{0,\infty}$.

The following result is a quantitative version of the martingale convergence
theorem.

######  Theorem 2.1 (Lépingle inequality).

For every $p\in[1,\infty)$, there exists a constant $C_{p}<\infty$ such that,
for every $r>2$, and every martingale $f$, we have

$\lVert V^{r}f\rVert_{L^{p}}\leq C_{p}\frac{r}{r-2}\lVert Mf\rVert_{L^{p}}.$
(2.2)

The main difficulty here is that the supremum in the definition of $V^{r}$ is
taken over not necessarily adapted partitions. We will remedy this by a greedy
selection algorithm, stopping as soon as we see a large jump. We have to use
stopping rules depending on a parameter $m$ in order to capture jumps of all
possible sizes.

###### Proof.

By the monotone convergence theorem, we may assume that $f$ becomes constant
after some time $N$.

Let $M_{t}:=\sup_{t^{\prime\prime}\leq t^{\prime}\leq t}\lvert
f_{t^{\prime}}-f_{t^{\prime\prime}}\rvert$. For each $m\in\mathbb{N}$, define
an adapted partition by

$\tau^{(m)}_{0}:=0,\quad\tau^{(m)}_{j+1}:=\inf\\{t\geq\tau^{(m)}_{j}\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lvert
f_{t}-f_{\tau_{j}}\rvert\geq 2^{-m}M_{t},M_{t}>0\\}.$ (2.3)

######  Claim.

Let $0\leq t^{\prime}<t<\infty$ and $m\geq 2$. Suppose that

$2<\lvert f_{t^{\prime}}-f_{t}\rvert/(2^{-m}M_{t})\leq 4.$ (2.4)

Then there exists $j$ with $t^{\prime}<\tau^{(m)}_{j}\leq t$ and

$\lvert f_{t^{\prime}}-f_{t}\rvert\leq 8\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert.$ (2.5)

###### Proof of the claim.

Let $j$ be the largest integer with $\tau^{\prime}:=\tau^{(m)}_{j}\leq t$. We
claim that $\tau^{\prime}>t^{\prime}$. Suppose for a contradiction that
$\tau^{\prime}<t^{\prime}$ (the case $\tau^{\prime}=t^{\prime}$ is similar but
easier). By the hypothesis (2.4) and the assumption that $t,t^{\prime}$ are
not stopping times, we obtain

$2\cdot 2^{-m}M_{t}<\lvert f_{t^{\prime}}-f_{t}\rvert\leq\lvert
f_{\tau^{\prime}}-f_{t^{\prime}}\rvert+\lvert
f_{\tau^{\prime}}-f_{t}\rvert<2^{-m}M_{t^{\prime}}+2^{-m}M_{t}\leq 2\cdot
2^{-m}M_{t},$

a contradiction. This shows $\tau^{\prime}>t^{\prime}$.

It remains to verify (2.5). Assume that $M_{\tau^{\prime}}<M_{t}/2$. Then, for
some $\tau^{\prime\prime}\in(\tau,t]$, we have $\lvert
f_{\tau^{\prime}}-f_{\tau^{\prime\prime}}\rvert\geq M_{t}/2\geq
2^{-m}M_{\tau^{\prime\prime}}$, contradicting maximality of $\tau^{\prime}$.
It follows that

$\lvert f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert\geq
2^{-m}M_{\tau^{\prime}}\geq 2^{-m}M_{t}/2\geq\lvert
f_{t^{\prime}}-f_{t}\rvert/8.\qed$

Next, for any $0<\rho<r<\infty$, we will show the pathwise inequality

$V^{r}_{t}(f_{t})^{r}\leq
8^{\rho}\sum_{m=2}^{\infty}\bigl{(}2^{-(m-2)}M_{\infty}\bigr{)}^{r-\rho}\sum_{j=1}^{\infty}\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert^{\rho}.$ (2.6)

Let $(u_{l})$ be any increasing sequence. For each $l$ with $\lvert
f_{u_{l}}-f_{u_{l+1}}\rvert\neq 0$, let $m=m(l)\geq 2$ be such that

$2<\lvert f_{u_{l}}-f_{u_{l+1}}\rvert/(2^{-m}M_{u_{l+1}})\leq 4.$

Such $m$ exists because the distance is bounded by $M_{u_{l+1}}$.

Let $j=j(l)$ be given by the Claim above with $t^{\prime}=u_{l}$ and
$t=u_{l+1}$. Then

$\lvert f_{u_{l}}-f_{u_{l+1}}\rvert^{r}\leq 8^{\rho}\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert^{\rho}\cdot(4\cdot
2^{-m}M_{u_{l+1}})^{r-\rho}.$

Since each pair $(m,j)$ occurs for at most one $l$, this implies

$\sum_{l}\lvert f_{u_{l}}-f_{u_{l+1}}\rvert^{r}\leq 8^{\rho}\sum_{m,j}\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert^{\rho}\cdot(2^{-(m-2)}M_{\infty})^{r-\rho}.$

Taking the supremum over all increasing sequences $(u_{l})$, we obtain (2.6).

Since we assumed that $f_{n}$ becomes independent of $n$ for sufficiently
large $n$, we have

$M_{\infty}\leq V^{r}_{t}(f_{t})<\infty.$

Substituting this inequality in (2.6) and canceling $V^{r}_{t}(f_{t})^{r-2}$
on both sides, we obtain

$V^{r}_{t}(f_{t})^{\rho}\leq
8^{\rho}\sum_{m=2}^{\infty}2^{-(m-2)(r-\rho)}\sum_{j=1}^{\infty}\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert^{\rho}.$ (2.7)

By the optional sampling theorem, for each $m$, the sequence
$(f_{\tau^{(m)}_{j}})_{j}$ is a martingale with respect to the filtration
$(\mathcal{F}_{\tau^{(m)}_{j}})_{j}$. Let

$S_{(m)}:=\Bigl{(}\sum_{j=1}^{\infty}\lvert
f_{\tau^{(m)}_{j-1}}-f_{\tau^{(m)}_{j}}\rvert^{2}\Bigr{)}^{1/2}$

denote the square function of this martingale. Then (2.7) implies

$V^{r}f\leq
8\Bigl{(}\sum_{m=2}^{\infty}2^{-(m-2)(r-2)}S_{(m)}^{2}\Bigr{)}^{1/2}\leq
8\sum_{m=2}^{\infty}2^{-(m-2)(r-2)/2}S_{(m)}$

By Minkowski’s inequality, this implies

$\lVert V^{r}f\rVert_{L^{p}}\leq 8\sum_{m=2}^{\infty}2^{-(m-2)(r-2)/2}\lVert
S_{(m)}\rVert_{L^{p}}.$

Applying the BDG inequality for the martingales $(f_{\tau^{(m)}_{j}})_{j}$, we
obtain

$\displaystyle\lVert V^{r}f\rVert_{L^{p}}$
$\displaystyle\lesssim_{p}\sum_{m=2}^{\infty}2^{-(m-2)(r-2)/2}\lVert\sup_{j}\lvert
f_{\tau^{(m)}_{j}}\rvert\rVert_{L^{p}}$
$\displaystyle\lesssim\frac{r}{r-2}\lVert Mf\rVert_{L^{p}}.\qed$

######  Remark.

The dependence of the constant in Theorem 2.1 on the variation exponent $r$
can be improved to $\sqrt{\frac{r}{r-2}}$ using a vector-valued BDG
inequality. In fact, a slightly more careful version of the above argument
already shows this for $p\in[2,\infty)$.

###  2.1 Martingale convergence

######  Corollary 2.2 (Martingale convergence).

Let $f$ be a martingale with $\lVert f\rVert_{1}<\infty$. Then, for every
$r>2$, $V^{r}f$ is finite a.s. In particular, the sequence $(f_{n})_{n}$
converges a.s.

###### Proof.

Fixing $\lambda<\infty$ and $N\in\mathbb{N}$, consider the stopping time

$\tau:=\inf\\{n\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lvert
f_{n}\rvert>\lambda\\}.$

Then $Mf^{\tau\wedge N}\leq\lambda\vee\lvert f_{\tau\wedge N}\rvert$. Since
$\tau\wedge N$ is a bounded stopping time, by the optional sampling theorem,
we have

$\lVert Mf^{\tau\wedge N}\rVert_{1}\leq\lambda+\lVert f_{\tau\wedge
N}\rVert_{1}\leq\lambda+\lVert f\rVert_{1}.$

By Lépingle’s inequality, for $r>2$, we obtain

$\lVert V^{r}f^{\tau\wedge N}\rVert_{1}\lesssim\lambda+\lVert f\rVert_{1}.$

Since the right-hand side does not depend on $N$ and by the monotone
convergence theorem, we obtain

$\lVert V^{r}f^{\tau}\rVert_{1}\lesssim\lambda+\lVert f\rVert_{1}.$

In particular, the function $V^{r}f^{\tau}$ is finite a.s. On the set
$\\{Mf<\lambda\\}$, we have $f=f^{\tau}$, so that $V^{r}f$ is finite a.s. on
this set.

Taking the union over $\lambda\in\mathbb{N}$, we find that $V^{r}f$ is finite
a.s. on the set $\\{Mf<\infty\\}$. By Doob’s maximal inequality, this set has
full probability. ∎

We have shown that, for an $L^{1}$ martingale $f$, the limit
$f_{\infty}:=\lim_{n\to\infty}f_{n}$ exists a.s. We recall the condition under
which this convergence also holds in $L^{1}$. We start with an example that
shows that this is not always the case.

######  Example 2.3.

Consider $\Omega=[0,1]$ with the dyadic filtration. The sequence

$f_{n}=2^{n}\mathbf{1}_{[0,2^{-n}]}$

is a martingale and converges pointwise a.s. to $0$. On the other hand,
$\mathbb{E}f_{n}=1\neq 0=\mathbb{E}0$.

######  Definition 2.4.

A set $F$ of real random variables is called _uniformly integrable_ if

$\lim_{\lambda\to\infty}\sup_{f\in F}\mathbb{E}(\mathbf{1}_{\lvert
f\rvert>\lambda}\lvert f\rvert)=0.$

######  Example 2.5.

If $f\in L^{1}$, then the singleton $\\{f\\}$ is uniformly integrable.

Uniform integrability is useful because it is the minimal hypothesis under
which an analogue of the dominated convergence theorem holds. We recall this
analogue.

######  Proposition 2.6 (Dominated convergence).

For every sequence of random variables $f=(f_{n})$, the following are
equivalent:

  1. 1.

$f$ is Cauchy in $L^{1}$,

  2. 2.

$f$ is uniformly integrable and Cauchy in probability, that is,

$(\forall\varepsilon>0)(\exists N\in\mathbb{N})(\forall m,n\geq N)\mu\\{\lvert
f_{n}-f_{m}\rvert>\varepsilon\\}<\varepsilon.$

In relation with martingales, it is important that uniform integrability is
preserved under conditional expectation. This can be seen using the following
characterization.

######  Lemma 2.7.

Let $F$ be a set of random variables. The following are equivalent.

  1. 1.

$F$ is uniformly integrable.

  2. 2.

$\sup_{f\in F}\mathbb{E}\lvert f\rvert<\infty$ and

$\lim_{\mu(A)\to 0}\sup_{f\in F}\mathbb{E}(\mathbf{1}_{A}\lvert f\rvert)=0.$
(2.8)

###### Proof.

Without loss of generality, all r.v. in $F$ are positive.

$\implies$: For every $\lambda>0$, we have

$\mathbb{E}f\leq\lambda+\mathbb{E}(\mathbf{1}_{f>\lambda}f).$

Taking the supremum over $f\in F$ on both sides and choosing $\lambda$ such
that

$\sup_{f\in F}\mathbb{E}(\mathbf{1}_{f>\lambda}f)<\infty,$

we see that $\sup_{f\in F}\mathbb{E}\lvert f\rvert<\infty$.

Moreover,

$\mathbb{E}(\mathbf{1}_{A}f)\leq\mu(A)\lambda+\mathbb{E}(\mathbf{1}_{f>\lambda}f).$

Taking first the supremum over $f\in F$, and then $\lim_{\mu(A)\to 0}$, we
obtain

$\lim_{\mu(A)\to 0}\sup_{f\in F}\mathbb{E}(\mathbf{1}_{A}f)\leq\sup_{f\in
F}\mathbb{E}(\mathbf{1}_{f>\lambda}f).$

Taking $\lim_{\lambda\to 0}$, we obtain (2.8).

$\impliedby$: We have $\mu\\{f>\lambda\\}\leq\lambda^{-1}\mathbb{E}f$. Taking
the supremum over $f\in F$, we obtain

$\sup_{f\in F}\mu\\{f>\lambda\\}\leq\lambda^{-1}\sup_{f\in F}\mathbb{E}f.$

The RHS converges to $0$ as $\lambda\to\infty$. The second hypothesis now
implies that $F$ is uniformly integrable. ∎

######  Lemma 2.8.

Let $F\subset L^{1}(\Omega,\mathcal{F})$ be uniformly integrable. Then the set

$\\{\mathbb{E}(f|\mathcal{F}^{\prime})\nonscript\>|\allowbreak\nonscript\>\mathopen{}f\in
F,\mathcal{F}^{\prime}\subseteq\mathcal{F}\text{ sub-$\sigma$-algebra}\\}$

is uniformly integrable.

###### Proof.

We verify the characterization above. We have

$\mathbb{E}\lvert\mathbb{E}(f|\mathcal{F}^{\prime})\rvert\leq\mathbb{E}\mathbb{E}(\lvert
f\rvert|\mathcal{F}^{\prime})=\mathbb{E}\lvert f\rvert.$

This verifies the first condition.

Fixing $\lambda>0$, we also have

$\mathbb{E}(\mathbf{1}_{A}\lvert\mathbb{E}(f|\mathcal{F}^{\prime})\rvert)\leq\mathbb{E}(\mathbf{1}_{A}\mathbb{E}(\lvert
f\rvert|\mathcal{F}^{\prime}))=\mathbb{E}(\mathbb{E}(\mathbf{1}_{A}|\mathcal{F}^{\prime})\lvert
f\rvert)\leq\lambda\mathbb{E}\lvert
f\rvert+\mathbb{E}(\mathbf{1}_{\mathbb{E}(\mathbf{1}_{A}|\mathcal{F}^{\prime})>\lambda}\lvert
f\rvert)$

Since
$\mu\\{\mathbb{E}(\mathbf{1}_{A}|\mathcal{F}^{\prime})>\lambda\\}\leq\lambda^{-1}\mu(A)$
and using (2.8) for the set $F$, we obtain

$\lim_{\mu(A)\to 0}\sup_{f\in
F,\mathcal{F}^{\prime}\subseteq\mathcal{F}}\mathbb{E}(\mathbf{1}_{A}\lvert\mathbb{E}(f|\mathcal{F}^{\prime})\rvert)\leq\lambda\mathbb{E}\lvert
f\rvert.$

Since $\lambda>0$ was arbitrary, this implies (2.8) for the set of conditional
expectations. ∎

The above results can now be summarized as follows.

######  Proposition 2.9 (Martingale closure).

For every martingale $f=(f_{n})_{n\in\mathbb{N}}$, the following are
equivalent.

  1. 1.

The set $\\{f_{n}\\}_{n\in\mathbb{N}}$ is uniformly integrable.

  2. 2.

The sequence $(f_{n})_{n\in\mathbb{N}}$ converges in $L^{1}$.

  3. 3.

There exists $f_{\infty}\in L^{1}$ such that, for every $n\in\mathbb{N}$, we
have $f_{n}=\mathbb{E}_{n}f_{\infty}$.

If the above conditions hold, then we can take
$f_{\infty}=\lim_{n\to\infty}f_{n}$ a.s. and in $L^{1}$.

######  Remark.

We have seen that, for a martingale $f$, we have

$Mf\in L^{1}\implies f\text{ is uniformly integrable}\implies f\in L^{1}.$

The second implication cannot be reversed by Example 2.3. The following
example shows that the first implication also cannot be reversed, and also
that $M$ is not bounded on $L^{1}$.

######  Example 2.10.

Let $\Omega=[0,1]$ with the dyadic filtration and

$f_{\infty}:=\sum_{m=0}^{\infty}(m+1)^{-2}2^{m}\mathbf{1}_{[2^{-m-1},2^{-m}]}$

and $f_{n}:=\mathbb{E}_{n}f_{\infty}$. Then, for $x\in(2^{-l-1},2^{-l})$, we
have

$Mf(x)=f_{l}(x)=2^{l}\sum_{m=l}^{\infty}\int_{0}^{2^{-l}}(m+1)^{-2}2^{m}\mathbf{1}_{[2^{-m-1},2^{-m}]}(y)\mathop{}\\!\mathrm{d}y\\\
=2^{l-1}\sum_{m=l}^{\infty}(m+1)^{-2}\sim(l+1)^{-1}2^{l}.$

We see that $Mf\not\in L^{1}$.

##  3 Vector-valued inequalities

######  Remark.

The norm on a Banach space $X$ will be sometimes denoted by $Xf:=\lVert
f\rVert_{X}$. This is consistent with operator notation, in that every norm is
also a sublinear operator.

######  Lemma 3.1.

Let $X$ be a Banach space. Let $T$ be a subadditive operator that maps
$X$-valued martingales starting at $0$ to $\mathbb{R}_{\geq 0}$-valued
increasing adapted processes starting at $0$ such that, for every stopping
time $\tau$, we have

$Tf_{\tau}=T(f^{\tau})_{\tau},$

and for every time $t$ we have

$Tf_{t}\leq Tf_{t-1}+Xdf_{t}.$

Let $U$ be another operator satisfying the same properties as $T$.

Assume that, for some $r\in(1,\infty)$ and every $X$-valued martingale $f$, we
have

$\mathbb{P}\\{Tf>\lambda\\}\lesssim\lambda^{-r}(L^{r}Uf)^{r}.$ (3.1)

Then, for every $q\in[1,\infty)$ and every $X$-valued martingale $f$, we have

$L^{q}Tf\lesssim_{q}L^{q}Uf+L^{q}MXdf.$ (3.2)

The formulation is flexible enough to apply to any combination
$\\{T,U\\}=\\{XM,XS\\}$, and also some other operators, such as martingale
transforms or $r$-variation norms.

###### Proof.

Consider first the case $q=1$. We use the Davis decomposition
$f=f^{\mathrm{pred}}+f^{\mathrm{bv}}$. Of the two possible generalizations of
the Davis decomposition to $X$-valued martingales, we use the one in which the
absolute value is replaced by the $X$-norm (the alternative would be to make a
decomposition for each $k$ separately). Specifically, we have

$Xdf^{\mathrm{pred}}_{n}\leq
2MXdf_{n-1},\quad\mathbb{E}\sum_{n}Xdf^{\mathrm{bv}}_{n}\leq 2\mathbb{E}MXdf.$

For $\lambda>0$, define the stopping time

$\tau:=\inf\\{t\nonscript\>|\allowbreak\nonscript\>\mathopen{}Uf^{\mathrm{pred}}_{t}>\lambda\text{
or }MXdf_{t}>\lambda\\}.$

We claim that

$Uf^{\mathrm{pred}}_{\tau}\leq Uf^{\mathrm{pred}}\wedge 3\lambda.$ (3.3)

Indeed, the first bound is trivial. The second bound is trivial if
$\tau=\infty$, so assume $\tau\in(0,\infty)$. Then, by properties of the Davis
decomposition, we have

$Uf^{\mathrm{pred}}_{\tau}\leq
Uf^{\mathrm{pred}}_{\tau-1}+Xdf^{\mathrm{pred}}_{\tau}\leq\lambda+2MXdf_{\tau-1}\leq
3\lambda.$

Also,

$\displaystyle\\{Tf^{\mathrm{pred}}>\lambda\\}$
$\displaystyle\subseteq\\{Tf^{\mathrm{pred}}_{\tau}>\lambda\\}\cup\\{\tau<\infty\\}$
$\displaystyle\subseteq\\{Tf^{\mathrm{pred}}_{\tau}>\lambda\\}\cup\\{Uf^{\mathrm{pred}}>\lambda\\}\cup\\{MXdf_{\tau}>\lambda\\}.$

By the layer cake formula,

$\displaystyle\lVert Tf^{\mathrm{pred}}\rVert_{L^{1}}$
$\displaystyle=\int_{0}^{\infty}\mathbb{P}\\{Tf^{\mathrm{pred}}>\lambda\\}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\leq\int_{0}^{\infty}\mathbb{P}\\{Tf^{\mathrm{pred}}_{\tau}>\lambda\\}\mathop{}\\!\mathrm{d}\lambda+\int_{0}^{\infty}\mathbb{P}\\{Uf^{\mathrm{pred}}>\lambda\\}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\quad+\int_{0}^{\infty}\mathbb{P}\\{MXdf>\lambda\\}\mathop{}\\!\mathrm{d}\lambda=:I+II+III.$

The term $III$ is easy to bound by the layer cake formula.

In the term $II$, we use the properties of the Davis decomposition to estimate

$\displaystyle II$ $\displaystyle=\lVert Uf^{\mathrm{pred}}\rVert_{L^{1}}$
$\displaystyle\leq\lVert Uf\rVert_{L^{1}}+\lVert
Uf^{\mathrm{bv}}\rVert_{L^{1}}$ $\displaystyle\leq\lVert
Uf\rVert_{L^{1}}+\lVert\sum_{n}Xdf^{\mathrm{bv}}_{n}\rVert_{L^{1}}$
$\displaystyle\leq\lVert Uf\rVert_{L^{1}}+2\lVert MXdf\rVert_{L^{1}}.$

Using the hypothesis (3.1) and (3.3), we bound the first term by

$\displaystyle I$ $\displaystyle\lesssim\int_{0}^{\infty}\lambda^{-r}\lVert
Uf^{\mathrm{pred}}_{\tau}\rVert_{L^{r}}^{r}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\leq\int_{0}^{\infty}\lambda^{-r}\lVert Uf^{\mathrm{pred}}\wedge
3\lambda\rVert_{L^{r}}^{r}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=\mathbb{E}\int_{0}^{\infty}\min\bigl{(}\lambda^{-r}(Uf^{\mathrm{pred}})^{r},3^{r}\bigr{)}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\lesssim\mathbb{E}Uf^{\mathrm{pred}}=II,$

and we reuse the previously established estimate for $II$.

We have shown (3.2) for $q=1$, and we will now extend this claim to $q>1$. In
doing so, we may replace $Uf$ by $(Uf+MXdf)/2$. This has the effect that we
may omit the second summand on the RHS of (3.2).

For $q>1$, we use the Garsia-Neveu lemma. For a martingale $f$, let
$\prescript{\tau}{}{\\!}f:=f-f^{\tau}$ be the martingale $f$ started at
(stopping) time $\tau$.

For any $t\in\mathbb{N}$ and $B\in\mathcal{F}_{t}$, we can apply the case
$q=1$ to the martingale $\mathbf{1}_{B}\prescript{t}{}{\\!}f$, and obtain

$\mathbb{E}_{t}T\prescript{t}{}{\\!}f\lesssim\mathbb{E}_{t}U\prescript{t}{}{\\!}f.$

Note that

$Tf_{\infty}-Tf_{t}=Tf_{\infty}-T(f^{t})_{\infty}\leq
T(f-f^{t})_{\infty}=T\prescript{t}{}{\\!}f_{\infty},$
$U\prescript{t}{}{\\!}f_{\infty}\leq(Uf+Uf^{t})_{\infty}\leq 2Uf_{\infty},$

so the conditions of Corollary 1.12 hold for $A_{t}=Tf_{(t-1)\vee 0}$ and
$\xi\sim Uf_{\infty}$. Corollary 1.12 now shows the claim. ∎

Monotonicity of $Uf$ is used in estimates for both $I$ and $II$ above.

######  Corollary 3.2 ($\ell^{r}$ valued BDG inequality).

For every $k\in\mathbb{N}$, let $f_{k}=(f_{k,n})_{n}$ be a real-valued
martingale starting at $0$. Then, for every $q,r\in[1,\infty)$, we have

$L^{q}\ell^{r}_{k}Mf_{k}\lesssim_{q}L^{q}\ell^{r}_{k}Sf_{k}.$

###### Proof.

For $r=q$, this follows from the scalar-valued Doob’s inequality and Fubini.
For general $q$, we apply Lemma 3.1 with $X=\ell^{r}$. ∎

Toy example: better constant in Lépingle.

###  3.1 Weighted and vector-valued Doob inequalities

Let $w\in L^{1}(\Omega,\mathcal{F}_{\infty})$ be a positive function, that we
will call a _weight_. We write $w_{k}:=\mathbb{E}_{k}w$,
$w^{*}:=\sup_{k}w_{k}$, and $wB:=\int_{B}w\mathop{}\\!\mathrm{d}\mu$ for
$B\in\mathcal{F}_{\infty}$.

######  Lemma 3.3.

Let $f$ be an $\mathbb{R}_{\geq 0}$-valued submartingale (recall that this
means $f_{k}\leq E_{k}f_{n}$ for $k\leq n$). Then, for every $\lambda>0$ and
$n\in\mathbb{N}$, we have

$\lambda
w\\{Mf_{n}>\lambda\\}\leq\int_{\\{Mf_{n}>\lambda\\}}f_{n}Mw_{n}\mathop{}\\!\mathrm{d}\mu.$

###### Proof.

For a fixed $\lambda$, define the stopping time

$\tau:=\inf\\{k\nonscript\>|\allowbreak\nonscript\>\mathopen{}f_{k}>\lambda\\}.$

Then, $\\{Mf_{n}>\lambda\\}=\\{\tau\leq n\\}$. Hence,

$\displaystyle\lambda w\\{Mf_{n}>\lambda\\}$ $\displaystyle=\lambda\sum_{k\leq
n}w\\{\tau=k\\}$ $\displaystyle\leq\sum_{k\leq
n}\int_{\\{\tau=k\\}}f_{k}w_{k}\mathop{}\\!\mathrm{d}\mu$ (submartingale
property) $\displaystyle\leq\sum_{k\leq
n}\int_{\\{\tau=k\\}}f_{n}w_{k}\mathop{}\\!\mathrm{d}\mu$
$\displaystyle\leq\int_{\\{Mf_{n}>\lambda\\}}f_{n}w^{*}\mathop{}\\!\mathrm{d}\mu.\qed$

We recall that, for a function $f$ on a measure space $(\Omega,\mu)$ and
$p\in[1,\infty)$, the $L^{p,\infty}$ quasinorm is defined by

$\lVert f\rVert_{L^{p,\infty}}:=\sup_{\lambda>0}\lambda\mu\\{\lvert
f\rvert>\lambda\\}^{1/p}.$

For $p=\infty$, we have $\lVert f\rVert_{L^{p,\infty}}=\lVert
f\rVert_{L^{\infty}}$.

######  Theorem 3.4 (Marcinkiewicz/real interpolation, see e.g. [Gra14]).

Let $(\Omega,\mu)$ and $(\tilde{\Omega},\tilde{\mu})$ be $\sigma$-finite
measure spaces. Let $0<p_{0}<p_{1}\leq\infty$. Let

$T:L^{p_{0}}(\Omega)+L^{p_{1}}(\Omega)\to L^{0}(\tilde{\Omega})$

be a subadditive operator such that, for every function $f$, we have

$\lVert Tf\rVert_{L^{p_{j},\infty}(\tilde{\Omega})}\leq A_{j}\lVert
f\rVert_{L^{p_{j}}(\Omega)},\quad j=0,1.$

Then, for every $p\in(p_{0},p_{1})$, we have

$\lVert Tf\rVert_{L^{p}(\tilde{\Omega})}\lesssim_{p}\lVert
f\rVert_{L^{p}(\Omega)}.$

######  Lemma 3.5.

Let $p\in(1,\infty)$, $f$ be a submartingale, and $w$ a weight. Then,

$\lVert Mf\rVert_{L^{p}(w)}\lesssim_{p}\lVert f\rVert_{L^{p}(w^{*})}$

###### Proof.

We will show the following more precise estimate:

$\lVert Mf_{N}\rVert_{L^{p}(w)}\lesssim_{p}\lVert
f_{N}\rVert_{L^{p}(Mw_{N})},$ (3.4)

from which the claim follows by letting $N\to\infty$. The main advantage of
this formulation is that the right-hand side increases in $N$, which would not
be the case if we would use the weight $w^{*}$ instead of $Mw_{N}$.

Lemma 3.3 implies in particular

$\lVert Mf_{N}\rVert_{L^{1,\infty}(w)}\lesssim\lVert
f_{N}\rVert_{L^{1}(Mw_{N})}.$

The estimate

$\lVert Mf_{N}\rVert_{L^{\infty}(w)}\lesssim\lVert
f_{N}\rVert_{L^{\infty}(Mw_{N})}$

is easy to see. By real interpolation, these two bounds imply (3.4). ∎

######  Proposition 3.6 (Vector-valued maximal inequality).

Let $p\in(1,\infty)$ and $r\in(1,\infty]$. Then, for any sequence of
martingales $f_{k}=(f_{k,n})_{n}$, we have

$L^{p}\ell^{r}_{k}Mf_{k}\lesssim_{q,r}L^{p}\ell^{r}_{k}f_{k}.$

###### Proof.

The case $r=\infty$ follows from the scalar case, because $\ell^{\infty}f_{k}$
is a submartingale.

The case $p=r$ also follows from the scalar case and Fubini.

For $1<r<p<\infty$, let $\rho:=(p/r)^{\prime}=p/(p-r)$. Then, for some weight
$w\in L^{\rho}$ with $\lVert w\rVert_{\rho}\leq 1$, we have

$\displaystyle\lVert\ell^{r}_{k}Mf_{k}\rVert_{p}^{r}$
$\displaystyle\lesssim\Bigl{(}\mathbb{E}\bigl{(}\sum_{k}(Mf_{k})^{r}\bigr{)}^{p/r}\Bigr{)}^{r/p}$
$\displaystyle=\mathbb{E}\sum_{k}(Mf_{k})^{r}w$
$\displaystyle\lesssim\mathbb{E}\sum_{k}\lvert f_{k}\rvert^{r}w^{*}$

and we conclude using Hölder’s inequality and boundedness of the maximal
operator on $L^{\rho}$.

For $1<p<r<\infty$, let $s\in(1,p)$. Then, for some sequence of weights with

$\lVert\ell^{(r/s)^{\prime}}_{k}w_{k}\rVert_{(p/s)^{\prime}}\sim 1,$

we have

$\displaystyle\lVert\ell^{r}_{k}Mf_{k}\rVert_{p}^{s}$
$\displaystyle=\lVert\ell^{r/s}_{k}(Mf_{k})^{s}\rVert_{p/s}$
$\displaystyle\sim\int\sum_{k}w_{k}(Mf_{k})^{s}$
$\displaystyle\lesssim\int\sum_{k}w_{k}^{*}\lvert f_{k}\rvert^{s}$
$\displaystyle\leq\lVert\ell^{(r/s)^{\prime}}_{k}w_{k}^{*}\rVert_{(p/s)^{\prime}}\lVert\ell^{r/s}_{k}\lvert
f_{k}\rvert^{s}\rVert_{p/s}.$

Since $(r/s)^{\prime}<(p/s)^{\prime}$, from the previously shown case, we
obtain

$\lVert\ell^{(r/s)^{\prime}}_{k}w_{k}^{*}\rVert_{(p/s)^{\prime}}\lesssim\lVert\ell^{(r/s)^{\prime}}_{k}w_{k}\rVert_{(p/s)^{\prime}}\sim
1.\qed$

######  Remark.

Most of this section is from [Hyt+16, Section 3].

##  4 Rough paths

In this section, we discuss how to define $\int a\mathop{}\\!\mathrm{d}g$ for
not very regular functions $a,g$. The intended application is that $g$ is a
sample path of a martingale, which only has bounded $r$-variation for $r>2$,
so that we cannot use Stieltjes integration.

The function $a$ is assumed to have the same regularity as $g$, because we
want the theory to be suitable for solving equations like
$f(t)=\int_{0}^{t}a(s,f(s))\mathop{}\\!\mathrm{d}g(s)$.

###  4.1 Young integration with Hölder functions

We begin with a criterion for convergence of Riemann-like sums. Let
$\Delta:=\\{(s,t)\nonscript\>|\allowbreak\nonscript\>\mathopen{}0\leq s\leq
t\leq T\\}$. For any $\Xi:\Delta\to\mathbb{R}$ and a partition
$0=\pi_{0}<\dotsc<\pi_{J}=T$, write

$\mathcal{I}^{\pi}\Xi_{0,T}:=\sum_{j=1}^{J}\Xi_{\pi_{j-1},\pi_{j}}.$

We will discuss sufficient conditions for the convergence of these sums along
the directed (by set inclusion) set of partitions.

For a mapping $\Xi:\Delta\to E$ ($E$ a vector space), we write

$\delta\Xi_{s,u,t}:=\Xi_{s,t}-\Xi_{s,u}-\Xi_{u,t}.$

A _control_ is a function $\omega:\Delta\to[0,\infty)$ that is superadditive
in the sense that

$\omega(s,t)+\omega(t,u)\leq\omega(s,u)$

for all $s\leq t\leq u$. This implies in particular that $\omega(t,t)=0$ for
all $t$.

######  Lemma 4.1.

Let $\omega$ be a control and $\theta>1$. Let $(E,\lvert\cdot\rvert)$ be a
Banach space and $\Xi:\Delta\to E$ such that, for every $0\leq s\leq t\leq
u\leq T$, we have

$\lvert\delta\Xi_{s,t,u}\rvert\leq\omega(s,u)^{\theta}.$ (4.1)

Then, for every partition $0=\pi_{0}\leq\dotsc\leq\pi_{J}=T$, we have

$\lvert\Xi_{0,T}-\mathcal{I}^{\pi}\Xi_{0,T}\rvert\lesssim(\theta-1)^{-1}\omega(0,T)^{\theta}.$

###### Proof.

By induction on the partition size $J$, we will show the bound

$\lvert\Xi_{0,T}-\mathcal{I}^{\pi}\Xi_{0,T}\rvert\leq\sum_{k=1}^{J-1}(2/k)^{\theta}\omega(0,T)^{\theta}.$

For $J=1$, we have $\mathcal{I}^{\pi}\Xi_{0,T}=\Xi_{0,T}$, which serves as the
induction base.

Suppose that the claim is known for all partitions of size $J$ and let $\pi$
be a partition of size $J+1$. By superadditivity of $\omega$, we have

$\sum_{k=0}^{J-1}\omega(\pi_{k},\pi_{k+2})=\sum_{k\leq J-1\text{
even}}\omega(\pi_{k},\pi_{k+2})+\sum_{k\leq J-1\text{
odd}}\omega(\pi_{k},\pi_{k+2})\leq 2\omega(0,T).$

Hence, there exists $k$ with $\omega(\pi_{k},\pi_{k+2})\leq 2\omega(0,T)/J$.
Let $\pi^{\prime}:=\pi\setminus\\{\pi_{k+1}\\}$. Then,

$\lvert\mathcal{I}^{\pi}\Xi_{0,T}-\mathcal{I}^{\pi^{\prime}}\Xi_{0,T}\rvert=\lvert\delta\Xi_{\pi_{k},\pi_{k+1},\pi_{k+2}}\rvert\leq\omega(\pi_{k},\pi_{k+2})^{\theta}\leq(2/J)^{\theta}\omega(0,T).$

Applying the inductive hypothesis to $\pi^{\prime}$, we obtain the claim. ∎

######  Proposition 4.2 (Sewing).

Let $\omega$ be a control such that

$\lim_{\pi}\sup_{j}\omega(\pi_{j},\pi_{j+1})=0.$ (4.2)

Let $\Xi:\Delta\to E$ be such that (4.1) holds. Then, the limit

$\mathcal{I}\Xi_{0,T}:=\lim_{\pi}\mathcal{I}^{\pi}\Xi_{0,T}$

exists and satisfies

$\lvert\mathcal{I}\Xi_{0,T}-\Xi_{0,T}\rvert\lesssim(\theta-1)^{-1}\omega(0,T)^{\theta}.$

###### Proof.

For any partitions $\pi\subseteq\pi^{\prime}$, by Lemma 4.1, we have

$\displaystyle\lvert\mathcal{I}^{\pi}\Xi_{s,t}-\mathcal{I}^{\pi^{\prime}}\Xi_{s,t}\rvert$
$\displaystyle\leq\sum_{j}\lvert\Xi_{\pi_{j},\pi_{j+1}}-\mathcal{I}^{\pi^{\prime}}\Xi_{\pi_{j},\pi_{j+1}}\rvert$
$\displaystyle\lesssim\sum_{j}\omega(\pi_{j},\pi_{j+1})^{\theta}$
$\displaystyle\lesssim\sup_{j}\omega(\pi_{j},\pi_{j+1})^{\theta-1}\sum_{j}\omega(\pi_{j},\pi_{j+1})$
$\displaystyle\lesssim\omega(s,t)\sup_{j}\omega(\pi_{j},\pi_{j+1})^{\theta-1}.$

The hypothesis (4.2) implies that $\pi\mapsto\mathcal{I}^{\pi}\Xi_{s,t}$ is a
Cauchy net. ∎

######  Example 4.3 (Young integral).

Let $a,g:[0,T]\to\mathbb{R}$ be $\alpha$-Hölder functions and

$\Xi_{s,t}:=a_{s}(g_{t}-g_{s}).$

Then,

$\displaystyle\delta\Xi_{s,t,u}$ $\displaystyle=\Xi_{s,u}-\Xi_{s,t}-\Xi_{t,u}$
$\displaystyle=a_{s}(g_{u}-g_{s})-a_{s}(g_{t}-g_{s})-a_{t}(g_{u}-g_{t})$
$\displaystyle=a_{s}(g_{u}-g_{t})+a_{s}(g_{t}-g_{s})-a_{s}(g_{t}-g_{s})-a_{t}(g_{u}-g_{t})$
$\displaystyle=(a_{s}-a_{t})(g_{u}-g_{t}).$

In particular, $\Xi_{s,t,u}\lesssim\lvert s-t\rvert^{\alpha}\lvert
u-t\rvert^{\alpha}\leq\lvert u-s\rvert^{2\alpha}$. Hence, provided that
$2\alpha>1$, the hypothesis of Proposition 4.2 holds with $\theta=2\alpha$ and
$\omega(s,t)\sim\lvert t-s\rvert$.

###  4.2 Young integration with $V^{r}$ functions

If the function $g:[0,T]\to\mathbb{R}$ has bounded $r$-variation, then

$\omega_{g,r}(s,t):=(V^{r}g_{s,t})^{r}=\sup_{s\leq\pi_{0}\leq\dotsb\leq\pi_{J}\leq
t}\sum_{j=1}^{J}\lvert g_{\pi_{j}}-g_{\pi_{j-1}}\rvert^{r}$

is a control. If both $a,g\in V^{r}$, then, in the situation of Example 4.3,
we have

$\lvert\delta\Xi_{s,t,u}\rvert\leq\omega_{a,r}(s,t)^{1/r}\omega_{g,r}(t,u)^{1/r}.$
(4.3)

This implies in particular (4.1) with $\omega=\omega_{a,r}+\omega_{g,r}$ and
$\theta=2/r$. However, (4.2) only holds for this control if $a,g$ are also
continuous. In order to integrate functions with jumps, we will show a version
of the sewing lemma under the condition (4.3).

######  Definition 4.4.

If $f$ is a function defined on a suitable interval, we write

$f(t+):=\lim_{t^{\prime}\to t,t^{\prime}>t}f(t^{\prime}),\quad
f(t-):=\lim_{t^{\prime}\to t,t^{\prime}<t}f(t^{\prime}),$

provided that the respective limits exist.

######  Lemma 4.5.

Let $\omega$ be a control. Then, for every $\varepsilon>0$, there exists a
partition $\pi$ such that

$\max_{j}\bigl{(}\omega(\pi_{j-1}+,\pi_{j})\wedge\omega(\pi_{j-1},\pi_{j}-)\bigr{)}\leq\varepsilon.$
(4.4)

###### Proof.

We select the partition greedily starting with $\pi_{0}=0$. It will be clear
from the construction that the claim (4.4) holds. If $\pi_{j}$ has been
already selected and $\pi_{j}<T$, we distinguish two cases.

Case 1: if $\omega(\pi_{j},\pi_{j}+)<\varepsilon$, then we let

$\pi_{j+1}:=\sup\\{t>\pi_{j}\nonscript\>|\allowbreak\nonscript\>\mathopen{}\omega(\pi_{j},t)<\varepsilon\\}.$

Then $\omega(\pi_{j},\pi_{j+1}-)\leq\varepsilon$.

Case 2: if $\omega(\pi_{j},\pi_{j}+)\geq\varepsilon$, then we choose any
$\pi_{j+1}\in(\pi_{j},T]$ with $\omega(\pi_{j}+,\pi_{j+1})\leq\varepsilon$.

To see that such $\pi_{j+1}$ exists, suppose for a contradiction that, for
every $t\in(\pi_{j},T]$, we have $\omega(\pi_{j}+,t)>\varepsilon$. Choose
recursively $t_{0}=T$ and, given $t_{k}$, let $t_{k+1}\in(\pi_{j},t_{k})$ be
such that $\omega(t_{k+1},t_{k})>\varepsilon$. By superadditivity, we obtain

$\omega(\pi_{j},T)\geq\sum_{k}\omega(t_{k+1},t_{k})=+\infty,$

a contradiction.

The selection of $\pi_{j}$’s ends after finitely many steps, because otherwise
we would have $\omega(\pi_{j},\pi_{j+1}+)\geq\varepsilon$ for every $j$, which
in turn implies $\omega(\pi_{j},\pi_{j+2})\geq\varepsilon$, and summing over
even $j$ we obtain a contradiction with the superadditivity of $\omega$. ∎

######  Theorem 4.6 (Sewing with jumps).

Let $\omega_{1,n},\omega_{2,n}$ be controls and $\alpha_{1,n},\alpha_{2,n}\geq
0$ with $\alpha_{1,n}+\alpha_{2,n}>1$ for all $n\in\\{1,\dotsc,N\\}$.

Let $\Xi:\Delta\to E$. Assume

$\lvert\delta\Xi_{s,u,t}\rvert\leq\sum_{n=1}^{N}\omega_{1,n}^{\alpha_{1,n}}(s,u)\omega_{2,n}^{\alpha_{2,n}}(u,t),$

Then the following net limit exists,

$\mathcal{I}\Xi_{0,T}:=\lim_{\pi}\mathcal{I}^{\pi}\Xi_{0,T},$

and one has the following estimate:

$\lvert\mathcal{I}\Xi_{0,T}-\Xi_{0,T}\rvert\lesssim\sum_{n=1}^{N}\omega_{1,n}^{\alpha_{1,n}}(0,T-)\omega_{2,n}^{\alpha_{2,n}}(0+,T),$

with $C$ depending only on $\min_{n}\alpha_{1,n}+\alpha_{2,n}$.

###### Proof.

Let

$\theta:=\min_{n}\alpha_{1,n}+\alpha_{2,n}>1$

and

$\omega(s,t):=\sum_{n=1}^{N}\omega_{1,n}^{\alpha_{1,n}/\theta}(s,t-)\omega_{2,n}^{\alpha_{2,n}/\theta}(s+,t).$

Note that the functions $(s,t)\mapsto\omega_{1,n}(s,t-)$ and
$(s,t)\mapsto\omega_{2,n}(s+,t)$ are also controls. The function $\omega$ is a
control, because, by Hölder’s inequality,

$\omega_{1,n}^{\alpha_{1,n}/\theta}(s,u)\omega_{2,n}^{\alpha_{2,n}/\theta}(s,u)\geq(\omega_{1,n}(s,t)+\omega_{1,n}(t,u))^{\alpha_{1,n}/\theta}(\omega_{2,n}(s,t)+\omega_{2,n}(t,u))^{\alpha_{2,n}/\theta}\\\
\geq\omega_{1,n}(s,t)^{\alpha_{1,n}/\theta}\omega_{2,n}(s,t)^{\alpha_{2,n}/\theta}+\omega_{1,n}(t,u)^{\alpha_{1,n}/\theta}\omega_{2,n}(t,u)^{\alpha_{2,n}/\theta}.$

It follows from Lemma 4.5 applied to the control
$\sum_{n}(\omega_{1,n}+\omega_{2,n})$ that (4.2) holds. Since
$\lvert\delta\Xi_{s,t,u}\rvert\leq\omega(s,u)^{\theta}$, we can apply
Proposition 4.2. ∎

Theorem 4.6 allows us to integrate under the hypothesis (4.3), but only for
$r<2$. By the law of the iterated logarithm, paths of the Brownian motion are
not in $V^{r}$ for any $r\leq 2$.

There are two options to refine the above reasoning in order to obtain an
integration theory suitable for martingales.

  * •

The classical stochastic integration uses orthogonality between martingale
increments. We will return to that.

  * •

Rough integration theory replaces Riemann–Stieltjes sums by higher order
quadrature schemes. Higher order quadrature improves the order of convergence
(under suitable regularity assumptions). In the rough setting, it improves
from having no convergence to having convergence.

Another motivation for us to look at the rough integration theory is that it
informs us about the estimates that we will want when we return to stochastic
integration.

###  4.3 Rough paths

Throughout this section, we fix some $r\in[2,3)$.

An $r$-rough path (with values in $\mathbb{R}^{d}$) consists of

$X:[0,T]\to\mathbb{R}^{d},\quad\mathbb{X}:\Delta\to\mathbb{R}^{d\times d}$

such that $X\in V^{r}$, $\mathbb{X}\in V^{r/2}$, where

$V^{r}\mathbb{X}_{s,t}=\sup_{s\leq u_{0}\leq\dotsb\leq u_{J}\leq
t}\Bigl{(}\sum_{j}\lvert\mathbb{X}_{u_{j},u_{j+1}}\rvert^{r}\Bigr{)}^{1/r}.$

and _Chen’s relation_ holds for $s<t<u$:

$\mathbb{X}_{s,u}=\mathbb{X}_{s,t}+\mathbb{X}_{t,u}+(X_{t}-X_{s})\otimes(X_{u}-X_{t}).$
(4.5)

######  Remark.

If $X$ is sufficiently regular for the integral to be defined, one can take

$\mathbb{X}_{s,t}:=\int_{s}^{t}(X_{u-}-X_{s})\otimes\mathop{}\\!\mathrm{d}X_{u}.$

Next, we introduce the spaces in which one can state and solve differential
equations driven by rough paths.

######  Definition 4.7 (Controlled path).

Let $\mathbf{X}=(X,\mathbb{X})$ be an $r$-rough path in $\mathbb{R}^{d}$. An
_$\mathbf{X}$ -controlled $r$-rough path_ in a vector space $E$ consists of
$Y:[0,T]\to E$, $Y^{\prime}:[0,T]\to L(\mathbb{R}^{d},E)$ such that
$Y^{\prime}\in V^{r}$ and

$R^{\mathbf{Y},\mathbf{X}}_{s,t}:=\delta Y_{s,t}-Y^{\prime}_{s}\delta
X_{s,t}\in V^{r/2}.$

Although $R^{\mathbf{Y},\mathbf{X}}$ depends on all of $Y,Y^{\prime},X$, it is
commonly abbreviated to $R^{Y}$, since the other dependencies are usually
unambiguous.

It may be helpful to think about the scalar case $d=d^{\prime}=1$, and we will
use the scalar notation. All arguments will be, however, formulated in such a
way that they work in the vector-valued case, which is important e.g. because
one may wish to incorporate time as an additional coordinate in $\mathbf{X}$.

######  Example 4.8.

$(X,1)$ is an $X$-controlled path.

######  Lemma 4.9 (Rough integral).

Let $\mathbf{X}=(X,\mathbb{X})$ be an $r$-rough path and
$\mathbf{Y}=(Y,Y^{\prime})$ an $\mathbf{X}$-controlled path in
$L(\mathbb{R}^{d},\mathbb{R}^{d^{\prime}})$. Let $\Xi_{s,t}:=Y_{s}\delta
X_{s,t}+Y_{s}^{\prime}\mathbb{X}_{s,t}$. Then,

$\int_{0}^{T}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}:=\lim_{\pi}\mathcal{I}^{\pi}\Xi_{0,T}$

exists and satisfies

$\Big{\lvert}\int_{s}^{t}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}-\Xi_{s,t}\Big{\rvert}\lesssim
V^{r/2}R_{s,t-}V^{r}X_{s+,t}+V^{r}Y^{\prime}_{s,t-}V^{r/2}\mathbb{X}_{s+,t}.$

###### Proof.

$\begin{split}\delta\Xi_{s,t,u}&=Y_{s}\delta
X_{s,u}+Y_{s}^{\prime}\mathbb{X}_{s,u}-(Y_{s}\delta
X_{s,t}+Y_{s}^{\prime}\mathbb{X}_{s,t})-(Y_{t}\delta
X_{t,u}+Y_{t}^{\prime}\mathbb{X}_{t,u})\\\ &=Y_{s}\delta
X_{t,u}+Y_{s}^{\prime}(\mathbb{X}_{s,u}-\mathbb{X}_{s,t})-(Y_{t}\delta
X_{t,u}+Y_{t}^{\prime}\mathbb{X}_{t,u})\\\ &=(Y_{s}-Y_{t})\delta
X_{t,u}+Y_{s}^{\prime}(\mathbb{X}_{t,u}+\delta X_{s,t}\otimes\delta
X_{t,u})-Y_{t}^{\prime}\mathbb{X}_{t,u}\\\ &=(Y_{s}-Y_{t})\delta
X_{t,u}+(Y_{s}^{\prime}-Y_{t}^{\prime})\mathbb{X}_{t,u}+Y_{s}^{\prime}(\delta
X_{s,t}\otimes\delta X_{t,u})\\\ &=-R^{\mathbf{Y},\mathbf{X}}_{s,t}\delta
X_{t,u}+(Y_{s}^{\prime}-Y_{t}^{\prime})\mathbb{X}_{t,u}.\end{split}$ (4.6)

Hence, we can apply Theorem 4.6. ∎

######  Example 4.10.

$\int_{0}^{T}(X-X_{0},1)\otimes\mathop{}\\!\mathrm{d}\mathbf{X}=\mathbb{X}_{0,T}$.
Indeed, in this case $\delta\Xi=0$ in (4.6).

This gives even a $V^{r/3}$ approximation of the integral. However, we drop
the second order term and only use a $V^{r/2}$ approximation, since we want to
run the iteration in the simplest possible space for the given regularity.

####  4.3.1 Rough differential equations

We consider an initial value problem that is informally described by
$Y_{0}=y_{0}$, $\mathop{}\\!\mathrm{d}Y=\varphi(Y)\mathop{}\\!\mathrm{d}X$. A
rigorous formulation is the following: given a rough path $\mathbf{X}$ in
$\mathbb{R}^{d}$ and a function $\varphi:\mathbb{R}^{d^{\prime}}\to
L(\mathbb{R}^{d},\mathbb{R}^{d^{\prime}})$, we are looking for an
$\mathbf{X}$-controlled path $\mathbf{Y}$ such that

$Y_{T}=y_{0}+\int_{0}^{T}\varphi(\mathbf{Y})\mathbf{X}.$ (4.7)

We still have to define what we mean by $\varphi(\mathbf{Y})$.

We will find that this problem has a unique solution if $\varphi\in
C^{2,1}_{b}$, the space of twice differentiable functions with
$\varphi,D\varphi,D^{2}\varphi$ bounded and $D^{2}\varphi$ Lipschitz.

The proof proceeds by a fixed point argument. First, we need to identify a
metric space in which this argument will be run. We abbreviate $\lVert
X\rVert_{r}:=V^{r}X_{0,T}$.

####  4.3.2 Operations on controlled paths

We have deliberately omitted the condition $Y\in V^{r}$ from the definition of
the controlled path, because it is implicit in the other conditions.

######  Lemma 4.11 (Implicit bound).

Let $\mathbf{X}$ be a rough path and $\mathbf{Y}$ an $\mathbf{X}$-controlled
path. Then

$\lVert Y\rVert_{r}\leq\lVert Y^{\prime}\rVert_{\sup}\lVert X\rVert_{r}+\lVert
R^{Y}\rVert_{r/2}.$

######  Lemma 4.12 (Integration).

Let $\mathbf{X}$ be a rough path and $\mathbf{Y}$ an $\mathbf{X}$-controlled
path. Let $Z_{t}:=\int_{0}^{t}Y\mathop{}\\!\mathrm{d}\mathbf{X}$ and
$Z_{t}^{\prime}:=Y_{t}$. Then, $\mathbf{Z}=(Z,Z^{\prime})$ is an
$\mathbf{X}$-controlled path, and it satisfies

$\lVert R^{Z}\rVert_{r/2}\lesssim\lVert R^{Y}\rVert_{r/2}\lVert
X\rVert_{r}+\lVert Y^{\prime}\rVert_{r}\lVert\mathbb{X}\rVert_{r/2}+\lVert
Y^{\prime}\rVert_{\sup}\lVert\mathbb{X}\rVert_{r/2}.$ (4.8)

###### Proof.

$\displaystyle R^{Z}_{s,t}$ $\displaystyle=\delta Z_{s,t}-Z^{\prime}_{s}\delta
X_{s,t}$
$\displaystyle=\int_{0}^{t}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}-\int_{0}^{s}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}-Y_{s}\delta
X_{s,t}$
$\displaystyle=\int_{s}^{t}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}-Y_{s}\delta
X_{s,t}-Y^{\prime}_{s}\mathbb{X}_{s,t}+Y^{\prime}_{s}\mathbb{X}_{s,t}.$

The last term corresponds to the last term in the conclusion, and the
remaining difference is estimated by Lemma 4.9. ∎

######  Lemma 4.13 (Composition with a smooth function).

Let $\mathbf{X}$ be a rough path and $\mathbf{Y}$ an $\mathbf{X}$-controlled
path. Suppose $\varphi\in C^{1,1}_{b}$. Then
$\varphi(\mathbf{Y}):=(\varphi(Y),D\varphi(Y)Y^{\prime})$ is an
$\mathbf{X}$-controlled path, and we have

$\lVert\varphi(\mathbf{Y})^{\prime}\rVert_{r}=\lVert
D\varphi(Y)Y^{\prime}\rVert_{r}\leq\lVert D\varphi\rVert_{\sup}\lVert
Y^{\prime}\rVert_{r}+\lVert D\varphi\rVert_{Lip}\lVert Y\rVert_{r}\lVert
Y^{\prime}\rVert_{\sup},$ (4.9) $\lVert R^{\varphi(Y)}\rVert_{r/2}\leq\lVert
D\varphi\rVert_{\sup}\lVert R^{Y}\rVert_{r/2}+\frac{1}{2}\lVert
D\varphi\rVert_{Lip}\lVert Y\rVert_{r}^{2}.$ (4.10)

Interestingly, this estimate does not depend on $\mathbf{X}$.

###### Proof.

Adding and subtracting $D\varphi(Y_{t})Y^{\prime}_{s}$, we obtain

$D\varphi(Y_{t})Y^{\prime}_{t}-D\varphi(Y_{s})Y^{\prime}_{s}=D\varphi(Y_{t})(Y^{\prime}_{t}-Y^{\prime}_{s})+(D\varphi(Y_{t})-D\varphi(Y_{s}))Y^{\prime}_{s}.$
(4.11)

This implies the first estimate.

By Taylor’s formula,

$\displaystyle\lvert R^{\varphi(Y)}_{s,t}\rvert$
$\displaystyle=\lvert\varphi(Y_{t})-\varphi(Y_{s})-D\varphi(Y_{s})Y_{s}^{\prime}X_{s,t}\rvert$
$\displaystyle\leq\lvert\varphi(Y_{t})-\varphi(Y_{s})-D\varphi(Y_{s})Y_{s,t}\rvert+\lvert
D\varphi(Y_{s})R^{Y}_{s,t}\rvert$ $\displaystyle\leq\frac{1}{2}\lVert
D\varphi\rVert_{Lip}\lvert Y_{s,t}\rvert^{2}+\lVert
D\varphi\rVert_{\sup}\lvert R^{Y}_{s,t}\rvert,$

which implies second estimate. ∎

The above discussion allows us to constrct a space in which the equation
(4.23) makes sense.

######  Lemma 4.14 (Solution space).

For any $\varphi\in C^{1,1}_{b}$ and any $A\in(0,\infty)$, there exists
$\varepsilon>0$ such that if $\lVert
X\rVert_{r}+\lVert\mathbb{X}\rVert_{r/2}<\varepsilon$, then the set of
$\mathbf{X}$-controlled paths

$\mathcal{Y}=\mathcal{Y}(\mathbf{X},A):=\\{\mathbf{Y}\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lVert
Y^{\prime}\rVert_{r}\leq A,\lVert R^{Y}\rVert_{r/2}\leq A^{2},\lVert
Y\rVert_{r}\leq A/\lVert\varphi\rVert_{Lip},\lVert
Y^{\prime}\rVert_{\sup}\leq\lVert\varphi\rVert_{\sup}\\}$ (4.12)

is invariant under the mapping

$\operatorname{Step}:\mathbf{Y}\mapsto(y_{0}+\int_{0}^{\cdot}\varphi(\mathbf{Y})\mathop{}\\!\mathrm{d}\mathbf{X},\varphi(Y))$
(4.13)

for any $y_{0}\in\mathbb{R}^{d^{\prime}}$.

###### Proof.

Implicit constants in this proof are allowed to depend on the $C^{1,1}_{b}$
norm of $\varphi$.

Suppose $\mathbf{Y}\in\mathcal{Y}$. Direct estimates show

$\lVert\varphi(Y)\rVert_{r}\leq\lVert\varphi\rVert_{Lip}\lVert Y\rVert_{r}\leq
A,\quad\lVert\varphi(Y)\rVert_{\sup}\leq\lVert\varphi\rVert_{\sup}.$

By Lemma 4.13, we have

$\lVert\varphi(\mathbf{Y})^{\prime}\rVert_{r}\lesssim A,\quad\lVert
R^{\varphi(\mathbf{Y})}\rVert_{r/2}\lesssim A^{2}.$ (4.14)

By Lemma 4.12, we obtain

$\lVert
R^{\int_{0}^{\cdot}\varphi(\mathbf{Y})\mathop{}\\!\mathrm{d}\mathbf{X}}\rVert_{r/2}\lesssim\varepsilon(A^{2}+A+1).$

By Lemma 4.11, this implies

$\lVert\int_{0}^{\cdot}\varphi(\mathbf{Y})\mathop{}\\!\mathrm{d}\mathbf{X}\rVert_{r}\lesssim\varepsilon(A^{2}+A+1).$

Choosing $\varepsilon$ sufficiently small, we obtain the claim. ∎

######  Remark.

Global bounds on $\varphi$ can be replaced by growth conditions, and then
$\varepsilon$ would also depend on $y_{0}$ and these growth conditions. There
is also a “rough Gronwall lemma” for concatenating local solutions in such a
setting.

####  4.3.3 Contractive iteration

In this section, we will show that the iteration (4.13) is contractive with
respect to a suitable metric on the space (4.12). This implies existence and
uniqueness of solutions.

For the later purpose of proving continuous dependence of the solution on
data, the estimates will involve two rough paths
$\mathbf{X},\tilde{\mathbf{X}}$ and paths $\mathbf{Y},\tilde{\mathbf{Y}}$
controlled by the respective rough paths. We abbreviate

$\Delta Y=Y-\tilde{Y},\quad\Delta
Y^{\prime}=Y^{\prime}-\tilde{Y}^{\prime},\quad\Delta
R^{Y}=R^{Y}-R^{\tilde{Y}},\quad\Delta
X=X-\tilde{X},\quad\Delta\mathbb{X}=\mathbb{X}-\tilde{\mathbb{X}},\\\
\quad\Delta\varphi(Y)=\varphi(Y)-\varphi(Y^{\prime}),\quad\Delta
R^{\varphi(Y)}=R^{\varphi(Y)}-R^{\varphi(\tilde{Y})}.$

######  Lemma 4.15 (Stability of composition).

Suppose $\varphi\in C^{2,1}_{b}$. Let $\mathbf{X},\tilde{\mathbf{X}}$ be rough
paths, $\mathbf{Y}$ a $\mathbf{X}$-controlled path, and $\tilde{\mathbf{Y}}$ a
$\tilde{\mathbf{X}}$-controlled path. Then,

$\lVert\varphi(Y)-\varphi(\tilde{Y})\rVert_{r}\leq\lVert
D\varphi\rVert_{\sup}\lVert\Delta Y\rVert_{r}+\lVert
D\varphi\rVert_{Lip}\lVert\Delta Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r},$
(4.15)
$\lVert\varphi(Y)^{\prime}-\varphi(\tilde{Y})^{\prime}\rVert_{r}\leq\lVert
D\varphi\rVert_{\sup}\lVert\Delta Y^{\prime}\rVert_{r}+\lVert
D\varphi\rVert_{Lip}\lVert Y\rVert_{r}\lVert\Delta Y^{\prime}\rVert_{\sup}\\\
+\lVert D\varphi\rVert_{Lip}\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}^{\prime}\rVert_{r}+\lVert
D^{2}\varphi\rVert_{\sup}\lVert\Delta
Y^{\prime}\rVert_{r}\lVert\tilde{Y}^{\prime}\rVert_{\sup}+\lVert
D^{2}\varphi\rVert_{Lip}\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}\lVert\tilde{Y}^{\prime}\rVert_{\sup},$
(4.16) $\lVert R^{\varphi(Y)}-R^{\varphi(\tilde{Y})}\rVert_{r/2}\leq\lVert
D\varphi\rVert_{\sup}\lVert\Delta R^{Y}\rVert_{r/2}+\lVert
D\varphi\rVert_{Lip}\lVert\Delta Y\rVert_{\sup}\lVert
R^{\tilde{Y}}\rVert_{r/2}\\\ +\frac{1}{2}\lVert
D^{2}\varphi\rVert_{\sup}(\lVert
Y\rVert_{r}+\lVert\tilde{Y}\rVert_{r})\lVert\Delta
Y\rVert_{r}+\frac{1}{2}\lVert D^{2}\varphi\rVert_{Lip}\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}^{2}.$ (4.17)

Lemma 4.15 almost recovers Lemma 4.13 upon setting $\tilde{\mathbf{Y}}=0$.

###### Proof.

In order to show (4.15), we write

$(\varphi(Y_{t})-\varphi(Y_{s}))-(\varphi(\tilde{Y}_{t})-\varphi(\tilde{Y}_{s}))\\\
=\int_{0}^{1}D\varphi(Y_{s}+r(Y_{t}-Y_{s}))(Y_{t}-Y_{s})\mathop{}\\!\mathrm{d}r-\int_{0}^{1}D\varphi(\tilde{Y}_{s}+r(\tilde{Y}_{t}-\tilde{Y}_{s}))(\tilde{Y}_{t}-\tilde{Y}_{s})\mathop{}\\!\mathrm{d}r\\\
=\int_{0}^{1}\Bigl{(}D\varphi(Y_{s}+r(Y_{t}-Y_{s}))(Y_{t}-Y_{s}-(\tilde{Y}_{t}-\tilde{Y}_{s}))\\\
+(D\varphi(Y_{s}+r(Y_{t}-Y_{s}))-D\varphi(\tilde{Y}_{s}+r(\tilde{Y}_{t}-\tilde{Y}_{s})))(\tilde{Y}_{t}-\tilde{Y}_{s})\Bigr{)}\mathop{}\\!\mathrm{d}r,$
(4.18)

and estimate

$\displaystyle\lvert
D\varphi(Y_{s}+r(Y_{t}-Y_{s}))-D\varphi(\tilde{Y}_{s}+r(\tilde{Y}_{t}-\tilde{Y}_{s}))\rvert$
$\displaystyle\leq\lVert
D\varphi\rVert_{Lip}\lvert(Y_{s}+r(Y_{t}-Y_{s}))-(\tilde{Y}_{s}+r(\tilde{Y}_{t}-\tilde{Y}_{s}))\rvert$
$\displaystyle\leq\lVert D\varphi\rVert_{Lip}(r\lvert
Y_{t}-\tilde{Y}_{t}\rvert+(1-r)\lvert Y_{s}-\tilde{Y}_{s}\rvert)$
$\displaystyle\leq\lVert D\varphi\rVert_{Lip}\lVert\Delta Y\rVert_{\sup}.$

In order to show (4.16), we write

$(D\varphi(Y_{t})Y^{\prime}_{t}-D\varphi(\tilde{Y}_{t})\tilde{Y}^{\prime}_{t})-(D\varphi(Y_{s})Y^{\prime}_{s}-D\varphi(\tilde{Y}_{s})\tilde{Y}^{\prime}_{s})\\\
=D\varphi(Y_{t})((Y^{\prime}_{t}-\tilde{Y}^{\prime}_{t})-(Y^{\prime}_{s}-\tilde{Y}^{\prime}_{s}))+(D\varphi(Y_{t})-D\varphi(Y_{s}))(Y^{\prime}_{s}-\tilde{Y}^{\prime}_{s})\\\
+(D\varphi(Y_{t})-D\varphi(\tilde{Y}_{t}))(\tilde{Y}^{\prime}_{t}-\tilde{Y}^{\prime}_{s})+((D\varphi(Y_{t})-D\varphi(Y_{s}))-(D\varphi(\tilde{Y}_{t})-D\varphi(\tilde{Y}_{s})))\tilde{Y}^{\prime}_{s}.$

The first 3 terms contribute the first 3 terms to (4.16). The 4-fold
difference in the last term can be written as in (4.18) with $D\varphi$ in
place of $\varphi$, which gives the last 2 terms in (4.16).

Now we show (4.17). By inserting $D\varphi(Y_{s})Y_{s,t}$ and
$D\varphi(\tilde{Y}_{s})\tilde{Y}_{s,t},$ one has

$\begin{split}R^{\varphi(Y)}_{s,t}-R^{\varphi(\tilde{Y})}_{s,t}&=\varphi(Y)_{s,t}-D\varphi(Y_{s})Y_{s}^{\prime}X_{s,t}-(\varphi(\tilde{Y})_{s,t}-D\varphi(\tilde{Y}_{s})\tilde{Y}_{s}^{\prime}\tilde{X}_{s,t})\\\
&=\varphi(Y)_{s,t}-D\varphi(Y_{s})Y_{s,t}-(\varphi(\tilde{Y})_{s,t}-D\varphi(\tilde{Y}_{s})\tilde{Y}_{s,t})\\\
&+D\varphi(Y_{s})R^{Y}_{s,t}-D\varphi(\tilde{Y}_{s})R^{\tilde{Y}}_{s,t}\end{split}$
(4.19)

By Taylor’s formula, the first line in (5.12) equals

$\int_{0}^{1}(1-r)\bigl{(}D^{2}\varphi(Y_{s}+rY_{s,t})Y_{s,t}^{\otimes
2}-D^{2}\varphi(\tilde{Y}_{s}+r\tilde{Y}_{s,t})\tilde{Y}_{s,t}^{\otimes
2}\bigr{)}\mathop{}\\!\mathrm{d}r.$

The integrand can be written in the form

$D^{2}\varphi(Y_{s}+rY_{s,t})(Y_{s,t}^{\otimes 2}-\tilde{Y}_{s,t}^{\otimes
2})+(D^{2}\varphi(Y_{s}+rY_{s,t})-D^{2}\varphi(\tilde{Y}_{s}+r\tilde{Y}_{s,t}))\tilde{Y}_{s,t}^{\otimes
2},$

and this contributes the last line to (4.17).

The second line in (5.12) equals

$\displaystyle
D\varphi(Y_{s})R^{Y}_{s,t}-D\varphi(\tilde{Y}_{s})R^{\tilde{Y}}_{s,t}$
$\displaystyle\lesssim
D\varphi(Y_{s})(R^{Y}_{s,t}-R^{\tilde{Y}}_{s,t})+(D\varphi(Y_{s})-D\varphi(\tilde{Y}_{s}))R^{\tilde{Y}}_{s,t}$

which completes the proof. ∎

######  Lemma 4.16 (Stability of rough integration).

Let $\mathbf{X},\tilde{\mathbf{X}}$ be rough paths, $\mathbf{Y}$ a
$\mathbf{X}$-controlled path, and $\tilde{\mathbf{Y}}$ a
$\tilde{\mathbf{X}}$-controlled path. Define $\mathbf{Z},\tilde{\mathbf{Z}}$
as in Lemma 4.12. Then,

$\lVert\Delta R^{Z}\rVert_{r/2}\lesssim_{r}\lVert\Delta
Y^{\prime}\rVert_{\sup}\lVert\mathbb{X}\rVert_{r/2}+\lVert\Delta
Y^{\prime}\rVert_{r}\lVert\mathbb{X}\rVert_{r/2}+\lVert\Delta
R^{Y}\rVert_{r/2}\lVert X\rVert_{r}\\\
+\lVert\tilde{Y}^{\prime}\rVert_{\sup}\lVert\Delta\mathbb{X}\rVert_{r/2}+\lVert\tilde{Y}^{\prime}\rVert_{r}\lVert\Delta\mathbb{X}\rVert_{r/2}+\lVert
R^{\tilde{Y}}\rVert_{r/2}\lVert\Delta X\rVert_{r}.$ (4.20)

Note that (4.20) with $\tilde{\mathbf{Y}}=0$ recovers (4.8).

###### Proof.

Let
$\Xi_{s,t}:=Y_{s}X_{s,t}+Y^{\prime}_{s}\mathbb{X}_{s,t}-(\tilde{Y}_{s}\tilde{X}_{s,t}+\tilde{Y}^{\prime}_{s}\tilde{\mathbb{X}}_{s,t}).$
Then one has,

$\displaystyle\lvert
R^{I_{\mathbf{X}}(Y)}_{s,t}-R^{I_{\tilde{\mathbf{X}}}(\tilde{Y})}_{s,t}\rvert$
$\displaystyle=\lvert\int_{s}^{t}\mathbf{Y}\mathop{}\\!\mathrm{d}\mathbf{X}-Y_{s}X_{s,t}-(\int_{s}^{t}\tilde{\mathbf{Y}}\mathop{}\\!\mathrm{d}\tilde{\mathbf{X}}-\tilde{Y}_{s}\tilde{X}_{s,t})\rvert$
$\displaystyle\leq\lvert
Y^{\prime}_{s}\mathbb{X}_{s,t}-\tilde{Y}^{\prime}_{s}\tilde{\mathbb{X}}_{s,t}\rvert+\lvert
I(\Xi)_{s,t}-\Xi_{s,t}\rvert.$

We estimate the first term by

$\lvert\Delta
Y^{\prime}_{s}\mathbb{X}_{s,t}\rvert+\lvert\tilde{Y}^{\prime}_{s}\Delta\mathbb{X}_{s,t}\rvert.$

We estimate the last term by the sewing lemma (Theorem 4.6) with (as the
calculation (4.6) shows)

$\delta(\Xi)_{\tau,u,\nu}=R^{Y}_{\tau,u}\delta X_{u,\nu}+\delta
Y^{\prime}_{\tau,u}\mathbb{X}_{u,\nu}-(R^{\tilde{Y}}_{\tau,u}\delta\tilde{X}_{u,\nu}+\delta\tilde{Y}^{\prime}_{\tau,u}\tilde{\mathbb{X}}_{u,\nu}).$

and

$\lvert\delta(\Xi)_{\tau,u,\nu}\rvert\leq\lvert\Delta\delta
Y^{\prime}_{\tau,u}\mathbb{X}_{u,\nu}\rvert+\lvert\delta\tilde{Y}^{\prime}_{\tau,u}\Delta\mathbb{X}_{u,\nu}\rvert+\lvert\Delta
R^{Y}_{\tau,u}\delta X_{u,\nu}\rvert+\lvert R^{\tilde{Y}}_{\tau,u}\Delta\delta
X_{u,\nu}\rvert.$

∎

######  Lemma 4.17 (Stability of implicit bound).

$\lVert Y-\tilde{Y}\rVert_{r}\leq\lVert\Delta Y^{\prime}\rVert_{\sup}\lVert
X\rVert_{r}+\lVert\tilde{Y}^{\prime}\rVert_{\sup}\lVert\Delta
X\rVert_{r}+\lVert\Delta R^{Y}\rVert_{r/2}.$

###### Proof.

This follows from writing

$Y_{s,t}-\tilde{Y}_{s,t}=(Y^{\prime}_{s}X_{s,t}+R^{Y}_{s,t})-(\tilde{Y}^{\prime}_{s}\tilde{X}_{s,t}+R^{\tilde{Y}}_{s,t})\\\
=(Y^{\prime}_{s}-\tilde{Y}_{s})X_{s,t}+\tilde{Y}_{s}(X_{s,t}-\tilde{X}_{s,t})+(\Delta
R^{Y}_{s,t}).\qed$

######  Lemma 4.18 (Contractivity of the iteration).

For every $A<\infty$ and $\varphi\in C^{2,1}_{b}$, there esists
$\varepsilon>0$ such that, for every rough path $\mathbf{X}$ with $\lVert
X\rVert_{r}<\varepsilon$ and $\lVert\mathbb{X}\rVert_{r/2}<\varepsilon$, the
map (4.13) is strictly contractive on the set

$\\{Y\in\mathcal{Y}(\mathbf{X},A)\nonscript\>|\allowbreak\nonscript\>\mathopen{}Y_{0}=y_{0},Y^{\prime}_{0}=\varphi(y_{0})\\}$
(4.21)

with respect to the metric

$d(\mathbf{Y},\tilde{\mathbf{Y}})=\max(\lVert\Delta
R^{Y}\rVert_{r/2},\lVert\Delta Y^{\prime}\rVert_{r},2(\lVert
D\varphi\rVert_{\sup}+A\lVert D\varphi\rVert_{Lip})\lVert\Delta Y\rVert_{r}).$
(4.22)

###### Proof.

The implicit constants here are allowed to depend on $A$ in the definition of
$\mathcal{Y}$.

Suppose $\mathbf{Y},\tilde{\mathbf{Y}}\in\mathcal{Y}(\mathbf{X},A)$ with
$d(\mathbf{Y},\tilde{\mathbf{Y}})=\alpha$ and $Y_{0}=\tilde{Y}_{0}=y_{0}$.
Then,

$\lVert\Delta Y\rVert_{\sup}\leq\lVert\Delta Y\rVert_{r}\lesssim\alpha.$

By Lemma 4.15, we obtain

$\lVert\varphi(Y)-\varphi(\tilde{Y})\rVert_{r}\leq\lVert
D\varphi\rVert_{\sup}\lVert\Delta Y\rVert_{r}+\lVert
D\varphi\rVert_{Lip}\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}\leq\alpha/2.$
$\lVert\varphi(Y)^{\prime}-\varphi(\tilde{Y})^{\prime}\rVert_{r}\lesssim_{\varphi}\lVert\Delta
Y^{\prime}\rVert_{r}+\lVert Y\rVert_{r}\lVert\Delta Y^{\prime}\rVert_{\sup}\\\
+\lVert\Delta Y\rVert_{\sup}\lVert\tilde{Y}^{\prime}\rVert_{r}+\lVert\Delta
Y^{\prime}\rVert_{r}\lVert\tilde{Y}^{\prime}\rVert_{\sup}+\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}\lVert\tilde{Y}^{\prime}\rVert_{\sup}\lesssim\alpha(1+A),$
$\lVert R^{\varphi(Y)}-R^{\varphi(\tilde{Y})}\rVert_{r/2}\lesssim\lVert\Delta
R^{Y}\rVert_{r/2}+\lVert\Delta Y\rVert_{\sup}\lVert
R^{\tilde{Y}}\rVert_{r/2}\\\ +(\lVert
Y\rVert_{r}+\lVert\tilde{Y}\rVert_{r})\lVert\Delta Y\rVert_{r}+\lVert\Delta
Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}^{2}\lesssim\alpha(1+A^{2}).$

Let $\mathbf{Z}=\operatorname{Step}(\mathbf{Y})$,
$\tilde{\mathbf{Z}}=\operatorname{Step}(\tilde{\mathbf{Y}})$. Then

$\lVert\Delta
Z^{\prime}\rVert_{r}=\lVert\varphi(Y)-\varphi(\tilde{Y})\rVert_{r}\leq\alpha/2.$

By Lemma 4.16,

$\lVert\Delta
R^{Z}\rVert_{r/2}\lesssim\varepsilon(\lVert\Delta\varphi(Y)^{\prime}\rVert_{\sup}+\lVert\Delta\varphi(Y)^{\prime}\rVert_{r}+\lVert\Delta
R^{\varphi(Y)}\rVert_{r/2})\lesssim\alpha\varepsilon.$

By Lemma 4.17, this implies

$\lVert\Delta Z\rVert_{r}\lesssim\lVert\Delta Z^{\prime}\rVert_{\sup}\lVert
X\rVert_{r}+\lVert\Delta R^{Z}\rVert_{r/2}\leq\lVert\Delta
Z^{\prime}\rVert_{r}\varepsilon+C\varepsilon\alpha\lesssim\varepsilon\alpha.$

Choosing $\varepsilon$ small enough, we obtain the claim. ∎

######  Theorem 4.19 (Existence of solutions).

For every $\varphi\in C^{2,1}_{b}$, there exist $A<\infty$ and $\varepsilon>0$
such that, for every rough path $\mathbf{X}$ with $\lVert
X\rVert_{r}<\varepsilon$ and $\lVert\mathbb{X}\rVert_{r/2}<\varepsilon$ and
every $y_{0}$, there exists a unique $\mathbf{X}$-controlled path
$\mathbf{Y}\in\mathcal{Y}(\mathbf{X},A)$ such that, for every $t\in[0,T]$, we
have

$\mathbf{Y}_{t}=y_{0}+\int_{0}^{t}\varphi(\mathbf{Y})\mathop{}\\!\mathrm{d}\mathbf{X},\quad
Y^{\prime}_{t}=\varphi(Y_{t}).$ (4.23)

###### Proof.

Let $A,\varepsilon$ be as in Lemma 4.18. Define a sequence of controlled paths
by

$(\mathbf{Y}_{0})_{t}=y_{0},\quad(\mathbf{Y}_{0})^{\prime}_{t}=0,\quad(\mathbf{Y}_{j+1})=\operatorname{Step}(\mathbf{Y}_{j}).$

For $j\geq 1$, the paths $(\mathbf{Y}_{j})$ are elements of (4.21). It follows
from Lemma 4.18 that this sequence is Cauchy with respect to the metric
(4.22), and its limit solves the RDE (4.23). Uniqueness of the solution
follows from strict contractivity of (4.13). ∎

One can also show that the solution $\mathbf{Y}$ in Theorem 4.19 is unique
among all $\mathbf{X}$-controlled paths (not only those in
$\mathcal{Y}(\mathbf{X},A)$). To this end, we note that, given any two
$\mathbf{X}$-controlled solutions, they are in $\mathcal{Y}(\mathbf{X},A)$ for
some large $A$. One then has to subdivide the time interval into smaller
intervals, on each of which $\mathbf{X}$ has sufficiently small variation norm
to apply Lemma 4.18. Large jumps of $\mathbf{X}$ have to be handled
separately. We omit the tedious details.

####  4.3.4 Lipschitz dependence on data

######  Lemma 4.20 (Local Lipschitz dependence of RDE solution on the data).

For every $\varphi\in C^{2,1}_{b}$ and every $A<\infty$, there exists
$\varepsilon>0$ such that if $\mathbf{X},\tilde{\mathbf{X}}$ are rough paths
with norms $<\varepsilon$, $\mathbf{Y}\in\mathcal{Y}(\mathbf{X},A)$ is a
solution of (4.23) with initial datum $y_{0}$, and
$\tilde{\mathbf{Y}}\in\mathcal{Y}(\tilde{\mathbf{X}},A)$ is a solution of
(4.23) with initial datum $\tilde{y}_{0}$ such that $\lvert
y_{0}-\tilde{y}_{0}\rvert\leq\varepsilon$ and rough path $\tilde{\mathbf{X}}$,
then

$\max(\lVert\Delta R^{Y}\rVert_{r/2},\lVert\Delta
Y^{\prime}\rVert_{r},\lVert\Delta
Y\rVert_{r})\lesssim_{r,\varphi,A}\max(\lVert
X-\tilde{X}\rVert_{r},\lVert\mathbb{X}-\tilde{\mathbb{X}}\rVert_{r/2},\lvert
y_{0}-\tilde{y}_{0}\rvert).$

###### Proof.

Let $\beta$ be the RHS of the conclusion and

$\alpha:=\max(\lVert\Delta R^{Y}\rVert_{r/2},\lVert\Delta
Y^{\prime}\rVert_{r},2(\lVert D\varphi\rVert_{\sup}+A\lVert
D\varphi\rVert_{Lip}+1)\lVert\Delta Y\rVert_{r}).$

We may assume $\alpha>\beta$, since otherwise the conclusion already holds. In
this case, we have

$\lVert\Delta Y\rVert_{\sup}\leq\lvert y_{0}-\tilde{y}_{0}\rvert+\lVert\Delta
Y\rVert_{r}\lesssim\alpha,\quad\lVert\Delta
Y^{\prime}\rVert_{\sup}\leq\lVert\varphi\rVert_{Lip}\lVert\Delta
Y\rVert_{\sup}\lesssim\alpha.$

By Lemma 4.15, we obtain

$\lVert\varphi(Y)^{\prime}-\varphi(\tilde{Y})^{\prime}\rVert_{r}\lesssim\alpha(1+A),\quad\lVert\Delta
R^{\varphi}\rVert_{r/2}\lesssim\alpha(1+A^{2}),$

A simple bound is

$\lVert\varphi(\tilde{Y})\rVert_{r}\leq\lVert\varphi\rVert_{Lip}\lVert\tilde{Y}\rVert_{r}\leq
A.$

As in (5.15), we have

$\lVert R^{\varphi(\tilde{\mathbf{Y}})}\rVert_{r/2}\lesssim A^{2}.$

Inserting all these bounds into Lemma 4.16, we obtain

$\lVert\Delta
R^{Y}\rVert_{r/2}\lesssim\varepsilon(\lVert\Delta\varphi(Y)^{\prime}\rVert_{\sup}+\lVert\Delta\varphi(Y)^{\prime}\rVert_{r}+\lVert\Delta
R^{\varphi(Y)}\rVert_{r/2})\\\
+\beta(\lVert\varphi(\tilde{Y})^{\prime}\rVert_{\sup}+\lVert\varphi(\tilde{Y})\rVert_{r}+\lVert
R^{\varphi(\tilde{Y})}\rVert_{r/2})\\\
\lesssim\varepsilon\alpha(1+A^{2})+\beta(1+A^{2}).$

By Lemma 4.15, we have

$\lVert\Delta
Y^{\prime}\rVert_{r}=\lVert\varphi(Y)-\varphi(\tilde{Y})\rVert_{r}\leq\lVert
D\varphi\rVert_{\sup}\lVert\Delta Y\rVert_{r}+\lVert
D\varphi\rVert_{Lip}\lVert\Delta Y\rVert_{\sup}\lVert\tilde{Y}\rVert_{r}\\\
\leq(\lVert D\varphi\rVert_{\sup}+A\lVert D\varphi\rVert_{Lip})\lVert\Delta
Y\rVert_{r}+A\lVert D\varphi\rVert_{Lip}\lvert
y_{0}-\tilde{y}_{0}\rvert\leq\alpha/2+C\beta.$

Moreover, by Lemma 4.17, we have

$\displaystyle\lVert\Delta Y\rVert_{r}$ $\displaystyle\leq\lVert\Delta
Y^{\prime}\rVert_{\sup}\lVert
X\rVert_{r}+\lVert\tilde{Y}^{\prime}\rVert_{\sup}\lVert\Delta
X\rVert_{r}+\lVert\Delta R^{Y}\rVert_{r/2}.$ $\displaystyle\leq
C\alpha\varepsilon+A\beta+C(\varepsilon\alpha+\beta).$

Inserting these bounds into the definition of $\alpha$, we obtain

$\alpha\leq\max(C(\varepsilon\alpha+\beta),\alpha/2+C\beta,C(\varepsilon\alpha+\beta)),$

where $C$ depends on $r,\varphi,A$. If $C\varepsilon\leq 1/2$, then this
implies $\alpha\leq C\beta$. ∎

######  Remark.

Rough paths were introduced in [Lyo98] and controlled paths in [Gub04]. For a
long time, the theory concentrated on Hölder continuous paths; a good
exposition of this case is in the book [FH20]. The treatment of $V^{r}$ paths
is adapted from [FZ18].

##  5 Martingale transforms

The main result of this section, Theorem 5.4, is a bound for discrete time
versions of the Itô integral.

We denote $\ell^{p}$ norms by

$\ell^{p}_{k}a_{k}:=(\sum_{k\in\mathbb{N}}\lvert a_{k}\rvert^{p})^{1/p}.$

In order to simplify notation, we only consider martingales $g$ with
$g_{0}=0$.

###  5.1 Davis decomposition

We will use the following $L^{q}$ bound for the Davis decomposition
(constructed in Lemma 1.9).

######  Lemma 5.1 (Davis decomposition in $L^{q}$).

For every martingale $(f_{n})$ with values in a Banach space $X$, there exists
a decomposition $f=f^{\mathrm{pred}}+f^{\mathrm{bv}}$ as a sum of two
martingales adapted to the same filtration with $f^{\mathrm{pred}}_{0}=0$ such
that the differences of $f^{\mathrm{pred}}$ have predictable majorants:

$Xdf^{\mathrm{pred}}_{n}\leq 2MXdf_{n-1}$ (5.1)

and $f^{\mathrm{bv}}$ has bounded variation, in an integral sense for every
$q\in[1,\infty)$:

$L^{q}\sum_{k\leq n}Xdf^{\mathrm{bv}}_{k}\leq(q+1)L^{q}MXdf_{n}.$ (5.2)

###### Proof.

Recall from the construction in Lemma 1.9 that

$df^{\mathrm{bv}}_{n}=d\tilde{h}_{n}-\mathbb{E}_{n-1}(d\tilde{h}_{n}),$

where $\tilde{h}_{n}$ is a process such that

$Xd\tilde{h}_{n}=MXdf_{n}-MXdf_{n-1}.$

Therefore,

$L^{q}\sum_{k\leq n}Xdf^{\mathrm{bv}}_{k}\leq L^{q}\sum_{k\leq
n}Xd\tilde{h}_{k}+L^{q}\sum_{k\leq n}\mathbb{E}_{k-1}Xd\tilde{h}_{k}.$

Using Lemma 1.15 in the second summand, we obtain the claim. ∎

######  Lemma 5.2.

Let $1\leq q<\infty$, $X$ be a Banach function space, elements of which are
$\mathbb{R}$-valued maps $x(\cdot)$, and $(f_{n})$ a martingale with values in
$X$. Then for $f^{\mathrm{pred}}$ given by Lemma 5.1 we have

$\lVert\lVert Sf^{\mathrm{pred}}\rVert_{X}\rVert_{L^{q}}\leq(q+2)\lVert\lVert
Sf\rVert_{X}\rVert_{L^{q}},$

where the square function is given by

$\lVert Sf\rVert_{X}:=\lVert\ell^{2}_{n}(df_{n}(\cdot))\rVert_{X}$

######  Remark.

We will apply this with $X=\ell^{r}$, i.e. $r$-summable series, viewed as maps
from $\mathbb{N}\to\mathbb{R}$, with the usual Banach structure.

###### Proof.

Using (5.2) we estimate

$\displaystyle\lVert\lVert Sf^{\mathrm{pred}}\rVert_{X}\rVert_{L^{q}}$
$\displaystyle\leq\lVert\lVert Sf\rVert_{X}\rVert_{L^{q}}+\lVert\lVert
Sf^{\mathrm{bv}}\rVert_{X}\rVert_{L^{q}}$ $\displaystyle\leq\lVert\lVert
Sf\rVert_{X}\rVert_{L^{q}}+\lVert\lVert\sum_{n}\lvert
d_{n}f^{\mathrm{bv}}\rvert\rVert_{X}\rVert_{L^{q}}$
$\displaystyle\leq\lVert\lVert Sf\rVert_{X}\rVert_{L^{q}}+\lVert\sum_{n}\lVert
d_{n}f^{\mathrm{bv}}\rVert_{X}\rVert_{L^{q}}$ $\displaystyle\leq\lVert\lVert
Sf\rVert_{X}\rVert_{L^{q}}+(q+1)\lVert\sup_{n}\lVert
d_{n}f\rVert_{X}\rVert_{L^{q}}$ $\displaystyle\lesssim\lVert\lVert
Sf\rVert_{X}\rVert_{L^{q}}.\qed$

###  5.2 Vector-valued maximal paraproduct estimate

We call a process $(F_{s,t})_{s\leq t}$ depending on two time variables
_adapted_ if $F_{s,t}$ is $\mathcal{F}_{t}$-measurable for every $s\leq t$.

For an adapted process $(F_{s,t})$ and a martingale $(g_{n})$, we define

$\Pi(F,g)_{s,t}:=\sum_{s<j\leq t}F_{s,j-1}dg_{j}$ (5.3)

Note that $\Pi(F,g)_{s,\cdot}$ only depends on $(F_{s,\cdot})$.

######  Proposition 5.3.

Let $0<q,q_{1}\leq\infty$, $1\leq q_{0},r,r_{0}<\infty$, $1\leq
r_{1}\leq\infty$. Assume $1/q=1/q_{0}+1/q_{1}$ and $1/r=1/r_{0}+1/r_{1}$.
Then, for any martingales $(g^{(k)}_{n})_{n}$, any adapted sequences
$(F^{(k)}_{s,t})_{s\leq t}$, and any stopping times
$\tau_{k}^{\prime}\leq\tau_{k}$ with $k\in\mathbb{Z}$, we have

$\big{\lVert}\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k)})_{\tau_{k}^{\prime},t}\rvert\big{\rVert}_{q}\leq
C_{q_{0},q_{1},r_{0},r_{1}}\big{\lVert}\ell^{r_{1}}_{k}\sup_{\tau_{k}^{\prime}\leq
t<\tau_{k}}\lvert
F^{(k)}_{\tau_{k}^{\prime},t}\rvert\big{\rVert}_{q_{1}}\lVert\ell^{r_{0}}_{k}Sg^{(k)}_{\tau_{k}^{\prime},\tau_{k}}\rVert_{q_{0}},$
(5.4)

where $Sg_{s,t}:=(\delta(Sg)^{2})^{1/2}_{s,t}=\bigl{(}\sum_{j=s+1}^{t}\lvert
dg_{j}\rvert^{2}\bigr{)}^{1/2}$.

###### Proof of Proposition 5.3.

We may replace each $g^{(k)}$ by the martingale

$\tilde{g}^{(k)}_{n}:=g^{(k)}_{n\wedge\tau_{k}}-g^{(k)}_{n\wedge\tau_{k}^{\prime}}$
(5.5)

without changing the value of either side of (5.4).

Consider first $q\geq 1$. For each $k$, the sequence

$h^{(k)}_{t}:=\begin{cases}0,&t<\tau_{k}^{\prime},\\\
\Pi(F^{(k)},g^{(k)})_{\tau_{k}^{\prime},t},&t\geq\tau_{k}^{\prime},\end{cases}$

is a martingale. We may also assume $F_{\tau_{k}^{\prime},t}=0$ if
$t\not\in[\tau_{k}^{\prime},\tau_{k})$. By the $\ell^{r}$ valued BDG
inequality (Corollary 3.2), we can estimate

$\displaystyle LHS\leavevmode\nobreak\ \eqref{eq:vv-pprod}$
$\displaystyle\lesssim\big{\lVert}\ell^{r}_{k}\lvert
Sh^{(k)}\rvert\big{\rVert}_{q}$
$\displaystyle=\big{\lVert}\ell^{r}_{k}\ell^{2}_{j}\lvert
F_{\tau_{k}^{\prime},j-1}^{(k)}dg^{(k)}_{j}\rvert\big{\rVert}_{q}$
$\displaystyle\leq\big{\lVert}\ell^{r}_{k}MF^{(k)}\ell^{2}_{j}\lvert
dg^{(k)}_{j}\rvert\big{\rVert}_{q}$
$\displaystyle\leq\lVert\ell^{r_{1}}_{k}MF^{(k)}\rVert_{q_{1}}\big{\lVert}\ell^{r_{0}}_{k}Sg^{(k)}\big{\rVert}_{q_{0}}.$

Here and later, we abbreviate $MF^{(k}:=\sup_{j}\lvert
F_{\tau_{k}^{\prime},j}^{(k}\rvert$.

Consider now $q<1$. By homogeneity, we may assume

$\big{\lVert}\ell^{r_{1}}_{k}MF^{(k)}\big{\rVert}_{q_{1}}=\big{\lVert}\ell^{r_{0}}_{k}Sg^{(k)}\big{\rVert}_{q_{0}}=1,$
(5.6)

and we have to show

$\big{\lVert}\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k)})_{\tau_{k}^{\prime},t}\rvert\big{\rVert}_{q}\lesssim
1.$

We use the Davis decomposition $g=g^{\mathrm{pred}}+g^{\mathrm{bv}}$ (Lemma
5.1 with $X=\ell^{r_{0}}$). The contribution of the bounded variation part is
estimated as follows:

$\displaystyle\lVert\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k),\mathrm{bv}})_{\tau_{k}^{\prime},t}\rvert\rVert_{q}$
$\displaystyle\leq\lVert\ell^{r}_{k}\sum_{j}\lvert
F^{(k)}_{\tau_{k}^{\prime},j-1}\rvert\cdot\lvert
dg^{(k),\mathrm{bv}}_{j}\rvert\rVert_{q}$
$\displaystyle\leq\lVert\ell^{r_{1}}_{k}MF^{(k)}\rVert_{q_{1}}\lVert\ell^{r_{0}}_{k}\Bigl{(}\sum_{j}\lvert
dg^{(k),\mathrm{bv}}_{j}\rvert\Bigr{)}\rVert_{q_{0}}$
$\displaystyle\leq\lVert\ell^{r_{1}}_{k}MF^{(k)}\rVert_{q_{1}}\lVert\sum_{j}\ell^{r_{0}}_{k}\lvert
dg^{(k),\mathrm{bv}}_{j}\rvert\rVert_{q_{0}}$
$\displaystyle\lesssim\lVert\ell^{r_{1}}_{k}MF^{(k)}\rVert_{q_{1}}\lVert\sup_{j}\ell^{r_{0}}_{k}\lvert
dg^{(k)}_{j}\rvert\rVert_{q_{0}}$
$\displaystyle\leq\lVert\ell^{r_{1}}_{k}MF^{(k)}\rVert_{q_{1}}\lVert\ell^{r_{0}}_{k}Sg^{(k)}\rVert_{q_{0}},$

where we used (5.2) in the penultimate step.

It remains to consider the part $g^{\mathrm{pred}}$ with predictable bounds
for jumps. We use the layer cake formula in the form

$\int
f^{q}=\int_{0}^{\infty}\mathbb{P}\\{f^{q}>\lambda\\}\mathop{}\\!\mathrm{d}\lambda=\int_{0}^{\infty}\mathbb{P}\\{f>\lambda^{1/q}\\}\mathop{}\\!\mathrm{d}\lambda.$

By the layer cake formula, we have

$\big{\lVert}\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k),\mathrm{pred}})_{\tau_{k}^{\prime},t}\rvert\big{\rVert}_{q}^{q}\\\
=\int_{0}^{\infty}\mathbb{P}\\{\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k),\mathrm{pred}})_{\tau_{k}^{\prime},t}\rvert>\lambda^{1/q}\\}\mathop{}\\!\mathrm{d}\lambda.$
(5.7)

Fix some $\lambda>0$ and define a stopping time

$\tau:=\inf\Big{\\{}t\nonscript\>\Big{|}\allowbreak\nonscript\>\mathopen{}M(\ell^{r_{0}}_{k}dg^{(k)})_{t}>c\lambda^{1/q_{0}}\text{
or }\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}}_{t}>c\lambda^{1/q_{0}}\text{ or
}\ell^{r_{1}}_{k}\sup_{0<j\leq t}\lvert
F_{\tau_{k}^{\prime},j}^{(k)}\rvert>\lambda^{1/q_{1}}\Big{\\}}.$ (5.8)

Define stopped martingales
$\tilde{g}^{(k)}_{t}:=g^{(k),\mathrm{pred}}_{t\wedge\tau}$ and adapted
processes

$\tilde{F}^{(k)}_{t^{\prime},t}:=F^{(k)}_{t^{\prime},t\wedge\tau-1}.$

Then, on the set $\\{\tau=\infty\\}$, we have

$\Pi(F^{(k)},g^{(k),\mathrm{pred}})_{\tau_{k}^{\prime},t}=\Pi(\tilde{F}^{(k)},\tilde{g}^{(k)})_{\tau_{k}^{\prime},t}\quad\text{for
all }k,t.$

Hence,

$\begin{split}&\\{\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(F^{(k)},g^{(k),\mathrm{pred}})_{\tau_{k}^{\prime},t}\rvert>\lambda^{1/q}\\}\\\
\subset&\\{\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(\tilde{F}^{(k)},\tilde{g}^{(k)})_{\tau_{k}^{\prime},t}\rvert>\lambda^{1/q}\\}\\\
&\cup\\{\ell^{r_{0}}_{k}Sg^{(k)}>\lambda^{1/q_{0}}\\}\cup\\{\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}}>\lambda^{1/q_{0}}\\}\\\
&\cup\\{\ell^{r_{1}}_{k}MF^{(k)}>\lambda^{1/q_{1}}\\}\end{split}$ (5.9)

The contributions of the latter three terms to (5.7) are $\lesssim 1$ by (5.6)
and Lemma 5.2. It remains to handle the first term.

By construction, we have
$\ell^{r_{1}}_{k}M\tilde{F}^{(k)}\leq\lambda^{1/q_{1}}$, and due to (5.1) we
also have $\ell^{r_{0}}_{k}S\tilde{g}^{(k)}\leq\lambda^{1/q_{0}}$, provided
that the absolute constant $c$ in (5.8) is small enough. Choose an arbitrary
exponent $\tilde{q}$ with $q_{0}<\tilde{q}<\infty$. By the already known case
of the Proposition with $(q_{0},q_{1})$ replaced by $(\tilde{q},\infty)$, we
obtain

$\begin{split}&\mathbb{P}\\{\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(\tilde{F}^{(k)},\tilde{g}^{(k)})_{\tau_{k}^{\prime},t}\rvert>\lambda^{1/q}\\}\\\
&\leq\lambda^{-\tilde{q}/q}\lVert\ell^{r}_{k}\sup_{\tau_{k}^{\prime}\leq
t\leq\tau_{k}}\lvert\Pi(\tilde{F}^{(k)},\tilde{g}^{(k)})_{\tau_{k}^{\prime},t}\rvert\rVert_{\tilde{q}}^{\tilde{q}}\\\
&\lesssim_{\tilde{q}}\lambda^{-\tilde{q}/q}\lVert\ell^{r_{1}}_{k}M\tilde{F}^{(k)}\rVert_{\infty}^{\tilde{q}}\lVert\ell^{r_{0}}_{k}S\tilde{g}^{(k)}\rVert_{\tilde{q}}^{\tilde{q}}\\\
&\leq\lambda^{-\tilde{q}/q_{0}}\lVert\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}}\wedge\lambda^{1/q_{0}}\rVert_{\tilde{q}}^{\tilde{q}}.\end{split}$
(5.10)

This estimate no longer depends on the stopping time $\tau$. Integrating the
right-hand side of (5.10) in $\lambda$, we obtain

$\displaystyle\int_{0}^{\infty}\lambda^{-\tilde{q}/q_{0}}\lVert\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}}\wedge\lambda^{1/q_{0}}\rVert_{\tilde{q}}^{\tilde{q}}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle=\mathbb{E}\int_{0}^{\infty}\bigl{(}\lambda^{-\tilde{q}/q_{0}}(\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}})^{\tilde{q}}\wedge
1\bigr{)}\mathop{}\\!\mathrm{d}\lambda$
$\displaystyle\sim\mathbb{E}(\ell^{r_{0}}_{k}Sg^{(k),\mathrm{pred}})^{q_{0}}$
$\displaystyle\sim 1,$

where we used $\tilde{q}>q_{0}$, Lemma 5.2 with $X=\ell^{r_{0}}$, and the
assumption (5.6). ∎

We will soon need a version of Proposition 5.3 with a supremum in both time
arguments of the paraproduct. Nice estimates of such form are only available
under some structural assumptions on $F$. In this course, we only consider
$F=\delta f$.

######  Theorem 5.4.

Let $q,q_{0},q_{1},r,r_{1}$ be as in Proposition 5.3 with $r_{0}=2$, that is,

$0<q,q_{1}\leq\infty,1\leq q_{0},r<\infty,1\leq
r_{1}\leq\infty,1/q=1/q_{0}+1/q_{1},1/r=1/2+1/r_{1}.$

Let $f$ be an adapted process, $g$ a martingale, and $\tau$ an adapted
partition. Then, we have

$\big{\lVert}\ell^{r}_{k}\sup_{\tau_{k-1}\leq s\leq
t\leq\tau_{k}}\lvert\Pi(\delta
f,g)_{s,t}\rvert\big{\rVert}_{q}\lesssim\big{\lVert}\ell^{r_{1}}_{k}\sup_{\tau_{k-1}\leq
t<\tau_{k}}\lvert\delta f_{\tau_{k-1},t}\rvert\big{\rVert}_{q_{1}}\lVert
Sg\rVert_{q_{0}}.$ (5.11)

###### Proof.

For any $s\leq t\leq u$, the sums (5.3) satisfy the relation

$\displaystyle\delta\Pi(F,g)_{s,t,u}$
$\displaystyle=\Pi_{s,u}(F,g)-\Pi_{s,t}(F,g)-\Pi_{t,u}(F,g)$
$\displaystyle=\sum_{s<j\leq u}F_{s,j-1}dg_{j}-\sum_{s<j\leq
t}F_{s,j-1}dg_{j}-\sum_{t<j\leq u}F_{t,j-1}dg_{j}$
$\displaystyle=\sum_{t<j\leq u}(F_{s,j-1}-F_{t,j-1})dg_{j}.$

In case of $F=\delta f$, the right-hand side becomes

$\sum_{t<j\leq u}(f_{t}-f_{s})dg_{j}=(f_{t}-f_{s})(g_{u}-g_{t}).$

Therefore, for any $\tau_{k}^{\prime}\leq s\leq t$, we can estimate

$\lvert\Pi_{s,t}(\delta f,g)\rvert\leq\lvert\Pi_{\tau_{k}^{\prime},t}(\delta
f,g)\rvert+\lvert\Pi_{\tau_{k}^{\prime},s}(\delta f,g)\rvert+\lvert\delta
f_{\tau_{k}^{\prime},s}\delta g_{s,t}\rvert.$

In the first two terms, we apply Proposition 5.3 with
$\tau_{k}^{\prime}=\tau_{k-1}$, $f^{(k)}=f$, $g^{(k)}=g$ for each $k$. In the
last term, by Hölder’s inequality, we have

$\big{\lVert}\ell^{r}_{k}\sup_{\tau_{k-1}\leq s<t\leq\tau_{k}}\lvert\delta
f_{\tau_{k}^{\prime},s}\delta
g_{s,t}\rvert\big{\rVert}_{q}\leq\big{\lVert}\ell^{r_{1}}_{k}\sup_{\tau_{k-1}\leq
s<t\leq\tau_{k}}\lvert\delta
f_{\tau_{k}^{\prime},s}\rvert\big{\rVert}_{q_{1}}\big{\lVert}\ell^{2}_{k}\sup_{\tau_{k-1}\leq
s<t\leq\tau_{k}}\lvert\delta g_{s,t}\rvert\big{\rVert}_{q_{0}}.$

In the former norm, we observe that the dependence on $t$ disappears. In the
latter norm, we use the $\ell^{2}$ valued BDG inequality (Corollary 3.2) with
the martingales $h^{(k)}=\prescript{\tau_{k-1}}{}{\\!}g^{\tau_{k}}$. ∎

###  5.3 Stopping time construction

In this section, we estimate the $r$-variation of a two-parameter function by
square function-like objects, like we did this for one-parameter functions in
the proof of Lépingle’s inequality.

For an adapted process $(\Pi_{s,t})_{s\leq t}$, let

$\Pi^{*}_{n^{\prime\prime}}:=\sup_{0\leq n<n^{\prime}\leq
n^{\prime\prime}}\lvert\Pi_{n,n^{\prime}}\rvert,\quad\Pi^{*}:=\Pi^{*}_{\infty}.$

######  Lemma 5.5.

For any discrete time adapted process $(\Pi_{s,t})_{s<t}$, there exist adapted
partitions $\tau^{(m)}_{j}$ such that, for every $0<\rho<r<\infty$, we have

$\sup_{\begin{subarray}{c}l_{\max},\\\
u_{0}<\dotsb<u_{l_{\max}}\end{subarray}}\sum_{l=1}^{l_{\max}}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{r}\leq\frac{(\Pi^{*})^{r}}{1-2^{-r}}+2^{\rho}\sum_{m=0}^{\infty}(2^{-m}\Pi^{*})^{r-\rho}\sum_{j=1}^{\infty}\Bigl{(}\sup_{\tau^{(m)}_{j-1}\leq
t<\tau^{(m)}_{j}}\lvert\Pi_{t,\tau^{(m)}_{j}}\rvert\Bigr{)}^{\rho}.$ (5.12)

###### Proof of Lemma 5.5.

For $m\in\mathbb{N}$, define stopping times

$\tau^{(m)}_{0}:=0,$

and then, for $j\geq 0$, allowing values in $\mathbb{N}\cup\\{\infty\\}$,

$\tau^{(m)}_{j+1}:=\inf\Big{\\{}t>\tau^{(m)}_{j}\nonscript\>\Big{|}\allowbreak\nonscript\>\mathopen{}\sup_{\tau^{(m)}_{j}\leq
t^{\prime}<t}\lvert\Pi_{t^{\prime},t}\rvert>2^{-m-1}\Pi^{*}_{t}\Big{\\}}.$
(5.13)

Fix $\omega\in\Omega$ and let $(u_{l})_{l=0}^{l_{\max}}$ be a finite strictly
increasing sequence. Consider $0<\rho<r<\infty$ and split

$\sum_{l=1}^{l_{\max}}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{r}=\sum_{m=0}^{\infty}\sum_{l\in
L(m)}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{r},$ (5.14)

where

$L(m):=\big{\\{}l\in\\{1,\dotsc,l_{\max}\\}\nonscript\>|\allowbreak\nonscript\>\mathopen{}2^{-m-1}\Pi^{*}_{u_{l}}<\lvert\Pi_{u_{l-1},u_{l}}\rvert\leq
2^{-m}\Pi^{*}_{u_{l}}\\}.$ (5.15)

In (5.14), we only omitted vanishing summands, since
$|\Pi_{u_{l-1},u_{l}}|\leq\Pi^{*}_{u_{l}}$. Let also
$L^{\prime}(m):=L(m)\setminus\\{\sup L(m)\\}$. Using (5.15), we obtain

$\sum_{l=1}^{l_{\max}}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{r}\leq\sum_{m=0}^{\infty}(2^{-m}\Pi^{*})^{r-\rho}\sum_{l\in
L^{\prime}(m)}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{\rho}+\sum_{m=0}^{\infty}(2^{-m}\Pi^{*})^{r}.$
(5.16)

######  Claim.

For every $l\in L(m)$, there exists $j$ s.t.
$\tau_{j}^{(m)}\in(u_{l-1},u_{l}]$.

###### Proof of the claim.

Let $j$ be maximal with $\tau^{(m)}_{j}\leq u_{l-1}$. Since $l\in L(m)$, by
definition (5.15), we have

$\lvert\Pi_{u_{l-1},u_{l}}\rvert>2^{-m-1}\Pi^{*}_{u_{l}}.$

By the definition of stopping times (5.13), we obtain $\tau^{(m)}_{j+1}\leq
u_{l}$. ∎

Fix $m$. For each $l\in L^{\prime}(m)$, let $j(l)$ be the largest $j$ such
that $\tau^{(m)}_{j}\in(u_{l-1},u_{l}]$. Then all $j(l)$ are distinct, and,
since $l\neq\max L(m)$, the claim shows that $\tau^{(m)}_{j(l)+1}<\infty$.
Furthermore, by (5.15), the monotonicity of $t\mapsto\Pi^{*}_{t}$, and the
fact that the infimum in the definition (5.13) of stopping times is in fact a
minimum unless it is infinite, we have

$\lvert\Pi_{u_{l-1},u_{l}}\rvert\leq 2^{-m}\Pi^{*}_{u_{l}}\leq
2^{-m}\Pi^{*}_{\tau^{(m)}_{j(l)+1}}\leq 2\sup_{\tau^{(m)}_{j(l)}\leq
t^{\prime}<\tau^{(m)}_{j(l)+1}}\lvert\Pi_{t^{\prime},\tau^{(m)}_{j(l)+1}}\rvert.$
(5.17)

Since all $j(l)$ are distinct, this implies

$\sum_{l\in L^{\prime}(m)}\lvert\Pi_{u_{l-1},u_{l}}\rvert^{\rho}\leq
2^{\rho}\sum_{j=1}^{\infty}\sup_{\tau^{(m)}_{j-1}\leq
t^{\prime}<\tau^{(m)}_{j}}\lvert\Pi_{t^{\prime},\tau^{(m)}_{j}}\rvert^{\rho}.$

Substituting this into (7.5), we conclude the proof of Lemma 5.5. ∎

######  Corollary 5.6.

Let $(\Pi_{s,t})_{s\leq t}$ be an adapted process with $\Pi_{t,t}=0$ for all
$t$. Then, for every $0<\rho<r<\infty$ and $q\in(0,\infty]$, we have

$\lVert
V^{r}\Pi\rVert_{L^{q}}\lesssim\sup_{\tau}\Big{\lVert}\Bigl{(}\sum_{j=1}^{\infty}\bigl{(}\sup_{\tau_{j-1}\leq
t<t^{\prime}\leq\tau_{j}}\lvert\Pi_{t,t^{\prime}}\rvert\bigr{)}^{\rho}\Bigr{)}^{1/\rho}\Big{\rVert}_{L^{q}},$
(5.18)

where the supremum is taken over all adapted partitions $\tau$.

###### Proof.

By the monotone convergence theorem, we can restrict the times in the
definition of $V^{r}$ to a finite set, and then apply Lemma 5.5.

The term $\Pi^{*}$ is of the form on the right-hand side of (5.18) with
$\tau_{1}=\infty$. Therefore, the claim follows from the triangle inequality
in $L^{q}$ (if $q\geq 1$), $q$-convexity of $L^{q}$ (if $q<1$), and Hölder’s
inequality. ∎

######  Theorem 5.7.

Let

$0<q,q_{1}\leq\infty,1\leq q_{0}<\infty,1\leq
r_{1}\leq\infty,1/q=1/q_{0}+1/q_{1},1/r<1/2+1/r_{1}.$

Let $f$ be an adapted process and $g$ a martingale. Then, we have

$\lVert V^{r}_{k}\Pi(\delta f,g)\rVert_{q}\lesssim\lVert
V^{r_{1}}f\rVert_{q_{1}}\lVert Sg\rVert_{q_{0}}.$ (5.19)

###### Proof.

Combine Corollary 5.6 and Theorem 5.4. ∎

######  Remark.

This section is mostly copied from [FZ23].

##  6 Itô integration

In this section, we discuss integration with respect to a càdlàg martingale in
continuous time, generalizing the paraproduct $\Pi$ from the previous section.
We consider stochastic processes adapted to some filtration
$(\mathcal{F}_{t})_{t\in\mathbb{R}_{\geq 0}}$ indexed by positive real times.
We begin with some convenient regularity assumptions. These assumptions are
not restrictive, in the sense that, for any martingale, one can reparametrize
time and change the filtration in such a way that they are satisfied, but we
will not discuss this, since our focus is on more quantitative issues.

A filtration $(\mathcal{F}_{t})_{t\in\mathbb{R}_{\geq 0}}$ is called _right-
continuous_ if, for every $t\geq 0$, we have

$\mathcal{F}_{t}=\bigcap_{t^{\prime}>t}\mathcal{F}_{t^{\prime}}.$

For a function $f:R_{\geq 0}\to E$ with values in a metric space $E$, the left
and right limits at a point $t$ are denoted by

$f_{t-}:=\lim_{s\to t,s<t}f_{s},\quad f_{t+}:=\lim_{s\to t,s>t}f_{s},$

if they exist. A function $f:\mathbb{R}_{\geq 0}\to E$ is called _càdlàg_ (for
“continue à droite, limite à gauche”, “right continuous with left limits”;
some authors use the English abbreviation “rcll”) if $f_{t-}$ exists for every
$t>0$, and $f_{t}=f_{t+}$ for every $t\geq 0$. A stochastic process
$g:\Omega\times\mathbb{R}_{\geq 0}\to E$ is called càdlàg if every path
$g(\omega,\cdot)$ is a càdlàg function.

######  Theorem 6.1 (Regularization, see e.g. [Kal21, Theorem 9.28]).

Let $\mathcal{F}=(\mathcal{F}_{t})_{t\in\mathbb{R}_{\geq 0}}$ be a right-
continuous filtration. If $g$ is a martingale with respect to $\mathcal{F}$,
then there exists a càdlàg martingale $\tilde{g}$ with respect to
$\mathcal{F}$ such that, for every $t\geq 0$, we have

$g(\cdot,t)=\tilde{g}(\cdot,t)\quad\text{a.e.}$

For an adapted partition $\pi$, we write

$\lfloor
t,\pi\rfloor:=\max\\{s\in\pi\nonscript\>|\allowbreak\nonscript\>\mathopen{}s\leq
t\\},\quad 0\leq t<\infty.$ (6.1)

For a càdlàg adapted process $f=(f_{t})_{t\geq 0}$, a càdlàg martingale
$g=(g_{t})_{t\geq 0}$, and an adapted partition $\pi$, we consider the
following Riemann–Stieltjes sums:

$\Pi^{\pi}(f,g)_{t,t^{\prime}}:=\sum_{t<\pi_{j}<t^{\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge t^{\prime}},\quad 0\leq
t\leq t^{\prime}<\infty.$ (6.2)

In each summand, the integrand $f$ is evaluated at the left endpoint of the
interval $[\pi_{j},\pi_{j+1}]$ on which we consider the increment of the
integrator $g$ (and also at $\lfloor t,\pi\rfloor$, which is even further to
the left). This is the distinguishing feature of the Itô integral that makes
it a martingale in the $t^{\prime}$ variable.

Unlike in Riemann(–Stieltjes) integration with a bounded variation integrator,
evaluating $t$ at other points in general produces different results (e.g.
Stratonovich integral, where $f$ is averaged over $\pi_{j}$ and $\pi_{j+1}$).

Most classical treatments involve sums

$\sum_{\pi_{j}<t^{\prime}}f_{\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}},$

which corerspond to fixing $t=0$, but this obscures the observation of the
natural regularity of the Itô integral that fits nicely with rough path
theory.

For an adpated process $f$ and an adapted partition $\pi$, we write

$f^{(\pi)}_{t}:=f_{\lfloor t,\pi\rfloor}.$ (6.3)

######  Proposition 6.2.

Let $0<q_{1}\leq\infty$, $1\leq q_{0}<\infty$, and $0<r,p_{1}\leq\infty$.
Suppose

$1/r<1/p_{1}+1/2,\quad 1/q=1/q_{0}+1/q_{1}.$ (6.4)

Let $(f_{t})$ be a càdlàg adapted process and $(g_{t})$ a càdlàg martingale.

Then, for every adapted partition $\pi$, we have the estimate

$\big{\lVert}V^{r}\Pi^{\pi}(f,g)\big{\rVert}_{L^{q}}\lesssim\lVert
V^{p_{1}}f^{(\pi)}\rVert_{L^{q_{1}}}\lVert V^{\infty}g\rVert_{L^{q_{0}}}.$
(6.5)

###### Proof of Proposition 6.2.

Since $\Pi^{\pi}(F,g)_{t,t^{\prime}}$ is càdlàg in both $t$ and $t^{\prime}$,
we have

$V^{r}\Pi^{\pi}(F,g)=\lim_{n\to\infty}\sup_{l_{\max},u_{0}<\dotsb<u_{l_{\max}},u_{l}\in\pi^{(n)}}\Bigl{(}\sum_{l=1}^{l_{\max}}\lvert\Pi^{\pi}(F,g)_{u_{l-1},u_{l}}\rvert^{r}\Bigr{)}^{1/r},$

where $\pi^{(n)}=\pi\cup 2^{-n}\mathbb{N}$. By the monotone convergence
theorem, it suffices to consider a fixed $\pi^{(n)}$, as long as the bound
does not depend on $n$.

For any adapted partitions $\pi\subseteq\tau$, we have

$\begin{split}\Pi^{\pi}(f,g)_{t,t^{\prime}}&=\sum_{k:t<\pi_{k}<t^{\prime}}\delta
f_{\lfloor t,\pi\rfloor,\pi_{k}}\delta g_{\pi_{k},\pi_{k+1}\wedge
t^{\prime}}\\\ &=\sum_{k:t<\pi_{k}<t^{\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{k}}\sum_{l:\pi_{k}\leq\tau_{l}<\pi_{k+1}\wedge
t^{\prime}}\delta g_{\tau_{l},\tau_{l+1}\wedge t^{\prime}}\\\
&=\sum_{k:t<\pi_{k}<t^{\prime}}\sum_{l:\pi_{k}\leq\tau_{l}<\pi_{k+1}\wedge
t^{\prime}}\delta f_{\lfloor t,\pi\rfloor,\lfloor\tau_{l},\pi\rfloor}\delta
g_{\tau_{l},\tau_{l+1}\wedge t^{\prime}}\\\
&=\sum_{l:t<\tau_{l}<t^{\prime}}\delta f^{(\pi)}_{\lfloor
t,\tau\rfloor,\tau_{l}}\delta g_{\tau_{l},\tau_{l+1}\wedge t^{\prime}}\\\
&=\Pi^{\tau}(f^{(\pi)},g)_{t,t^{\prime}},\end{split}$ (6.6)

where $f^{(\pi)}$ is given by (6.3). Define discrete time processes
$f^{(\pi)}_{\tau},g_{\tau}$ by

$(f^{(\pi)}_{\tau})_{j}=f^{(\pi)}_{\tau_{j}},\quad(g_{\tau})_{j}=g_{\tau_{j}}.$

Then, we have

$\displaystyle\Pi^{\pi}(f,g)_{\tau_{j},\tau_{j^{\prime}}}$
$\displaystyle=\Pi^{\tau}(f^{(\pi)},g)_{\tau_{j},\tau_{j^{\prime}}}$
$\displaystyle=\sum_{l:\tau_{j}<\tau_{l}<\tau_{j^{\prime}}}\delta
f^{(\pi)}_{\lfloor\tau_{j},\tau\rfloor,\tau_{l}}\delta
g_{\tau_{l},\tau_{l+1}\wedge\tau_{j^{\prime}}}$
$\displaystyle=\sum_{l:j<l<j^{\prime}}\delta
f^{(\pi)}_{\tau_{j},\tau_{l}}\delta g_{\tau_{l},\tau_{l+1}}$
$\displaystyle=\Pi(f^{(\pi)}_{\tau},g_{\tau})_{j,j^{\prime}},$

where the last line is the discrete time paraproduct defined in (5.3). By
Theorem 5.7 and the BDG inequality (Corollary 1.13) for the discrete time
martingale $g_{\tau}$, we obtain

$\lVert V^{r}\Pi(f^{(\pi)}_{\tau},g_{\tau})\rVert_{q}\lesssim\lVert
V^{r_{1}}f^{(\pi)}_{\tau}\rVert_{q_{1}}\lVert
Sg_{\tau}\rVert_{q_{0}}\lesssim\lVert
V^{r_{1}}f^{(\pi)}_{\tau}\rVert_{q_{1}}\lVert
V^{\infty}g_{\tau}\rVert_{q_{0}}\leq\lVert V^{r_{1}}f\rVert_{q_{1}}\lVert
V^{\infty}g\rVert_{q_{0}}.\qed$

######  Lemma 6.3.

Let $(f_{t})_{t\geq 0}$ be a càdlàg adapted process. Suppose that
$V^{p_{1}}f\in L^{q_{1}}$ for some $p_{1},q_{1}\in(0,\infty]$. Then, for every
$\tilde{p}_{1}\in(p_{1},\infty)\cup\\{\infty\\}$, we have

$\lim_{\pi}\lVert V^{\tilde{p}_{1}}(f-f^{(\pi)})\rVert_{L^{q_{1}}}=0.$

###### Proof.

We have $V^{p_{1}}f^{(\pi)}\leq V^{p_{1}}f$ and, by Hölder’s inequality,

$V^{\tilde{p}_{1}}(f-f^{(\pi)})\leq
V^{p_{1}}(f-f^{(\pi)})^{1-\theta}V^{\infty}(f-f^{(\pi)})^{\theta}$

with some $\theta\in(0,1]$, so it suffices to consider $\tilde{p}_{1}=\infty$.

Let $\varepsilon>0$ and define a sequence of stopping times recursively,
starting with $\pi_{0}:=0$, by

$\pi_{j+1}:=\inf\Big{\\{}t>\pi_{j}\nonscript\>\Big{|}\allowbreak\nonscript\>\mathopen{}\lvert\delta
f_{\pi_{j},t}\rvert\geq\varepsilon\Big{\\}}.$

Since $f$ is càdlàg, the infimum is either $+\infty$ or a minimum, so that
this is indeed a stopping time. Also by the càdlàg assumption, this sequence
of stopping times is strictly monotonically increasing in the sense that
$\pi_{j}<\infty\implies\pi_{j}<\pi_{j+1}$. Moreover, if
$T:=\sup_{j}\pi_{j}<\infty$, then the left limit $f_{T-}$ does not exist,
contradicting the càdlàg hypothesis. Therefore, $\pi_{j}\to\infty$, so that
$\pi$ is an adapted partition.

Then, by (4.18), for any adapted partition $\pi^{\prime}\supseteq\pi$ and
$s\leq t$, we have

$\displaystyle\lvert\delta f_{s,t}-\delta f^{(\pi^{\prime})}_{s,t}\rvert$
$\displaystyle\leq\lvert f_{t}-f_{\lfloor t,\pi^{\prime}\rfloor}\rvert+\lvert
f_{s}-f_{\lfloor s,\pi^{\prime}\rfloor}\rvert$ $\displaystyle\leq
2\varepsilon+2\varepsilon.\qed$

######  Theorem 6.4 (Itô integral).

In the situation of Proposition 6.2, suppose that the right-hand side of (6.8)
is finite. Then

$\Pi(f,g):=\lim_{\pi}\Pi^{\pi}(f,g)$ (6.7)

exists in $L^{q}(\Omega,V^{r})$, satisfies the bound

$\big{\lVert}V^{r}\Pi(f,g)\big{\rVert}_{L^{q}}\lesssim\lVert
V^{p_{1}}f\rVert_{L^{q_{1}}}\lVert V^{\infty}g\rVert_{L^{q_{0}}},$ (6.8)

and, for any $0\leq t\leq t^{\prime}\leq t^{\prime\prime}<\infty$, _Chen’s
relation_

$\Pi(f,g)_{t,t^{\prime\prime}}-\Pi(f,g)_{t,t^{\prime}}-\Pi(f,g)_{t^{\prime},t^{\prime\prime}}=\delta
f_{t,t^{\prime}}\delta g_{t^{\prime},t^{\prime\prime}}.$ (6.9)

The limit (6.7) is called the _Itô integral_ (with integrand $f$ and
integrator $g$).

###### Proof of Theorem 6.4.

By the Cauchy criterion for net convergence, the existence of the limit (6.7)
will follow if we can show that

$\lim_{\pi}\sup_{\tau\supseteq\pi}\big{\lVert}V^{r}(\Pi^{\pi}(f,g)-\Pi^{\tau}(f,g))\big{\rVert}_{L^{q}}=0.$
(6.10)

To this end, we use that, by (6.6), we have

$\Pi^{\pi}(f,g)-\Pi^{\tau}(f,g)=\Pi^{\tau}(f^{(\pi)}-f^{(\tau)},g).$

Let $\tilde{p}_{1}\in(p_{1},\infty]\cup\\{\infty\\}$ be such that
$1/r<1/\tilde{p}_{1}+1/2$. By Proposition 6.2 with $f$ replaced by
$f^{(\pi)}-f^{(\tau)}$, we obtain

$\displaystyle\big{\lVert}V^{r}\Pi^{\tau}(f^{(\pi)}-f^{(\tau)},g)\big{\rVert}_{L^{q}}$
$\displaystyle\lesssim\lVert
V^{\tilde{p}_{1}}(f^{(\pi)}-f^{(\tau)})^{(\tau)}\rVert_{L^{q_{1}}}\lVert
V^{\infty}g\rVert_{L^{q_{0}}}$

This converges to $0$ by Lemma 6.3.

In order to show the Chen relation (6.9), we first show that the corresponding
relation holds pointwise for the discretized paraproducts $\Pi^{\pi}$. Indeed,
by definition (6.2), for $t\leq t^{\prime}\leq t^{\prime\prime}$, we have

$\Pi^{\pi}(f,g)_{t,t^{\prime\prime}}-\Pi^{\pi}(f,g)_{t,t^{\prime}}-\Pi^{\pi}(f,g)_{t^{\prime},t^{\prime\prime}}$
(6.11) $=\sum_{t<\pi_{j}<t^{\prime\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime\prime}}-\sum_{t<\pi_{j}<t^{\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge t^{\prime}}\\\
-\sum_{t^{\prime}<\pi_{j}<t^{\prime\prime}}\delta f_{\lfloor
t^{\prime},\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime\prime}}$ $=\sum_{t<\pi_{j}<t^{\prime}}f_{\lfloor
t,\pi\rfloor,\pi_{j}}(\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime\prime}}-\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}})+\sum_{t<\pi_{j}=t^{\prime}<t^{\prime\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge t^{\prime\prime}}\\\
+\sum_{t^{\prime}<\pi_{j}<t^{\prime\prime}}(\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}-\delta f_{\lfloor t^{\prime},\pi\rfloor,\pi_{j}})\delta
g_{\pi_{j},\pi_{j+1}\wedge t^{\prime\prime}}$
$=\sum_{t<\pi_{j}<t^{\prime}}f_{\lfloor t,\pi\rfloor,\pi_{j}}\delta
g_{\pi_{j+1}\wedge t^{\prime},\pi_{j+1}\wedge
t^{\prime\prime}}+\sum_{t<\pi_{j}=t^{\prime}<t^{\prime\prime}}\delta
f_{\lfloor t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime\prime}}\\\ +\delta f_{\lfloor t,\pi\rfloor,\lfloor
t^{\prime},\pi\rfloor}\sum_{t^{\prime}<\pi_{j}<t^{\prime\prime}}\delta
g_{\pi_{j},\pi_{j+1}\wedge t^{\prime\prime}}$

All summands except possibly the one with $\pi_{j}<t^{\prime}<\pi_{j+1}$ in
the first sum vanish, and it follows that

$\dots=\delta f_{\lfloor t,\pi\rfloor,\lfloor
t^{\prime},\pi\rfloor}\Bigl{(}\sum_{t<\pi_{j}<t^{\prime}<\pi_{j+1}}\delta
g_{\pi_{j+1}\wedge t^{\prime},\pi_{j+1}\wedge t^{\prime\prime}}\\\
+\sum_{t<\pi_{j}=t^{\prime}<t^{\prime\prime}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime\prime}}+\sum_{t^{\prime}<\pi_{j}<t^{\prime\prime}}\delta
g_{\pi_{j},\pi_{j+1}\wedge t^{\prime\prime}}\Bigr{)}$ $=\delta f_{\lfloor
t,\pi\rfloor,\lfloor t^{\prime},\pi\rfloor}\Bigl{(}\sum_{t<\pi_{j}\leq
t^{\prime}<\pi_{j+1}\wedge t^{\prime\prime}}\delta
g_{t^{\prime},\pi_{j+1}\wedge
t^{\prime\prime}}+\sum_{t^{\prime}<\pi_{j}<t^{\prime\prime}}\delta
g_{\pi_{j},\pi_{j+1}\wedge t^{\prime\prime}}\Bigr{)}=\delta
f^{(\pi)}_{t,t^{\prime}}\delta g_{t^{\prime},t^{\prime\prime}}.$

By the already known conclusion (6.7), the process (6.11) converges to the
left-hand side of (6.9). By Lemma 6.3, the last expression in the above chain
of equalities converges to the right-hand side of (6.9). ∎

###  6.1 Quadratic covariation

######  Proposition 6.5.

Let $q_{0},q_{1}\in[1,\infty)$, $1/q=1/q_{0}+1/q_{1}$, and $r>1$. For càdlàg
martingales $f,g$ with $V^{\infty}f\in L^{q_{0}}$, $V^{\infty}g\in L^{q_{1}}$,
let

$[f,g]^{\pi}_{t,t^{\prime}}:=\sum_{\lfloor t,\pi\rfloor\leq\pi_{j}<\lfloor
t^{\prime},\pi\rfloor}\delta f_{\pi_{j},\pi_{j+1}}\delta
g_{\pi_{j},\pi_{j+1}}.$

Then,

$[f,g]_{t,t^{\prime}}:=\lim_{\pi}[f,g]^{\pi}_{t,t^{\prime}}$ (6.12)

exists in $L^{q}(V^{r})$. Moreover, we have the integration by parts formula

$\delta f_{t,t^{\prime}}\delta
g_{t,t^{\prime}}=\Pi(f,g)_{t,t^{\prime}}+\Pi(g,f)_{t,t^{\prime}}+[f,g]_{t,t^{\prime}}.$
(6.13)

The process (6.12) is called the _quadratic covariation_ of $f$ and $g$. Here
are a few facts about it that easily follow from the definition:

$\forall t\leq t^{\prime}\leq
t^{\prime\prime},\quad[f,g]_{t,t^{\prime\prime}}=[f,g]_{t,t^{\prime}}+[f,g]_{t^{\prime},t^{\prime\prime}}\quad\text{a.s.}$

To see this, note that the same identity holds for $[\cdot,\cdot]^{\pi}$ for
any adapted partition $\pi$ that contains the times
$t,t^{\prime},t^{\prime\prime}$.

$\forall t\leq
t^{\prime},\quad\mathbb{E}_{t}[f,f]_{t,t^{\prime}}=\mathbb{E}_{t}\lvert\delta
f_{t,t^{\prime}}\rvert^{2}.$

To see this, note that the same identity holds for $[\cdot,\cdot]^{\pi}$ for
any adapted partition $\pi$ that contains the times $t,t^{\prime}$, because,
by the optinal sampling theorem, $\delta f_{\pi_{j},\pi_{j+1}}$ are martingale
increments, and therefore orthogonal in $L^{2}$.

###### Proof of Proposition 6.5.

First, we compute

$\displaystyle\delta f_{\lfloor t,\pi\rfloor,t^{\prime}}\delta g_{\lfloor
t,\pi\rfloor,t^{\prime}}-\Pi^{\pi}(f,g)_{t,t^{\prime}}-\Pi^{\pi}(g,f)_{t,t^{\prime}}$
$\displaystyle=\bigl{(}\sum_{\lfloor t,\pi\rfloor\leq\pi_{k}<t^{\prime}}\delta
f_{\pi_{k},\pi_{k+1}\wedge t^{\prime}}\bigr{)}\bigl{(}\sum_{\lfloor
t,\pi\rfloor\leq\pi_{j}<t^{\prime}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}}\bigr{)}$ $\displaystyle-\sum_{\lfloor
t,\pi\rfloor<\pi_{j}<t^{\prime}}\delta f_{\lfloor t,\pi\rfloor,\pi_{j}}\delta
g_{\pi_{j},\pi_{j+1}\wedge t^{\prime}}-\sum_{\lfloor
t,\pi\rfloor<\pi_{k}<t^{\prime}}\delta g_{\lfloor t,\pi\rfloor,\pi_{k}}\delta
f_{\pi_{k},\pi_{k+1}\wedge t^{\prime}}$ $\displaystyle=\sum_{\lfloor
t,\pi\rfloor\leq\pi_{j}<t^{\prime}}\delta f_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}}\delta g_{\pi_{j},\pi_{j+1}\wedge t^{\prime}}$
$\displaystyle+\sum_{\lfloor t,\pi\rfloor\leq\pi_{k}<\pi_{j}<t^{\prime}}\delta
f_{\pi_{k},\pi_{k+1}\wedge t^{\prime}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}}-\sum_{\lfloor t,\pi\rfloor<\pi_{j}<t^{\prime}}\delta f_{\lfloor
t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},\pi_{j+1}\wedge t^{\prime}}$
$\displaystyle+\sum_{\lfloor t,\pi\rfloor\leq\pi_{j}<\pi_{k}<t^{\prime}}\delta
f_{\pi_{k},\pi_{k+1}\wedge t^{\prime}}\delta g_{\pi_{j},\pi_{j+1}\wedge
t^{\prime}}-\sum_{\lfloor t,\pi\rfloor<\pi_{k}<t^{\prime}}\delta g_{\lfloor
t,\pi\rfloor,\pi_{k}}\delta f_{\pi_{k},\pi_{k+1}\wedge t^{\prime}}.$

Each of the last two lines vanishes identically.

In particular, replacing $t^{\prime}$ by $\lfloor t^{\prime},\pi\rfloor$, we
obtain

$\delta f_{\lfloor t,\pi\rfloor,\lfloor t^{\prime},\pi\rfloor}\delta
g_{\lfloor t,\pi\rfloor,\lfloor
t^{\prime},\pi\rfloor}-\Pi^{\pi}(f,g)_{t,\lfloor
t^{\prime},\pi\rfloor}-\Pi^{\pi}(g,f)_{t,\lfloor
t^{\prime},\pi\rfloor}=\sum_{\lfloor t,\pi\rfloor\leq\pi_{j}<\lfloor
t^{\prime},\pi\rfloor}\delta f_{\pi_{j},\pi_{j+1}}\delta
g_{\pi_{j},\pi_{j+1}}.$ (6.14)

Note that

$\Pi^{\pi}(f,g)_{t,\lfloor
t^{\prime},\pi\rfloor}=\Pi^{\pi}(f,g)_{t,t^{\prime}}-\sum_{\pi_{j}<t^{\prime}<\pi_{j+1}}\delta
f_{\lfloor t,\pi\rfloor,\pi_{j}}\delta g_{\pi_{j},t^{\prime}},$ (6.15)

where the sum consists of either $0$ or $1$ summands. For any inrcreasing
sequence $u_{0}\leq\dotsc\leq u_{K}$, we have In particular, with any
$\alpha\in(0,r-1)$, we have

$\displaystyle\sum_{k=0}^{K-1}\Big{\lvert}\sum_{\pi_{j}<u_{k+1}<\pi_{j+1}}\delta
f_{\lfloor u_{k},\pi\rfloor,\pi_{j}}\delta
g_{\pi_{j},u_{k+1}}\Big{\rvert}^{r}$ $\displaystyle=\sum_{k:\lfloor
u_{k},\pi\rfloor<\lfloor u_{k+1},\pi\rfloor}\lvert\delta f_{\lfloor
u_{k},\pi\rfloor,\lfloor u_{k+1},\pi\rfloor}\delta g_{\lfloor
u_{k+1},\pi\rfloor,u_{k+1}}\rvert^{r}$
$\displaystyle\leq\Bigl{(}\sum_{k:\lfloor u_{k},\pi\rfloor<\lfloor
u_{k+1},\pi\rfloor}\lvert\delta f_{\lfloor u_{k},\pi\rfloor,\lfloor
u_{k+1},\pi\rfloor}\rvert^{2r}\Bigr{)}^{1/2}\Bigl{(}\sum_{k:\lfloor
u_{k},\pi\rfloor<\lfloor u_{k+1},\pi\rfloor}\lvert\delta g_{\lfloor
u_{k+1},\pi\rfloor,u_{k+1}}\rvert^{2r}\Bigr{)}^{1/2}$
$\displaystyle\leq(V^{2r}f)^{r}\sup_{t}\lvert\delta g_{\lfloor
t,\pi\rfloor,t}\rvert^{\alpha}(V^{2(r-\alpha)}g)^{r-\alpha}.$

This expression no longer depends on the sequence $(u_{k})$. By Lépingle’s and
Hölder’s inequalities, this is bounded in $L^{q}$, and, taking into account
Lemma 6.3, this converges to $0$ in $L^{q}$. Hence, the second term on the
right-hand side of (6.15) converges to $0$ in $L^{q}V^{r}$. Theorem 6.4 and
Lépingle’s inequality for $f$ now imply

$\lim_{\pi}\Pi^{\pi}(f,g)_{t,\lfloor
t^{\prime},\pi\rfloor}=\Pi(f,g)_{t,t^{\prime}}\quad\text{in }L^{q}V^{r}.$

Writing (6.14) as

$[f,g]^{\pi}_{t,t^{\prime}}=\delta f^{(\pi)}_{t,t^{\prime}}\delta
g^{(\pi)}_{t,t^{\prime}}-\Pi^{\pi}(f,g)_{t,\lfloor
t^{\prime},\pi\rfloor}-\Pi^{\pi}(g,f)_{t,\lfloor t^{\prime}\rfloor},$ (6.16)

we see that the first summand on the right-hand side converges to $\delta
f\delta g$ in $L^{q}V^{r}$ by Lemma 6.3, and the remaining summands to Itô
integrals by the above discussion. Hence, the left hand side converges, as was
claimed in (6.12). The identity (6.13) is the limit of the above equality. ∎

##  7 A sharp inequality for the square function

The proof of the BDG inequality that we have seen in this notes was quite
indirect: we started with some $L^{2}$ identities and used the Davis
decomposition to lower the $L^{p}$ exponent. In this section, we will take a
look at a more direct method for proving martingale inequalities. In this
special case, it will yield an inequality with an optimal constant.

This section follows [Bur02].

######  Theorem 7.1.

Let $(f_{n})_{n\in\mathbb{N}}$ be a real-valued martingale. Then, for every
$N\in\mathbb{N}$, we have

$\mathbb{E}\Bigl{(}3\lvert f_{0}\rvert+\sum_{n=1}^{N}\frac{\lvert
df_{n}\rvert^{2}}{f^{*}_{n}}\Bigr{)}\leq\mathbb{E}\Bigl{(}2f^{*}_{N}+\frac{\lvert
f_{N}\rvert^{2}}{f^{*}_{N}}\Bigr{)}.$ (7.1)

We will see later that the inequality (7.1) implies the Davis inequality for
the martingale square function with the sharp constant. In applications, one
replaces (7.1) by the slightly weaker inequality

$\mathbb{E}\Bigl{(}3\lvert f_{0}\rvert+\sum_{n=1}^{N}\frac{\lvert
df_{n}\rvert^{2}}{f^{*}_{n}}\Bigr{)}\leq 3\mathbb{E}f^{*}_{N}.$

However, the preculiar form of the right-hand side of (7.1) permits to show
this result _by induction on $N$_.

###  7.1 A Bellman function

The inductive step in the proof of Theorem 7.1 is usually stated as a
concavity property of a special function. Functions used in such arguments are
called “Bellman functions”; many more examples can be found in the books
[Osę12, VV20].

Thoroughout this section,

$\mathcal{D}:=\\{(x,t,z)\in\mathbb{R}\times\mathbb{R}_{\geq
0}\times\mathbb{R}_{\geq
0}\nonscript\>|\allowbreak\nonscript\>\mathopen{}\lvert x\rvert\leq z\\}.$

We define $U:\mathcal{D}\to\mathbb{R}$ by

$U(x,y,m):=y-\frac{\lvert x\rvert^{2}+(\gamma-1)m^{2}}{m},$

where $\gamma=3$. The main feature of this function is the following concavity
property.

######  Proposition 7.2.

For any $x,h\in\mathbb{R}$ and $y,m\in\mathbb{R}_{\geq 0}$ with $\lvert
x\rvert\leq m$, we have

$U(x+h,y+\frac{\lvert h\rvert^{2}}{(\lvert x+h\rvert\vee m)},\lvert
x+h\rvert\vee m)\leq U(x,y,m)-\frac{2(x)h}{m}.$ (7.2)

###### Proof of Theorem 7.1 assuming Proposition 7.2.

Using (7.2) with

$x=f_{n},\quad y=\tilde{S}_{n}:=\gamma\lvert
f_{0}\rvert+\sum_{j=1}^{n}\frac{\lvert df_{j}\rvert^{2}}{f^{*}_{j}},\quad
m=f^{*}_{n},\quad h=df_{n+1},$

we obtain

$U(f_{n+1},\tilde{S}_{n+1},f^{*}_{n+1})\leq
U(f_{n},\tilde{S}_{n},f^{*}_{n})-\frac{2f_{n}df_{n+1}}{f^{*}_{n}}.$ (7.3)

By conditional independence, we have

$\mathbb{E}(\frac{2f_{n}df_{n+1}}{f^{*}_{n}}\lvert\mathcal{F}_{n})=\frac{2f_{n}}{f^{*}_{n}}\mathbb{E}(df_{n+1}\rvert\mathcal{F}_{n})=0.$

Taking expectations, we obtain

$\mathbb{E}U(f_{n+1},\tilde{S}_{n+1},f^{*}_{n+1})\leq\mathbb{E}U(f_{n},\tilde{S}_{n},f^{*}_{n}).$

Iterating this inequality, we obtain

$\mathbb{E}\Bigl{(}3\lvert f_{0}\rvert+\sum_{n=1}^{N}\frac{\lvert
df_{n}\rvert^{2}}{(f^{*}_{n})}-\frac{\lvert
f_{N}\rvert^{2}}{f^{*}_{N}}-2f^{*}_{N}\Bigr{)}=\mathbb{E}U(f_{N},\tilde{S}_{N},f^{*}_{N})\leq\mathbb{E}U(f_{0},\tilde{S}_{0},f^{*}_{0})=0.\qed$

######  Remark.

The above proof in fact shows the pathwise inequality

$3\lvert f_{0}\rvert+\sum_{n=1}^{N}\frac{\lvert
df_{n}\rvert^{2}}{(f^{*}_{n})}\leq 2f^{*}_{N}+\frac{\lvert
f_{N}\rvert^{2}}{f^{*}_{N}}-\sum_{n=1}^{N}\frac{2(f_{n})df_{n+1}}{(f^{*}_{n})}.$

###### Proof of Proposition 7.2.

If $\lvert x+h\rvert\leq m$, then

$\displaystyle U(x+h,y+\frac{\lvert h\rvert^{2}}{(\lvert x+h\rvert\vee
m)},\lvert x+h\rvert\vee m)$ $\displaystyle=(y+\frac{\lvert
h\rvert^{2}}{m})-\frac{\lvert x+h\rvert^{2}+(\gamma-1)m^{2}}{m}$
$\displaystyle=y+\frac{\lvert h\rvert^{2}}{m}-\frac{\lvert
x\rvert^{2}+2xh+\lvert h\rvert^{2}+(\gamma-1)m^{2}}{m}$
$\displaystyle=y-\frac{\lvert x\rvert^{2}+(\gamma-1)m^{2}}{m}-\frac{2xh}{m}$
$\displaystyle=U(x,y,m)-\frac{2(x)h}{m}.$

If $\lvert x+h\rvert>m$, then we need to show

$(y+\frac{\lvert h\rvert^{2}}{\lvert x+h\rvert})-\frac{\lvert