# The norm of linear extension operators for $C^{m-1,1}(\mathbb{R}^{n})$

J. Carruth, A. Frei-Pearson, A. Israel

###### Abstract

Fix integers $m\geq 2$, $n\geq 1$. We prove the existence of a bounded linear
extension operator for $C^{m-1,1}(\mathbb{R}^{n})$ with operator norm at most
$\exp(\gamma D^{k})$, where $D:=\binom{m+n-1}{n}$ is the number of
multiindices of length $n$ and order at most $m-1$, and $\gamma,k>0$ are
absolute constants (independent of $m,n,E$). Upper bounds on the norm of this
operator are relevant to basic questions about fitting a smooth function to
data. Our results improve on a previous construction of extension operators of
norm at most $\exp(\gamma D^{k}2^{D})$. Along the way, we establish a
finiteness theorem for $C^{m-1,1}(\mathbb{R}^{n})$ with improved bounds on the
involved constants.

##  1 Introduction

Fix $m\geq 1$, $n\geq 1$. We let $C^{m}(\mathbb{R}^{n})$ denote the Banach
space of all $m$-times continuously differentiable functions
$F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ whose partial derivatives up to order
$m$ are bounded functions on $\mathbb{R}^{n}$. We equip
$C^{m}(\mathbb{R}^{n})$ with a standard norm:

$\|F\|_{C^{m}(\mathbb{R}^{n})}:=\sup_{x\in\mathbb{R}^{n}}\max_{|\alpha|\leq
m}|\partial^{\alpha}F(x)|.$

Here, for a multiindex
$\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathbb{N}^{n}$, we write
$|\alpha|:=\sum_{j}\alpha_{j}$ to denote the order of $\alpha$. We write
$\partial^{\alpha}F(x)=\partial^{\alpha_{1}}_{1}\cdots\partial^{\alpha_{n}}_{n}F(x)$
for the $\alpha^{\text{th}}$ partial derivative of a function $F\in
C^{m}_{loc}(\mathbb{R}^{n})$. We also define
$\alpha!:=\prod_{i=1}^{n}\alpha_{i}!$.

The following problem goes back to Whitney [28, 29, 30]. Let $E$ be an
arbitrary subset of $\mathbb{R}^{n}$. Given a function
$f:E\rightarrow\mathbb{R}$, determine whether there exists a function $F\in
C^{m}(\mathbb{R}^{n})$ with $F=f$ on $E$.

Whitney’s problem was solved by C. Fefferman in 2006 [16].111The Whitney
problem has a long history with contributions by many authors; below, we
discuss some of the most relevant to our work. For a more complete history see
[18] and the references therein. In a remarkable series of papers, Fefferman
posed and solved a variety of related problems. In three of these papers [17,
19, 20], two of them joint with B. Klartag, the authors connected this work to
the practical problem of computing a $C^{m}$ interpolant for a given set of
data.

Suppose now that $E$ is a finite subset of $\mathbb{R}^{n}$. We define the
trace norm of a function $f:E\rightarrow\mathbb{R}$ by

$||f||_{C^{m}(E)}:=\inf\\{||F||_{C^{m}(\mathbb{R}^{n})}:F=f\text{ on }E\\}.$

A function $F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is an _interpolant_ of $f$
if $F=f$ on $E$. Given $C\geq 1$, a function $F\in C^{m}(\mathbb{R}^{n})$ is a
_$C$ -optimal interpolant_ of $f$ provided that $F=f$ on $E$ and
$||F||_{C^{m}(\mathbb{R}^{n})}\leq C||f||_{C^{m}(E)}$. That is, $F$ is an
interpolant of $f$ with $C^{m}$ norm that is within a factor of $C$ of the
optimal value. In [17, 19, 20], Fefferman and Klartag proved the following
theorem.

######  Theorem 1.1.

Fix $m\geq 1$, $n\geq 1$. Let $E\subseteq\mathbb{R}^{n}$ be a finite set with
cardinality $\\#(E)=N$ and fix $f:E\rightarrow\mathbb{R}$. There exists an
algorithm that computes a $C$-optimal interpolant $F\in C^{m}(\mathbb{R}^{n})$
of $f$. Specifically, the algorithm takes as input $(E,f,m)$ and performs
$C_{1}N\log N$ units of one-time work, on an idealized (von Neumann) computer
with $C_{2}N$ units of memory. Given $x\in\mathbb{R}^{n}$, the computer
responds to a query by returning the values of $\partial^{\alpha}F(x)$ for all
$\alpha$ with $|\alpha|\leq m$, where $F$ is a $C$-optimal interpolant of $f$.
The algorithm requires $C_{3}\log N$ computer operations to answer a query.
The constants $C,C_{1},C_{2},C_{3}$ depend only on $m$ and $n$.

For details on the model of computation, including an explanation of the terms
“one-time work”, “query”, or what it means to “compute” a function on
$\mathbb{R}^{n}$, see [17, 19, 20].

We note that (1) the running time of the algorithm in Theorem 1.1 likely has
optimal dependence on $N=\\#(E)$ and (2) this is the only known algorithm for
solving the $C^{m}$ interpolation problem for arbitrary finite sets
efficiently in $N$. Therefore, at least in theory, this algorithm could have
widespread practical application.

Unfortunately, the constant $C$ in Theorem 1.1 grows rapidly with $m$ and $n$,
rendering the algorithm impractical for real-world applications. While $C$ is
not computed explicitly in [17, 19, 20], an examination of the arguments in
those papers shows that one must take $C$ to have order of magnitude at least
$\exp(\gamma D^{k}2^{D})$ for some real number $\gamma>0$ and integer $k>0$;
here $D:={m+n-1\choose n}$ denotes the dimension of the vector space of
polynomials in $n$ variables of degree at most $m-1$. In other words, the
optimality guarantees on the interpolant produced by this algorithm
deteriorate rapidly as $n$ and $m$ grow. Any practical version of Theorem 1.1
will have to address this issue. There is considerable interest in finding
such an algorithm; see [11].

The proof of Theorem 1.1 is based on a _finiteness theorem_ for
$C^{m-1,1}(\mathbb{R}^{n})$. This theorem is the source of the double
exponential dependence on $D$ of the constant $C$ in Theorem 1.1. Next, we
state this result.

We let $C^{m-1,1}(\mathbb{R}^{n})$ denote the space of all $(m-1)$-times
differentiable functions $F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ whose
$(m-1)^{\text{rst}}$ order partial derivatives are Lipschitz continuous on
$\mathbb{R}^{n}$. We equip this space with a seminorm:

$||F||_{C^{m-1,1}(\mathbb{R}^{n})}:=\sup_{x,y\in\mathbb{R}^{n}}\left(\sum_{|\alpha|=m-1}\frac{(\partial^{\alpha}F(x)-\partial^{\alpha}F(y))^{2}}{|x-y|^{2}}\right)^{1/2}.$

Given a ball $B\subseteq\mathbb{R}^{n}$, we write $C^{m-1,1}(B)$ for the
corresponding space of $C^{m-1,1}$ functions $F:B\rightarrow\mathbb{R}$.

######  Theorem 1.2.

[Finiteness theorem for $C^{m-1,1}(\mathbb{R}^{n})$ – see [15]]

Let $m\geq 2,n\geq 1$. There exist constants $k^{\\#},C^{\\#}$ depending on
$m$ and $n$ such that the following holds.

Let $f:E\rightarrow\mathbb{R}$, $E\subseteq\mathbb{R}^{n}$ an arbitrary set.
Suppose that for every finite subset $S\subseteq E$ with cardinality
$\\#(S)\leq k^{\\#}$ there exists a function $F^{S}\in
C^{m-1,1}(\mathbb{R}^{n})$ satisfying $F^{S}=f$ on $S$ and
$\|F^{S}\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq 1$.

Then there exists a function $F\in C^{m-1,1}(\mathbb{R}^{n})$ with $F=f$ on
$E$ and $\|F\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq C^{\\#}$.

The finiteness theorem was first proved in the case $m=2$, $n\geq 1$ by
Shvartsman [24]; in this case, it was shown that one can take $k^{\\#}=3\cdot
2^{n-1}$ and $C^{\\#}=A\exp(\gamma n)$, where $A,\gamma>0$ are absolute
constants (independent of $n$). Further, Shvartsman [23] proves that the value
$k^{\\#}=3\cdot 2^{n-1}$ is the smallest possible when $m=2$. In other words,
if $k^{\\#}<3\cdot 2^{n-1}$ then the finiteness theorem fails to hold for any
$C^{\\#}>1$.

Theorem 1.2 was conjectured to hold for any $m\geq 2$, $n\geq 1$ by Brudnyi
and Shvartsman in [5].

In [15], Fefferman proved the conjecture of Brudnyi and Shvartsman. He showed
that Theorem 1.2 holds for any $m\geq 2$, $n\geq 1$ with
$k^{\\#}\leq(D+1)^{3\cdot 2^{D}}$. He did not state an explicit bound on the
value of $C^{\\#}$, but one can check that his proof gives
$C^{\\#}\leq\exp(\gamma D^{k}2^{D})$ for absolute constants $\gamma,k>0$
(independent of $m,n$).

Note that in the case $m=2$, Fefferman’s result implies Shvartsman’s with the
caveat that Shvartman’s result holds for smaller $k^{\\#}$, $C^{\\#}$. Indeed,
if $m=2$, then $D=(n+1)$; therefore Shvartsman’s result implies that the
finiteness theorem holds with $k^{\\#}=3\cdot 2^{D-2}$ and
$C^{\\#}=A\exp(\gamma D)$.

The constant $C$ in Theorem 1.1 inherits its double exponential dependence on
$D$ from the constant $C^{\\#}$ in Theorem 1.2. This leads us to pose the
following problem.

######  Problem 1.

Is it possible to improve the dependence of the constant $C^{\\#}$ in Theorem
1.2 on $D=\binom{m+n-1}{n}$?

Progress on Problem 1 is not possible by optimizing the constants in each line
of Fefferman’s proof of Theorem 1.2. Without going into detail, his proof is
by induction, and it produces a $C^{\\#}$ which is exponential in the number
of induction steps. The number of induction steps is equal to $2^{D}$, leading
to the double exponential dependence of $C^{\\#}$ on $D$. Thus, lowering the
constant $C^{\\#}$ requires new ideas.

In a joint work [6] with B. Klartag, we gave a new proof of Theorem 1.2 which
avoided Fefferman’s induction scheme. Our proof relied on semialgebraic
geometry and compactness arguments, however, and therefore it did not give an
effective bound on $C^{\\#}$. In this paper, we replace the qualitative
arguments of [6] with quantitative ones and improve the dependence of
$C^{\\#}$ on $D$ in Theorem 1.2 to exponential in a power of $D$.
Specifically, we prove the following theorem.

######  Theorem 1.3.

There exist absolute constants $\gamma>0$ and $k\geq 1$, independent of $m$
and $n$, such that the finiteness theorem for $C^{m-1,1}(\mathbb{R}^{n})$
(Theorem 1.2) holds with $C^{\\#}=\exp(\gamma D^{k})$ and $k^{\\#}=\exp(\gamma
D^{k})$.

In [14], Fefferman showed that his proof of Theorem 1.2 can be modified to
produce a $C^{\\#}$-optimal interpolant $F$ that depends linearly on the data
$f$. This property is crucial in getting from Theorem 1.2 to the algorithm in
Theorem 1.1. Our proof also has this property. Specifically, the next theorem
is a byproduct of the proof of Theorem 1.3.

Given an arbitrary set $E\subseteq\mathbb{R}^{n}$ (not necessarily finite), we
let $C^{m-1,1}(E)$ denote the space of all restrictions to $E$ of functions in
$C^{m-1,1}(\mathbb{R}^{n})$, equipped with the standard trace seminorm:

$\|f\|_{C^{m-1,1}(E)}:=\inf\\{\|F\|_{C^{m-1,1}(\mathbb{R}^{n})}:F=f\mbox{ on
}E\\}\quad(f\in C^{m-1,1}(E)).$

######  Theorem 1.4.

There exist absolute constants $\gamma>0$ and $k\geq 1$, independent of $m$
and $n$, such that the following holds. Given $E\subset\mathbb{R}^{n}$, there
exists a linear map $T:C^{m-1,1}(E)\rightarrow C^{m-1,1}(\mathbb{R}^{n})$
satisfying $Tf|_{E}=f$ and $||Tf||_{C^{m-1,1}(\mathbb{R}^{n})}\leq
C^{\\#}||f||_{C^{m-1,1}(E)}$ for all $f\in C^{m-1,1}(E)$, where
$C^{\\#}=\exp(\gamma D^{k})$.

While the constant $C^{\\#}$ in Theorems 1.3 and 1.4 is still too large to
give rise to a practical algorithm for $C^{m}$ interpolation, this marks the
first progress on Problem 1 since Fefferman’s proof of Theorem 1.2.

Theorem 1.3 shows that the constant $C^{\\#}$ in the finiteness theorem can be
taken to be exponential in a power of $D$. We do not know whether this is the
optimal dependence—little is known about lower bounds for $C^{\\#}$. Trivially
one has the lower bound $C^{\\#}\geq 1$. One might hope that for any
$C^{\\#}>1$ there exists some $k^{\\#}$ sufficiently large depending on
$C^{\\#}$ such that the Finiteness Theorem holds. This is true when $m=1$ (see
[12]), but not in general. In [13], Fefferman and Klartag show that there
exists a constant $c_{0}>0$ such that Theorem 1.2 does not hold for
$C^{\\#}<1+c_{0}$ for any $k^{\\#}$ when $m=n=2$. It would be interesting to
obtain a lower bound on $C^{\\#}$ that grows with $n$ or $m$.

A loose inspection of our proof indicates that it is sufficient to take the
power $k=8$ in Theorem 1.3. In the case $m=2$ we know that this is not
sharp—Shvartsman’s work shows that Theorem 1.3 holds with $k=1$ when $m=2$
(see the discussion of Theorem 1.2 above).

While this paper is concerned with upper bounds on the constant $C^{\\#}$,
there is also interest in understanding the dependence of the constant
$k^{\\#}$ on $m$ and $n$. Bierstone and Milman, in [3], and Shvartsman, in
[25], independently showed that the Finiteness Theorem holds with
$k^{\\#}=2^{D}$ and $C^{\\#}$ as in Fefferman’s proof of Theorem 1.2, i.e.
$C^{\\#}=\exp(\gamma D^{k}2^{D})$ for absolute constants $\gamma,k>0$. Our
proof gives $k^{\\#},C^{\\#}\leq\exp(\hat{\gamma}D^{\hat{k}})$ for absolute
constants $\hat{\gamma},\hat{k}$. We would be interested to know whether the
Finiteness Theorem holds with $k^{\\#}=2^{D}$ and
$C^{\\#}\leq\exp(\hat{\gamma}D^{\hat{k}})$ simultaneously.

We remark that, by standard arguments, Theorem 1.4 implies the analogous
theorem for $C^{m}(\mathbb{R}^{n})$ when $E$ is a _finite_ subset of
$\mathbb{R}^{n}$. Fefferman proved the analogue of Theorem 1.4 for
$C^{m}(\mathbb{R}^{n})$ when $E$ is compact; the argument is significantly
more complicated (see [9]). It would be interesting to understand the norm of
linear extension operators $T:C^{m}(E)\rightarrow C^{m}(\mathbb{R}^{n})$ for
$E$ compact.

We will now sketch the proof of Theorem 1.3, highlighting the new ideas in the
argument. Small modifications to this argument enable us to obtain the
existence of a linear extension operator $T:C^{m-1,1}(E)\rightarrow
C^{m-1,1}(\mathbb{R}^{n})$ with improved bounds on the operator norm, as in
Theorem 1.4.

By a compactness argument, it suffices to prove the finiteness theorem for a
finite set $E$ in $\mathbb{R}^{n}$. Note that the constants $C^{\\#}$ and
$k^{\\#}$ in the finiteness theorem are to be chosen independent of $E$. In
the following, constants written $C$, $C^{\\#}$, etc., are assumed to depend
only on $m$ and $n$. We write
$\|\varphi\|=\|\varphi\|_{C^{m-1,1}(\mathbb{R}^{n})}$ for the $C^{m-1,1}$
seminorm of a function $\varphi\in C^{m-1,1}(\mathbb{R}^{n})$.

Fix a finite set $E\subseteq\mathbb{R}^{n}$ and function
$f:E\rightarrow\mathbb{R}$. We assume the data $(E,f)$ satisfies the
hypotheses of the finiteness theorem; namely, we assume the following
_finiteness hypothesis_ is valid:

$\mathcal{(FH)}\quad\left\\{\begin{aligned} &\mbox{for any subset }S\subseteq
E\mbox{ with }\\#(S)\leq k^{\\#}\\\ &\mbox{there exists a }C^{m-1,1}\mbox{
function }F^{S}:\mathbb{R}^{n}\rightarrow\mathbb{R}\\\ &\mbox{satisfying
}F^{S}=f\mbox{ on }S\mbox{ and }\|F^{S}\|\leq 1.\end{aligned}\right.$

We assume $k^{\\#}$ in the finiteness hypothesis is a sufficiently large
constant determined by $m$ and $n$.

To prove the finiteness theorem, we will construct an $F\in
C^{m-1,1}(\mathbb{R}^{n})$ satisfying $F=f$ on $E$ and $\|F\|\leq C^{\\#}$ for
a constant $C^{\\#}$ determined by $m$ and $n$. That is, we will construct an
interpolant $F\in C^{m-1,1}(\mathbb{R}^{n})$ of $f$ with $C^{m-1,1}$-seminorm
at most $C^{\\#}$.

Let $\mathcal{P}$ be the vector space of real-valued polynomials on
$\mathbb{R}^{n}$ of degree $\leq m-1$. Write $J_{x}(\varphi)$ to denote the
$(m-1)^{\text{rst}}$ order Taylor polynomial at $x$ of a function $\varphi\in
C^{m-1,1}(\mathbb{R}^{n})$, defined by

$J_{x}(\varphi)(z):=\sum_{|\alpha|\leq
m-1}(\partial^{\alpha}\varphi(x)/\alpha!)(z-x)^{\alpha}.$

We call $J_{x}(\varphi)\in\mathcal{P}$ the $(m-1)$-jet of $\varphi$ at $x$. We
define a ring product $\odot_{x}$ on $\mathcal{P}$ by defining
$P\odot_{x}Q=J_{x}(PQ)$ for $P,Q\in\mathcal{P}$. We write $\mathcal{R}_{x}$
for the ring $(\mathcal{P},\odot_{x})$.

Fefferman’s papers on the Whitney extension problem (e.g., [8, 9, 10, 14, 15,
16]) introduce and make extensive use of a family of convex subsets
$\sigma(x)\subseteq\mathcal{P}$, indexed by $x\in E$. Informally, the set
$\sigma(x)$ measures the freedom in choosing the $(m-1)$-jet $J_{x}(F)$ for an
interpolant $F\in C^{m-1,1}(\mathbb{R}^{n})$ of $f$. Let

$\sigma(x):=\\{J_{x}(\varphi):\varphi|_{E}=0,\;\|\varphi\|\leq
1\\}\subseteq\mathcal{P}.$

Note that if $J_{x}(F_{1})=P_{1}$ and $J_{x}(F_{2})=P_{2}$ for two different
interpolants $F_{1},F_{2}$ of $f$, and if $\|F_{1}\|\leq M$ and $\|F_{2}\|\leq
M$ for some $M>0$, then $P_{1}-P_{2}$ belongs to $2M\sigma(x)$. Indeed,
$\varphi:=F_{1}-F_{2}$ satisfies $\varphi|_{E}=0$ and $\|\varphi\|\leq 2M$;
hence, $P_{1}-P_{2}=J_{x}(\varphi)\in 2M\sigma(x)$. Thus, the (dilates) of
$\sigma(x)$ can be used to control the freedom in the choice of $J_{x}(F)$ for
an interpolant $F$ of $f$ on $E$ of bounded seminorm.

A key idea in Fefferman’s proof of the finiteness theorem is to index an
interpolation problem by a _label_ 222A label is a multi-index set
$\mathcal{A}=\\{\alpha_{1},\cdots,\alpha_{L}\\}$ with each $\alpha_{i}$ a
multiindex of order at most $m-1$. $\mathcal{A}$ which records information on
the “large coordinate directions” in the set $\sigma(x)$. Fefferman introduces
an order relation $<$ on labels, which can be used to sort interpolation
problems according to their “difficulty”. By a divide and conquer approach, he
decomposes an interpolation problem with a given label $\mathcal{A}$ into a
family of easier interpolation problems with smaller labels
$\mathcal{A}^{\prime}<\mathcal{A}$. The proof is organized as an induction on
the label assigned to a given interpolation problem. For details, see [15].

In a joint work [6] with B. Klartag, we gave a coordinate-free proof of the
finiteness theorem. To accomplish this we explained how to replace the notion
of a label in Fefferman’s inductive scheme by the notion of a _DTI subspace_.
We record information on the large directions in $\sigma(x)$ by specifying
that a DTI subspace is _transverse_ to $\sigma(x)$. We mimic Fefferman’s
divide and conquer strategy. However, one crucial difference is that our proof
is organized as an induction with respect to an integer-valued quantity called
the _complexity_ of $E$. Roughly speaking, the complexity of $E$ measures how
often the geometry of the set $\sigma(x)$ changes dramatically as one applies
a rescaling transformation about a fixed point $x\in E$.

Let $V$ be a subspace of $\mathcal{P}$. We say that $V$ is _dilation-and-
translation-invariant_ , or _DTI_ , provided that (1) $V$ is _dilation-
invariant_ , i.e., $P(\cdot/\delta)\in V$ for all $P\in V$, $\delta>0$ and (2)
$V$ is _translation-invariant_ , i.e., $P(\cdot-h)\in V$ for all $P\in V$,
$h\in\mathbb{R}^{n}$. These conditions on $V$ can be reformulated as follows:
A subspace $V$ is dilation-invariant provided that
$V=\bigoplus_{i=0}^{m-1}V_{i}$, where
$V_{i}\subseteq\mathcal{P}_{i}:=\mathrm{span}\\{x^{\alpha}:|\alpha|=i\\}$ is a
_homogeneous subspace_ of $\mathcal{P}$, for $i=0,1,2,\dots,m-1$. Further, a
subspace $V$ is translation-invariant if and only if the orthogonal complement
$V^{\perp}$ of $V$ with respect to a natural inner product333This claim is
valid, e.g., for the inner product $\langle
P,Q\rangle^{\prime}:=\sum_{|\alpha|\leq
m-1}\frac{1}{\alpha!}\partial^{\alpha}P(0)\partial^{\alpha}Q(0)$ for
$P,Q\in\mathcal{P}$; see Lemma 3.11 of [6]. We will make use of another inner
product $\langle\cdot,\cdot\rangle$ on $\mathcal{P}$ later in the paper. on
$\mathcal{P}$ satisfies that $V^{\perp}$ is an ideal in the ring of
$(m-1)$-jets $\mathcal{R}_{0}=(\mathcal{P},\odot_{0})$ based at $x=0$. It
follows that the DTI subspaces $V$ are orthogonal to those ideals $I$ in
$\mathcal{R}_{0}$ which admit a direct sum decomposition into homogeneous
subspaces.

We assign a _DTI label_ $V$ to the set $E$ at position $x\in E$ and scale
$\delta>0$ provided that $V$ is a DTI subspace of $\mathcal{P}$, while
$\sigma(x)$ and $V$ satisfy a quantitative _transversality condition_ at
$(x,\delta)$. Roughly speaking, the transversality condition states that the
“big directions” in $\sigma(x)$ do not make a small angle with $V$, and the
intersection $V\cap\sigma(x)$ is suitably small. Here, to make sense of
angles, we equip the vector space $\mathcal{P}$ with a suitable inner product
$\langle\cdot,\cdot\rangle_{x,\delta}$. See Definition 7.6 for the precise
statement of the transversality condition.

We associate to a point $x\in E$ a sequence of DTI subspaces

$V_{1},V_{2},\dots,V_{L}$

and lengthscales

$\delta_{1}>\delta_{2}>\dots>\delta_{L}$

such that $V_{\ell}$ is a DTI label assigned to $E$ at position $x$ and scale
$\delta_{\ell}$ $(\ell\leq L)$, and $V_{\ell}$ is not a DTI label assigned to
$E$ at position $x$ and scale $\delta_{\ell+1}<\delta_{\ell}$ ($\ell<L$). We
denote by $\mathcal{C}(E)$ the supremal length of any such sequence associated
to any $x\in E$. By convention, $\mathcal{C}(E)=0$ if $E=\emptyset$. Borrowing
notation from our earlier work [6], we refer to the quantity $\mathcal{C}(E)$
as the _complexity_ of $E$. It is evident from the definition that complexity
is locally monotone with respect to inclusion, in the sense that
$\mathcal{C}(E\cap B)\geq\mathcal{C}(E\cap B^{\prime})$ whenever
$B^{\prime}\subseteq B\subseteq\mathbb{R}^{n}$. To construct an extension $F$
of $f$ of bounded $C^{m-1,1}$ norm, we proceed by induction on
$\mathcal{C}(E)$.

The base case of the induction corresponds to the case $\mathcal{C}(E)=0$. If
$\mathcal{C}(E)=0$ it easily follows that $E$ is the empty set, whence it is
trivially true that there exists an extension of $f$ on $E$ of bounded
$C^{m-1,1}$ seminorm.

For the induction step, we assume the induction hypothesis that the finiteness
theorem is true for any data $(\widetilde{E},\widetilde{f})$ satisfying that
$\mathcal{C}(\widetilde{E})<L_{0}$ for fixed $L_{0}\geq 1$. We then fix data
$(E,f)$ satisfying the hypotheses of the finiteness theorem, with
$\mathcal{C}(E)=L_{0}$. To complete the induction step we must construct an
interpolant $F$ of $f$ with $\|F\|\leq C$.

Fix a closed ball $B_{0}\subseteq\mathbb{R}^{n}$ with $E\subseteq B_{0}$ and
$\operatorname{diam}(B_{0})=\operatorname{diam}(E)$. We define a cover of
$B_{0}$ by a family $\mathcal{W}$ of closed balls in $\mathbb{R}^{n}$; thus,
$B_{0}\subseteq\bigcup_{B\in\mathcal{W}}B$. We construct the cover
$\mathcal{W}$ to have the following properties: First, $\mathcal{C}(E\cap
B)<\mathcal{C}(E)=L_{0}$ for all $B\in\mathcal{W}$. On the other hand,
$\mathcal{C}(E\cap 100B)=\mathcal{C}(E)=L_{0}$ for all $B\in\mathcal{W}$.
Finally, the cover $\mathcal{W}$ has _good geometry_ in the sense that for
every $B\in\mathcal{W}$ we have $B\cap B^{\prime}\neq\emptyset$ for at most
$C$ balls $B^{\prime}\in\mathcal{W}$; also, if $B\cap B^{\prime}\neq\emptyset$
for $B,B^{\prime}\in\mathcal{W}$ then $\operatorname{diam}(B)$ and
$\operatorname{diam}(B^{\prime})$ differ by a factor of at most $K$. Here,
$C=C(n)$ and $K=K(n)$ are appropriate dimensional constants.

Evidently, it is sufficient to construct an interpolant $F$ of $f$ on $B_{0}$,
satisfying $\|F\|_{C^{m-1,1}(B_{0})}\leq C$. For then, it is trivial to extend
$F$ to all of $\mathbb{R}^{n}$, while not increasing the $C^{m-1,1}$-seminorm
by more than a constant factor.

By the induction hypothesis applied to the set $\widetilde{E}=E\cap B$, for
each $B\in\mathcal{W}$ there exists a _local interpolant_ $F_{B}$ of $f$ on
$E\cap B$ satisfying two conditions: (local interpolation) $F_{B}=f$ on $E\cap
B$ and (bounded seminorm) $\|F_{B}\|\leq M$ for all $B\in\mathcal{W}$. Here,
$M$ will be a constant determined by $m$,$n$ and the induction index $L_{0}$.
So $\\{F_{B}\\}_{B\in\mathcal{W}}$ is a family of local interpolants
associated to the balls in the cover $\mathcal{W}$. We define

$F=\sum_{B\in\mathcal{W}}F_{B}\theta_{B}\mbox{ on }B_{0},$

where $\\{\theta_{B}\\}_{B\in\mathcal{W}}$ is a partition of unity on $B_{0}$
(thus, $\sum_{B}\theta_{B}=1$ on $B_{0}$), while each $\theta_{B}$ is
supported on $B$, $\theta_{B}\equiv 1$ near the center of $B$, and each
partition function $\theta_{B}$ satisfies the derivative bounds
$\|\partial^{\alpha}\theta_{B}\|_{L^{\infty}}\leq
C\operatorname{diam}(B)^{-|\alpha|}$ for $|\alpha|\leq m$. Such a partition of
unity is guaranteed to exist by the covering and good geometry properties of
$\mathcal{W}$. Evidently, since $F_{B}=f$ on $E\cap B$ for all
$B\in\mathcal{W}$, we have $F=f$ on $E$. We hope to prove that
$\|F\|_{C^{m-1,1}(B_{0})}\leq\widetilde{C}M$ for a constant $\widetilde{C}$
determined by $m$ and $n$. Unfortunately, there is no reason to expect this to
be true, given that the $F_{B}$ were chosen independently of one another. By
following the ideas in [6] (inspired by analogous ideas in [15]), we construct
local interpolants $F_{B}$ which are compatible with one another – to enforce
these compatibility conditions, we modify by a small additive correction
function the $F_{B}$ specified above. We now state the extra compatibility
conditions on the $F_{B}$. First we establish the existence of a DTI subspace
$V$ that is transverse to $\sigma(x)$ for each $x\in E$ at some scale
$\delta>0$. Then fix an appropriate jet $P_{0}\in\mathcal{P}$ (determined by
the data $(f,E)$) and specify that $J_{x_{B}}F_{B}\in P_{0}+V$ for every
$B\in\mathcal{W}$; here $x_{B}$ is a specified point of $B$. Essentially, the
compatibility conditions state that $J_{x_{B}}F_{B}$ belongs to the same coset
of $V$ for every $B\in\mathcal{W}$. These are the extra conditions required of
the local interpolants $F_{B}$, beyond those stated before. For a family of
local interpolants $F_{B}$ satisfying the aforementioned conditions, we can
prove that
$\|F\|_{C^{m-1,1}(B_{0})}\leq\widetilde{C}(m,n)\max_{B}\|F_{B}\|\leq\widetilde{C}(m,n)M$
for the $F$ defined before. Since $F$ is an interpolant of $f$, this completes
the induction step. As a final remark, we note that to carry out the above
modification step and prove the existence of local solutions $F_{B}$
satisfying the extra compatibility conditions, it is required to bring in the
finiteness hypothesis $\mathcal{(FH)}$ and certain convex sets
$\Gamma_{\ell}(x,f,M)$ (these being sometimes referred to as
$\mathcal{K}_{f}(x;k,M)$ in Fefferman’s work). We spare the details in this
sketch.

Thus we have shown, by induction on $\mathcal{C}(E)$, that there exists an
extension $F$ of $f$ with norm at most $\widetilde{C}^{\mathcal{C}(E)}$, where
$\widetilde{C}$ is a fixed constant determined by $m$ and $n$. To see this,
note that the bound on the norm of the extension $F$ increases by a factor of
$\widetilde{C}=\widetilde{C}(m,n)$ at each step of the induction proof.

To conclude the proof of the finiteness theorem, we must demonstrate that the
complexity $\mathcal{C}(E)$ is bounded uniformly for all finite subsets
$E\subseteq\mathbb{R}^{n}$. We define the worst-case complexity $L_{\max}$ by

$L_{\max}:=\sup_{E\subseteq\mathbb{R}^{n}}\mathcal{C}(E),$

where the supremum is over finite sets $E\subseteq\mathbb{R}^{n}$. In [6], we
demonstrated that $L_{\max}$ is bounded by a constant $C(D)$ determined by
$D=\binom{n+m-1}{n}$. Our proof used semialgebraic geometry, resulting in poor
dependence $C(D)\gtrsim\exp(\exp(D))$. Also in [6], we conjectured that

$L_{\max}\lesssim\operatorname{poly}(D).$ (1)

The first main technical result of this paper, Proposition 2.11, establishes
the conjecture (1). More specifically, in Section 4, we prove that
$L_{\max}\leq 4mD^{2}$.

By our discussion above, we can construct an extension $F$ of
$f:E\rightarrow\mathbb{R}$ with
$||F||_{C^{m-1,1}(\mathbb{R}^{n})}\leq\widetilde{C}^{L_{\max}}$ for any finite
set $E\subseteq\mathbb{R}^{n}$. Combining this with (1) gives
$||F||_{C^{m-1,1}(\mathbb{R}^{n})}\leq\widetilde{C}^{\operatorname{poly}(D)}$.
Therefore to establish Theorem 1.3 it just remains to show that

$\widetilde{C}\lesssim\exp(\operatorname{poly}(D)).$ (2)

Indeed, (2) follows from a careful bookkeeping of various constants appearing
in the proof, and our second main technical result, Proposition 2.9, which we
prove in Section 5.

This completes our sketch of the proof of Theorem 1.3.

To establish Theorem 1.4, we show that our construction can be modified so
that, for a fixed set $E$, the extension $F$ depends linearly on the data $f$.

We finish the introduction by describing the content of Sections 6-10 in more
detail.

Section 6 contains the statement of our main extension theorem for finite sets
$E\subseteq\mathbb{R}^{n}$.

Section 7 contains the definitions of the convex sets $\sigma(x)$ and their
variants, and gives results on the basic properties of these sets.

Section 8 contains additional technical results (many borrowed from [14])
needed for the proof of Theorem 1.4.

Sections 9–10 contain the main analytic ingredients of the paper, including
the Main Decomposition Lemma (Lemma 10.2), which is the apparatus used to
decompose the extension problem for $(E,f)$ into easier subproblems.

Finally, Section 11 contains the proof of the extension theorem for finite
$E$, and the proofs of the theorems from the introduction (Theorems 1.3 and
1.4).

The notation and terminology in the previous discussion is not necessarily
used in the rest of the paper. This discussion captures the spirit of the
proof of our theorems, but some of the definitions given above are simplified
for ease of explanation. In particular, the phrase “DTI label” does not appear
in the remainder of the paper, nor in our earlier work [6]. Furthermore, the
definition of complexity and the description of the properties of the cover
$\mathcal{W}$ are presented somewhat differently than in the main body of the
paper – for instance, certain technical constants have been obscured in the
above discussion to simplify the exposition.

####  1.0.1 Acknowledgements

We are grateful to the participants of the 14th Whitney Problems Workshop for
their interest in this work. We are particularly grateful to Charles Fefferman
and Bo’az Klartag, for providing valuable comments on an early draft of this
paper. We are also grateful to the National Science Foundation and the Air
Force Office of Scientific Research for their generous financial
support.444The first-named author acknowledges the support of AFOSR grant
FA9550-19-1-0005. The third-named author acknowledges the support of NSF grant
DMS-1700404 and AFOSR grant FA9550-19-1-0005. Last, we would like to thank the
anonymous referee, whose feedback led to improvements in the paper.

##  2 Notation and preliminaries

Fix $m\geq 2$, $n\geq 1$ throughout the paper. Let $D:=\binom{m+n-1}{n}$.

We write $B(x,r)=\\{z\in\mathbb{R}^{n}:|z-x|\leq r\\}$ for the closed ball of
radius $r$ and center $x$ in $\mathbb{R}^{n}$.

Given a ball $B\subseteq\mathbb{R}^{n}$ and $\lambda>0$, let $\lambda B$
denote the ball with the same center as $B$ and radius equal to $\lambda$
times the radius of $B$.

For any finite set $S$, write $\\#(S)$ to denote the number of elements of
$S$. If $S$ is infinite, we put $\\#(S)=\infty$.

Let
$\mathcal{M}:=\\{\alpha=(\alpha_{1},\alpha_{2},\dots,\alpha_{n}):|\alpha|=\alpha_{1}+\alpha_{2}+\dots+\alpha_{n}\leq
m-1\\}$ be the set of all multiindices of length $n$ and order at most $m-1$.
Then $\\#(\mathcal{M})=D$.

###  2.1 Convention on constants

By an “absolute constant” we mean a numerical constant whose value is
independent of $m$ and $n$.

Given quantities $A,B\geq 0$, we write $A=O(B)$ to indicate that $A\leq\gamma
B$ for an absolute constant $\gamma>0$. We write $\operatorname{poly}(x)$ to
denote a polynomial $\operatorname{poly}(x)=\sum_{k=0}^{d}a_{k}x^{k}$ with
coefficients $a_{k}$ and maximum degree $d$ given by absolute constants.
Similarly, we write $\operatorname{poly}(x,y)$ to denote a polynomial in two
variables with coefficients and maximum degree given by absolute constants.

We say that $C>0$ is a _controlled constant_ if $C$ depends only on $m$, $n$
and both $1/C$ and $C$ are $O(\exp(\operatorname{poly}(D)))$. Note that the
product of $O(\operatorname{poly}(D))$ many controlled constants is again a
controlled constant.

Provided $m\geq 2$, the binomial coefficient $D=\binom{m+n-1}{n}$ satisfies
$\max\\{m,n\\}\leq D$. So, if both $C$ and $1/C$ are
$O(\exp(\operatorname{poly}(m,n)))$ then $C$ is a controlled constant.

We say that two quantities $X,Y\geq 0$ are _equivalent up to a controlled
constant_ if $C^{-1}Y\leq X\leq CY$ for a controlled constant $C$.

###  2.2 Function spaces $C^{m-1,1}$ and $\dot{C}^{m}$

Let $G\subseteq\mathbb{R}^{n}$ be a convex domain with nonempty interior. We
write $C^{m-1,1}(G)$ to denote the space of all $(m-1)$-times differentiable
functions $F:G\rightarrow\mathbb{R}$ whose $(m-1)$-st order partial
derivatives are Lipschitz continuous on $G$, equipped with the seminorm

$\|F\|_{C^{m-1,1}(G)}:=\sup_{x,y\in
G}\left(\sum_{|\alpha|=m-1}\frac{(\partial^{\alpha}F(x)-\partial^{\alpha}F(y))^{2}}{|x-y|^{2}}\right)^{1/2}.$
(3)

We define the space $\dot{C}^{m}(G)$ to consist of all $m$-times continuously
differentiable functions $F:G\rightarrow\mathbb{R}$ whose $m$-th order partial
derivatives are uniformly bounded on $G$, equipped with the seminorm

$\|F\|_{\dot{C}^{m}(G)}:=\sup_{z\in G}\max_{|\beta|=m}|\partial^{\beta}F(z)|.$
(4)

Let $F\in\dot{C}^{m}(G)$. Given a multiindex $\alpha$ with $|\alpha|=m-1$, the
Mean Value Theorem implies that the difference quotient
$|\partial^{\alpha}F(x)-\partial^{\alpha}F(y)|/|x-y|$ is bounded by
$\sup_{z\in[x,y]}|\nabla\partial^{\alpha}F(z)|$, where $[x,y]$ is the line
segment connecting $x$ and $y$ (contained in $G$). The latter quantity is
bounded by $\sqrt{n}\cdot\|F\|_{\dot{C}^{m}(G)}$. Therefore, if
$F\in\dot{C}^{m}(G)$ then $F\in C^{m-1,1}(G)$ and

$\|F\|_{C^{m-1,1}(G)}\leq C\|F\|_{\dot{C}^{m}(G)},$ (5)

for a controlled constant $C$.

We write $C^{m-1}_{loc}(\mathbb{R}^{n})$ to denote the space of all functions
$F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $F\in\dot{C}^{m-1}(B(0,R))$
for any $R>0$.

###  2.3 Jet space

Let $\mathcal{P}$ denote the vector space of all polynomials on
$\mathbb{R}^{n}$ of degree at most $m-1$. Then $\mathcal{P}$ admits a basis of
monomials,
$\mathfrak{V}_{x}:=\\{m_{\alpha,x}(z):=(z-x)^{\alpha}:\alpha\in\mathcal{M}\\}$
for any $x\in\mathbb{R}^{n}$. In particular,
$\dim(\mathcal{P})=\\#(\mathcal{M})=D$.

Given $x\in\mathbb{R}^{n}$ and $F\in C^{m-1}_{loc}(\mathbb{R}^{n})$, let
$J_{x}(F)\in\mathcal{P}$ denote the $(m-1)$-jet of $F$ at $x$, given by

$J_{x}(F)(z):=\sum_{|\alpha|\leq
m-1}(\partial^{\alpha}F(x)/\alpha!)\cdot(z-x)^{\alpha}.$

We endow $\mathcal{P}$ with a product $\odot_{x}$ (“jet multiplication at
$x$”) defined by $P\odot_{x}Q=J_{x}(P\cdot Q)$ for $P,Q\in\mathcal{P}$. We
write $\mathcal{R}_{x}$ to denote the ring $(\mathcal{P},\odot_{x})$ of
$(m-1)$-jets at $x$. We write $\odot=\odot_{0}$ for the jet product at $x=0$.

Note that if $F,G\in C^{m-1}_{loc}(\mathbb{R}^{n})$ then $J_{x}(F\cdot
G)=J_{x}(F)\odot_{x}J_{x}(G)$. That is,
$J_{x}:C^{m-1}_{loc}(\mathbb{R}^{n})\rightarrow\mathcal{R}_{x}$ is a ring
isomorphism.

We often use the notation $\mathcal{P}$ and $\mathcal{R}_{x}$ interchangeably.
We shall use $\mathcal{P}$ when the ring structure of the jet space is
irrelevant to the intended application.

####  2.3.1 Translations and dilations

The jet space $\mathcal{P}$ inherits the structure of translations and
dilations from $\mathbb{R}^{n}$. Specifically, we let
$\tau^{h}:\mathcal{P}\rightarrow\mathcal{P}$ ($h\in\mathbb{R}^{n}$) and
$\tau_{x,\delta}:\mathcal{P}\rightarrow\mathcal{P}$ ($x\in\mathbb{R}^{n}$,
$\delta>0$) be translation and dilation operators defined by

$\displaystyle\tau^{h}(P)(z):=P(z-h),\mbox{ and}$ (6)
$\displaystyle\tau_{x,\delta}(P)(z):=\delta^{-m}P(x+\delta\cdot(z-x))\qquad(P\in\mathcal{P}).$

####  2.3.2 Inner products and norms

Let $x\in\mathbb{R}^{n}$. We define the inner product $\langle P,Q\rangle_{x}$
of $P,Q\in\mathcal{P}$ by

$\langle P,Q\rangle_{x}:=\sum_{|\alpha|\leq
m-1}\partial^{\alpha}P(x)\partial^{\alpha}Q(x)/(\alpha!)^{2}.$

The corresponding norm $|P|_{x}$ of $P\in\mathcal{P}$ is given by

$|P|_{x}:=\sqrt{\langle P,P\rangle_{x}}=\sqrt{\sum_{|\alpha|\leq
m-1}(\partial^{\alpha}P(x))^{2}/(\alpha!)^{2}}.$

The purpose of the $1/(\alpha!)^{2}$ factor in the above expressions is to
ensure the monomials $m_{\alpha,x}(z):=(z-x)^{\alpha}$ have unit length, i.e.,
$|m_{\alpha,x}|_{x}=1$ for $|\alpha|\leq m-1$.

For $x\in\mathbb{R}^{n}$, $\delta>0$, we define the _scaled inner product_
$\langle P,Q\rangle_{x,\delta}$ of $P,Q\in\mathcal{P}$ by

$\displaystyle\langle P,Q\rangle_{x,\delta}$
$\displaystyle:=\langle\tau_{x,\delta}(P),\tau_{x,\delta}(Q)\rangle_{x}$
$\displaystyle=\sum_{|\alpha|\leq
m-1}\frac{1}{(\alpha!)^{2}}\delta^{2(|\alpha|-m)}\partial^{\alpha}P(x)\cdot\partial^{\alpha}Q(x).$

The associated _scaled norm_ $|P|_{x,\delta}$ of $P\in\mathcal{P}$ is

$|P|_{x,\delta}:=\sqrt{\langle
P,P\rangle_{x,\delta}}=\biggl{(}\sum_{|\alpha|\leq
m-1}\frac{1}{(\alpha!)^{2}}\cdot(\delta^{|\alpha|-m}\cdot\partial^{\alpha}P(x))^{2}\biggr{)}^{\frac{1}{2}}.$

The closed unit ball for the scaled norm $|\cdot|_{x,\delta}$ is denoted by

$\mathcal{B}_{x,\delta}:=\biggl{\\{}P:\;|P|_{x,\delta}\leq
1\biggr{\\}}\subseteq\mathcal{P}.$

For fixed $x$ the monomial basis
$\mathfrak{V}_{x}:=\\{m_{\alpha,x}:|\alpha|\leq m-1\\}$ is orthogonal in
$\mathcal{P}$, with respect to the scaled inner product
$\langle\cdot,\cdot\rangle_{x,\delta}$ for any $\delta>0$. The monomial basis
$\mathfrak{V}_{x}$ is orthonormal in $\mathcal{P}$ only for the inner product
$\langle\cdot,\cdot\rangle_{x}=\langle\cdot,\cdot\rangle_{x,1}$.

For any $\delta\geq\rho>0$, and $P\in\mathcal{P}$,

$\left(\frac{\rho}{\delta}\right)^{m}\cdot|P|_{x,\rho}\leq|P|_{x,\delta}\leq\left(\frac{\rho}{\delta}\right)\cdot|P|_{x,\rho}.$
(7)

Therefore,

$\left(\frac{\delta}{\rho}\right)\mathcal{B}_{x,\rho}\subseteq\mathcal{B}_{x,\delta}\subseteq\left(\frac{\delta}{\rho}\right)^{m}\mathcal{B}_{x,\rho}.$
(8)

In particular,

$|P|_{x,\delta}\leq|P|_{x,\rho},\;\mbox{and}\;\mathcal{B}_{x,\rho}\subseteq\mathcal{B}_{x,\delta}\mbox{
for }\delta\geq\rho>0.$ (9)

Observe that $|P|_{x,\delta}=|\tau_{x,\delta}P|_{x}$ for $P\in\mathcal{P}$. It
follows that

$\tau_{x,r}\mathcal{B}_{x,\delta}=\mathcal{B}_{x,\delta/r}.$ (10)

Note that $\langle\cdot,\cdot\rangle_{x,1}=\langle\cdot,\cdot\rangle_{x}$ and
$|\cdot|_{x,1}=|\cdot|_{x}$ for $x\in\mathbb{R}^{n}$. When $x=0$, we write
$\langle P,Q\rangle=\langle P,Q\rangle_{0,1}$ and $|P|=|P|_{0,1}$ for the
_standard inner product and norm_ on $\mathcal{P}$. Write
$\mathcal{B}=\mathcal{B}_{0,1}$ to denote the closed unit ball for the
standard norm on $\mathcal{P}$.

Unless stated otherwise, we equip $\mathcal{P}$ by default with the standard
norm and inner product.

We write
$\mathcal{P}_{i}=\mathrm{span}\\{x^{\alpha}:|\alpha|=i\\}\subseteq\mathcal{P}$
to denote the subspace of homogeneous polynomials of degree $i$.

We require bounds on the norm of a product of polynomials. These bounds are
sometimes referred to in the literature as Bombieri inequalities. Recall that
$\odot$ is the jet product at $x=0$.

######  Lemma 2.1.

Let $C_{b}:=(m+1)!$. Then

$\displaystyle|P\odot Q|\leq C_{b}|P|\cdot|Q|\quad(P,Q\in\mathcal{P})$ (11)
$\displaystyle|P\odot Q|\geq
C_{b}^{-1}|P|\cdot|Q|\quad(P\in\mathcal{P}_{i},Q\in\mathcal{P}_{j},\;i+j<m).$
(12)

###### Proof.

We use two inequalities from [2], stated below in (13). Our standard norm on
$\mathcal{P}$ is given by $|P|=\sqrt{\sum c_{\alpha}^{2}}$ if $P=\sum
c_{\alpha}x^{\alpha}$. In [2] this is called the _$2$ -norm_ and denoted by
$|P|_{2}$. From [2] (see Proposition 1.B.3 and Theorem 1.1), the following
holds: If $P\in\mathcal{P}_{i}$ and $Q\in\mathcal{P}_{j}$ for $i+j<m$, then

$((i+j)!)^{-1/2}|P||Q|\leq|P\cdot Q|\leq 2^{(i+j)/2}|P||Q|.$ (13)

Note that $P\cdot Q=P\odot Q$ if $P\in\mathcal{P}_{i}$ and
$Q\in\mathcal{P}_{j}$ for $i+j<m$; else, if $P\in\mathcal{P}_{i}$ and
$Q\in\mathcal{P}_{j}$ for $i+j\geq m$ then $P\odot Q=0$. Therefore, the left-
hand inequality in (13) implies (12).

Now let $P,Q\in\mathcal{P}$. Write $P=\sum_{i<m}P_{i}$ and $Q=\sum_{i<m}Q_{i}$
for $P_{i},Q_{i}\in\mathcal{P}_{i}$. Then $|P|=\sqrt{\sum|P_{i}|^{2}}$ and
$|Q|=\sqrt{\sum|Q_{i}|^{2}}$ by orthogonality of the homogeneous subspaces
$\mathcal{P}_{i}$. Also, $P\odot Q=\sum_{i+j<m}P_{i}\cdot Q_{j}$. By the
triangle inequality, and the right-hand inequality in (13),

$|P\odot Q|\leq\sum_{i+j<m}|P_{i}\cdot Q_{j}|\leq
2^{m/2}\cdot\sum_{i+j<m}|P_{i}|\cdot|Q_{j}|\leq
2^{m/2}\left(\sum_{i<m}|P_{i}|\right)\cdot\left(\sum_{j<m}|Q_{j}|\right).$

By Cauchy-Schwartz,
$\sum_{i<m}|P_{i}|\leq\sqrt{m}\sqrt{\sum_{i<m}|P_{i}|^{2}}$, and similarly for
the $Q_{i}$. Hence,

$|P\odot Q|\leq
m2^{m/2}\sqrt{\sum_{i<m}|P_{i}|^{2}}\sqrt{\sum_{i<m}|Q_{i}|^{2}}=m2^{m/2}|P||Q|.$
(14)

Observe that $m2^{m/2}\leq(m+1)!$. Thus, (14) implies (11). ∎

######  Proposition 2.2 (Taylor’s Theorem).

Let $G$ be a convex domain with nonempty interior. There exists a controlled
constant $C_{T}\geq 1$ such that, for all $F\in C^{m-1,1}(G)$, $x,y\in G$, and
$\delta\geq|x-y|$,

$|J_{x}F-J_{y}F|_{x,\delta}\leq C_{T}\|F\|_{C^{m-1,1}(G)}.$ (15)

###### Proof.

Taylor’s theorem implies that if $F\in C^{m-1,1}(G)$, $x,y\in G$, and
$|\beta|\leq m-1$ then

$|\partial^{\beta}(J_{x}F-J_{y}F)(x)|\leq
C\cdot\|F\|_{C^{m-1,1}(G)}\cdot|x-y|^{m-|\beta|}.$

for a controlled constant $C$. Thus, for $\delta\geq|x-y|$, we obtain:

$\delta^{|\beta|-m}|\partial^{\beta}(J_{x}F-J_{y}F)(x)|\leq
C\cdot\|F\|_{C^{m-1,1}(G)}.$

Now square both sides of the above inequality, divide by $(\beta!)^{2}$, sum
over $\beta$ with $|\beta|\leq m-1$, and take the square root, to obtain (15).
∎

####  2.3.3 The Classical Whitney Extension Theorem

We make use of the classical Whitney Extension Theorem for $(m-1)$-jets. We
state the result here in a convenient form for later use.

Let $E\subseteq\mathbb{R}^{n}$. Suppose we are given a family of polynomials
$P_{x}\in\mathcal{P}$, indexed by $x\in E$. We use the notation
$P_{\bullet}:E\to\mathcal{P}$ to denote the polynomial-valued map
$P_{\bullet}:x\mapsto P_{x}$. We refer to $P_{\bullet}$ as a _Whitney field_
on $E$. Endow the space of Whitney fields with a seminorm
$\|P_{\bullet}\|_{\mathcal{P}(E)}:=\sup\\{|P_{x}-P_{y}|_{x,|x-y|}:x,y\in
E,\;x\neq y\\}$. We let
$\mathcal{P}(E):=\\{P_{\bullet}:E\rightarrow\mathcal{P}:\|P_{\bullet}\|_{\mathcal{P}(E)}<\infty\\}$.

######  Proposition 2.3 (Classical Whitney Extension Theorem).

There exists a linear map $T:\mathcal{P}(E)\to C^{m-1,1}(\mathbb{R}^{n})$ such
that $\|T(P_{\bullet})\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq
C_{Wh}\|P_{\bullet}\|_{\mathcal{P}(E)}$, and $J_{x}T(P_{\bullet})=P_{x}$ for
all $x\in E$, and all $P_{\bullet}\in\mathcal{P}(E)$. Here, $C_{Wh}$ is a
controlled constant.

We refer the reader to [7], where it is proven that the classical Whitney
extension theorem holds with the constant $C_{Wh}=C_{m}n^{5m/2}$, for a
constant $C_{m}$ determined by $m$. The proof in [7] does not give an explicit
bound on $C_{m}$, but by inspection of the proof one can see that $C_{m}$ is a
polynomial function of $m!$. Therefore, $C_{Wh}$ is controlled.

We now state an elementary consequence of the Whitney extension theorem: We
can extend a $C^{m-1,1}$ function on a convex domain
$G\subseteq\mathbb{R}^{n}$ to all of $\mathbb{R}^{n}$, with control on the
$C^{m-1,1}$ seminorm of the extension.

######  Lemma 2.4.

Let $G$ be a convex domain in $\mathbb{R}^{n}$ with nonempty interior. Let
$F\in C^{m-1,1}(G)$. Then there exists a function $\widehat{F}\in
C^{m-1,1}(\mathbb{R}^{n})$ with $\widehat{F}|_{G}=F$ and
$\|\widehat{F}\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq C\|F\|_{C^{m-1,1}(G)}$, for a
controlled constant $C\geq 1$. Furthermore, $\widehat{F}$ can be taken to
depend linearly on $F$.

###### Proof.

Given $F\in C^{m-1,1}(G)$, define a Whitney field
$P_{\bullet}\in\mathcal{P}(G)$ by $P_{x}=J_{x}F$ for $x\in G$ (note that
$J_{x}F$ is well-defined for $x\in G$ by the hypothesis that $G$ has nonempty
interior). By Taylor’s theorem (Proposition 2.2),
$\|P_{\bullet}\|_{\mathcal{P}(G)}\leq C_{T}\|F\|_{C^{m-1,1}(G)}$. Let
$T:\mathcal{P}(G)\to C^{m-1,1}(\mathbb{R}^{n})$ be as in the classical Whitney
extension theorem, and set $\widehat{F}:=T(P_{\bullet})$. Then $\widehat{F}$
depends linearly on $F$. Because $J_{x}\widehat{F}=P_{x}=J_{x}F$ for all $x\in
G$, we have $\widehat{F}|_{G}=F$. Furthermore,

$\|\widehat{F}\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq
C_{Wh}\|P_{\bullet}\|_{\mathcal{P}(G)}\leq C_{Wh}C_{T}\|F\|_{C^{m-1,1}(G)}.$

This completes the proof of the lemma with $C=C_{T}C_{Wh}$. ∎

####  2.3.4 Graded decomposition of the jet space

Given $x\in\mathbb{R}^{n}$, the jet space $\mathcal{R}_{x}\simeq\mathcal{P}$
admits a _graded decomposition_ into homogeneous vector subspaces.
Specifically,

$\mathcal{R}_{x}=\bigoplus_{i=0}^{m-1}\mathcal{R}_{x}^{i},\quad\mbox{where}\;\mathcal{R}_{x}^{i}:=\operatorname{span}\\{m_{x,\alpha}(z):=(z-x)^{\alpha}:|\alpha|=i\\}.$

Note that $\tau_{x,\delta}(P)=\delta^{i-m}P$ for $P\in\mathcal{R}_{x}^{i}$ –
thus, $\mathcal{R}_{x}^{i}$ is homogeneous of order $i-m$ with respect to the
dilations $\tau_{x,\delta}$ ($\delta>0$). The subspaces $\mathcal{R}_{x}^{i}$
are pairwise orthogonal with respect to the inner product
$\langle\cdot,\cdot\rangle_{x,\delta}$ (any $\delta>0$). Furthermore,
$\operatorname{span}(\mathcal{R}_{i}^{x}\odot_{x}\mathcal{R}_{j}^{x})=\mathcal{R}_{i+j}^{x}$
if $i+j<m$, and $\mathcal{R}_{i}^{x}\odot_{x}\mathcal{R}_{j}^{x}=\\{0\\}$ if
$i+j\geq m$.

####  2.3.5 Dilation and translation invariant subspaces

Let $V$ be a subspace of $\mathcal{P}$. We say $V$ is _translation invariant_
if $\tau^{h}(P)\in V$ for all $P\in V$, $h\in\mathbb{R}^{n}$. Let
$x_{0}\in\mathbb{R}^{n}$. We say $V$ is _dilation invariant at $x_{0}$_ if
$\tau_{x_{0},\delta}(P)\in V$ for all $P\in V$, $\delta>0$. For the
definitions of the translations $\tau^{h}$ and dilations $\tau_{x,\delta}$,
see (6).

Note that $V$ is dilation invariant at $x_{0}$ if and only if $V$ admits a
decomposition

$V=\bigoplus_{i=0}^{m-1}V_{i}^{x_{0}},$

for subspaces $V_{i}^{x_{0}}\subseteq\mathcal{R}_{x_{0}}^{i}$ ($0\leq i\leq
m-1$).

We say $V$ is _DTI_ (_dilation-and-translation-invariant_) if $V$ is both
translation invariant and dilation invariant at $x_{0}$ for some
$x_{0}\in\mathbb{R}^{n}$. If $V$ is DTI then $V$ is dilation invariant at $x$
for all $x\in\mathbb{R}^{n}$, due to the identity
$\tau_{x,\delta}=\tau^{x-x_{0}}\tau_{x_{0},\delta}\tau^{x_{0}-x}$.

A special class of DTI subspaces arises by looking at the span of monomials in
$\mathcal{P}$. Given $\mathcal{A}\subseteq\mathcal{M}$, let
$V_{\mathcal{A}}:=\operatorname{span}\\{x^{\alpha}:\alpha\in\mathcal{A}\\}$.

######  Definition 2.5.

A set $\mathcal{A}\subseteq\mathcal{M}$ is _monotonic_ provided that if
$\alpha\in\mathcal{A}$, $\beta\in\mathcal{M}$, and
$\alpha+\beta\in\mathcal{M}$, then $\alpha+\beta\in\mathcal{A}$.

######  Lemma 2.6.

Let $\mathcal{A}\subseteq\mathcal{M}$. Then the following are equivalent:

  * (i)

$\mathcal{A}$ is monotonic.

  * (ii)

$V_{\mathcal{A}}$ is an ideal in the ring
$\mathcal{R}_{0}=(\mathcal{P},\odot)$.

  * (iii)

$V_{\mathcal{M}\setminus\mathcal{A}}$ is a DTI subspace.

###### Proof.

Recall that $\odot=\odot_{0}$ is the “jet product at $x=0$”. Note that
$V_{\mathcal{A}}$ is an ideal in $\mathcal{R}_{0}$ if and only if
$x^{\beta}\odot P\in V_{\mathcal{A}}$ for every polynomial $P$ in a basis for
$V_{\mathcal{A}}$ and every $\beta\in\mathcal{M}$. Thus, $V_{\mathcal{A}}$ is
an ideal if and only if $x^{\beta}\odot x^{\alpha}\in V_{\mathcal{A}}$ for all
$\beta\in\mathcal{M}$ and $\alpha\in\mathcal{A}$. Observe that $x^{\beta}\odot
x^{\alpha}=0$ if $|\beta|+|\alpha|\geq m$, and else, $x^{\beta}\odot
x^{\alpha}=x^{\alpha+\beta}$ if $|\alpha|+|\beta|\leq m-1$. Thus,
$V_{\mathcal{A}}$ is an ideal if and only if $\alpha+\beta\in\mathcal{A}$
whenever $\alpha\in\mathcal{A}$, $\beta\in\mathcal{M}$, $|\alpha|+|\beta|\leq
m-1$. Therefore, $V_{\mathcal{A}}$ is an ideal if and only if $\mathcal{A}$ is
monotonic, establishing the equivalence of (i) and (ii).

It remains to establish the equivalence of (i) and (iii). Evidently,
$V=V_{\mathcal{M}\setminus\mathcal{A}}$ is dilation invariant at $x_{0}=0$ due
to the fact that $V$ is spanned by monomials based at $x_{0}=0$. Therefore it
suffices to show that $V$ is translation invariant if and only if
$\mathcal{A}$ is monotonic.

Suppose $\mathcal{A}$ is monotonic. By linearity it suffices to show that
$\tau^{h}P\in V$ for any element $P$ in the basis
$\\{x^{\gamma}\\}_{\gamma\in\mathcal{M}\setminus\mathcal{A}}$ for $V$. Fix
$\gamma\in\mathcal{M}\setminus\mathcal{A}$ and $h\in\mathbb{R}^{n}$, and use
the binomial identity to write

$\tau^{h}[x^{\gamma}]=(x-h)^{\gamma}=\sum_{\begin{subarray}{c}\gamma_{1},\gamma_{2}\in\mathcal{M}\\\
\gamma_{1}+\gamma_{2}=\gamma\end{subarray}}c_{\gamma_{1}\gamma_{2}}x^{\gamma_{1}}h^{\gamma_{2}}.$

Since $\mathcal{A}$ is monotonic, and
$\gamma\in\mathcal{M}\setminus\mathcal{A}$, we have
$\gamma_{1}\in\mathcal{M}\setminus\mathcal{A}$ if
$\gamma=\gamma_{1}+\gamma_{2}$. Consequently, each term
$c_{\gamma_{1}\gamma_{2}}x^{\gamma_{1}}h^{\gamma_{2}}$ in the above sum
belongs to $V$. By linearity, $\tau^{h}[x^{\gamma}]\in V$ for any
$\gamma\in\mathcal{M}\setminus\mathcal{A}$. Thus, $V$ is translation
invariant.

Next we suppose $V$ is translation invariant and show that $\mathcal{A}$ is
monotonic. Note the identity $\partial_{x_{i}}P=\lim_{h\rightarrow
0}h^{-1}(P-\tau^{he_{i}}(P))$ where $e_{i}\in\mathbb{R}^{n}$ is the $i$’th
coordinate vector. Because $V$ is translation invariant, this identity implies
that $\partial_{x_{i}}P\in V$ for any $P\in V$. Therefore,
$\partial^{\beta}P\in V$ for $P\in V$ and any multiindex $\beta$. For sake of
contradiction suppose that $\mathcal{A}$ is not monotonic. Then there exist
$\alpha\in\mathcal{A}$, $\beta\in\mathcal{M}$ with
$\alpha+\beta\in\mathcal{M}\setminus\mathcal{A}$. Thus, $x^{\alpha+\beta}\in
V$. Consequently, $\partial^{\beta}x^{\alpha+\beta}\in V$. Note that
$\partial^{\beta}x^{\alpha+\beta}=cx^{\alpha}$ for $c\in\mathbb{R}$, $c\neq
0$. Thus, $x^{\alpha}\in V$, implying that
$\alpha\in\mathcal{M}\setminus\mathcal{A}$, a contradiction.

This completes the proof of the lemma.

∎

####  2.3.6 Whitney convexity

A subset $\Omega$ of a vector space is _symmetric_ provided that
$v\in\Omega\implies-v\in\Omega$.

Given $x\in\mathbb{R}^{n}$, we denote $X\odot_{x}Y:=\\{P\odot_{x}Q:P\in
X,\;Q\in Y\\}$ for subsets $X,Y\subseteq\mathcal{R}_{x}$.

The next definition plays a key role in the theory of $C^{m-1,1}$ extension.

######  Definition 2.7 (Whitney convexity).

Let $x\in\mathbb{R}^{n}$, and let $\Omega\subseteq\mathcal{R}_{x}$ be a closed
symmetric convex set. We say that $\Omega$ is $A$-Whitney convex at $x$ if
$(\Omega\cap\mathcal{B}_{x,\delta})\odot_{x}\mathcal{B}_{x,\delta}\subseteq
A\delta^{m}\Omega$ for all $\delta>0$. If $\Omega$ is $A$-Whitney convex at
$x$ for some $A<\infty$, then we say that $\Omega$ is Whitney convex at $x$.

The Whitney coefficient $w_{x}(\Omega)$ of $\Omega$ at $x$ is the infimum of
all $A>0$ such that $\Omega$ is $A$-Whitney convex at $x$. If no finite $A$
exists, then $w_{x}(\Omega):=+\infty$.

###  2.4 Main technical results

Here, we state the new technical results of this paper. The second result will
be used to affirm a conjecture from the introduction of [6]. Sections 3, 4 and
5 are dedicated to the proofs of these results,

Fix $x\in\mathbb{R}^{n}$. We equip the jet space
$\mathcal{R}_{x}=(\mathcal{P},\odot_{x})$ with the inner product
$\langle\cdot,\cdot\rangle_{x}$ and norm $|\cdot|_{x}$; see Section 2.3.2.
Then $\mathcal{R}_{x}$ is a finite-dimensional Hilbert space, with
$\dim(\mathcal{R}_{x})=D=\binom{m+n-1}{n}$. Let $\mathcal{B}_{x}$ be the unit
ball of $\mathcal{R}_{x}$. We let $\Pi_{V}:\mathcal{R}_{x}\rightarrow V$
denote the orthogonal projection map on a subspace
$V\subseteq\mathcal{R}_{x}$.

######  Definition 2.8.

Let $V$ be a subspace of $\mathcal{R}_{x}$, let $\Omega$ be a closed symmetric
convex subset of $\mathcal{R}_{x}$, and let $R\geq 1$. Say that $\Omega$ is
$R$-transverse to $V$ at $x$ if $\Omega\cap V\subseteq R\mathcal{B}_{x}$ and
$\Pi_{V^{\perp}}(\Omega\cap\mathcal{B}_{x})\supseteq R^{-1}\mathcal{B}_{x}\cap
V^{\perp}$. Here, $V^{\perp}$ is the orthogonal complement of $V$ with respect
to the inner product $\langle\cdot,\cdot\rangle_{x}$ on $\mathcal{R}_{x}$.

Obviously, we can state a corresponding definition of transversalty in a
general finite-dimensional Hilbert space. We do so in Definition 3.7.

We now note a couple of trivial properties of $R$-transversality for the
unfamiliar reader.

  * •

If $\Omega$ is $R$-transverse to $V$, then $\Omega$ is $R^{\prime}$-transverse
to $V$ for any $R^{\prime}\geq R$.

  * •

If $\Omega=V^{\perp}$, then $\Omega$ is $R$-transverse to $V$ for any $R\geq
1$.

Our first technical result is as follows:

######  Proposition 2.9.

Let $x\in\mathbb{R}^{n}$, $A\geq 1$, and $\Omega\subseteq\mathcal{R}_{x}$ be
given. Suppose that $\Omega$ is $A$-Whitney convex at $x$. Then there is a DTI
subspace $V\subseteq\mathcal{R}_{x}$ such that $\Omega$ is $R_{0}$-transverse
to $V$ at $x$. Here, $R_{0}$ is a constant determined by $m$, $n$, and $A$ of
the form $R_{0}=\exp(\operatorname{poly}(D)\log(A))$.

We write $l(I)\leq r(I)$ to denote the left and right endpoints of a compact
interval $I\subseteq\mathbb{R}$, respectively. If $I$ and $J$ are compact
intervals, we write $I>J$ if $l(I)>r(J)$. We write $I>0$ if $l(I)>0$.

######  Definition 2.10.

Let $x\in\mathbb{R}^{n}$. Given a closed symmetric convex set
$\Omega\subseteq\mathcal{R}_{x}$, $\delta>0$, and real numbers
$1<R<R^{*}<\infty$, we define the quantity
$\mathcal{C}_{x}(\Omega,R,R^{*},\delta)$ to be the supremum of all integers
$K$ such that there exist subspaces $V_{k}\subseteq\mathcal{R}_{x}$ and
compact intervals $I_{k}\subseteq(0,\delta]$ ($k=1,2,\dots,K$) such that the
following conditions hold:

  * •

$I_{1}>I_{2}>I_{3}>\cdots>I_{K}>0$

  * •

For all $k$, $\tau_{x,r(I_{k})}\Omega$ is $R$-transverse to $V_{k}$ at $x$.

  * •

For all $k$, $\tau_{x,l(I_{k})}\Omega$ is not $R^{*}$-transverse to $V_{k}$ at
$x$.

  * •

For all $k$, $V_{k}$ is dilation invariant at $x$.

We refer to $\mathcal{C}_{x}(\Omega,R,R^{*},\delta)$ as the pointwise
complexity of $\Omega$ at $x$ at scale below $\delta$ with parameters
$(R,R^{*})$.

If $\delta=\infty$, we set
$\mathcal{C}_{x}(\Omega,R,R^{*})=\mathcal{C}_{x}(\Omega,R,R^{*},\infty)$,
which we refer to as the pointwise complexity of $\Omega$ at $x$ with
parameters $(R,R^{*})$.

Our second technical result provides a bound on the pointwise complexity of a
general closed symmetric convex subset of $\mathcal{R}_{x}$.

######  Proposition 2.11.

Let $x\in\mathbb{R}^{n}$, $\delta>0$, $R\geq 16$, and $R^{*}\geq
D^{2D+1/2}R^{4D}$ be given. Then $\mathcal{C}_{x}(\Omega,R,R^{*},\delta)\leq
4mD^{2}$ for any closed symmetric convex set $\Omega\subseteq\mathcal{R}_{x}$.

###  2.5 Elementary tools and techniques

This section contains elementary lemmas on polynomial inequalities and cutoff
functions. Many of these results were proven in [6] via compactness arguments.
Here we give direct proofs that yield explicit constants.

####  2.5.1 Properties of polynomial norms

We present inequalities for polynomial norms used throughout the paper.

######  Lemma 2.12.

[cf. Lemma 2.1, part (i) in [6]] Let $x,y\in\mathbb{R}^{n}$ and $\delta>0$.
Suppose $|x-y|\leq\eta\delta$ for $0\leq\eta\leq 1$. Then for any
$P\in\mathcal{P}$,

$|P|_{y,\delta}^{2}\leq(1+C\eta)|P|_{x,\delta}^{2}$

for a controlled constant $C$.

###### Proof.

By Taylor’s theorem, for any $\alpha$ with $|\alpha|\leq m-1$ we have

$\partial^{\alpha}P(y)=\sum_{\gamma:|\alpha+\gamma|<m}\frac{1}{\gamma!}(\partial^{\alpha+\gamma}P)(x)\cdot(y-x)^{\gamma}.$

Therefore

$\begin{split}|P|_{y,\delta}^{2}&=\sum_{|\alpha|<m}\frac{\delta^{2(|\alpha|-m)}}{\alpha!}\left(\partial^{\alpha}P(x)+\sum_{\begin{subarray}{c}\gamma>0:\\\
|\alpha+\gamma|<m\end{subarray}}\frac{1}{\gamma!}(\partial^{\alpha+\gamma}P)(x)\cdot(y-x)^{\gamma}\right)^{2}\\\
&=|P|_{x,\delta}^{2}+(\mathrm{R})\end{split}$ (16)

where

$\displaystyle(\mathrm{R})$
$\displaystyle=\sum_{|\alpha|<m}\frac{\delta^{2(|\alpha|-m)}}{\alpha!}\sum_{\begin{subarray}{c}\gamma_{1}>0,\gamma_{2}>0:\\\
|\alpha+\gamma_{1}|<m\\\
|\alpha+\gamma_{2}|<m\end{subarray}}\frac{1}{\gamma_{1}!\gamma_{2}!}(\partial^{\alpha+\gamma_{1}}P)(x)(\partial^{\alpha+\gamma_{2}}P)(x)(y-x)^{\gamma_{1}+\gamma_{2}}$
$\displaystyle+\sum_{|\alpha|<m}\frac{\delta^{2(|\alpha|-m)}}{\alpha!}2(\partial^{\alpha}P)(x)\cdot\sum_{\begin{subarray}{c}\gamma>0:\\\
|\alpha+\gamma|<m\end{subarray}}\frac{1}{\gamma!}(\partial^{\alpha+\gamma}P)(x)\cdot(y-x)^{\gamma}$
$\displaystyle\leq
2\sum_{|\alpha|<m}\frac{\delta^{2(|\alpha|-m)}}{\alpha!}\sum_{\begin{subarray}{c}\gamma_{1}\geq
0,\gamma_{2}>0:\\\ |\alpha+\gamma_{1}|<m\\\
|\alpha+\gamma_{2}|<m\end{subarray}}\frac{1}{\gamma_{1}!\gamma_{2}!}(\partial^{\alpha+\gamma_{1}}P)(x)(\partial^{\alpha+\gamma_{2}}P)(x)\cdot(y-x)^{\gamma_{1}+\gamma_{2}}.$

Now use the trivial bound

$|\partial^{\alpha+\gamma}P(x)|\leq|P|_{x,\delta}\frac{(\alpha+\gamma)!}{\delta^{|\alpha+\gamma|-m}}$

and the hypothesis $|y-x|\leq\delta\eta$ to get that

$|(\mathrm{R})|\leq
2\sum_{|\alpha|<m}\frac{\delta^{2(|\alpha|-m)}}{\alpha!}\sum_{\begin{subarray}{c}\gamma_{1}\geq
0,\gamma_{2}>0:\\\ |\alpha+\gamma_{1}|<m\\\
|\alpha+\gamma_{2}|<m\end{subarray}}\frac{(\alpha+\gamma_{1})!(\alpha+\gamma_{2})!}{\gamma_{1}!\gamma_{2}!}\frac{\eta^{\gamma_{1}+\gamma_{2}}}{\delta^{2(|\alpha|-m)}}|P|_{x,\delta}^{2}.$

Using the fact that the number of multiindices appearing in each of the above
sums is bounded by $D$, and the hypothesis $\eta<1$, we see that

$|(\mathrm{R})|\leq D^{3}\left((m-1)!\right)^{2}\eta|P|_{x,\delta}^{2}.$ (17)

Combining (16) and (17) proves the lemma, with
$C=D^{3}\left((m-1)!\right)^{2}$. ∎

######  Lemma 2.13.

[cf. Lemma 2.1, part (ii) in [6]] Let $x\in\mathbb{R}^{n}$ and
$0<\rho\leq\delta$. Then there exists a controlled constant $C$ such that for
any $P,Q\in\mathcal{P}$,

$|P\odot_{x}Q|_{x,\rho}\leq C\delta^{m}|P|_{x,\delta}|Q|_{x,\rho}.$

###### Proof.

By translating and rescaling, we reduce matters to the case $x=0$, $\rho=1$.
For $\delta\geq 1$, we have $\delta^{m}|P|_{0,\delta}\geq|P|_{0,1}$, by (7).
Thus, it suffices to prove the bound $|P\odot_{0}Q|_{0,1}\leq
C|P|_{0,1}|Q|_{0,1}$. This inequality is a consequence of Lemma 2.1. ∎

######  Lemma 2.14.

[cf. Lemma 2.1, part (iii) in [6]] Let $x,y\in\mathbb{R}^{n}$ and
$\delta,\rho>0$. Assume that $|x-y|\leq\rho\leq\delta$. Then there exists a
controlled constant $C$ such that for any $P,Q\in\mathcal{P}$,

$|(P\odot_{y}Q)-(P\odot_{x}Q)|_{x,\rho}\leq
C\delta^{m}|P|_{x,\delta}|Q|_{x,\delta}.$

###### Proof.

By translating and rescaling, we reduce matters to the case $x=0$, $\delta=1$.
Write $|\cdot|=|\cdot|_{0,1}$ for the standard norm on $\mathcal{P}$. Fix
$P,Q\in\mathcal{P}$ with $|P|\leq 1$, $|Q|\leq 1$. Then
$P(z)=\sum_{|\alpha|\leq m-1}c_{\alpha}z^{\alpha}$ and
$Q(z)=\sum_{|\alpha|\leq m-1}d_{\alpha}z^{\alpha}$, with $|c_{\alpha}|$,
$|d_{\alpha}|$ each bounded by a controlled constant. Our task is to show that
$|P\odot_{y}Q-P\odot_{0}Q|_{0,\rho}\leq C$ for $|y|\leq\rho\leq 1$.

Let $B=B(0,1)$ be the closed unit ball in $\mathbb{R}^{n}$ of radius $1$
centered at $0$.

Let $F(z)=P(z)Q(z)$. Then $F$ is a polynomial of degree at most $2m-2$ of the
form

$F(z)=\sum_{|\alpha|\leq 2m-2}f_{\alpha}z^{\alpha},\;\;|f_{\alpha}|\leq
C,\;\;C\mbox{ controlled}.$

Each of the monomial functions $z\mapsto z^{\alpha}$ is in $\dot{C}^{m}(B)$
with $\dot{C}^{m}$ seminorm bounded by a controlled constant. Thus,
$\|F\|_{\dot{C}^{m}(B)}\leq C^{\prime}$ for a controlled constant
$C^{\prime}$. Using (5), we deduce that $F$ is in $C^{m-1,1}(B)$ and
$\|F\|_{C^{m-1,1}(B)}\leq C$ for a controlled constant $C$.

By Taylor’s theorem (15), we have

$|(P\odot_{0}Q)-(P\odot_{y}Q)|_{0,\rho}=|J_{0}F-J_{y}F|_{0,\rho}\leq
C_{T}\|F\|_{C^{m-1,1}(B)}\leq C_{T}C$

for $1\geq\rho\geq|y|$. This completes the proof of the lemma.

∎

######  Lemma 2.15 (cf. Lemma 2.2 of [6]).

Fix polynomials $P_{x},Q_{x},R_{x}$ and $P_{y},Q_{y},R_{y}$ in $\mathcal{P}$,
for $|x-y|\leq\rho\leq\delta$. Suppose that $P_{x},P_{y}\in
M_{0}\mathcal{B}_{x,\delta}$, $Q_{x},Q_{y}\in M_{1}\mathcal{B}_{x,\delta}$,
and $R_{x},R_{y}\in M_{2}\mathcal{B}_{x,\delta}$. Also suppose that
$P_{x}-P_{y}\in M_{0}\mathcal{B}_{x,\rho}$, $Q_{x}-Q_{y}\in
M_{1}\mathcal{B}_{x,\rho}$, and $R_{x}-R_{y}\in M_{2}\mathcal{B}_{x,\delta}$.
Then

$|P_{x}\odot_{x}Q_{x}\odot_{x}R_{x}-P_{y}\odot_{y}Q_{y}\odot_{y}R_{y}|_{x,\rho}\leq
C\delta^{2m}M_{0}M_{1}M_{2},$

where $C$ is a controlled constant.

###### Proof.

This lemma is identical to Lemma 2.2 in [6] with the additional claim that the
constant $C$ is controlled. To see that this is true, we examine the proof of
Lemma 2.2 in [6]. Note that $C$ is a product of a finite number (independent
of $D$) of the constants appearing in Lemma 2.1 in [6]. Lemmas 2.12, 2.13, and
2.14 of this paper show that we can take these constants to be controlled. ∎

######  Lemma 2.16 (cf. equation (2.4) of [6]).

If $|x-y|\leq\lambda\delta$ for $\lambda\geq 1$, then for any
$P\in\mathcal{P}$,

$|P|_{y,\delta}\leq C^{\prime}\lambda^{m-1}|P|_{x,\delta}$

for a controlled constant $C^{\prime}$. Consequently,

$\mathcal{B}_{x,\delta}\subseteq
C^{\prime}\lambda^{m-1}\mathcal{B}_{y,\delta}.$

###### Proof.

Apply (7) twice and Lemma 2.12 to get:

$|P|_{y,\delta}\leq\lambda^{m}|P|_{y,\lambda\delta}\leq(1+C)\lambda^{m}|P|_{x,\lambda\delta}\leq(1+C)\lambda^{m-1}|P|_{x,\delta},$

where $C$ is the controlled constant from Lemma 2.12. ∎

####  2.5.2 Whitney Covers and Partitions of Unity

######  Lemma 2.17.

For any ball $B\subseteq\mathbb{R}^{n}$ and any $0<r<1$ there exists a cutoff
function $\theta\in C^{m}(\mathbb{R}^{n})$ with $\theta\equiv 0$ on
$\mathbb{R}^{n}\backslash B$, $\theta\equiv 1$ on $(1-r)B$,
$\|\partial^{\alpha}\theta\|_{L^{\infty}(\mathbb{R}^{n})}\leq
C_{\theta,1}(r)\operatorname{diam}(B)^{-|\alpha|}$ for any $|\alpha|\leq m$,
where $C_{\theta,1}(r):=9\frac{(4m)^{4m}}{r^{m}}$.

###### Proof.

By translating and rescaling it suffices to construct $\theta$ supported on
the unit ball $B=\\{x:|x|\leq 1\\}$.

Let $\psi:\mathbb{R}\rightarrow\mathbb{R}_{+}$ be given by
$\psi(x)=e^{-x^{-1}}e^{-(1-x)^{-1}}$ for $x\in(0,1)$, and $\psi(x)=0$ for
$x\notin(0,1)$. Evidently, $\psi\in C^{\infty}(\mathbb{R})$, and
$\psi^{(k)}(0)=\psi^{(k)}(1)=0$ for all $k\geq 0$. By the product rule, for
$x\in(0,1)$, $\psi^{(k)}(x)$ is the sum of at most $2^{k}$ terms of the form
$f_{i,j}(x)=\frac{d^{i}}{dx^{i}}(e^{-x^{-1}})\frac{d^{j}}{dx^{j}}(e^{-(1-x)^{-1}})$
with $i+j=k$. By induction on $i$, $\frac{d^{i}}{dx^{i}}(e^{-x^{-1}})$ is the
sum of at most $2^{i}$ terms of the form $h_{w,r,s}(x)=wx^{-2s-r}e^{-x^{-1}}$
for integers $r,s$ with $r+s=i$, and real $w$ with $|w|\leq(2s+r)^{r}$. Using
the bound $t^{K}e^{-t}\leq K^{K}$ ($t,K>0$), we find
$|h_{w,r,s}(x)|\leq|w|(2s+r)^{2s+r}\leq(2s+r)^{2s+2r}\leq(2i)^{2i}$, and thus
$|\frac{d^{i}}{dx^{i}}(e^{-x^{-1}})|\leq 2^{i}(2i)^{2i}=8^{i}i^{2i}$ for
$x>0$. Similarly, $|\frac{d^{j}}{dx^{j}}(e^{-(1-x)^{-1}})|\leq 8^{j}j^{2j}$
for $x<1$. Thus, $|f_{i,j}(x)|\leq 8^{i+j}\max\\{i,j\\}^{2(i+j)}$ for
$x\in(0,1)$. We deduce that $\|\psi^{(k)}\|_{L^{\infty}(\mathbb{R})}\leq
2^{k}8^{k}k^{2k}=(4k)^{2k}$ for $k\geq 0$.

Note that
$\gamma:=\int_{-\infty}^{\infty}\psi(t)dt\geq\frac{1}{3}e^{-2/3}\geq\frac{1}{9}$.
Now, let

$\upsilon(x):=\gamma^{-1}\int_{-\infty}^{x}\psi(t)dt.$

Then $\upsilon(t)=0$ for $t\leq 0$, $\upsilon(t)=1$ for $t\geq 1$, and
$\upsilon^{(k)}(0)=\upsilon^{(k)}(1)=0$ for $k\geq 1$. Finally,
$\|\upsilon^{(k)}\|_{L^{\infty}(\mathbb{R})}\leq 9\cdot(4k)^{2k}$ for $k\geq
0$; here, our convention is that $0^{0}=1$.

For $0<\eta<1$ let $\varphi_{\eta}:\mathbb{R}^{+}\rightarrow\mathbb{R}$ given
by $\varphi_{\eta}(t)=v((1-t)/(1-\eta))$. Then

  1. 1.

$\varphi_{\eta}(t)=1$ for $t\leq\eta$,

  2. 2.

$\varphi_{\eta}(t)=0$ for $t\geq 1$,

  3. 3.

$\|\varphi_{\eta}^{(k)}\|_{L^{\infty}(\mathbb{R}^{+})}\leq
9\cdot\frac{(4k)^{2k}}{(1-\eta)^{k}}$ for $k\geq 0$.

Define $\theta:\mathbb{R}^{n}\rightarrow\mathbb{R}$ by
$\theta(x):=\varphi_{(1-r)^{2}}(|x|^{2})$. Note that $\theta(x)\equiv 0$ for
$|x|\geq 1$ due to property 2 of $\varphi_{\eta}$. Furthermore,
$\theta(x)\equiv 1$ for $|x|\leq 1-r$, by property 1 of $\varphi_{\eta}$. By
induction on $|\alpha|$, using the product and chain rules, we establish the
following claim: For $0<|\alpha|\leq m$, the function
$\partial^{\alpha}\theta(x)$ is a sum of at most $2^{|\alpha|-1}\leq 2^{m}$
terms of the form
$h_{j,\beta}(x)=C_{j,\beta}\varphi_{(1-r)^{2}}^{(j)}(|x|^{2})\cdot x^{\beta}$
for integers $j\leq m$, multindices $\beta$ with $|\beta|\leq m$, and
constants $C_{j,\beta}$ satisfying $|C_{j,\beta}|\leq m^{|\alpha|}\leq m^{m}$.
If $|x|\leq 1$ then $|x^{\beta}|\leq 1$. Property 3 of $\varphi_{\eta}$
implies that, for $|x|\leq 1$ and $|\alpha|\leq m$:

$|\partial^{\alpha}\theta(x)|\leq 2^{m}\cdot m^{m}\cdot
9\frac{(4m)^{2m}}{(1-(1-r)^{2})^{m}}\leq 9\frac{(4m)^{3m}}{(2r-r^{2})^{m}}\leq
9\frac{(4m)^{3m}}{r^{m}}.$

(We use $2r-r^{2}\geq r$.) Because $\theta(x)\equiv 0$ for $|x|\geq 1$, we
conclude that

$\|\partial^{\alpha}\theta\|_{L^{\infty}(\mathbb{R}^{n})}\leq
9\frac{(4m)^{3m}}{r^{m}}\leq
9\frac{(4m)^{3m}}{r^{m}}2^{m}\operatorname{diam}(B)^{-|\alpha|}\leq
9\frac{(4m)^{4m}}{r^{m}}\operatorname{diam}(B)^{-|\alpha|}.$

This completes the proof of the lemma.

∎

######  Definition 2.18.

A finite collection $\mathcal{W}$ of closed balls is a _Whitney cover_ of a
ball $\widehat{B}\subseteq\mathbb{R}^{n}$ if  (1) $\mathcal{W}$is a cover of
$\widehat{B}$,  (2) the collection of third-dilates
$\\{\frac{1}{3}B:B\in\mathcal{W}\\}$ is pairwise disjoint, and  (3)
$\operatorname{diam}(B_{1})/\operatorname{diam}(B_{2})\in[1/8,8]$for all balls
$B_{1},B_{2}\in\mathcal{W}$ with
$\frac{6}{5}B_{1}\cap\frac{6}{5}B_{2}\neq\emptyset$.

######  Lemma 2.19 (Bounded Overlap of Whitney Covers).

If $\mathcal{W}$ is a Whitney cover of $\widehat{B}$ then
$\\#\\{B\in\mathcal{W}:x\in\frac{6}{5}B\\}\leq 100^{n}$ for all
$x\in\mathbb{R}^{n}$.

###### Proof.

See Lemma 2.14 of [6] for the proof. ∎

######  Lemma 2.20 (Partitions of Unity adapted to Whitney Covers – cf. Lemma
2.15 of [6]).

If $\mathcal{W}$ is a Whitney cover of $\widehat{B}$, then for each
$B\in\mathcal{W}$ there exists a non-negative $C^{\infty}$ function
$\theta_{B}:\widehat{B}\rightarrow[0,\infty)$ such that

  1. 1.

$\theta_{B}=0$ on $\widehat{B}\backslash\frac{6}{5}B$.

  2. 2.

$|\partial^{\alpha}\theta_{B}(x)|\leq C\operatorname{diam}(B)^{-|\alpha|}$ for
all $|\alpha|\leq m$ and $x\in\widehat{B}$.

  3. 3.

$\sum_{B\in\mathcal{W}}\theta_{B}=1$ on $\widehat{B}$.

Here, $C$ is a controlled constant.

###### Proof.

Use Lemma 2.17 to obtain a function
$\psi_{B}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ for each $B\in\mathcal{W}$
satisfying  (1) $\operatorname{supp}(\psi_{B})\subseteq\frac{6}{5}B$, (2)
$\psi_{B}=1$on $B$, and  (3) $\|\partial^{\alpha}\psi_{B}\|_{L^{\infty}}\leq
C\operatorname{diam}(B)^{-|\alpha|}$for all $|\alpha|\leq m$ , for a
controlled constant $C$.

Set $\Psi:=\sum_{B\in\mathcal{W}}\psi_{B}$ and define

$\theta_{B}(x):=\psi_{B}(x)/\Psi(x),\quad x\in\widehat{B}.$ (18)

Since each point in $\widehat{B}$ belongs to some $B\in\mathcal{W}$, $\Psi\geq
1$ on $\widehat{B}$ and thus $\theta_{B}$ is well-defined on $\widehat{B}$.
Property 1 follows from the fact that $\psi_{B}$ is supported on
$\frac{6}{5}B$. Property 3 follows because
$\sum_{B\in\mathcal{W}}\theta_{B}=\sum_{\mathcal{B}\in\mathcal{W}}\psi_{B}/\Psi=1$
on $\widehat{B}$.

Property 2 is valid if $x\in\widehat{B}\backslash\frac{6}{5}B$ since then
$J_{x}(\theta_{B})=0$. Now fix $x\in\frac{6}{5}B\cap\widehat{B}$. If
$\psi_{B^{\prime}}(x)\neq 0$ for some $B^{\prime}$, then
$x\in\frac{6}{5}B^{\prime}$, so
$\frac{6}{5}B\cap\frac{6}{5}B^{\prime}\neq\emptyset$, and hence,
$\operatorname{diam}(B)/\operatorname{diam}(B^{\prime})\in[\frac{1}{8},8]$ by
definition of Whitney covers. By Lemma 2.19, the cardinality of
$\mathcal{W}_{x}:=\\{B^{\prime}\in\mathcal{W}:x\in\frac{6}{5}B^{\prime}\\}$ is
$\leq 100^{n}$. Therefore,

$|\partial^{\alpha}\Psi(x)|\leq\sum_{B^{\prime}\in
W_{x}}|\partial^{\alpha}\psi_{B^{\prime}}(x)|\leq\sum_{B^{\prime}\in
W_{x}}C\operatorname{diam}(B^{\prime})^{-|\alpha|}\leq
C^{\prime}\operatorname{diam}(B)^{-|\alpha|}$ (19)

for controlled constants $C$, $C^{\prime}$. Given (19) and the fact that
$\Psi\geq 1$ on $\widehat{B}$, by repeated application of the quotient rule we
obtain $|\partial^{\gamma}(1/\Psi(x))|\leq
C^{\prime\prime}\operatorname{diam}(B)^{-|\gamma|}$ for $|\gamma|\leq m$ for a
controlled constant $C^{\prime\prime}$. By application of the product rule to
(18), we see that $|\partial^{\alpha}\theta_{B}(x)|$ is bounded above by a sum
of $2^{|\alpha|}$ terms of the form

$|\partial^{\beta}\psi_{B}(x)|\cdot|\partial^{\gamma}(1/\Psi(x))|,\mbox{ where
}\beta+\gamma=\alpha.$

Given $|\partial^{\beta}\psi_{B}(x)|\leq C\operatorname{diam}(B)^{-|\beta|}$
we conclude that $|\partial^{\alpha}\theta_{B}(x)|\leq
C^{\prime\prime\prime}\operatorname{diam}(B)^{-|\alpha|}$ for a controlled
constant $C^{\prime\prime\prime}$. This finishes the proof of property 2. ∎

######  Lemma 2.21 (Gluing lemma – cf. Lemma 2.16 of [6]).

Fix a Whitney cover $\mathcal{W}$ of $\widehat{B}$, a partition of unity
$\\{\theta_{B}\\}_{B\in\mathcal{W}}$ as in Lemma 2.20, and points
$x_{B}\in\frac{6}{5}B$ for each $B\in\mathcal{W}$. Suppose
$\\{F_{B}\\}_{B\in\mathcal{W}}$ is a collection of functions in
$C^{m-1,1}(\mathbb{R}^{n})$ with the following properties:

  * •

$\|F_{B}\|_{C^{m-1,1}(\mathbb{R}^{n})}\leq M_{0}$

  * •

$F_{B}=f$ on $E\cap\frac{6}{5}B$.

  * •

$|J_{x_{B}}F_{B}-J_{x_{B^{\prime}}}F_{B^{\prime}}|_{x_{B},\operatorname{diam}(B)}\leq
M_{0}$ whenever $\frac{6}{5}B\cap\frac{6}{5}B^{\prime}\neq\emptyset$.

Let $F=\sum_{B\in\mathcal{W}}\theta_{B}F_{B}$. Then $F\in
C^{m-1,1}(\widehat{B})$ with $F=f$ on $E\cap\widehat{B}$ and
$\|F\|_{C^{m-1,1}(\widehat{B})}\leq CM_{0}$, where $C$ is a controlled
constant.

###### Proof.

We sketch the proof, following the proof of Lemma 2.16 in [6], which is
identical to Lemma 2.21 but without the claim that $C$ is a controlled
constant.

See the proof of Lemma 2.16 of [6] for verification that $F=f$ on
$E\cap\widehat{B}$.

The proof of Lemma 2.16 of [6] then goes on to show that

$|J_{x}(F)-J_{y}(F)|_{x,|x-y|}\leq CM_{0}$ (20)

whenever $x,y\in\widehat{B}$ with
$|x-y|\leq\delta_{\min}:=\frac{1}{100}\min\\{\operatorname{diam}(B):B\in\mathcal{W}\\}$.
By definition of the $|\cdot|_{x,\delta}$-norm, (20) implies the local
Lipschitz condition:

$|\partial^{\alpha}F(x)-\partial^{\alpha}F(y)|\leq
C^{\prime}M_{0}|x-y|\quad\mbox{for
}|\alpha|=m-1,\;x,y\in\widehat{B},\;|x-y|\leq\delta_{\min}.$

Then by the triangle inequality, the Lipschitz constant of
$\partial^{\alpha}F$ on all of $\widehat{B}$ is $\leq C^{\prime}M_{0}$, for
each $|\alpha|=m-1$. Therefore, $\|F\|_{C^{m-1,1}(\widehat{B})}\leq
C^{\prime\prime}M_{0}$, as desired.

All that remains is to show that $C$ in (20) is a controlled constant. From
the proof in [6], we note that $C$ is a sum or product of finitely many
(independent of $m,n$) of the constants $C_{T}$ (appearing in Taylor’s
theorem), $100^{n}$, $4^{m}$, and the constants in Lemmas 2.2, 2.15, and
equation (2.4) of [6]. By Lemmas 2.15, 2.20, and 2.16 of the present paper, we
see that each of the last three of these constants is controlled. $C_{T}$ is
controlled by Proposition 2.2. Thus, $C$ is a controlled constant. ∎

##  3 Geometry in the Grassmanian

Let $(X,\langle\cdot,\cdot\rangle)$ be a real finite-dimensional Hilbert
space, and set $d:=\dim X$. Denote the norm on $X$ by
$|\cdot|=\sqrt{\langle\cdot,\cdot\rangle}$, and let $\mathcal{B}=\\{x\in
X:|x|\leq 1\\}$ be the unit ball of $X$. Write $\mathcal{K}(X)$ for the
collection of all closed, convex, symmetric subsets of $X$. Recall that a
subset $\Omega\subseteq X$ is symmetric if $v\in\Omega\implies-v\in\Omega$.

###  3.1 Tools from linear and multilinear algebra

Here we present a few tools and pieces of terminology from multilinear
algebra.

For $0\leq k\leq\dim(X)$, let $\bigwedge^{k}X$ be the $k$’th exterior power of
$X$. We refer to elements of $\bigwedge^{k}X$ as tensors. If
$v_{1},v_{2},\dots,v_{k}\in X$ then $v_{1}\wedge v_{2}\wedge\dots\wedge
v_{k}\in\bigwedge^{k}X$ is called a _pure tensor_. Every tensor is a finite
linear combination of pure tensors. We specify a Hilbert space structure on
$\bigwedge^{k}X$ as follows. Let $e_{1},\dots,e_{d}$ be an orthonormal basis
for $X$. For $1\leq i_{1}<\dots<i_{k}\leq d$, $1\leq j_{1}<\dots<j_{k}\leq d$,
let
$\langle\bigwedge_{\ell=1}^{k}e_{i_{\ell}},\bigwedge_{\ell=1}^{k}e_{j_{\ell}}\rangle$
be $1$ if $i_{\ell}=j_{\ell}$ for all $\ell$, and $0$ otherwise. We extend
this inner product to all of $\bigwedge^{k}X$ by bilinearity. Then
$\\{\bigwedge_{\ell=1}^{k}e_{i_{\ell}}:1\leq i_{1}<\dots<i_{k}\leq d\\}$ is an
orthonormal basis for $\bigwedge^{k}X$. Write $\langle\cdot,\cdot\rangle$ and
$|\cdot|$ for the inner product and associated norm on $\bigwedge^{k}X$. This
inner product can be defined in a basis-independent manner as the unique
bilinear mapping obeying the identity

$\left\langle\bigwedge_{i=1}^{k}v_{i},\bigwedge_{i=1}^{k}w_{i}\right\rangle=\det(\langle
v_{i},w_{j}\rangle)_{1\leq i,j\leq k},\mbox{ for all
}v_{1},\dots,v_{k},w_{1},\dots,w_{k}\in X.$

In particular, the Hilbert space structure on $\bigwedge^{k}X$ is independent
of the choice of orthonormal basis for $X$.

Let $V$ be a $k$-dimensional subspace of $X$, and fix a basis
$\\{v_{j}\\}_{1\leq j\leq k}$ for $V$. We set $\omega_{V}:=v_{1}\wedge
v_{2}\wedge\dots\wedge v_{k}\in\bigwedge^{k}X$. We call $\omega_{V}$ a
_representative form_ for $V$. The next remark implies that the representative
forms associated to different choices of basis for $V$ are scalar multiples of
one another.

######  Remark 3.1.

If $\\{\hat{v}_{j}\\}_{1\leq j\leq k}$ and $\\{v_{j}\\}_{1\leq j\leq k}$ are
two bases for $V$ then
$\hat{v}_{1}\wedge\hat{v}_{2}\wedge\dots\wedge\hat{v}_{k}=\det(A)\cdot
v_{1}\wedge v_{2}\wedge\dots\wedge v_{k}$, where
$A=(A_{ij})\in\mathbb{R}^{k\times k}$ is the change-of-basis matrix defined by
the relations $\hat{v}_{i}=\sum_{j}A_{ij}v_{j}$ ($i=1,2,\dots,k$).

The eigenvalues of a self-adjoint operator $T:X\rightarrow X$ will be written
in descending order:
$\lambda_{1}(T)\geq\lambda_{2}(T)\geq\cdots\geq\lambda_{d}(T)$ ($d=\dim X$).

Let $X_{0}$, $X_{1}$ be $k$-dimensional Hilbert spaces. We denote the singular
values of a linear transformation $T:X_{0}\rightarrow X_{1}$ by
$\sigma_{1}(T)\geq\sigma_{2}(T)\geq\dots\geq\sigma_{k}(T)\geq 0$. The squared
singular values of $T$ are eigenvalues of $T^{*}T$, or $TT^{*}$
(equivalently), i.e.,

$\sigma_{\ell}(T)=\sqrt{\lambda_{\ell}(T^{*}T)}=\sqrt{\lambda_{\ell}(TT^{*})}\mbox{
for }\ell=1,2,\dots,k.$

The extremal singular values $\sigma_{1}(T)$ and $\sigma_{k}(T)$ are related
to the operator norms of $T$ and $T^{-1}$. First, $\sigma_{1}(T)=\|T\|_{op}$.
Also, $\sigma_{k}(T)>0$ if and only if $T:X_{0}\rightarrow X_{1}$ is
invertible, and then $\sigma_{k}(T)=\|T^{-1}\|_{op}^{-1}$. This implies the
following description:

$\sigma_{k}(T)=\sup\\{\eta\geq 0:\|Tx\|_{X_{1}}\geq\eta\|x\|_{X_{0}}\mbox{ for
all }x\in X_{0}\\}.$ (21)

Finally, $\sigma_{k}(T)$ has a description in terms of the images of balls
under $T$. Let $\mathcal{B}_{X_{j}}:=\\{x\in X_{j}:\|x\|_{X_{j}}\leq 1\\}$ be
the unit ball of $X_{j}$ ($j\in\\{0,1\\}$). Then

$\sigma_{k}(T)=\sup\\{\eta\geq
0:T(\mathcal{B}_{X_{0}})\supseteq\eta\mathcal{B}_{X_{1}}\\}.$ (22)

###  3.2 Angles between subspaces

Let $G(k,X)$ be the Grassmannian of $k$-dimensional subspaces of $X$ ($1\leq
k\leq d$). Note that $G(k,X)\subseteq\mathcal{K}(X)$.

Given $V,W\in G(k,X)$, the _maximum principal angle_
$\theta_{\max}(V,W)\in[0,\frac{\pi}{2}]$ between $V$ and $W$ is defined by

$\theta_{\max}(V,W):=\arccos\left(\inf\left\\{\frac{|\Pi_{W}v|}{|v|}:v\in
V,v\neq 0\right\\}\right).$ (23)

Here and below, we write $\Pi_{W}:X\rightarrow W$ to denote the orthogonal
projection operator on a subspace $W$ of $X$. Below we show that
$\theta_{\max}(\cdot,\cdot)$ is symmetric. In fact,
$\theta_{\max}(\cdot,\cdot)$ is a metric on the Grassmanian $G(k,X)$. For a
further discussion of principal angles, see [22] and the references therein.

Given $V,W\in G(k,X)$, let

$\angle(V,W)=\arccos\left(\frac{|\langle\omega_{V},\omega_{W}\rangle|}{|\omega_{V}|\cdot|\omega_{W}|}\right).$
(24)

By Remark 3.1, the quantity $\angle(V,W)$ is independent of the choice of
representative forms for $V$ and $W$.

The quantities $\angle(V,W)$ and $\theta_{\max}(V,W)$ are related to singular
values of the projection operator $T_{V\rightarrow
W}:=\Pi_{W}|_{V}:V\rightarrow W$. In fact, we have the identities

$\displaystyle\cos(\angle(V,W))=\sigma_{1}(T_{V\rightarrow
W})\sigma_{2}(T_{V\rightarrow W})\dots\sigma_{k}(T_{V\rightarrow W}),$ (25)
$\displaystyle\cos(\theta_{\max}(V,W))=\sigma_{k}(T_{V\rightarrow W}).$

The first identity can easily be seen to be true by computing (24) using
principal vectors for $V$ and $W$, and using the singular value
characterization of principal angles; see [21] for details. The second
identity follows from (21) and (23).

######  Lemma 3.2.

Fix $\eta>0$. If $X_{0}$ and $H$ are subspaces of $X$ such that
$|\Pi_{H}x|\geq\eta|x|$ for all $x\in X_{0}$, then
$\dim(X_{0})=\dim(\Pi_{H}X_{0})$ and
$\cos(\theta_{\max}(X_{0},\Pi_{H}X_{0}))\geq\eta$.

###### Proof.

Condition $\dim(X_{0})=\dim(\Pi_{H}X_{0})$ holds as
$\Pi_{H}|_{X_{0}}:X_{0}\rightarrow X$ is injective. As
$\Pi_{\Pi_{H}X_{0}}=\Pi_{H}$ on $X_{0}$, $|\Pi_{\Pi_{H}X_{0}}(x)|\geq\eta|x|$
for $x\in X_{0}$ by the lemma’s hypothesis. The bound
$\cos(\theta_{\max}(X_{0},\Pi_{H}X_{0}))\geq\eta$ is a consequenece of the
definition (23). ∎

######  Lemma 3.3.

Let $W$ and $V$ be subspaces of $X$ of equal dimension. Then the following
conditions are equivalent.

  1. 1.

$\cos(\theta_{\max}(V,W))\geq\eta$

  2. 2.

$|\Pi_{W}(v)|\geq\eta|v|$ for all $v\in V$.

  3. 3.

$|\Pi_{V}(w)|\geq\eta|w|$ for all $w\in W$.

  4. 4.

$\cos(\theta_{\max}(W,V))\geq\eta$

###### Proof.

The equivalence of conditions 1 and 2 is immediate from the definition (23).
The equivalence of conditions 3 and 4 follows for the same reason.

We prove the equivalence of conditions 2 and 3 by duality. Let
$T_{V\rightarrow W}:=\Pi_{W}|_{V}$ and $T_{W\rightarrow V}:=\Pi_{V}|_{W}$.
Condition 2 is equivalent to the claim that $T_{V\rightarrow W}$ is invertible
and $\|T_{V\rightarrow W}^{-1}\|_{op}\leq\eta^{-1}$. Similarly, condition 3 is
equivalent to the claim that $T_{W\rightarrow V}$ is invertible and
$\|T_{W\rightarrow V}^{-1}\|_{op}\leq\eta^{-1}$. Since $T_{V\rightarrow W}$ is
the adjoint of $T_{W\rightarrow V}$, we obtain the equivalence of conditions 2
and 3. ∎

From the equivalence of conditions 1 and 4 of Lemma 3.3, we learn that

$\theta_{\max}(V,W)=\theta_{\max}(W,V)\quad\mbox{for }V,W\in G(k,X).$ (26)

See Section 2 of [22] for a proof of the following result.

######  Lemma 3.4.

If $V$ and $W$ are subspaces of $X$ of equal dimension then
$\cos(\theta_{\max}(V,W))=\cos(\theta_{\max}(V^{\perp},W^{\perp}))$.

Thanks to (25), we have the following result.

######  Lemma 3.5.

Let $V,W\in G(k,X)$. Then

$\cos(\theta_{\max}(V,W))^{k}\leq\cos(\angle(V,W))\leq\cos(\theta_{\max}(V,W)).$
(27)

###  3.3 Transversality

Recall that $\mathcal{K}(X)$ denotes the set of all closed, convex, symmetric
subsets of $X$. Given $\Omega\in\mathcal{K}(X)$ and $a>0$, let
$a\cdot\Omega:=\\{a\cdot x:x\in\Omega\\}$. Given a function $T:X\rightarrow
X$, let $T(\Omega):=\\{T(x):x\in\Omega\\}$.

We start with an elementary lemma. Given $A\subseteq X$ and a subspace $V$ in
$X$, let $A/V$ denote the subset $\\{a+V:a\in A\\}$ of the quotient space
$X/V=\\{x+V:x\in X\\}$.

######  Lemma 3.6.

Let $A,B\subseteq X$. Then $A/V\subseteq B/V$ if and only if
$\Pi_{V^{\perp}}A\subseteq\Pi_{V^{\perp}}B$.

###### Proof.

Note that $A/V\subseteq B/V$ if and only if for every $a\in A$ there exists
$b\in B$ such that $a-b\in V$.

Suppose $A/V\subseteq B/V$. Fix an arbitrary $x\in\Pi_{V^{\perp}}A$. Let $a\in
A$ with $x=\Pi_{V^{\perp}}a$. Because $A/V\subseteq B/V$, there exists $b\in
B$ so that $a-b\in V$. Then $\Pi_{V^{\perp}}b=\Pi_{V^{\perp}}a$. Thus,
$x\in\Pi_{V^{\perp}}B$. So, we’ve shown
$\Pi_{V^{\perp}}A\subseteq\Pi_{V^{\perp}}B$.

Conversely, suppose $\Pi_{V^{\perp}}A\subseteq\Pi_{V^{\perp}}B$. Fix $a\in A$.
Let $x=\Pi_{V^{\perp}}a$. Because $\Pi_{V^{\perp}}A\subseteq\Pi_{V^{\perp}}B$,
there exists $b\in B$ with $x=\Pi_{V^{\perp}}b$. But then
$\Pi_{V^{\perp}}(a-b)=x-x=0$. So, $a-b\in V$. This proves $A/V\subseteq B/V$.
∎

We now introduce the concept of transversality in the Hilbert space $X$.

######  Definition 3.7.

Let $\Omega\in\mathcal{K}(X)$, let $V\subseteq X$ be a subspace, and let
$R\geq 1$. Then $\Omega$ is _$R$ -transverse_ to $V$ if

$\displaystyle\Omega\cap V\subseteq R\cdot\mathcal{B}$ (28)
$\displaystyle\Pi_{V^{\perp}}(\Omega\cap\mathcal{B})\supseteq
R^{-1}\cdot\mathcal{B}\cap V^{\perp}.$ (29)

In particular, if $X=\mathcal{R}_{x}$, then $\Omega\subseteq X$ is
$R$-transverse to $V$ if and only if it is $R$-transverse to $V$ at $x$ (see
Definition 2.8).

Using Lemma 3.6, we obtain an equivalent formulation of transversality used in
our previous work [6]. This will allow us to later borrow results from [6].

######  Corollary 3.8.

Let $\Omega\in\mathcal{K}(X)$ and let $V$ be a subspace of $X$. Then $\Omega$
is $R$-transverse to $V$ if and only if (A) $\Omega\cap V\subseteq
R\cdot\mathcal{B}$, and (B) $(\Omega\cap\mathcal{B})/V\supseteq
R^{-1}\cdot\mathcal{B}/V$.

The notion of transversality between a pair of subspaces (i.e., when $\Omega$
is a subspace) admits an equivalent formulation in terms of principal angles.

######  Lemma 3.9.

Let $W,V$ be subspaces of $X$, and $R\geq 1$. Then $W$ is $R$-transverse to
$V$ if and only if $\dim(W)=\dim(V^{\perp})$ and
$\cos(\theta_{\max}(W,V^{\perp}))\geq R^{-1}$.

###### Proof.

When $\Omega=W$ is a subspace, condition (28) is equivalent to the assertion
that $W\cap V=\\{0\\}$. Thus, from (28), (29), $W$ is $R$-transverse to $V$ if
and only if (a) $W\cap V=\\{0\\}$ and (b)
$\Pi_{V^{\perp}}(W\cap\mathcal{B})\supseteq R^{-1}\cdot\mathcal{B}\cap
V^{\perp}$.

Note that condition (a) implies $\dim(W)\leq\dim(V^{\perp})$.

Note that condition (b) implies that $T_{W\rightarrow
V^{\perp}}:=\Pi_{V^{\perp}}|_{W}:W\rightarrow V^{\perp}$ is surjective. Hence,
condition (b) implies $\dim(W)\geq\dim(V^{\perp})$.

Hence, if $W$ is $R$-transverse to $V$ then $\dim(W)=\dim(V^{\perp})$, and
condition (b) is then equivalent to the inequality $\sigma_{k}(T_{W\rightarrow
V^{\perp}})\geq R^{-1}$ (see (22)), which is equivalent to the inequality
$\cos(\theta_{\max}(W,V^{\perp}))\geq R^{-1}$ (see (25)).

On the other hand, suppose $\dim(W)=\dim(V^{\perp})$ and
$\cos(\theta_{\max}(W,V^{\perp}))\geq R^{-1}$. Thus,
$\sigma_{k}(T_{W\rightarrow V^{\perp}})\geq R^{-1}$, which implies condition
(b) above (again, see (25) and (22)). In particular, $T_{W\rightarrow
V^{\perp}}:W\rightarrow V^{\perp}$ is surjective. As
$\dim(W)=\dim(V^{\perp})$, we have that $T_{W\rightarrow V^{\perp}}$ is
injective. Thus, $\\{0\\}=\ker(\Pi_{V^{\perp}}|_{W})=W\cap V$, which gives
condition (a) above. So $W$ is $R$-transverse to $V$.

This completes the proof of the lemma. ∎

By Lemma 3.9 and Lemma 3.4, we have the following result.

######  Corollary 3.10.

Let $W$ and $V$ be subspaces of $X$, and let $R\geq 1$. Then $W$ is
$R$-transverse to $V$ if and only if $V$ is $R$-transverse to $W$.

######  Lemma 3.11.

Let $W,V$ be subspaces of $X$, and let $r>0$ and $R\geq 1$. If $W$ is
$R$-transverse to $V$ then $(W+r\mathcal{B})\cap V\subseteq Rr\mathcal{B}$.

###### Proof.

Fix $x\in(W+r\mathcal{B})\cap V$. As $x\in W+r\mathcal{B}$, we have
$|\Pi_{W^{\perp}}x|=\operatorname{dist}(x,W)\leq r$. Observe that
$|\Pi_{W^{\perp}}x|\geq R^{-1}|x|$ due to the condition
$\theta_{\max}(V,W^{\perp})\geq R^{-1}$ and since $x\in V$. Thus, $|x|\leq
Rr$, so $x\in Rr\mathcal{B}$, as desired. ∎

######  Lemma 3.12.

Let $\Omega\in\mathcal{K}(X)$ and let $V$ be a subspace of $X$. Let
$T:X\rightarrow X$ be an invertible linear transformation satisfying either
(A) $|x|\leq|Tx|\leq M|x|$ for all $x\in X$, or (B) $M^{-1}|x|\leq|Tx|\leq|x|$
for all $x\in X$.

If $\Omega$ is $R$-transverse to $V$ then $T(\Omega)$ is $MR$-transverse to
$T(V)$.

###### Proof.

We suppose $\Omega$ is $R$-transverse to $V$, so (a) $\Omega\cap V\subseteq
R\mathcal{B}$ and (b) $R^{-1}\mathcal{B}/V\subseteq(\Omega\cap\mathcal{B})/V$.
Here we use the formulation of transversality given in Corollary 3.8.

If $T$ satisfies condition (A) then $\|T^{-1}\|_{op}\leq 1$ and
$\|T\|_{op}\leq M$, implying the set inclusions $\mathcal{B}\subseteq
T(\mathcal{B})$ and $T(\mathcal{B})\subseteq M\mathcal{B}$. By (a),

$T(\Omega)\cap T(V)=T(\Omega\cap V)\subseteq T(R\mathcal{B})\subseteq
MR\mathcal{B},$

and (b) implies that

$\displaystyle R^{-1}\mathcal{B}/T(V)$ $\displaystyle\subseteq
R^{-1}T(\mathcal{B})/T(V)\subseteq
T(\Omega\cap\mathcal{B})/T(V)=(T(\Omega)\cap T(\mathcal{B}))/T(V)$
$\displaystyle\subseteq(T(\Omega)\cap M\mathcal{B})/T(V)\subseteq
M(T(\Omega)\cap\mathcal{B})/T(V).$

Thus, $(MR)^{-1}\mathcal{B}/T(V)\subseteq(T(\Omega)\cap\mathcal{B})/T(V))$. We
deduce from the previous inclusions that $T(\Omega)$ is $MR$-transverse to
$V$.

If $T$ satisfies condition (B) then $T(\mathcal{B})\subseteq\mathcal{B}$ and
$M^{-1}\mathcal{B}\subseteq T(\mathcal{B})$, thus by (a),

$T(\Omega)\cap T(V)=T(\Omega\cap V)\subseteq T(R\mathcal{B})\subseteq
R\mathcal{B}\subseteq MR\mathcal{B},$

and by (b),

$\displaystyle M^{-1}R^{-1}\mathcal{B}/T(V)\subseteq
R^{-1}T(\mathcal{B})/T(V)\subseteq T(\Omega\cap\mathcal{B})/T(V)$
$\displaystyle=(T(\Omega)\cap T(\mathcal{B}))/T(V)$
$\displaystyle\subseteq(T(\Omega)\cap\mathcal{B})/T(V),$

so $T(\Omega)$ is $MR$-transverse to $T(V)$. ∎

##  4 Rescaling dynamics

Let $(X,\langle\cdot,\cdot\rangle)$ be a real Hilbert space of finite
dimension $d:=\dim(X)<\infty$. Write $|x|=\sqrt{\langle x,x\rangle}$ for the
norm of a vector $x\in X$. Let $\tau_{\delta}$ be a $1$-parameter family of
linear operators on $X$ of the following form. Fix $m\geq 1$. Suppose that $X$
admits a direct sum decomposition

$X=\bigoplus\limits_{\nu=1}^{m}X_{\nu},$ (30)

for pairwise orthogonal subspaces $X_{\nu}\subseteq X$. Let
$\tau_{\delta}:X\rightarrow X$ satisfy

$\tau_{\delta}|_{X_{\nu}}=\delta^{-\nu}\cdot\mathrm{id}|_{X_{\nu}}\quad(\delta>0).$
(31)

In this description of $\tau_{\delta}$ we allow that $X_{\nu}=\\{0\\}$ for
certain $\nu$.

######  Definition 4.1.

We refer to a tuple $\mathcal{X}=(X,\tau_{\delta})_{\delta>0}$ satisfying
(30), (31) as a _Hilbert dilation system_. A Hilbert dilation system
$\mathcal{X}$ is said to be _simple_ provided that $\dim(X_{\nu})\in\\{0,1\\}$
for all $\nu=1,2,\dots,m$.

######  Definition 4.2.

A subspace $V\subseteq X$ is _dilation-invariant_ , or DI, if
$\tau_{\delta}V=V$ for all $\delta>0$. If $V$ is DI then

$V=\bigoplus\limits_{\nu=1}^{m}V_{\nu},\quad\mbox{with }V_{\nu}=V\cap
X_{\nu}\subseteq X_{\nu}.$

The _signature_ of a DI subspace $V\subseteq X$ is the quantity

$\operatorname{sgn}(V)=\sum_{\nu=1}^{m}\nu\cdot\dim(V\cap X_{\nu}).$

We note that

$\left\\{\begin{aligned} &V\mbox{ dilation-invariant}\implies V^{\perp}\mbox{
dilation-invariant, and}\\\
&\operatorname{sgn}(V^{\perp})=\Sigma_{0}-\operatorname{sgn}(V),\;\;\Sigma_{0}:=\sum_{\nu=1}^{m}\nu\cdot\dim(X_{\nu}).\end{aligned}\right.$
(32)

We study the behavior of orbits of $\tau_{\delta}$ acting on the Grassmanian
$G(k,X)$. To do so, we will pass to the action of $\tau_{\delta}$ on the
$k$-fold exterior product $\bigwedge^{k}X$.

The linear transformation $\tau_{\delta}:X\rightarrow X$ induces a linear
transformation $\tau_{\delta}^{*}:\bigwedge^{k}X\rightarrow\bigwedge^{k}X$
defined by its action on the pure tensors:

$\tau_{\delta}^{*}(v_{1}\wedge v_{2}\wedge\dots\wedge
v_{k})=\tau_{\delta}(v_{1})\wedge\tau_{\delta}(v_{2})\wedge\dots\wedge\tau_{\delta}(v_{k}).$

If $V$ is a DI subspace of $X$ of dimension $k$, and
$\omega_{V}\in\bigwedge^{k}X$ is a representative form for $V$ (i.e.,
$\omega_{V}$ is the tensor product of a basis for $V$), then

$\tau_{\delta}^{*}(\omega_{V})=\delta^{-\operatorname{sgn}(V)}\omega_{V}.$
(33)

Because all representative forms of $V$ are equivalent up to a scalar
multiple, it suffices to verify (33) for the form associated to a particular
basis for $V$. Because $V$ is DI it admits a basis of the form
$\\{e_{j}\\}_{j=1}^{k}$, with $e_{j}\in X_{i_{j}}$ for each $j$. Consider the
representative form $\omega_{V}=e_{1}\wedge\dots\wedge e_{k}$ for $V$. Note
that $\tau_{\delta}(e_{j})=\delta^{-i_{j}}e_{j}$ for all $j$, and
$\operatorname{sgn}(V)=\sum_{j=1}^{k}i_{j}$. Thus, by definition of
$\tau_{\delta}^{*}$,

$\tau_{\delta}^{*}(\omega_{V})=\tau_{\delta}(e_{1})\wedge\dots\wedge\tau_{\delta}(e_{k})=\delta^{-\operatorname{sgn}(V)}e_{1}\wedge\dots\wedge
e_{k},$

giving (33).

###  4.1 Quantitative stabilization for the action of a simple Hilbert
dilation system on the Grassmanian

Let $H\in G(k,X)$. The parametrized family of subspaces
$(\tau_{\delta}H)_{\delta>0}$ is an _orbit_ of $\tau_{\delta}$ in $G(k,X)$.
The orbit $\tau_{\delta}H$ converges to a subspace $H_{0}\in G(k,X)$ in the
Grassmanian topology in the limit as $\delta\rightarrow 0^{+}$, and
furthermore the limit subspace $H_{0}$ is dilation invariant (see the proof of
Lemma 3.12 in [6]).

The main result of this section is a quantitative bound on the distance of the
orbit $\tau_{\delta}H$ to the set of dilation invariant subspaces when
$\delta$ varies in a compact interval $I$. Specifically, we have:

######  Proposition 4.3.

Let $(X,\tau_{\delta})_{\delta>0}$ be a simple Hilbert dilation system. Let
$H\in G(k,X)$, $1\leq k\leq d=\dim(X)$. Fix $\eta\in(0,1/2)$ and a compact
interval $I\subseteq(0,\infty)$ with
$\frac{r(I)}{l(I)}\geq\left(\frac{2^{d}}{\eta}\right)^{dk+2}$. There exist
$\delta\in I$ and a dilation invariant subspace $V\in G(k,X)$ satisfying
$\cos(\theta_{\max}(\tau_{\delta}H,V))\geq 1-\eta$.

The rest of Section 4.1 is devoted to proving Proposition 4.3. The restriction
that $(X,\tau_{\delta})_{\delta>0}$ is simple ($\dim X_{\nu}\in\\{0,1\\}$
$\forall\nu$) will be in place for the rest of this section. We expect it is
possible to prove a variant of Proposition 4.3 without this restriction, but
the arguments are likely more involved and the constants are slightly worse.
Anyway, the above version is sufficient for the needed application in Section
5.

We introduce notation to be used in the proof. We order the indices $\nu$ for
which $\dim X_{\nu}=1$ in an increasing sequence:
$1\leq\nu_{1}<\nu_{2}<\dots<\nu_{d}\leq m$. For $j=1,2,\dots,d$, let $e_{j}\in
X$ be a unit vector spanning $X_{\nu_{j}}$. Then:

$\displaystyle X\mbox{ admits an orthonormal basis
}\\{e_{1},e_{2},\dots,e_{d}\\}\mbox{ with}$ (34)
$\displaystyle\tau_{\delta}(e_{j})=\delta^{-\nu_{j}}e_{j}\qquad(j=1,2,\dots,d,\;\delta>0),\mbox{
and}$
$\displaystyle\nu_{1},\dots,\nu_{d}\in\mathbb{N},\;\;1\leq\nu_{1}<\dots<\nu_{d}\leq
m.$

Let $[d]:=\\{1,2,\dots,d\\}$. Given $S\subseteq[d]$, let
$V_{S}:=\operatorname{span}\\{e_{j}:j\in S\\}$. Note that a subspace
$V\subseteq X$ is dilation invariant if and only if $V=V_{S}$ for some
$S\subseteq[d]$.

For $S\subseteq[d]$ with $\\#(S)=k$, let $\omega_{S}:=\bigwedge_{j\in
S}e_{j}\in\bigwedge^{k}X$ be a representative form for $V_{S}\subseteq X$.
Note that $\\{\omega_{S}:S\subseteq[d],\;\\#(S)=k\\}$ is an orthonormal basis
for $\bigwedge^{k}X$. See Section 3.1 for a discussion of the Hilbert space
structure on $\bigwedge^{k}X$.

Given $S\subseteq[d]$, define
$c(S):=(c_{1}(S),c_{2}(S),\dots,c_{d}(S))\in\\{0,1,2,\dots,d\\}^{d}$ by

$c_{\ell}(S):=\\#\\{j\in S:j\leq\ell\\}\qquad(\ell=1,2,\dots,d).$

By definition,

$c_{\ell}(S)=\dim(V_{S}\cap X_{\leq\ell})\mbox{, where
}X_{\leq\ell}:=\operatorname{span}\\{e_{1},e_{2},\dots,e_{\ell}\\}.$ (35)

Also, observe that

$S\neq S^{\prime}\implies c_{\ell}(S)\neq c_{\ell}(S^{\prime})\mbox{ for some
}\ell=1,2,\dots,d.$ (36)

Let $A,B:X\rightarrow X$ be linear operators. Then we write $A\geq B$ to mean
that $(A-B)$ is positive semidefinite.

######  Lemma 4.4.

Let $H\in G(k,X)$ for $1\leq k\leq d$. Suppose $\epsilon\in(0,1/2)$,
$\delta,\delta^{\prime}>0$ and $S,S^{\prime}\subseteq[d]$ satisfy
$\\#(S)=\\#(S^{\prime})=k$, $\delta\geq\frac{1}{\epsilon^{2}}\delta^{\prime}$,

$\frac{|\langle\omega_{\tau_{\delta}H},\omega_{S}\rangle|}{|\omega_{\tau_{\delta}H}|}\geq\epsilon,\mbox{
and
}\;\;\frac{|\langle\omega_{\tau_{\delta^{\prime}}H},\omega_{S^{\prime}}\rangle|}{|\omega_{\tau_{\delta^{\prime}}H}|}\geq\epsilon.$

Then $c_{\ell}(S)\geq c_{\ell}(S^{\prime})$ for $\ell=1,2,\dots,d$.

###### Proof.

For sake of contradiction, let $H,\delta,\delta^{\prime},S,S^{\prime}$ be as
in the hypotheses of the lemma, and suppose that there exists $\ell\in[d]$
with $c_{\ell}(S)<c_{\ell}(S^{\prime})$. Without loss of generality,
$\delta^{\prime}=1$. Then $\delta\geq\frac{1}{\epsilon^{2}}$,
$|\langle\omega_{\tau_{\delta}H},\omega_{S}\rangle|\geq\epsilon|\omega_{\tau_{\delta}H}|$,
and $|\langle\omega_{H},\omega_{S^{\prime}}\rangle|\geq\epsilon|\omega_{H}|$.
Thanks to (24) and (27), we have

$\cos(\theta_{\max}(\tau_{\delta}H,V_{S}))\geq\cos(\angle(\tau_{\delta}H,V_{S}))=\frac{|\langle\omega_{\tau_{\delta}H},\omega_{S}\rangle|}{|\omega_{\tau_{\delta}H}|}\geq\epsilon,$
(37)

and similarly

$\cos(\theta_{\max}(H,V_{S^{\prime}}))\geq\epsilon.$ (38)

Consider the orthogonal subspaces

$X_{\leq\ell}:=\operatorname{span}\\{e_{j}:j\leq\ell\\},\;\;\;X_{>\ell}:=\operatorname{span}\\{e_{j}:j>\ell\\}.$

Then $X=X_{\leq\ell}\oplus X_{>\ell}$. If $\ell=d$, by convention
$X_{>\ell}=\\{0\\}$. By (35),

$\displaystyle\dim(V_{S}\cap X_{\leq\ell})=c_{\ell}(S),$ (39)
$\displaystyle\dim(V_{S^{\prime}}\cap X_{\leq\ell})=c_{\ell}(S^{\prime}).$
(40)

Let $\Pi_{\leq\ell}:=\Pi_{X_{\leq\ell}}$ and $\Pi_{>\ell}:=\Pi_{X_{>\ell}}$ be
the orthogonal projection operators associated to $X_{\leq\ell}$ and
$X_{>\ell}$, respectively.

From (38) and Lemma 3.3, $|\Pi_{H}(x)|\geq\epsilon|x|$ for all $x\in
V_{S^{\prime}}$. Set

$\widetilde{H}=\Pi_{H}(V_{S^{\prime}}\cap X_{\leq\ell})\subseteq H.$

Applying Lemma 3.2 to the subspace $X_{0}=V_{S^{\prime}}\cap X_{\leq\ell}$
gives

$\dim(\widetilde{H})=\dim(V_{S^{\prime}}\cap X_{\leq\ell})$ (41)

and $\cos(\theta_{\max}(\widetilde{H},V_{S^{\prime}}\cap
X_{\leq\ell}))\geq\epsilon$. The prior inequality implies, by Lemma 3.3,

$|\Pi_{\leq\ell}x|\geq|\Pi_{X_{\leq\ell}\cap
V_{S^{\prime}}}x|\geq\epsilon|x|\;\mbox{ for all }x\in\widetilde{H}.$ (42)

By the Pythagorean theorem and (42),

$|\Pi_{>\ell}x|=\sqrt{|x|^{2}-|\Pi_{\leq\ell}x|^{2}}\leq\sqrt{1-\epsilon^{2}}|x|\;\mbox{
for all }x\in\widetilde{H}.$ (43)

By the form of $\tau_{\delta}$ (see (34)) and because $\delta\geq 1$, we have
$\tau_{\delta}|_{X_{\leq\ell}}\geq\delta^{-\nu_{\ell}}\cdot\mathrm{id}|_{X_{\leq\ell}}$
and
$\tau_{\delta}|_{X_{>\ell}}\leq\delta^{-\nu_{\ell}-1}\cdot\mathrm{id}|_{X_{>\ell}}$.
Therefore, for $x\in\widetilde{H}$, (42) gives

$|\tau_{\delta}\Pi_{\leq\ell}x|\geq\delta^{-\nu_{\ell}}|\Pi_{\leq\ell}x|\geq\delta^{-\nu_{\ell}}\epsilon|x|,$

and (43) gives

$|\tau_{\delta}\Pi_{>\ell}x|\leq\delta^{-\nu_{\ell}-1}|\Pi_{>\ell}x|\leq\delta^{-\nu_{\ell}-1}\sqrt{1-\epsilon^{2}}|x|.$

Because $\tau_{\delta}$ fixes $X_{>\ell}$ and $X_{\leq\ell}$, the operators
$\tau_{\delta}$, $\Pi_{>\ell}$, $\Pi_{\leq\ell}$ all commute. Thus, combining
the above inequalities gives

$\frac{|\Pi_{>\ell}\tau_{\delta}x|}{|\tau_{\delta}x|}\leq\frac{|\tau_{\delta}\Pi_{>\ell}x|}{|\tau_{\delta}\Pi_{\leq\ell}x|}\leq\frac{1}{\delta}\frac{\sqrt{1-\epsilon^{2}}}{\epsilon}<\epsilon\quad\mbox{for
all }x\in\widetilde{H}\setminus\\{0\\}.$ (44)

where the last inequality uses the assumption that
$\delta\geq\frac{1}{\epsilon^{2}}$.

From (39), (40), (41), and the assumption $c_{\ell}(S)<c_{\ell}(S^{\prime})$,
we have that $\dim(\widetilde{H})>\dim(V_{S}\cap X_{\leq\ell})$. Thus, we can
find an $x\in\widetilde{H}\cap(V_{S}\cap X_{\leq\ell})^{\perp}$ with $x\neq
0$. Note that $(V_{S}\cap X_{\leq\ell})^{\perp}$ is spanned by a subcollection
of the basis $\\{e_{j}\\}$, and each $e_{j}$ is an eigenvector of
$\tau_{\delta}$. Thus, since $x\in(V_{S}\cap X_{\leq\ell})^{\perp}$, we have
$\tau_{\delta}x\in(V_{S}\cap X_{\leq\ell})^{\perp}$. Therefore, due to the
orthogonal decomposition $V_{S}=(V_{S}\cap X_{\leq\ell})\oplus(V_{S}\cap
X_{>\ell})$, we have $\Pi_{V_{S}}\tau_{\delta}x=\Pi_{V_{S}\cap
X_{>\ell}}\tau_{\delta}x$. We deduce that

$|\Pi_{V_{S}}\tau_{\delta}x|=|\Pi_{V_{S}\cap
X_{>\ell}}\tau_{\delta}x|\leq|\Pi_{>\ell}\tau_{\delta}x|.$

Since $x\in\widetilde{H}\setminus\\{0\\}$, (44) implies that
$|\Pi_{>\ell}\tau_{\delta}x|<\epsilon|\tau_{\delta}x|$. Thus,
$|\Pi_{V_{S}}\tau_{\delta}x|<\epsilon|\tau_{\delta}x|$ and
$\tau_{\delta}x\in\tau_{\delta}\widetilde{H}\subseteq\tau_{\delta}H$, which
implies $\cos(\theta_{\max}(\tau_{\delta}H,V_{S}))<\epsilon$. This contradicts
(37), completing the proof of Lemma 4.4. ∎

####  4.1.1 Proof of Proposition 4.3

Fix $H\in G(k,X)$, $\eta\in(0,1/2)$.

Fix a compact interval $I\subseteq(0,\infty)$ with
$\frac{r(I)}{l(I)}\geq\left(\frac{2^{d}}{\eta}\right)^{dk+2}$.

For ease of notation, let $H_{\delta}=\tau_{\delta}H$ and
$\omega_{\delta}=\omega_{H_{\delta}}=\tau_{\delta}^{*}\omega_{H}$ for
$\delta>0$.

We aim to show that there exist $\delta\in I$ and a $k$-dimensional dilation
invariant subspace $V\subseteq X$ with $\cos(\theta_{\max}(H_{\delta},V))\geq
1-\eta$.

Recall that every $k$-dimensional dilation invariant subspace $V\subseteq X$
has the form $V=V_{S}=\operatorname{span}\\{e_{j}:j\in S\\}$ for some
$S\subseteq[d]$ with $\\#(S)=k$, and that $\omega_{S}=\bigwedge_{j\in
S}e_{j}\in\bigwedge^{k}X$ is a representative form for $V_{S}$.

By (24), (27), it is enough to show that there exists $\delta\in I$ such that

$\cos(\angle(H_{\delta},V_{S}))=\frac{|\langle\omega_{\delta},\omega_{S}\rangle|}{|\omega_{\delta}|}\geq
1-\eta\quad\mbox{ for some }S\subseteq[d],\;\\#(S)=k.$ (45)

Let

$\epsilon=\sqrt{\eta/2^{d}}.$ (46)

Observe that if $k=\dim(H)=d$ then (45) is true with $S=[d]$ for any
$\delta\in I$. That’s because $\bigwedge^{d}X$ is one-dimensional, hence,
$\omega_{\delta}\in\operatorname{span}\\{\omega_{[d]}\\}$.

We may thus assume $d\geq 2$ and $1\leq k<d$.

We will then prove (45) by contradiction. For sake of contradiction, suppose
(45) fails for every $\delta\in I$.

Recall $\l(I)$ and $r(I)$ are the left and right endpoints of $I$. For $j\geq
0$, let $\delta_{j}:=\epsilon^{-2j}\cdot l(I)$. Let $J$ be the largest
positive integer such that $\delta_{J}\in I$. By assumption,
$r(I)/l(I)\geq\left(\frac{2^{d}}{\eta}\right)^{dk+2}=\epsilon^{-2(dk+2)}$,
thus

$J\geq dk+2.$ (47)

For each $j=0,1,\dots,J$ we claim that there exist distinct subsets $S_{j,1}$
and $S_{j,2}$ of $[d]$ of cardinality $k$, such that

$\frac{|\langle\omega_{\delta_{j}},\omega_{S_{j,\mu}}\rangle|}{|\omega_{\delta_{j}}|}\geq\epsilon\qquad(\mu=1,2).$
(48)

To see this, order the subsets of $[d]$ of cardinality $k$ in a sequence,
$S_{j,1},S_{j,2},\dots,S_{j,L}$, $L=\binom{d}{k}$, so that

$\ell\mapsto|\langle\omega_{\delta_{j}},\omega_{S_{j,\ell}}\rangle|\mbox{ is
non-increasing (for fixed }j\mbox{)}.$

Set
$a_{j,\ell}:=|\langle\omega_{\delta_{j}},\omega_{S_{j,\ell}}\rangle|/|\omega_{\delta_{j}}|$
for $\ell=1,\dots,L$. By assumption, (45) fails for $\delta=\delta_{j}$, thus

$a_{j,\ell}<1-\eta\quad\mbox{for all }\ell=1,\dots,L.$

Because
$\\{\omega_{S}:S\subseteq[d],\;\\#(S)=k\\}=\\{\omega_{S_{j,\ell}}:\ell=1,\dots,L\\}$
is an orthonormal basis for $\bigwedge^{k}X$, we have
$\sum_{\ell=1}^{L}a_{j,\ell}^{2}=1$. Since $\ell\mapsto a_{j,\ell}$ is non-
increasing,

$a_{j,1}\geq\sqrt{1/L}\geq\sqrt{1/2^{d}}>\epsilon.$

Since $a_{j,1}\leq 1-\eta$, we have

$\sum_{\ell=2}^{L}a_{j,\ell}^{2}=1-a_{j,1}^{2}\geq
1-(1-\eta)^{2}=2\eta-\eta^{2}\geq\eta.$

Thus, because $\ell\mapsto a_{j,\ell}$ is non-increasing, we have

$a_{j,2}\geq\sqrt{\eta/(L-1)}>\sqrt{\eta/L}>\sqrt{\eta/2^{d}}=\epsilon.$

As $a_{j,1},a_{j,2}\geq\epsilon$, we complete the proof of (48).

Let $\mu_{0}=1$, and for $1\leq j\leq J$, let $\mu_{j}\in\\{1,2\\}$ be such
that $S_{j,\mu_{j}}\neq S_{j-1,\mu_{j-1}}$. By definition, note that
$\delta_{j}=\delta_{j-1}/\epsilon^{2}$ for $j\geq 1$. Thus, using (48), we may
apply Lemma 4.4 to deduce that $c_{\ell}(S_{j,\mu_{j}})\geq
c_{\ell}(S_{j-1,\mu_{j-1}})$ for every $\ell=1,2,\dots,d$ and $j=1,2,\dots,J$.
Further, since $S_{j,\mu_{j}}\neq S_{j-1,\mu_{j-1}}$, for each $j$ this
inequality is strict for some $\ell$ (see (36)). It follows that

$\psi_{j}:=\sum\limits_{\ell=1}^{d}c_{\ell}(S_{j,\mu_{j}})>\psi_{j-1}\qquad(j=1,2,\dots,J).$
(49)

But note that

$0\leq c_{\ell}(S)=\\#\\{j\in S:j\leq\ell\\}\leq k$

for all $\ell=1,2,\dots,d$ and all $S\subseteq[d]$ with $\\#(S)=k$. Thus,

$0\leq\psi_{j}\leq dk\qquad(j=1,2,\dots,J).$

From this and (49) we deduce that $J\leq dk+1$. But this contradicts (47).
This completes the proof of (45) and finishes the proof of Proposition 4.3.

###  4.2 Monotonicity of the orbits of a Hilbert dilation system on the
Grassmanian.

Fix a Hilbert dilation system $\mathcal{X}=(X,\tau_{\delta})_{\delta>0}$. We
drop the assumption that $\mathcal{X}$ is simple. Thus,
$X=\bigoplus_{\nu=1}^{m}X_{\nu}$ and $\tau_{\delta}:X\rightarrow X$ has the
form $\tau_{\delta}|_{X_{\nu}}=\delta^{-\nu}\cdot\mathrm{id}|_{X_{\nu}}$, as
in (30), (31). Our next result describes a qualitative property of the
functions $\delta\mapsto\angle(\tau_{\delta}H,V)$ (for fixed $H,V$) that will
enter into the proof of Proposition 2.11. See (24) for the definition of the
quantity $\angle(V,W)$.

######  Lemma 4.5.

Let $H,V\subseteq X$ be subspaces with $\dim(H)=\dim(V)\geq 1$, such that $V$
is dilation invariant. Then the map $f(\delta)=\cos(\angle(\tau_{\delta}H,V))$
is unimodal: if $a<b<c$ and $f(b)<f(c)$, then $f(a)<f(b)$.

###### Proof.

Let $l=\dim(H)=\dim(V)\geq 1$. Fix representative forms
$\omega_{H},\omega_{V}\in\bigwedge^{l}X$ for $H$, $V$, respectively, with unit
norm, $|\omega_{H}|=|\omega_{V}|=1$. Then $\tau_{\delta}^{*}\omega_{H}$ is a
representative form for $\tau_{\delta}H$. So we have

$f(\delta)=\cos(\angle(\tau_{\delta}H,V))=\frac{|\langle\tau_{\delta}^{*}\omega_{H},\omega_{V}\rangle|}{|\tau_{\delta}^{*}\omega_{H}||\omega_{V}|}=\frac{|\langle\omega_{H},\tau_{\delta}^{*}\omega_{V}\rangle|}{|\tau_{\delta}^{*}\omega_{H}|}.$
(50)

Since $V$ is dilation invariant,
$\tau_{\delta}^{*}\omega_{V}=\delta^{-\operatorname{sgn}(V)}\omega_{V}$ (see
(33)). So the numerator of (50) is

$|\langle\omega_{H},\tau_{\delta}^{*}\omega_{V}\rangle|=\alpha\cdot\delta^{-\operatorname{sgn}(V)}\mbox{,
for }\alpha:=|\langle\omega_{H},\omega_{V}\rangle|\geq 0.$ (51)

To compute the denominator of (50), we fix a basis for $\bigwedge^{l}X$. Fix a
family of dilation invariant subspaces $U_{1},U_{2},\dots,U_{M}$, such that
the associated unit-norm representative forms
$\omega_{U_{1}},\omega_{U_{2}},\dots,\omega_{U_{M}}$ give an orthonormal basis
for $\bigwedge^{l}X$ ($M=\binom{d}{l}$). Then
$\tau_{\delta}^{*}\omega_{U_{i}}=\delta^{-\operatorname{sgn}(U_{i})}\omega_{U_{i}}$
by (33). So the denominator of (50) is

$\displaystyle|\tau_{\delta}^{*}\omega_{H}|=\sqrt{\sum_{i=1}^{M}\langle\tau_{\delta}^{*}\omega_{H},\omega_{U_{i}}\rangle^{2}}=\sqrt{\sum_{i=1}^{M}\langle\omega_{H},\tau_{\delta}^{*}\omega_{U_{i}}\rangle^{2}}$
(52)
$\displaystyle=\sqrt{\sum_{i=1}^{M}\langle\omega_{H},\omega_{U_{i}}\rangle^{2}\delta^{-2\operatorname{sgn}(U_{i})}}=\sqrt{\sum_{p=1}^{dm}\alpha_{p}\delta^{-2p}}$

for constants

$\alpha_{p}=\sum_{i\in[M],\operatorname{sgn}(U_{i})=p}\langle\omega_{H},\omega_{U_{i}}\rangle^{2}\geq
0,\qquad 1\leq p\leq dm.$

Here, we used that $1\leq\operatorname{sgn}(U)\leq dm$ for any dilation
invariant subspace $U\subseteq X$ with $\dim(U)\geq 1$. Not all of the
coefficients $\alpha_{p}$ are equal to zero, because $\\{\omega_{U_{i}}\\}$ is
an orthonormal basis for $\bigwedge^{\ell}X$. Combining (51) and (52), we have

$f(\delta)=\frac{\alpha\delta^{-\operatorname{sgn}(V)}}{\sqrt{\sum\limits_{p=1}^{dm}\alpha_{p}\delta^{-2p}}}.$
(53)

If $\alpha=0$ then $f=0$, and we obtain the desired conclusion because
constant functions are unimodal.

Now suppose $\alpha>0$, so that $f(\delta)>0$ for all $\delta$. Define
$g(\delta)=\log(f(\delta^{-1}))$. Then compute

$\displaystyle
g^{\prime}(\delta)=\operatorname{sgn}(V)\frac{1}{\delta}-\frac{\sum\limits_{p=1}^{dm}p\alpha_{p}\delta^{2p-1}}{\sum\limits_{p=1}^{dm}\alpha_{p}\delta^{2p}}=\frac{P(\delta)}{\sum\limits_{p=1}^{dm}\alpha_{p}\delta^{2p}},$
$\displaystyle
P(\delta)=\sum\limits_{p=1}^{dm}\alpha_{p}(\operatorname{sgn}(V)-p)\delta^{2p-1}.$

We now split the proof into two cases.

Case 1: $\alpha_{p}=0$ for all $p\neq\operatorname{sgn}(V)$. Then from the
above identities, $P\equiv 0$, and so $g^{\prime}\equiv 0$. So $g$ is
constant, and thus $f$ is constant, giving the desired result.

Case 2: $\alpha_{p}\neq 0$ for some $p\neq\operatorname{sgn}(V)$. If there
exist $r,q$ with $\alpha_{r}>0$, $\alpha_{q}>0$, $r<\operatorname{sgn}(V)<q$,
then the signs of the coefficients of $P(\delta)$ change exactly once;
otherwise they change $0$ times. By Descartes’ rule of signs, there is at most
one value of $\delta>0$ with $P(\delta)=0$, so at most one value of $\delta>0$
with $g^{\prime}(\delta)=0$. This leaves three options: $g$ is monotone, $g$
has one interior maximum and no interior minima, and $g$ has one interior
minimum and no interior maxima. The first two options imply that $g$ is
unimodal, hence $f$ is unimodal. The third option is impossible. To see this,
we exploit the assumption that $\alpha_{p}\neq 0$ for some
$p\neq\operatorname{sgn}(V)$. Therefore, from (53), either
$\lim_{\delta\rightarrow\infty}f(\delta)=0$ or $\lim_{\delta\rightarrow
0}f(\delta)=0$. Therefore, $g(\delta)\to-\infty$ for at least one of
$\delta\to 0$ or $\delta\to\infty$, ruling out that $g$ has one interior
minimum and no interior maxima.

This completes the proof of Lemma 4.5. ∎

###  4.3 Rescaling dynamics on the space of ellipsoids

We present further preparatory results to be used in the proofs of
Propositions 2.9 and 2.11.

Let $(X,\tau_{\delta})_{\delta>0}$ be a Hilbert dilation system. So, $X$ is a
real Hilbert space of dimension $d$ and $\tau_{\delta}:X\rightarrow X$ are
linear operators of the form (30),(31).

Given a set $\Omega\subseteq X$ and $T:X\rightarrow X$, we denote
$T\Omega=\\{T(x):x\in\Omega\\}$.

A (centered) _ellipsoid_ $\mathcal{E}\subseteq X$ is a set of the form

$\mathcal{E}=\left\\{\sum_{i=1}^{d}c_{i}\sigma_{i}v_{i}:\sum_{i=1}^{d}c_{i}^{2}\leq
1\right\\},$ (54)

where $\sigma_{1}\geq\dots\geq\sigma_{d}\geq 0$ and $\\{v_{1},\dots,v_{d}\\}$
is an orthonormal basis for $X$. We call $v_{1},\dots,v_{d}$ (normalized)
principal axis directions of $\mathcal{E}$, and $\sigma_{1},\dots,\sigma_{d}$
the principal axis lengths of $\mathcal{E}$. Denote
$\sigma_{j}(\mathcal{E})=\sigma_{j}$ for the $j$’th principal axis length of
$\mathcal{E}$. Principal axis lengths (but not directions) are uniquely
determined by $\mathcal{E}$.

Note that the intersection of an ellipsoid and a subspace is also an
ellipsoid. Further, the image of an ellipsoid under a linear transformation is
an ellipsoid.

Let $\mathcal{B}$ be the closed unit ball of $X$. If $A:X\rightarrow X$ is a
linear transformation then $A\mathcal{B}$ is an ellipsoid in $X$. Let
$\sigma_{1}\geq\dots\geq\sigma_{d}\geq 0$ be the singular values of $A$, let
$\\{v_{1},\dots,v_{d}\\}$ be left singular vectors of $A$, and let
$\\{w_{1},\dots,w_{d}\\}$ be right singular vectors of $A$. That is,
$\\{v_{i}\\}$ and $\\{w_{i}\\}$ are orthonormal bases for $X$, and
$Aw_{i}=\sigma_{i}v_{i}$ for all $i$. We express $\mathcal{B}$ in the form
$\\{\sum_{i}c_{i}w_{i}:\sum_{i}c_{i}^{2}\leq 1\\}$. Then

$A\mathcal{B}=\left\\{\sum_{i=1}^{d}c_{i}\sigma_{i}v_{i}:\sum_{i=1}^{d}c_{i}^{2}\leq
1\right\\}.$ (55)

So, the principal axis lengths of $A\mathcal{B}$ are the singular values of
$A$, and the principal axis directions of $A\mathcal{B}$ are corresponding
left singular vectors of $A$.

In particular, every ellipsoid $\mathcal{E}$ can be written as
$\mathcal{E}=A\mathcal{B}$ for some linear transformation $A:X\rightarrow X$.

Given an ellipsoid $\mathcal{E}\subseteq X$, let
$\mathcal{E}_{\delta}:=\tau_{\delta}\mathcal{E}$ for $\delta>0$. Then
$(\mathcal{E}_{\delta})_{\delta>0}$ is an _orbit_ of $\tau_{\delta}$ in the
space of ellipsoids. Our next result, Lemma 4.7, states that this orbit can be
approximated by an orbit in the Grassmanian $G(k,X)$ if a condition on the
$\mathcal{E}_{\delta}$ is met.

######  Definition 4.6.

Let $\epsilon\in(0,1/2)$, and let $\mathcal{E}\subseteq X$ be an ellipoid. Say
that $\mathcal{E}$ is $\epsilon$-degenerate if
$\sigma_{j}(\mathcal{E})\notin[\epsilon,\epsilon^{-1}]$ for all $j$. In other
words, $\mathcal{E}$ is $\epsilon$-degenerate if the length of every principal
axis of $\mathcal{E}$ is either less than $\epsilon$ or greater than
$\epsilon^{-1}$.

######  Lemma 4.7.

Let $\mathcal{E}$ be an ellipsoid in $X$, let $\epsilon\in(0,1/2)$, and let
$I\subseteq(0,\infty)$ be a compact interval. Let
$\mathcal{E}_{\delta}:=\tau_{\delta}\mathcal{E}$ for $\delta>0$. Suppose that
$\mathcal{E}_{\delta}$ is $\epsilon$-degenerate for all $\delta\in I$. Then
there exists a subspace $H\subseteq X$ such that, for all $\delta\in I$,

  1. (a)

$\mathcal{E}_{\delta}\subseteq\tau_{\delta}H+\epsilon\mathcal{B}$, and

  2. (b)

$\tau_{\delta}H\cap(\frac{1}{2\epsilon}\mathcal{B})\subseteq\mathcal{E}_{\delta}$.

###### Proof.

By rescaling, we may assume that $I$ has the form $I=[1,T]$ for $T\geq 1$.
Write $\mathcal{E}=A\mathcal{B}$ for a linear transformation $A:X\rightarrow
X$. Then $\mathcal{E}_{\delta}=A_{\delta}\mathcal{B}$, with
$A_{\delta}:=\tau_{\delta}A$. For $\delta>0$, consider the singular values of
$A_{\delta}$:

$\sigma_{1}(\delta)\geq\sigma_{2}(\delta)\geq\dots\geq\sigma_{d}(\delta)\geq
0,$ (56)

and let $\\{v_{1}(\delta),\dots,v_{d}(\delta)\\}$ be the associated left
singular vectors of $A_{\delta}$, which form an orthonormal basis for $X$. By
(55),

$\mathcal{E}_{\delta}=\left\\{\sum_{i=1}^{d}c_{i}\sigma_{i}(\delta)v_{i}(\delta):\sum_{i=1}^{d}c_{i}^{2}\leq
1\right\\}.$ (57)

The singular values of $A_{\delta}$ are the square roots of eigenvalues of
$A_{\delta}A_{\delta}^{*}$:

$\sigma_{j}(\delta)=\sqrt{\lambda_{j}(A_{\delta}A_{\delta}^{*})}=\sqrt{\lambda_{j}(\tau_{\delta}AA^{*}\tau_{\delta})}.$

The ordered tuple of eigenvalues
$(\lambda_{1}(B),\dots,\lambda_{d}(B))\in\mathbb{R}^{d}$ of a symmetric matrix
$B\in\mathbb{R}^{d\times d}$ is a continuous function of the entries of $B$.
It follows that $\delta\mapsto\sigma_{j}(\delta)$ is continuous for each $j$.
By the intermediate value theorem, and the assumption that
$\mathcal{E}_{\delta}$ is $\epsilon$-degenerate for each $\delta\in I$, there
exists $k\in\\{0,1,\dots,d\\}$ so that

$\displaystyle\sigma_{j}(\delta)>\epsilon^{-1}\mbox{ for }1\leq j\leq k,$ (58)
$\displaystyle\sigma_{j}(\delta)<\epsilon\mbox{ for }k<j\leq d\quad(\mbox{all
}\delta\in I).$ (59)

Let $H=\operatorname{span}\\{v_{j}(1):1\leq j\leq k\\}$. Thus, $H\in G(k,X)$
is spanned by the $k$ longest principle axes of $\mathcal{E}_{1}$, and
$H^{\perp}=\operatorname{span}\\{v_{j}(1):k<j\leq d\\}$ is spanned by the
$(d-k)$ shortest principal axes of $\mathcal{E}_{1}$.

Evidently, by (57), $\Pi_{H}\mathcal{E}_{1}=\mathcal{E}_{1}\cap H$ and
$\Pi_{H^{\perp}}\mathcal{E}_{1}=\mathcal{E}_{1}\cap H^{\perp}$. By the second
identity, a general element $x$ of $\Pi_{H^{\perp}}\mathcal{E}_{1}$ has the
form $x=\sum_{i>k}c_{i}\sigma_{i}(1)v_{i}(1)$, for coefficients $c_{i}$ with
$\sum_{i}c_{i}^{2}\leq 1$. By (59) for $\delta=1$, the fact that
$|v_{j}(1)|=1$ for all $j$, and the Pythagorean theorem, we deduce that
$|x|\leq\epsilon$ for any $x\in\Pi_{H^{\perp}}\mathcal{E}_{1}$. Thus,
$\Pi_{H^{\perp}}\mathcal{E}_{1}\subseteq\epsilon\mathcal{B}$. Thus, given that
$\mathcal{E}=\mathcal{E}_{1}$, we obtain

$\mathcal{E}\subseteq\Pi_{H}\mathcal{E}+\Pi_{H^{\perp}}\mathcal{E}=(\mathcal{E}\cap
H)+(\mathcal{E}\cap H^{\perp})\subseteq(\mathcal{E}\cap
H)+\epsilon\mathcal{B}.$

Thus, for $\delta\geq 1$,

$\tau_{\delta}\mathcal{E}\subseteq\tau_{\delta}((\mathcal{E}\cap
H)+\epsilon\mathcal{B})=\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H+\epsilon\tau_{\delta}\mathcal{B}\subseteq\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H+\epsilon\mathcal{B},$
(60)

where the last inclusion uses that $\|\tau_{\delta}\|_{op}\leq 1$ for
$\delta\geq 1$.

Note that (60) implies
$\tau_{\delta}\mathcal{E}\subseteq\tau_{\delta}H+\epsilon\mathcal{B}$ for
$\delta\geq 1$. This implies (a).

We next establish (b). For contradiction, suppose there exists
$\delta\in[1,T]$ with

$\tau_{\delta}H\cap\left((2\epsilon)^{-1}\mathcal{B}\right)\not\subseteq\tau_{\delta}\mathcal{E}.$
(61)

We regard $\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H$ as an ellipsoid in the
vector space $\tau_{\delta}H$. Let $\sigma\geq 0$ be the shortest principal
axis length of $\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H$ in
$\tau_{\delta}H$, and let $v\in\tau_{\delta}H$ be an associated unit-norm
principal axis direction. Then $\pm\sigma
v\in\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H$, and by (61),
$\sigma<\frac{1}{2\epsilon}$. Thus, if $U:=\tau_{\delta}H\cap v^{\perp}$, then
$\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H\subseteq
U+\frac{1}{2\epsilon}\mathcal{B}$. By (60),

$\tau_{\delta}\mathcal{E}\subseteq(\tau_{\delta}\mathcal{E}\cap\tau_{\delta}H)+\epsilon\mathcal{B}\subseteq
U+\left((2\epsilon)^{-1}+\epsilon\right)\mathcal{B}\subseteq
U+(3/4)\epsilon^{-1}\mathcal{B}.$ (62)

Given $\dim(\tau_{\delta}H)=k$, and $U$ has codimension $1$ in
$\tau_{\delta}H$, then $\dim(U)=k-1$. From (57) and (58),
$\tau_{\delta}\mathcal{E}$ contains a $k$-dimensional disk of radius
$\epsilon^{-1}$. Together with (62), these remarks lead to a contradiction. ∎

Let $\mathcal{E}$ be an ellipsoid in $X$. The next lemma guarantees that
$\tau_{\delta}\mathcal{E}$ is $\epsilon$-degenerate for “most”
$\delta\in(0,\infty)$. We write $r(I)$ and $l(I)$ for the right and left
endpoints of an interval $I\subseteq(0,\infty)$, respectively.

######  Lemma 4.8.

Let $d=\dim X$. Let $\mathcal{E}\subseteq X$ be an ellipsoid and let
$\epsilon\in(0,1/2)$. There exists a collection of closed intervals
$J_{1},J_{2},\dots,J_{d}\subseteq(0,\infty)$ such that
$\tau_{\delta}\mathcal{E}$ is $\epsilon$-degenerate for all
$\delta\notin\bigcup_{p=1}^{d}J_{p}$, and such that
$r(J_{p})/l(J_{p})\leq\frac{1}{\epsilon^{2}}$ for all $p$.

###### Proof.

Write $\mathcal{E}=A\mathcal{B}$ for a linear transformation $A:X\rightarrow
X$. For $\delta>0$, let
$\mathcal{E}_{\delta}=\tau_{\delta}\mathcal{E}=A_{\delta}\mathcal{B}$, with
$A_{\delta}=\tau_{\delta}A$. Let
$\sigma_{1}(\delta)\geq\sigma_{2}(\delta)\geq\dots\geq\sigma_{d}(\delta)\geq
0$ be the principal axis lengths of $\mathcal{E}_{\delta}$, given by the
singular values of $A_{\delta}$.

The $j$-th singular value $\sigma_{j}(\delta)$ of $A_{\delta}$ is given by
$\sigma_{j}(\delta)=\sqrt{\lambda_{j}(\delta)}$, where $\lambda_{j}(\delta)$
is the $j$-th eigenvalue of $A_{\delta}A_{\delta}^{*}$, i.e.,

$\lambda_{j}(\delta)=\lambda_{j}(\tau_{\delta}AA^{*}\tau_{\delta}).$

Here, we write the eigenvalues of $A_{\delta}A_{\delta}^{*}$ in decreasing
order,
$\lambda_{1}(\delta)\geq\lambda_{2}(\delta)\geq\dots\geq\lambda_{d}(\delta)\geq
0$, for each $\delta$. Let $\delta_{*}>0$. We claim that

$\lambda_{j}(\delta)\leq\left(\delta_{*}/\delta\right)^{2}\cdot\lambda_{j}(\delta_{*})\quad\mbox{for}\;j=1,2,\dots,d,\;\;\delta\geq\delta_{*}.$
(63)

Using that $A_{\delta}=(A_{\delta_{*}})_{\delta/\delta_{*}}$, we make the
substitution $A\leftarrow A_{\delta_{*}}$ and
$\delta\leftarrow\delta/\delta_{*}$ and reduce the proof of (63) to the case
$\delta_{*}=1$. By the min-max characterization of eigenvalues, for any
$\delta\geq 1$, with $B=AA^{*}$, we have

$\displaystyle\lambda_{j}(\delta)$ $\displaystyle=\sup\limits_{V\in
G(j,X)}\inf\limits_{x\in
V\setminus\\{0\\}}\frac{\langle\tau_{\delta}B\tau_{\delta}x,x\rangle}{|x|^{2}}$
$\displaystyle=\sup\limits_{V\in G(j,X)}\inf\limits_{x\in
V\setminus\\{0\\}}\frac{|\tau_{\delta}x|^{2}}{|x|^{2}}\frac{\langle
B\tau_{\delta}x,\tau_{\delta}x\rangle}{|\tau_{\delta}x|^{2}}$
$\displaystyle=\sup\limits_{\hat{V}\in
G(j,X)}\inf\limits_{\hat{x}\in\hat{V}\setminus\\{0\\}}\frac{|\hat{x}|^{2}}{|\tau_{\delta^{-1}}\hat{x}|^{2}}\frac{\langle
B\hat{x},\hat{x}\rangle}{|\hat{x}|^{2}}\leq\delta^{-2}\lambda_{j}(1).$

The last equality above makes use of the substitution $\hat{V}=\tau_{\delta}V$
and $\hat{x}=\tau_{\delta}x$. The last inequality holds because
$|\tau_{a}y|\geq a^{-1}|y|$ for $a\leq 1$, and by the min-max characterization
of the eigenvalue $\lambda_{j}(1)=\lambda_{j}(B)$. We have proven (63).

Let $J_{p}$ be the closure of the set
$\\{\delta\in(0,\infty):\sigma_{p}(\delta)\in[\epsilon,\epsilon^{-1}]\\}$ for
$p=1,2,\dots,d$. From (63) and
$\sigma_{p}(\delta)=\sqrt{\lambda_{p}(\delta)}$, we have
$\sigma_{p}(\delta)\leq(\delta_{*}/\delta)\sigma_{p}(\delta_{*})$ for
$\delta\geq\delta_{*}$. Thus, $\sigma_{p}$ is a decreasing function of
$\delta$, and if $\delta>\epsilon^{-2}\delta_{*}$ then
$\sigma_{p}(\delta)<\epsilon^{2}\sigma_{p}(\delta_{*})$. It follows that
$J_{p}$ is an interval and $r(J_{p})/l(J_{p})\leq\epsilon^{-2}$.

Finally, note, for $\delta\notin\bigcup_{p}J_{p}$, that
$\sigma_{p}(\delta)\notin[\epsilon,\epsilon^{-1}]$ for all $p$ (by definition
of the intervals $J_{p}$), thus, $\tau_{\delta}\mathcal{E}$ is
$\epsilon$-degenerate. ∎

###  4.4 Complexity

Given a Hilbert space $X$, we let $\mathcal{K}(X)$ denote the collection of
all closed, convex, symmetric subsets of $X$. Let
$\mathcal{B}\in\mathcal{K}(X)$ denote the unit ball of $X$. Given
$\Omega\in\mathcal{K}(X)$, $V$ a subspace of $X$, and $R\geq 1$, recall that
$\Omega$ is $R$-transverse to $V$ if (a) $\Omega\cap V\subseteq R\mathcal{B}$,
and (b) $\Pi_{V^{\perp}}(\Omega\cap\mathcal{B})\supseteq R^{-1}\mathcal{B}\cap
V^{\perp}$ (see Definition 3.7).

For an interval $I$, let $l(I)$ and $r(I)$ denote the left and right endpoints
of $I$, respectively. We say $I>J$ if $l(I)>r(J)$, and $I>0$ if $l(I)>0$.

######  Definition 4.9.

Let $\mathcal{X}=(X,\tau_{\delta})_{\delta>0}$ be a Hilbert dilation system.
For $\Omega\in\mathcal{K}(X)$, $R\in[1,\infty)$, $R^{*}\in(R,\infty)$, the
complexity of $\Omega$ with respect to $\mathcal{X}$ with parameters
$(R,R^{*})$, written
$\mathcal{C}_{\mathcal{X}}(\Omega,R,R^{*})=\mathcal{C}(\Omega,R,R^{*})$, is
the largest positive integer $K$ such that there exist compact intervals
$I_{1}>I_{2}>\dots>I_{K}>0$ in $(0,\infty)$ and dilation invariant subspaces
$V_{1},V_{2},\dots,V_{K}\subseteq X$ such that, for every $j$,
$\tau_{r(I_{j})}\Omega$ is $R$-transverse to $V_{j}$, and
$\tau_{l(I_{j})}\Omega$ is not $R^{*}$-transverse to $V_{j}$.

Fix a Hilbert dilation system $(X,\tau_{\delta})_{\delta>0}$. Thus,
$X=\bigoplus_{\nu=1}^{m}X_{\nu}$ and $\tau_{\delta}:X\rightarrow X$ is given
by $\tau_{\delta}|_{X_{\nu}}=\delta^{-\nu}\mathrm{id}|_{X_{\nu}}$. Let
$d:=\dim(X)$. Let $V$ be a dilation-invariant (DI) subspace of $X$ (see
Definition 4.2). Then $V$ has the form

$V=\bigoplus\limits_{\nu=1}^{m}V\cap X_{\nu}.$

Recall that the _signature_ of $V$ is defined by
$\operatorname{sgn}(V)=\sum\limits_{\nu=1}^{m}\nu\cdot\dim(V\cap X_{\nu})$.
Note that $0\leq\operatorname{sgn}(V)\leq md$ for any dilation-invariant
subspace $V$.

If $\Omega_{1},\Omega_{2}\in\mathcal{K}(X)$ satisfy
$\lambda^{-1}\Omega_{2}\subseteq\Omega_{1}\subseteq\lambda\Omega_{2}$ for
$\lambda\geq 1$ then we say that $\Omega_{1}$ and $\Omega_{2}$ are
$\lambda$-equivalent, and we write $\Omega_{1}\sim_{\lambda}\Omega_{2}$.

We now rephrase a classical theorem of F. John (see [1]) in terms of the
definitions just provided.

######  Proposition 4.10 (John’s theorem).

Given a compact $\Omega\in\mathcal{K}(X)$, there exists an ellipsoid
$\mathcal{E}\subseteq X$ such that $\Omega$ and $\mathcal{E}$ are
$\sqrt{d}$-equivalent.

######  Remark 4.11.

If $\Omega_{1}$ is $R$-transverse to $V$ and
$\Omega_{1}\sim_{\lambda}\Omega_{2}$ then $\Omega_{2}$ is $\lambda
R$-transverse to $V$. It follows that if $\Omega_{1}\sim_{\lambda}\Omega_{2}$
then $\mathcal{C}(\Omega_{1},R,R^{*})\leq\mathcal{C}(\Omega_{2},\lambda
R,\lambda^{-1}R^{*})$ provided that $R^{*}>\lambda^{2}R$ so the right-hand-
side is well-defined.

######  Lemma 4.12.

Fix $\xi>R\geq 1$. Suppose $\Omega_{1}$ is $R$-transverse to $V$ and
$\Omega_{1}\cap\xi\mathcal{B}=\Omega_{2}\cap\xi\mathcal{B}$. Then $\Omega_{2}$
is $R$-transverse to $V$.

###### Proof.

Given that $\Omega_{1}$ is $R$-transverse to $V$ and
$\Omega_{1}\cap\xi\mathcal{B}=\Omega_{2}\cap\xi\mathcal{B}$, we have

$\Omega_{2}\cap V\cap\xi\mathcal{B}=\Omega_{1}\cap
V\cap\xi\mathcal{B}\subseteq R\mathcal{B}.$

Since $\xi>R$, we deduce that $\Omega_{2}\cap V\subseteq R\mathcal{B}$.

Since $\Omega_{1}\cap\xi\mathcal{B}=\Omega_{2}\cap\xi\mathcal{B}$ for $\xi>1$,
we have $\Omega_{1}\cap\mathcal{B}=\Omega_{2}\cap\mathcal{B}$, thus

$R^{-1}\mathcal{B}\cap
V^{\perp}\subseteq\Pi_{V^{\perp}}(\Omega_{1}\cap\mathcal{B})=\Pi_{V^{\perp}}(\Omega_{2}\cap\mathcal{B}).$

So, $\Omega_{2}$ is $R$-transverse to $V$. ∎

The remainder of this section is devoted to the proof of the next result.

######  Proposition 4.13.

For any $\Omega\in\mathcal{K}(X)$, $\mathcal{C}(\Omega,R_{1},R_{2})\leq
4md^{2}$ provided that $R_{1}\geq 16$ and
$R_{2}\geq\max\\{(\sqrt{d})^{4m+1}R_{1}^{4m},(\sqrt{d})^{3d+1}R_{1}^{3d}\\}$.

Using John’s theorem, we shall reduce Proposition 4.13 to the following:

######  Proposition 4.14.

For any ellipsoid $\mathcal{E}\subseteq X$, $R\geq 16$ and
$R^{*}\geq\max\\{R^{4m},R^{3d}\\}$,

$\mathcal{C}(\mathcal{E},R,R^{*})\leq 4md^{2}.$

We will later give details on the reduction of Proposition 4.13 to Proposition
4.14. Next we make preparations for the proof of Proposition 4.14. Fix
$R,R^{*}$ and $\epsilon>0$ such that

$\displaystyle 16\leq R\leq\max\\{R^{3d},R^{4m}\\}\leq R^{*},$ (64)
$\displaystyle\epsilon\leq 1/(4R)\mbox{ and }R/R^{*}\leq\epsilon^{2m}.$

Note (64) is satisfied if $\epsilon=\frac{1}{4R}$, as then
$\frac{R}{R^{*}}\leq R^{1-4m}\leq R^{-3m}\leq(4R)^{-2m}=\epsilon^{2m}$.

The following result is the key ingredient in the proof of Proposition 4.14.

######  Proposition 4.15.

Let $R$, $R^{*}$, $\epsilon$ be as in (64). Let $\mathcal{E}$ be an ellipsoid
in $X$, and let $I=[\delta_{\min},\delta_{\max}]\subseteq(0,\infty)$. Suppose
that $\tau_{\delta}\mathcal{E}$ is $\epsilon$-degenerate for all $\delta\in
I$.

If there exist $\delta_{*}\in I$ and dilation invariant subspaces
$V,W\subseteq X$ such that

  1. 1.

$\tau_{\delta_{\max}}\mathcal{E}$ is $R$-transverse to $V,$

  2. 2.

$\tau_{\delta_{*}}\mathcal{E}$ is not $R^{*}$-transverse to $V,$ and

  3. 3.

$\tau_{\delta_{\min}}\mathcal{E}$ is $R$-transverse to $W$,

then $\operatorname{sgn}(V)>\operatorname{sgn}(W)$.

Before the proof of Proposition 4.15, we present two preparatory lemmas.

######  Lemma 4.16.

If $A,K,T\in\mathcal{K}(X)$, and $K\subseteq T$, then $(A+K)\cap
T\subseteq(A\cap 2T)+K$.

###### Proof.

Fix $x\in(A+K)\cap T$. Then $x=a+k$ for $a\in A$, $k\in K$. Note that
$a=x-k\in T+K\subseteq 2T$. Hence, $x=a+k\in(A\cap 2T)+K$. ∎

######  Lemma 4.17.

Under the hypotheses of Proposition 4.15, there exists a subspace $H\subseteq
X$ such that Conditions 1,2,3 of Proposition 4.15 hold with $H$ and $4R$ in
place of $\mathcal{E}$ and $R$, respectively.

###### Proof of Lemma 4.17.

By Lemma 4.7, there exists a subspace $H\subseteq X$ such that for all
$\delta\in I$,

  1. (a)

$\tau_{\delta}\mathcal{E}\subseteq\tau_{\delta}H+\epsilon\mathcal{B}$

  2. (b)

$\tau_{\delta}H\cap(\frac{1}{2\epsilon}\mathcal{B})\subseteq\tau_{\delta}\mathcal{E}$.

Using (b) for $\delta=\delta_{\max}$, the inequality
$R\leq\frac{1}{4\epsilon}$ (see (64)), and the condition that
$\tau_{\delta_{\max}}\mathcal{E}$ is $R$-transverse to $V$,

$(\tau_{\delta_{\max}}H\cap(2R\mathcal{B}))\cap
V\subseteq\tau_{\delta_{\max}}\mathcal{E}\cap V\subseteq R\mathcal{B},$

which implies that $\tau_{\delta_{\max}}H\cap V\subseteq R\mathcal{B}$.

Using the condition that $\tau_{\delta_{\max}}\mathcal{E}$ is $R$-transverse
to $V$, and (a) for $\delta=\delta_{\max}$,

$\displaystyle R^{-1}\mathcal{B}\cap V^{\perp}$
$\displaystyle\subseteq\Pi_{V^{\perp}}(\tau_{\delta_{\max}}\mathcal{E}\cap\mathcal{B}))$
$\displaystyle\subseteq\Pi_{V^{\perp}}((\tau_{\delta_{\max}}H+\epsilon\mathcal{B})\cap\mathcal{B}))$
$\displaystyle\subseteq\Pi_{V^{\perp}}(2(\tau_{\delta_{\max}}H\cap\mathcal{B})+\epsilon\mathcal{B}),$

where we used Lemma 4.16 for the last inclusion. Because
$\epsilon\leq\frac{1}{2R}$ and $\Pi_{V^{\perp}}\mathcal{B}=\mathcal{B}\cap
V^{\perp}$, it follows that

$R^{-1}\mathcal{B}\cap V^{\perp}\subseteq
2\Pi_{V^{\perp}}(\tau_{\delta_{\max}}H\cap\mathcal{B})+(1/2)R^{-1}\mathcal{B}\cap
V^{\perp}.$

We deduce that $\frac{1}{4}R^{-1}\mathcal{B}\cap
V^{\perp}\subseteq\Pi_{V^{\perp}}(\tau_{\delta_{\max}}H\cap\mathcal{B})$.

Therefore, we see that $\tau_{\delta_{\max}}H$ is $4R$-transverse to $V$.

Repeating the previous argument, using that $\tau_{\delta_{\min}}\mathcal{E}$
is $R$-transverse to $W$, and (a), (b) for $\delta=\delta_{\min}$, we see that
$\tau_{\delta_{\min}}H$ is $4R$-transverse to $W$.

Assume for sake of contradiction that $\tau_{\delta_{*}}H$ is
$R^{*}$-transverse to $V$. By Lemma 3.11,
$(\tau_{\delta_{*}}H+\epsilon\mathcal{B})\cap V\subseteq
R^{*}\epsilon\mathcal{B}\subseteq R^{*}\mathcal{B}$. Thus, by condition (a)
for $\delta=\delta_{*}$,

$\tau_{\delta_{*}}\mathcal{E}\cap
V\subseteq(\tau_{\delta_{*}}H+\epsilon\mathcal{B})\cap V\subseteq
R^{*}\mathcal{B}.$

Condition (b) for $\delta=\delta_{*}$ implies that
$\tau_{\delta_{*}}H\cap\mathcal{B}\subseteq\tau_{\delta_{*}}\mathcal{E}\cap\mathcal{B}$.
Thus,

$(R^{*})^{-1}\mathcal{B}\cap
V^{\perp}\subseteq\Pi_{V^{\perp}}(\tau_{\delta_{*}}H\cap\mathcal{B})\subseteq\Pi_{V^{\perp}}(\tau_{\delta_{*}}\mathcal{E}\cap\mathcal{B}),$

where the first inclusion uses the assumption that $\tau_{\delta_{*}}H$ is
$R^{*}$-transverse to $V$. Thus, $\tau_{\delta_{*}}\mathcal{E}$ is
$R^{*}$-transverse to $V$, contradicting the hypotheses on $\mathcal{E}$ and
$V$. ∎

###### Proof of Proposition 4.15.

By Lemma 4.17, there exists a subspace $H\subseteq X$ such that
$\tau_{\delta_{\max}}H$ is $4R$-transverse to $V$, $\tau_{\delta_{*}}H$ is not
$R^{*}$-transverse to $V$, and $\tau_{\delta_{\min}}H$ is $4R$-transverse to
$W$, where $\delta_{\min}\leq\delta_{*}\leq\delta_{\max}$. According to Lemma
3.9,

$\displaystyle\cos(\theta_{\max}(\tau_{\delta_{\max}}H,V^{\perp}))\geq(4R)^{-1}$
$\displaystyle\cos(\theta_{\max}(\tau_{\delta_{*}}H,V^{\perp}))\leq(R^{*})^{-1},$
$\displaystyle\cos(\theta_{\max}(\tau_{\delta_{\min}}H,W^{\perp}))\geq(4R)^{-1},$

with $\dim(V^{\perp})=\dim(W^{\perp})=\ell$, where $\ell:=\dim(H)$.

By Lemma 3.5, we then have

$\displaystyle\cos(\angle(\tau_{\delta_{\max}}H,V^{\perp}))\geq(4R)^{-\ell}$
(65)
$\displaystyle\cos(\angle(\tau_{\delta_{*}}H,V^{\perp}))\leq(R^{*})^{-1},$
(66)
$\displaystyle\cos(\angle(\tau_{\delta_{\min}}H,W^{\perp}))\geq(4R)^{-\ell}.$
(67)

Suppose for contradiction that
$\operatorname{sgn}(V)\leq\operatorname{sgn}(W)$. Then, by (32), we have
$\operatorname{sgn}(V^{\perp})\geq\operatorname{sgn}(W^{\perp})$. Now, let
$\alpha(\delta)=\frac{\cos(\angle(\tau_{\delta}H,V^{\perp}))}{\cos(\angle(\tau_{\delta}H,W^{\perp}))}$.
Let $\omega_{H}$, $\omega_{V^{\perp}}$, $\omega_{W^{\perp}}$ be representative
forms for $H$, $V^{\perp}$, and $W^{\perp}$, respectively. We then write

$\displaystyle\alpha(\delta)$
$\displaystyle=\frac{|\omega_{W^{\perp}}|\cdot|\langle\tau_{\delta}^{*}\omega_{H},\omega_{V^{\perp}}\rangle|}{|\omega_{V^{\perp}}|\cdot|\langle\tau_{\delta}^{*}\omega_{H},\omega_{W^{\perp}}\rangle|}$