# Optimal linear response for Markov Hilbert-Schmidt integral operators and
stochastic dynamical systems

Fadi Antown1, Gary Froyland1, and Stefano Galatolo2

1School of Mathematics and Statistics, University of New South Wales, Sydney
NSW 2052, Australia
2Department of Mathematics, University of Pisa, Via Buonarroti 1, 56127 Pisa,
Italy

###### Abstract

We consider optimal control problems for discrete-time random dynamical
systems, finding unique perturbations that provoke maximal responses of
statistical properties of the system. We treat systems whose transfer operator
has an $L^{2}$ kernel, and we consider the problems of finding (i) the
infinitesimal perturbation maximising the expectation of a given observable
and (ii) the infinitesimal perturbation maximising the spectral gap, and hence
the exponential mixing rate of the system. Our perturbations are either (a)
perturbations of the kernel or (b) perturbations of a deterministic map
subjected to additive noise. We develop a general setting in which these
optimisation problems have a unique solution and construct explicit formulae
for the unique optimal perturbations. We apply our results to a Pomeau-
Manneville map and an interval exchange map, both subjected to additive noise,
to explicitly compute the perturbations provoking maximal responses.

Keywords— Stochastic dynamical system, optimal linear response, transfer
operator, mixing rate, optimal control

[MSC2020]: 37H30, 47A55, 49J50, 49N05

##  1 Introduction

The statistical properties of the long-term behaviour of deterministic or
stochastic dynamical systems are strongly related to the properties of
invariant or stationary measures and to the spectral properties of the
associated transfer operator. When the dynamical system is perturbed it is
useful to understand and predict the response of the statistical properties of
the system through these objects. When such responses are differentiable, we
say that the system exhibits a _linear response_ to the class of
perturbations. To first order, this response can be described by a suitable
derivative expressing the infinitesimal rate of change in e.g. the natural
invariant measure or in the spectrum. Understanding the response of
statistical properties to perturbation has particular importance in
applications, including to climate science (see e.g. [25], [27] and the
references therein).

In the present paper we go beyond quantifying responses and address natural
problems concerning the _optimal response_ , namely which perturbations elicit
a _maximal_ response. For example, given an observation function, which
perturbation produces the greatest change in the expectation of this
observation, and which perturbation produces the greatest change in the rate
of convergence to equilibrium. Continuing the climate science application, one
may wish to know which small climate action (which perturbation) would produce
the greatest reduction in the average temperature (the expected observation
value). We note that by considering trajectories of a perturbed map and using
ergodicity, one may view the problem of maximising the response in the
expectation of an observation as an infinite-horizon optimal control problem,
averaging an observation along trajectories.

The linear response of dynamical systems is an area of intense research and we
present a brief overview of the literature that is related to the present
work. Early results concerning the response of invariant measures to the
perturbation of a deterministic system have been obtained by Ruelle [42] in
the uniformly hyperbolic case. More recently, these results have been extended
to several other situations in which one has some hyperbolicity and sufficient
regularity of the system and its perturbations. We refer the reader to the
survey [5] for an extended discussion of the literature about linear response
(and its failure) for deterministic systems.

The mathematical literature on linear response of invariant measures of
stochastic or random dynamical systems is more recent. In the framework of
continuous-time random processes and stochastic differential equations, linear
response results were proved in [27, 32]. Results related to the linear
response of the stationary measure for diffusion in random media appear in
[33, 24, 23, 13, 40]. In the discrete-time case, examples of linear response
for small random perturbations of uniformly hyperbolic deterministic systems
appeared in [26]. In [4] linear response results are given for random
compositions of expanding or non-uniformly expanding maps. In the paper [49]
the smoothness of the invariant measure response under suitable perturbations
is proved for a class of random diffeomorphisms, but no explicit formula is
given for the derivatives; an application to the smoothness of the rotation
number of Arnold circle maps with additive noise is presented. Systems
generated by the iteration of a deterministic map subjected to i.i.d. additive
random perturbations are one class of stochastic systems studied in the
present paper (see Section 6). The linear response of such systems is
considered systematically in [20] and linear response results are proved for
perturbations to the deterministic map or to the additive noise. These results
are used to by [39] to extend some results of [49] outside the diffeomorphism
case and applied to an idealized model of El Niño-Southern Oscillation, given
by a noninvertible circle map with additive noise. Higher derivative results
for the response of systems with additive noise are presented in [22].
Response results for random systems in the so-called quenched point of view
appeared recently in [43], [44] where the random composition of expanding maps
is considered using Hilbert cones techniques and in [11] where the random
composition of hyperbolic maps is considered by a transfer operator based
approach.

We remark that the addition of random perturbations is not necessarily
sufficient to guarantee a linear response. An i.i.d. composition of the
identity map and a rotation on the circle is considered in [19], and it is
shown that using observables with square-integrable first derivative, one only
has Hölder continuity of the response with respect to $C^{0}$ perturbations of
the circle rotation.

One can similarly consider the linear response of the dominant eigenvalues of
the transfer operator under perturbation. In the literature there are several
results describing the way eigenvalues and eigenvectors of suitable classes of
operators change when those operators are perturbed in some way, for example
classical results concerning compact operators subjected to analytic
perturbations [29], and quasi-compact Markov operators subjected to $C^{k}$
perturbations [28]. In specific classes of dynamics, differentiability of
isolated spectral data is demonstated in [26] for transfer operators of Anosov
maps where the map is subjected to $C^{k}$ perturbations and in [32] for
transfer operators arising from SDEs subjected to $C^{k}$ perturbations of the
drift.

Optimal linear response questions have been considered in the dynamical
setting of homogeneous (and inhomogeneous) finite-state Markov chains [1],
where explicit formulae are provided for the unique maximising perturbations
that (i) maximise the norm of the response, (ii) maximise the expectation of a
given observable, and (iii) maximise the spectral gap. The efficient Lagrange
multiplier approach developed in [1] for questions (ii) and (iii) will be
extended to the infinite-dimensional setting in the present paper. In
continuous time, [18] maximised the spectral gap of a numerical discretisation
of a periodically forced Fokker-Planck equation (perturbing the velocity field
to maximally speed up or slow down the exponential mixing rate). The same
problem is considered by [16], but for general aperiodic forcing over a finite
time, using the Lagrange multiplier approach of [1]. A non-spectral approach
to increasing mixing rates by optimal kernel perturbations in discrete time is
[15].

Related optimal control problems have been considered in [21] where the goal
was to find a minimal perturbation realising a specific response to the
invariant measure of a deterministic system (about the problem of finding an
infinitesimal perturbation realising a given response see also [30]). These
kinds of questions and other similar ones were also briefly considered in [20]
for random dynamical systems consisting of deterministic maps perturbed by
additive noise. Similar problems in the case of probabilistic cellular
automata were considered in [38].

The present work takes the point of view of [1], extending the theory to the
infinite-dimensional setting of stochastic integral operators, proving the
existence of unique optimal perturbations, deriving explicit formulae for
these optimal perturbations, and illustrating the formulae and their
conclusions via two topical examples. We consider the class of stochastic
dynamical systems with transfer operators representable by an $L^{2}$-compact,
integral operator, which includes deterministic systems perturbed by additive
noise. The transfer operator $L$ has the form

$Lf(x)=\int k(x,y)f(y)\ dy,$ (1)

where $k$ is a stochastic kernel; in the case of deterministic systems $T$
with additive noise, $k(x,y)=\rho(x-T(y))$, with $\rho$ a probability density
representing the distribution of the noise intensity (see Section 6). We
consider perturbations of two types: firstly, perturbations to the kernel $k$,
and secondly, perturbations to the map $T$.

An outline of the paper is as follows. In Section 2 we consider general
compact, integral-preserving operators $L:L^{2}\to L^{2}$ (see (3)) and state
general linear response statements for the normalised fixed points and the
leading eigenvalues of these operators (Theorem 2.2 and Proposition 2.6). In
Section 3, we derive response formulae for the normalised fixed points
(Corollary 3.5) and spectral values (Corollary 3.6) of operators of the form
(1), under perturbation of the kernel $k$. In Section 4 we consider the
problem of finding the perturbation that provokes a _maximal_ response in the
average of a given observable (General Problem 1) and the spectral gap
(General Problem 2). We show that if the feasible set of perturbations is
convex, an optimal solution exists, and that this optimum is unique if the
feasible set is strictly convex. In Section 5.1, using Lagrange multipliers we
derive an explicit formula for the unique optimal kernel perturbation that
maximises the expectation of an observable (Theorem 5.4). In section 5.2 we
prove an explicit formula for the perturbation that maximise the change in
spectral gap (and therefore the rate of mixing) of the system (Theorem 5.6).

In Section 6 we specialise our integral operators to annealed transfer
operators corresponding to deterministic maps $T$ with additive noise. For
these systems the kernel $k$ has the form $k(x,y)=\rho(x-T(y))$ for some
nonsingular transformation $T$, and we consider perturbations of the map $T$
directly. Response formulas for these perturbations are developed in
Proposition 6.3 and Proposition 6.6 for the invariant measure and the
dominating eigenvalues, respectively. In this framework we again prove
existence and uniqueness of the map perturbation maximising the derivative of
the expectation of an observation (Proposition 7.3) and then derive an
explicit formula for the extremiser (Theorem 7.4). Proposition 7.6 and Theorem
7.7 state results analogous to Proposition 7.3 and Theorem 7.4 for the
optimization of the spectral gap and mixing rate.

In section 8 we apply and illustrate the theoretical findings of this work on
the Pomeau-Manneville map and a weakly mixing interval exchange, each
perturbed by additive noise. For each map we numerically estimate (i) the
optimal stochastic perturbation (perturbing the kernel $k$) and (ii) the
optimal deterministic perturbation (perturbing the map $T$) that maximises the
derivatives of the expectation of an observable and the mixing rate. One of
the interesting lessons is that to maximally increase the mixing rate of the
noisy Pomeau-Manneville map, one should perturb the kernel (stochastic
perturbation) to move mass away from the indifferent fixed point or deform the
map to transport mass away from the fixed point (deterministic perturbation);
see Figures 4 and 7, respectively. Further numerical outcomes are discussed
and explained in Section 8.

##  2 Linear response for compact integral-preserving operators

In this section we introduce general response results for integral-preserving
compact operators. We consider both the response of the invariant measure to
the perturbations and the response of the dominant eigenvalues.

###  2.1 Existence of linear response for the invariant measure

In the following, we consider integral-preserving compact operators acting on
$L^{2}$, which are not necessarily positive. We will give a general linear
response statement for their invariant measures. In Section 3 we show how
these results can be applied to Hilbert-Schmidt integral operators, which will
later be transfer operators of suitable random dynamical systems.

Let $L^{2}([0,1])$ be the space of square-integrable functions over the unit
interval (considered with the Lebesgue measure $m$); for brevity, we will
denote it as simply $L^{2}$. We remark that the analysis in the rest of the
paper can be extended to manifolds, but we keep the setting simple so as not
to obscure the main ideas. Let us consider the space of zero-average functions

$V:=\bigg{\\{}f\in L^{2}~{}s.t.~{}~{}\int f\,dm=0\bigg{\\}}.$

######  Definition 2.1.

We say that an operator $L:L^{2}\rightarrow L^{2}$ has _exponential
contraction of the zero average space_ $V$ if there are $C\geq 0$ and
$\lambda<0$ such that $\forall g\in V$

$\|L^{n}g\|_{2}\leq Ce^{\lambda n}\|g\|_{2}$ (2)

for all $n\geq 0$.

For $\bar{\delta}>0$ and $\delta\in[0,\bar{\delta})$, we consider a family of
integral-preserving, compact operators $L_{\delta}:L^{2}\rightarrow L^{2}$; we
think of $L_{\delta}$ as perturbations of $L_{0}$. We say that $f_{\delta}\in
L^{2}$ is an _invariant function_ of $L_{\delta}$ if
$L_{\delta}f_{\delta}=f_{\delta}$. We will see that under natural assumptions,
the operators $L_{\delta}$, $\delta\in[0,\bar{\delta})$, have a family of
normalized invariant functions $f_{\delta}\in L^{2}$. Furthermore, for
suitable perturbations the invariant functions vary smoothly in $L^{2}$ and we
get an explicit formula for the resulting derivative
$\frac{df_{\delta}}{d\delta}$. We remark that since the operators we consider
are not necessarily positive, the invariant functions are not necessarily
positive.

######  Theorem 2.2 (Linear response for integral-preserving compact
operators).

Let us consider a family of compact operators $L_{\delta}:L^{2}\rightarrow
L^{2}$, with $\delta\in\left[0,\overline{\delta}\right)$, preserving the
integral: for each $g\in L^{2}$

$\int L_{\delta}g~{}dm=\int g~{}dm.$ (3)

Then,

  1. (I)

The operators have invariant functions in $L^{2}$: for each $\delta$ there is
$g_{\delta}\neq 0$ such that $L_{\delta}g_{\delta}=g_{\delta}$.

  2. (II)

Suppose $L_{0}$ also satisfies the following:

(A1) (mixing of the unperturbed operator) For every $g\in V$,

$\lim_{n\rightarrow\infty}\|L_{0}^{n}g\|_{2}=0.$

Under this assumption, the unperturbed operator $L_{0}$ has a unique
normalized invariant function $f_{0}$ such that $\int{f}_{0}\ dm=1$.
Furthermore, $L_{0}$ has exponential contraction of the zero average space
$V$.

  3. (III)

Suppose the family of operators $L_{\delta}$ also satisfy the following:

(A2) ($L_{\delta}$ are small perturbations and existence of derivative
operator at $f_{0}$) Suppose there is a $K\geq 0$ such that
$\left||L_{\delta}-L_{0}|\right|_{L^{2}\rightarrow L^{2}}\leq K\delta$ for
small $\delta$. Furthermore, suppose there exist $\hat{f}\in V$ such that

$\underset{\delta\rightarrow
0}{\lim}\frac{(L_{\delta}-L_{0})}{\delta}f_{0}=\hat{f}.$

Under these assumptions, the following hold:

    1. (a)

There exists a $\delta_{2}>0$ such that for each $0\leq\delta<\delta_{2}$, the
operators $L_{\delta}$ have unique invariant functions ${f}_{\delta}$ such
that $\int{f}_{\delta}dm=1.$

    2. (b)

The resolvent operator $({Id}-L_{0})^{-1}:V\rightarrow$ $V$ is continuous.

    3. (c)

$\lim_{\delta\rightarrow
0}\left\|\frac{f_{\delta}-f_{0}}{\delta}-({Id}-L_{0})^{-1}\hat{f}\right\|_{2}=0;$

thus, $({Id}-L_{0})^{-1}\hat{f}$ represents the first order term in the
perturbation of the invariant function for the family of systems $L_{\delta}$.

###### Proof.

##### Claim (I):

We start by proving the existence of the invariant functions $g_{\delta}$ for
the operators $L_{\delta}$. Since the operators are compact and integral
preserving, $L_{\delta}$ has an eigenvalue $1$ for each $\delta$. Indeed, let
us consider the adjoint operators $L^{*}_{\delta}:L^{2}\to L^{2}$ defined by
the duality relation $\langle L_{\delta}f,g\rangle=\langle
f,L^{*}_{\delta}g\rangle$ for all $f,g\in L^{2}.$ Because of the integral-
preserving assumption, we have $\langle
f,L^{*}_{\delta}\mathbf{1}\rangle=\langle L_{\delta}f,\mathbf{1}\rangle=\int
L_{\delta}f\ dm=\int f\ dm=\langle f,\mathbf{1}\rangle$.111We use the notation
$\mathbf{1}$ for the constant function and $\mathbf{1}_{A}$ for the indicator
function of the set $A$. This implies $L^{*}_{\delta}\mathbf{1}=\mathbf{1}$
and thus, $1$ is in the spectrum of $L^{*}_{\delta}$ and $L_{\delta}$. Since
$L_{\delta}$ is compact, its spectrum equals the eigenvalues and we have
nontrivial fixed points for the operators $L_{\delta}$.

##### Claim (III)(a) for $\delta=0$:

Now we prove the uniqueness of the normalized invariant function of $L_{0}$.
Above we proved that $L_{0}$ has some invariant function $g_{0}\neq 0$. The
mixing assumption $(A1)$ implies that $\int g_{0}\ dm\neq 0$; to see this, we
note that if $\int g_{0}\ dm=0$, then $g_{0}\in V$, and, by $(A1)$, $g_{0}$
cannot be a nontrivial fixed point of $L_{0}$. We claim that
$f_{0}=\frac{g_{0}}{\int g_{0}\ dm}$ is the unique normalized invariant
function for $L_{0}$. To see this, suppose there was a second normalized
invariant function $f^{\prime}_{0}$; then, $f^{\prime}_{0}-f_{0}$ would be an
invariant function in $V$, which is a contradiction.

##### Claim (II):

To show that $L_{0}$ has exponential contraction on $V$, we first note that
for $f\in L^{2}$, we can write $f=f_{0}\int f\ dm+[f-f_{0}\int f\ dm]$. Since
$[f-f_{0}\int f\ dm]\in V$, it follows from $(A1)$ that
$L_{0}^{n}f\to_{L^{2}}f_{0}\int f\ dm$. Thus, the spectrum of $L_{0}$ is
contained in the unit disk by the spectral radius theorem. Now suppose
$\lambda$ is in the spectrum of $L_{0}$ and $|\lambda|=1$. By the compactness
assumption, there is an eigenvector $f_{\lambda}$ for $\lambda$ and then we
have $||L_{0}^{n}(f_{\lambda})||_{2}=||f_{\lambda}||_{2}$. However,
$L_{0}^{n}(f_{\lambda})\to_{L^{2}}f_{0}\int f_{\lambda}\ dm$, which is not
possible unless $\lambda=1$. Hence, the spectrum of $L_{0}|_{V}$ is strictly
contained in the unit disk. Thus, by the spectral radius theorem, there is an
$n>0$ such that $||L_{0}^{n}|_{V}||_{L^{2}\rightarrow L^{2}}\leq\frac{1}{2}$
and we have exponential contraction of $L_{0}$ on $V$.

##### Claim (III)(a) for $\delta\in[0,\bar{\delta}]$:

From the assumption $||L_{0}-L_{\delta}||_{L^{2}\rightarrow L^{2}}\leq
K\delta$, we have for small enough $\delta$ that
$||L_{\delta}^{n}|_{V}||_{L^{2}\rightarrow L^{2}}\leq\frac{2}{3}$ and
therefore, $L_{\delta}$ is also mixing. We can apply the argument above to the
operators $L_{\delta}$ and obtain, for each small enough $\delta$, a unique
normalized invariant function $f_{\delta}$.

##### Claim (III)(b):

Using the exponential contraction of $L_{0}$ on $V$, we now show that
$(\text{Id}-L_{0})^{-1}:V\rightarrow V$ is continuous. Indeed, for $f\in V$,
we get $(\text{Id}-L_{0})^{-1}f=f+\sum_{n=1}^{\infty}L_{0}^{n}f$. Since
$L_{0}$ is exponentially contracting on $V$, and
$\sum_{n=1}^{\infty}Ce^{\lambda n}:=M<\infty,$ the sum
$\sum_{n=1}^{\infty}L_{0}^{n}f$ converges in $V$ with respect to the $L^{2}$
norm. The resolvent $(\text{Id}-L_{0})^{-1}:V\rightarrow V$ is then a
continuous operator and $||(\text{Id}-L_{0})^{-1}||_{V\rightarrow V}\leq 1+M.$
We remark that since $\hat{f}\in V,$ the resolvent can be computed at
$\hat{f}$.

##### Claim (III)(c):

Now we are ready to prove the linear response formula. Furthermore, we have

$\displaystyle\|f_{\delta}-f_{0}\|_{2}$ $\displaystyle\leq$
$\displaystyle\|L_{\delta}^{n}f_{\delta}-L_{0}^{n}f_{0}\|_{2}$
$\displaystyle\leq$
$\displaystyle\|L_{\delta}^{n}f_{0}-L_{0}^{n}f_{0}\|_{2}+\|L_{\delta}^{n}f_{\delta}-L_{\delta}^{n}f_{0}\|_{2}$
$\displaystyle\leq$
$\displaystyle\|L_{\delta}^{n}-L_{0}^{n}\|_{2}\|f_{0}\|_{2}+\|L_{\delta}^{n}|_{V}\|_{L^{2}\rightarrow
L^{2}}\|f_{\delta}-f_{0}\|_{2}$ $\displaystyle\leq$
$\displaystyle\|L_{\delta}^{n}-L_{0}^{n}\|_{2}\|f_{0}\|_{2}+\frac{2}{3}\|f_{\delta}-f_{0}\|_{2},$

from which we get $\|f_{\delta}-f_{0}\|_{2}\leq
3\|L_{\delta}^{n}-L_{0}^{n}\|_{L^{2}\rightarrow L^{2}}\|f_{0}\|_{2}$. Since
$||L_{0}-L_{\delta}||_{L^{2}\rightarrow L^{2}}\leq K\delta$, we have
$\|f_{\delta}-f_{0}\|_{2}\to 0$ as $\delta\to 0.$ Since $f_{0}$ and
$f_{\delta}$ are the invariant functions of $L_{0}$ and $L_{\delta}$, we have

$(\text{Id}-L_{0})\frac{f_{\delta}-f_{0}}{\delta}=\frac{1}{\delta}(L_{\delta}-L_{0})f_{\delta}.$

By applying the resolvent to both sides we obtain

$\displaystyle\frac{f_{\delta}-f_{0}}{\delta}$
$\displaystyle=(\text{Id}-L_{0})^{-1}\frac{L_{\delta}-L_{0}}{\delta}f_{\delta}$
$\displaystyle=(\text{Id}-L_{0})^{-1}\frac{L_{\delta}-L_{0}}{\delta}f_{0}+(\text{Id}-L_{0})^{-1}\frac{L_{\delta}-L_{0}}{\delta}(f_{\delta}-f_{0}).$

Moreover, from assumption $(A2)$, we have for sufficiently small $\delta$ that

$\left\|(\text{Id}-L_{0})^{-1}\frac{L_{\delta}-L_{0}}{\delta}(f_{\delta}-f_{0})\right\|_{2}\leq\|(\text{Id}-L_{0})^{-1}\|_{V\rightarrow
V}K\|f_{\delta}-f_{0}\|_{2}.$

Since we already proved that $\lim_{\delta\rightarrow
0}\|f_{\delta}-f_{0}\|_{2}=0$, we are left with

$\lim_{\delta\rightarrow
0}\frac{f_{\delta}-f_{0}}{\delta}=(\text{Id}-L_{0})^{-1}\hat{f}$

converging in the $L^{2}$ norm.

We remark that the strategy of proof of Theorem 2.2 is similar to the one of
Theorem 3 of [20] although the assumptions made are quite different, here we
consider a compact integral preserving operator on $L^{2}$, while in [20]
several norms are considered to allow low regularity perturbations and the
operator is required to be positive.

It is worth to remark that the above proof gives a description of the spectral
picture of $L_{0}$. By Theorem 2.2, if $L_{0}$ satisfies $(A1)$ then the
invariant function is unique, up to normalization; this shows that $1$ is a
simple eigenvalue. Furthermore, $L_{0}$ preserves the direct sum $L^{2}=$
span$\\{f_{0}\\}\oplus V$ and the spectrum of $L_{0}$ is strictly inside the
unit disk when $L_{0}$ is restricted to $V$. Hence, the spectrum of $L_{0}$ is
contained in the unit disk and there is a spectral gap.

######  Remark 2.3.

The mixing assumption in $(A1)$ is required only for the unperturbed operator
$L_{0}$. This assumption is satisfied, for example, if $L_{0}$ is an integral
operator and an iterate of this operator has a strictly positive kernel, see
Corollary 5.7.1 of [35]. Later in Remark 6.4 we show this assumption is
verified for a wide range of examples of stochastic dynamical systems.

###  2.2 Existence of linear response of the dominant eigenvalues

In this section, we consider the existence of linear response for the second
largest eigenvalues (in magnitude) and provide a formula for the linear
response. An important object needed to quantify linear response statements is
a “derivative” of the transfer operator with respect to the perturbation.

######  Definition 2.4.

We define $\dot{L}:L^{2}\to V$ as the unique linear operator satisfying

$\lim_{\delta\to
0}\left\|\frac{(L_{\delta}-L_{0})}{\delta}-\dot{L}\right\|_{L^{2}\to V}=0.$

Let $\mathcal{B}(L^{2})$ denote the space of bounded linear operators from the
Banach space $L^{2}$ to itself and $r(L)$ denote the spectral radius of an
operator $L$; we begin with the following definition.

######  Definition 2.5 ([28], Definition III.7).

Let $s\in\mathbb{N},s\geq 1$. We say that
$L\in\mathcal{B}(L^{2}([0,1],\mathbb{C}))$ has $s$ dominating simple
eigenvalues if there exists closed subspaces $E$ and $\tilde{E}$ such that

  1. 1.

$L^{2}([0,1],\mathbb{C})=E\oplus\tilde{E}$,

  2. 2.

$L(E)\subset E$, $L(\tilde{E})\subset\tilde{E}$,

  3. 3.

dim$(E)=s$ and $L|_{E}$ has $s$ geometrically simple eigenvalues
$\lambda_{i}$, $i=1,\dots,s$,

  4. 4.

$r(L|_{\tilde{E}})<\min\\{|\lambda_{i}|:i=1,\dots,s\\}$.

Adapting Theorem III.8 and Corollary III.11 of [28] to our situation, we can
now state a linear response result for these eigenvalues.

######  Proposition 2.6.

Let $L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow
L^{2}([0,1],{\mathbb{C}})$, where $\delta\in[0,\bar{\delta})=:I_{0}$, be
integral-preserving (see equation (3)) compact operators. Assume that the map
$\delta\mapsto L_{\delta}$ is in
$C^{1}(I_{0},\mathcal{B}(L^{2}([0,1],{\mathbb{C}})))$ and $L_{0}$ is mixing
(see $(A1)$ in Theorem 2.2). Then, $\lambda_{1,0}:=1\in\sigma(L_{0})$ and
$r(L_{0})=1$. Let $\mathcal{I}\subset\sigma(L_{0})\setminus\\{1\\}$ be the
eigenvalue(s) of maximal modulus strictly inside the unit disk; assume they
are geometrically simple and let $s:=|\mathcal{I}|+1$. Then there exists an
interval $I_{1}:=[0,\delta_{1})$, $I_{1}\subset I_{0}$ such that for
$\delta\in I_{1}$, $L_{\delta}$ has $s$ dominating simple eigenvalues. Thus,
there exists functions $e_{i,(\cdot)},\ \hat{e}_{i,(\cdot)}\in
C^{1}(I_{1},L^{2}([0,1],{\mathbb{C}}))$ and $\lambda_{i,(\cdot)}\in
C^{1}(I_{1},{\mathbb{C}})$ such that for $\delta\in I_{1}$ and $i,j=2,\dots,s$

  * (i)

$L_{\delta}e_{i,\delta}=\lambda_{i,\delta}e_{i,\delta}$,
$L^{*}_{\delta}\hat{e}_{i,\delta}=\lambda_{i,\delta}\hat{e}_{i,\delta}$,

  * (ii)

$\langle
e_{i,\delta},\hat{e}_{j,\delta}\rangle_{L^{2}([0,1],{\mathbb{C}})}=\delta_{i,j}$,
where $\delta_{i,j}$ is the Kronecker delta.

Furthermore, let $\dot{\lambda}_{i}\in{\mathbb{C}}$ satisfy

$\lim_{\delta\rightarrow
0}\bigg{|}\frac{\lambda_{i,\delta}-\lambda_{i,0}}{\delta}-\dot{\lambda}_{i}\bigg{|}=0,$

then

$\displaystyle\dot{\lambda}_{i}=\langle\hat{e}_{i,0},\dot{L}e_{i,0}\rangle_{L^{2}([0,1],{\mathbb{C}})},$
(4)

where $\dot{L}$ is as in Definition 2.4.

###### Proof.

From Theorem 2.2 and the discussion following it, $1\in\sigma(L_{0})$ and
$r(L_{0})=1$.

We now use Theorem III.8 in [28] to obtain the existence of linear response
and Corollary III.11 [28] to obtain the formula. We begin by verifying the two
hypotheses of Theorem III.8 [28]. We remark that our map $\delta\mapsto
L_{\delta}$ belonging to
$C^{1}([0,\bar{\delta}),\mathcal{B}(L^{2}([0,1],{\mathbb{C}})))$ can be
extended to a map
$C^{1}((-\bar{\delta},\bar{\delta}),\mathcal{B}(L^{2}([0,1],{\mathbb{C}})))$.

Doing so, hypothesis $(H1)$ of Theorem III.8 [28] is satisfied. Since
$r(L_{0})=1$, we just need to show that $L_{0}$ has $s$ dominating
eigenvalues. Since $L_{0}$ is a compact operator, the eigenvalues
$\lambda_{i,0}\in\mathcal{I}$ are isolated. Let $\Pi_{i}$ be the
eigenprojection onto the eigenspace of $\lambda_{i,0}$ and
$E_{i}:=\Pi_{i}(L^{2}([0,1],{\mathbb{C}}))$. Define the eigenspaces
$E:=\bigoplus_{i=1}^{s}E_{i}$ and
$\widetilde{E}:=(\text{Id}-\sum_{i=1}^{s}\Pi_{i})(L^{2}([0,1],{\mathbb{C}}))$.
We thus have:

  * (1)

$L^{2}([0,1],{\mathbb{C}})=E\oplus\widetilde{E}$.

  * (2)

$L_{0}\left(E\right)\subset E$ and $L_{0}(\widetilde{E})\subset\widetilde{E}$.

  * (3)

dim$\left(E\right)=s$ and $L_{0}|_{E}$ has $s$ simple eigenvalues
$\lambda_{1,0}\cup\mathcal{I}$. This point follows from the assumption that
the eigenvalues in $\mathcal{I}$ are geometrically simple and the fact that
$\lambda_{1,0}$ is simple (see Theorem 2.2).

  * (4)

$r(L_{0}|_{\widetilde{E}})<|\lambda_{i,0}|$ where
$\lambda_{i,0}\in\mathcal{I}$.

Thus, $L_{0}$ satisfies hypothesis (H2) of Theorem III.8 since it has $s$
dominating simple eigenvalues and $r(L_{0})=1$. Hence, from Theorem III.8
[28], the map $\delta\mapsto\lambda_{i,\delta}$ is differentiable at
$\delta=0$.

We can now apply the argument in Corollary III.11 [28] for $\lambda_{i,0}$ to
obtain (15) (the result and proof of Corollary III.11 [28] is for the top
eigenvalue, however the argument still holds for any eigenvalue
$\lambda_{i,0}$, $\in\mathcal{I}$ by changing the index value in the proof of
the corollary).

##  3 Application to Hilbert-Schmidt integral operators

In this section we apply the results of the previous section to Hilbert-
Schmidt integral operators and suitable perturbations. The operators we
consider are compact operators on $L^{2}([0,1],{\mathbb{R}})$ (or
$L^{2}([0,1],{\mathbb{C}})$); for brevity we will denote222We will also denote
$L^{p}:=L^{p}([0,1],{\mathbb{R}})$; this notation will not be used for
$L^{2}([0,1],{\mathbb{C}})$. $L^{2}:=L^{2}([0,1],{\mathbb{R}})$. To avoid
confusion we point out that in the following we will also consider the space
$L^{2}([0,1]^{2})$ of square integrable real functions on the unit square;
this space contains the kernels of the operators we consider.

Let $k\in L^{2}([0,1]^{2})$ and consider the operator $L:L^{2}\to L^{2}$
defined in the following way: for $f\in L^{2}$

$Lf(x)=\int k(x,y)f(y)dy;$ (5)

such an operator is called a Hilbert-Schmidt integral operator. Such operators
may represent the annealed transfer operators of systems perturbed by additive
noise (see Section 6).

We now list some well-known and basic facts about Hilbert-Schmidt integral
operators with kernels in $L^{2}([0,1]^{2})$:

  * •

The operator $L:L^{2}\rightarrow L^{2}$ is bounded and

$||Lf||_{2}\leq||k||_{L^{2}([0,1]^{2})}||f||_{2}$ (6)

(see Proposition 4.7 in II.§4 [8]).

  * •

If $k\in L^{\infty}([0,1]^{2})$, then

$||Lf||_{\infty}\leq||k||_{L^{\infty}([0,1]^{2})}||f||_{1}$ (7)

and the operator $L:L^{1}\rightarrow L^{\infty}$ is bounded. Furthermore,
$\|L\|_{L^{p}\rightarrow L^{\infty}}\leq\|k\|_{L^{\infty}([0,1]^{2})}$ for
$1\leq p\leq\infty$.

  * •

If for almost every $y\in[0,1]$ we have

$\int k(x,y)dx=1,$

then the Hilbert-Schmidt integral operator associated to the kernel $k$ is
integral preserving (satisfies (3)).

  * •

The operator $L:L^{2}\rightarrow L^{2}$ is compact (see [31]).

Combining the last two points, we have from Theorem 2.2 that such an operator
has an invariant function in $L^{2}$. Furthermore, for $k\in
L^{\infty}([0,1]^{2})$ we have an analogous result.

######  Lemma 3.1.

Let $L:L^{2}\rightarrow L^{2}$ be an integral operator, with integral-
preserving kernel $k\in L^{\infty}([0,1]^{2})$, that is mixing (satisfies
$(A1)$ of Theorem 2.2). Then, there exists a unique fixed point $f\in
L^{\infty}$ of $L$ satisfying $\int f\ dm=1$. Furthermore, if the kernel is
nonnegative, then $f$ is nonnegative.

###### Proof.

Since $k$ is an integral-preserving kernel, $L_{0}$ satisfies (3). Thus, we
can apply Theorem 2.2 to conclude that there exists a unique $f\in L^{2}$,
$\int f\ dm=1$, such that $Lf=f$. Noting that $k\in L^{\infty}([0,1]^{2})$, we
have from inequality (7) that $f\in L^{\infty}$.

We now assume $k$ is nonnegative. Let $k^{j}$ be the kernel of the operator
$L^{j}$. Since $k$ is an integral-preserving kernel, we have

$\displaystyle|k^{2}(x,y)|$ $\displaystyle=\bigg{|}\int
k(x,z)k(z,y)dz\bigg{|}\leq\int|k(x,z)k(z,y)|dz$
$\displaystyle\leq\|k\|_{L^{\infty}([0,1]^{2})}\int
k(z,y)dz=\|k\|_{L^{\infty}([0,1]^{2})};$

it easily follows that
$\|k^{j}\|_{L^{\infty}([0,1]^{2})}\leq\|k\|_{L^{\infty}([0,1]^{2})}$. Thus,
for any probability density $g\in L^{1}$, we have
$\|L^{j}g\|_{\infty}\leq\|k\|_{L^{\infty}([0,1]^{2})}$; thus, by Corollary
5.2.2 in [35], there exists a probability density $\hat{f}\in L^{1}$ such that
$L\hat{f}=\hat{f}$. Since $f$ is the unique invariant function with integral
$1$, we have $\hat{f}=f$; thus, $f$ is a probability density.

###  3.1 Characterising valid perturbations and the derivative of the transfer
operator

In this subsection we consider perturbations of integral-preserving Hilbert-
Schmidt integral operators such that assumption (A2) of Theorem 2.2 can be
verified and the derivative operator $\dot{L}$ computed. We begin, however, by
first characterizing the set of perturbations for which the integral
preserving property of the operators is preserved.

Consider the set $V_{\ker}$ of kernels having zero average in the $x$
direction, defined as

$V_{\ker}:=\bigg{\\{}k\in L^{2}([0,1]^{2}):\int
k(x,y)dx=0~{}for~{}a.e.~{}y\bigg{\\}}.$

######  Lemma 3.2.

Consider a kernel operator $A:L^{2}([0,1])\rightarrow L^{2}([0,1])$ defined by
$Af(x)=\int k(x,y)f(y)dy$. Then, the following are equivalent

  1. 1.

$A(L^{2}([0,1]))\subseteq V$,

  2. 2.

$k\in V_{\ker}$.

###### Proof.

Clearly, the second condition implies the first. For the other direction we
prove the contrapositive. If $\int k(x,y)dx\neq 0$ on a set of positive
measure, then for a small $\epsilon>0$ there is a set $S$ of positive measure
$m(S)>0$ such that $\int k(x,y)dx\geq\epsilon$ or $\int k(x,y)dx\leq-\epsilon$
for each $y\in S$. Suppose $\int k(x,y)dx\geq\epsilon$ in this set, consider
$f:=\mathbf{1}_{S}$ and $g:=Af.$ Then, $g(x)=\int k(x,y)\mathbf{1}_{S}(y)dy$
and we have $\int g(x)dx=\int_{S}\int k(x,y)dxdy\geq\epsilon\ m(S)$ and
$g\notin V$. The other case $\int k(x,y)dx\leq-\epsilon$ is analogous.

We now prove that $V_{\ker}$ is closed.

######  Lemma 3.3.

The set $V_{\ker}$ is a closed vector subspace of $L^{2}([0,1]^{2}).$

###### Proof.

The fact that $V_{\ker}$ is a vector space is trivial. For fixed $f\in
L^{2}([0,1])$, the set of $k\in L^{2}([0,1]^{2})$ such that $\int
k(x,y)f(y)dx\in V$ is closed. To see this, define the function
$K_{f}:L^{2}([0,1]^{2})\rightarrow L^{2}([0,1])$ as

$K_{f}(k)=\int k(x,y)f(y)dy.$ (8)

By (6), $K_{f}$ is continuous. Since $V$ is closed in $L^{2}([0,1])$, this
implies that $K_{f}^{-1}(V)$ is closed in $L^{2}([0,1]^{2}).$ Finally,
$V_{\ker}$ is closed in $L^{2}([0,1]^{2})$ because $V_{\ker}=\cap_{f\in
L^{2}([0,1])}K_{f}^{-1}(V)$.

We now introduce the type of perturbations which we will investigate
throughout the paper. Let $L_{\delta}:L^{2}\rightarrow L^{2}$ be a family of
integral operators, with kernels $k_{\delta}\in L^{2}([0,1]^{2})$, given by

$L_{\delta}f(x)=\int k_{\delta}(x,y)f(y)dy.$

######  Lemma 3.4.

Let $k_{\delta}\in L^{2}([0,1]^{2})$ for each $\delta\in[0,\bar{\delta}).$
Suppose that

$k_{\delta}=k_{0}+\delta\cdot\dot{k}+r_{\delta}$ (9)

where $\dot{k},\ r_{\delta}\in L^{2}([0,1]^{2})$ and
$||r_{\delta}||_{L^{2}([0,1]^{2})}=o(\delta).$ The bounded linear operator
$\dot{L}:L^{2}\to V$ defined by

$\dot{L}f(x):=\int\dot{k}(x,y)f(y)dy$ (10)

satisfies

$\lim_{\delta\rightarrow
0}\bigg{\|}\frac{L_{\delta}-L_{0}}{\delta}-\dot{L}\bigg{\|}_{L^{2}\to V}=0.$

If additionally the derivative of the map $\delta\mapsto k_{\delta}$ with
respect to $\delta$ varies continuously in a neighborhood of $\delta=0$, then
$\delta\mapsto L_{\delta}$ has a continuous derivative in a neighborhood of
$\delta=0$.

###### Proof.

By integral preservation of $L_{\delta}$ and the fact that $\dot{k}\in
L^{2}([0,1]^{2})$, one sees that $\dot{L}:L^{2}\to V$ and is bounded. By (9),

$\displaystyle\left\|\frac{L_{\delta}-L_{0}}{\delta}-\dot{L}\right\|_{L^{2}\to
V}$ $\displaystyle=$
$\displaystyle\sup_{\|f\|_{L^{2}}=1}\left\|\int\frac{k_{\delta}(x,y)-k_{0}(x,y)}{\delta}f(y)\
dy-\int\dot{k}(x,y)f(y)\ dy\right\|_{L^{2}}$ $\displaystyle=$
$\displaystyle\sup_{\|f\|_{L^{2}}=1}\left\|\int r_{\delta}(x,y)f(y)\
dy\right\|_{L^{2}}$ $\displaystyle\leq$
$\displaystyle\|r_{\delta}\|_{L^{2}([0,1]^{2})}=o(\delta).$

Proceeding similarly, one shows that if the map $\delta\mapsto k_{\delta}$ has
a continuous derivative with respect to $\delta$ in a neighborhood of
$\delta=0$, then $\delta\mapsto L_{\delta}$ has a continuous derivative.
Indeed we are supposing that for each $\delta\in[0,\overline{\delta})$ there
is $\dot{k}_{\delta}$ such that for small enough $h$

$k_{\delta+h}=k_{\delta}+h\cdot\dot{k}_{\delta}+r_{\delta,h}$

where $\dot{k}_{\delta},\ r_{\delta,h}\in L^{2}([0,1]^{2})$,
$||r_{\delta,h}||_{L^{2}([0,1]^{2})}=o(h)$ and furthermore
$\delta\mapsto\dot{k}_{\delta}$ is continuous. We have then by (6) that the
associated operators $\dot{L}_{\delta}$ defined as

$\dot{L}_{\delta}f(x):=\int\dot{k}_{\delta}(x,y)f(y)dy$ (11)

also varies in a continuous way as $\delta$ increases.

###  3.2 A formula for the linear response of the invariant measure and its
continuity

Now we apply Theorem 2.2 to Hilbert-Schmidt integral operators to obtain a
linear response formula for $L^{2}$ perturbations.

######  Corollary 3.5 (Linear response formula for kernel operators).

Suppose $L_{\delta}:L^{2}\rightarrow L^{2}$ are integral-preserving
(satisfying (3)) integral operators with stochastic kernels $k_{\delta}\in
L^{2}([0,1]^{2})$ as in (9). Suppose $L_{0}$ satisfies assumption $(A1)$ of
Theorem 2.2. Then $\dot{k}\in V_{\ker}$, the system has linear response for
this perturbation and an explicit formula for it is given by

$\lim_{\delta\rightarrow
0}\frac{f_{\delta}-f_{0}}{\delta}=(\text{Id}-L_{0})^{-1}\int\dot{k}(x,y)f_{0}(y)dy$
(12)

with convergence in $L^{2}.$

###### Proof.

Since $L_{\delta}$, $\delta\in[0,\bar{\delta})$, is integral preserving, we
have $(L_{\delta}-L_{0})(L^{2})\subset V$ and therefore, $k_{\delta}-k_{0}\in
V_{\ker}$ by Lemma 3.2, i.e. $\delta\dot{k}+r_{\delta}\in V_{\ker}$. Then, for
a.e. $y\in[0,1]$ and $\delta\not=0$, we have

$\displaystyle\bigg{|}\int\dot{k}(x,y)dx\bigg{|}\leq\frac{1}{\delta}\int|r_{\delta}(x,y)|dx\leq\frac{1}{\delta}\|r_{\delta}\|_{L^{2}([0,1]^{2})}.$

As $\delta\rightarrow 0$, the right hand side approaches $0$ and, since
$\int\dot{k}(x,y)dx$ is independent of $\delta$, we have
$\int\dot{k}(x,y)dx=0$ for a.e. $y\in[0,1]$, i.e. $\dot{k}\in V_{\ker}$.

Furthermore by $(\ref{perturb1})$ there is a $K\geq 0$ such that

$\left||L_{0}-L_{\delta}|\right|_{L^{2}\rightarrow L^{2}}\leq K\delta.$ (13)

Hence the family of operators satisfy the first part of assumption $(A2)$. The
second part of this assumption is established by the results of Lemma 3.4.

Since the operators $L_{\delta}$ are compact, integral preserving, and satisfy
assumptions $(A1)$ and $(A2)$ we can conclude by applying Theorem 2.2 to this
family of operators, obtaining

$\lim_{\delta\rightarrow
0}\left\|\frac{f_{\delta}-f_{0}}{\delta}-(\text{Id}-L_{0})^{-1}\int\dot{k}(x,y)f_{0}(y)dy\right\|_{2}=0.$

Now we show that the linear response of the invariant measure is continuous
with respect to the kernel perturbation. This will be used in Section 4 for
the proof of the existence of solutions of our main optimization problems.

Consider the transfer operator $L_{0}$, having a kernel $k_{0}\in
L^{2}([0,1]^{2})$, and a set of infinitesimal perturbations $P\subset
V_{\ker}$ of $k_{0}$. We will endow $P$ with the topology induced by its
inclusion in $L^{2}([0,1]^{2})$. Suppose $L_{\delta}$ is a perturbation of
$L_{0}$ satisfying the assumptions of Lemma 3.4. By Corollary 3.5, the linear
response will depend on the first-order term of the perturbation, $\dot{k}\in
P$, allowing us to define the function $R:P\rightarrow V$ by

$R(\dot{k}):=(\text{Id}-L_{0})^{-1}\int\dot{k}(x,y)f_{0}(y)dy.$ (14)

By (6) and the continuity of the resolvent operator it follows directly that
the response function
$R:(P,\|\cdot\|_{L^{2}([0,1]^{2})})\to(V,\|\cdot\|_{L^{2}})$ is continuous.

###  3.3 A formula for the linear response of the dominant eigenvalues and its
continuity

We apply Proposition 2.6 to Hilbert-Schmidt integral operators and obtain a
linear response formula for the dominant eigenvalues in the case of $L^{2}$
perturbations.

######  Corollary 3.6.

Suppose $L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow
L^{2}([0,1],{\mathbb{C}})$ are integral-preserving (satisfying (3)) integral
operators with kernels $k_{\delta}\in L^{2}([0,1]^{2})$ satisfying
$\delta\mapsto k_{\delta}\in C^{1}([0,\bar{\delta}),L^{2}([0,1]^{2}))$.
Suppose $L_{0}$ satisfies $(A1)$ of Theorem 2.2. Let
$\lambda_{0}\in{\mathbb{C}}$ be an eigenvalue of $L_{0}$ with the largest
magnitude strictly inside the unit circle and assume that $\lambda_{0}$ is
geometrically simple. Then, there exists $\dot{\lambda}\in\mathbb{C}$ such
that

$\lim_{\delta\rightarrow
0}\bigg{|}\frac{\lambda_{\delta}-\lambda_{0}}{\delta}-\dot{\lambda}\bigg{|}=0.$

Furthermore,

$\displaystyle\dot{\lambda}$
$\displaystyle=\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Re(\hat{e})(x)\Re(e)(y)+\Im(\hat{e})(x)\Im(e)(y)\right)dydx$
(15)
$\displaystyle\qquad+i\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Im(\hat{e})(x)\Re(e)(y)-\Re(\hat{e})(x)\Im(e)(y)\right)dydx,$

where $e\in L^{2}([0,1],\mathbb{C})$ is the eigenvector of $L_{0}$ associated
to the eigenvalue $\lambda_{0}$, $\hat{e}\in L^{2}([0,1],\mathbb{C})$ is the
eigenvector of $L_{0}^{*}$ associated to the eigenvalue $\lambda_{0}$ and
$\dot{L}$ is the operator in Lemma 3.4.

###### Proof.

Since $k_{\delta}\in L^{2}([0,1]^{2})$, the operator
$L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow L^{2}([0,1],{\mathbb{C}})$ is
compact; by assumption, it also satisfies (3). From Lemma 3.4, the map
$\delta\mapsto L_{\delta}$ is $C^{1}$. Hence, by Proposition 2.6, we have
$\dot{\lambda}=\langle\hat{e},\dot{L}e\rangle_{L^{2}([0,1],\mathbb{C})}$.
Finally, we compute

$\displaystyle\dot{\lambda}=\langle\hat{e},\dot{L}e\rangle_{L^{2}([0,1],\mathbb{C})}$
$\displaystyle=\int_{0}^{1}\hat{e}(x)\overline{\dot{L}e}(x)dx$
$\displaystyle=\int_{0}^{1}\hat{e}(x)\overline{\int_{0}^{1}\dot{k}(x,y)e(y)dy}dx$
$\displaystyle=\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\hat{e}(x)\bar{e}(y)dydx$
$\displaystyle=\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Re(\hat{e})(x)\Re(e)(y)+\Im(\hat{e})(x)\Im(e)(y)\right)dydx$
$\displaystyle\qquad+i\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Im(\hat{e})(x)\Re(e)(y)-\Re(\hat{e})(x)\Im(e)(y)\right)dydx.$

From the expression in the final line of the proof above, it is clear that if
we consider $\dot{\lambda}$ as a function of $\dot{k}$, the map
$\dot{\lambda}:(V_{\ker},\|\cdot\|)_{L^{2}([0,1]^{2})}\to\mathbb{C}$ is
continuous.

##  4 Optimal response: optimising the expectation of observables and mixing
rate

Having described the responses of our dynamical systems to perturbations, it
is natural to consider the optimisation problem of finding perturbations that
provoke _maximal_ responses. We consider the problems of finding the
infinitesimal perturbation that maximises the expectation of a given
observable and the infinitesimal perturbation that maximally enhances mixing.
In doing so, we extend the approach in [1] from the setting of finite-state
Markov chains to the integral operators considered in the present paper.

We show that at an abstract level these problems reduce to the optimization of
a linear continuous functional $\mathcal{J}$ on a convex set $P$ of feasible
perturbations; this problem has a solution and the solution is unique if the
set $P$ of allowed infinitesimal perturbations is strictly convex. The
convexity assumption on $P$ is natural because if two different perturbations
of the system are possible, then their convex combination (applying the two
perturbations with different intensities) will also be possible. After
introducing the abstract setting, we construct the objective functions for our
two optimal response problems and state general existence and uniqueness
results for the optima. Later, in Section 5 we focus on the construction of
the set of feasible perturbations and provide explicit formulae for the
maximising perturbations.

###  4.1 General optimisation setting, existence and uniqueness

We recall some general results (adapted for our purposes) on optimizing a
linear continuous function on convex sets; see also Lemma 6.2 [16]. The
abstract problem is to find $\dot{k}$ such that

$\mathcal{J}(\dot{k})=\max_{\dot{h}\in P}\mathcal{J}(\dot{h}),$ (16)

where $\mathcal{J}:\mathcal{H}\rightarrow{\mathbb{R}}$ is a continuous linear
function, $\mathcal{H}$ is a separable Hilbert space and
$P\subset\mathcal{H}$.

######  Proposition 4.1 (Existence of the optimal solution).

Let $P$ be bounded, convex, and closed in $\mathcal{H}$. Then, problem
considered at (16) has at least one solution.

###### Proof.

Since $P$ is bounded and $\mathcal{J}$ is continuous, we have that $\sup_{k\in
P}\mathcal{J}(k)<\infty$. Consider a maximizing sequence $k_{n}$ such that
$\lim_{n\rightarrow\infty}\mathcal{J}(k_{n})=\sup_{k\in P}\mathcal{J}(k)$.
Then, $k_{n}$ has a subsequence $k_{n_{j}}$ converging in the weak topology.
Since $P$ is strongly closed and convex in $\mathcal{H}$, we have that it is
weakly closed. This implies that
$\overline{k}:=\lim_{j\rightarrow\infty}k_{n_{j}}\in P.$ Also, since
$\mathcal{J}(k)$ is continuous and linear, it is continuous in the weak
topology. Then we have that
$\mathcal{J}(\overline{k})=\lim_{j\rightarrow\infty}\mathcal{J}(k_{n_{j}})=\sup_{k\in
P}\mathcal{J}(k)$ and we realise a maximum.

Uniqueness of the optimal solution will be provided by strict convexity of the
feasible set.

######  Definition 4.2.

We say that a convex closed set $A\subseteq\mathcal{H}$ is _strictly convex_
if for each pair $x,y\in A$ and for all $0<\gamma<1$, the points $\gamma
x+(1-\gamma)y\in\mathrm{int}(A)$, where the relative interior333The relative
interior of a closed convex set $C$ is the interior of $C$ relative to the
closed affine hull of $C$, see e.g. [7]. is meant.

######  Proposition 4.3 (Uniqueness of the optimal solution).

Suppose $P$ is closed, bounded, and strictly convex subset of $\cal{H}$, and
that $P$ contains the zero vector in its relative interior. If $\mathcal{J}$
is not uniformly vanishing on $P$ then the optimal solution to (16) is unique.

###### Proof.

Suppose that there are two distinct maxima $\dot{k}_{1},\dot{k}_{2}\in P$ with
$\mathcal{J}(\dot{k}_{1})=\mathcal{J}(\dot{k}_{2})=\alpha$. Let $0<\gamma<1$
and set $z=\gamma\dot{k}_{1}+(1-\gamma)\dot{k}_{2}$. By strict convexity of
$P$, $z\in\mathrm{int}(P)$, and by linearity of $\mathcal{J}$,
$\mathcal{J}(z)=\alpha$. Let $B_{r}(z)$ denote a (relative in $P$) open ball
of radius $r$ centred at $z$, with $r>0$ chosen small enough so that
$B_{r}(z)\subset\mathrm{int}(P)$. Because the zero vector lies in the relative
interior of $P$, and $\mathcal{J}$ does not uniformly vanish on $P$, there
exists a vector $v\in B_{r}(z)$ such that $\mathcal{J}(v)>0$. Now
$z+\frac{rv}{2\|v\|}\in\mathrm{int}(P)$ and
$\mathcal{J}(z+\frac{rv}{2\|v\|})>\alpha$, contradicting maximality of
$\dot{k}_{1}$.

In the following subsections we apply the general results of this section to
our specific optimisation problems.

###  4.2 Optimising the response of the expectation of an observable

Let $c\in L^{2}$ be a given observable. We consider the problem of finding an
infinitesimal perturbation that maximises the expectation of $c$. The
perturbations we consider are perturbations to the kernels of Hilbert-Schmidt
integral operators, of the form (9). If we denote the average of $c$ with
respect to the perturbed invariant density $f_{\delta}$ by

$\mathbb{E}_{c,\delta}:=\int c~{}f_{\delta}~{}dm,$

we have

$\frac{d\mathbb{E}_{c,\delta}}{d\delta}\bigg{|}_{\delta=0}=\lim_{\delta\rightarrow
0}\frac{\mathbb{E}_{c,\delta}-\mathbb{E}_{c,0}}{\delta}=\lim_{\delta\rightarrow
0}\int c~{}\frac{f_{\delta}-f_{0}}{\delta}~{}dm=\int c~{}R(\dot{k})~{}dm,$

where the last equality follows from Corollary 3.5.

The function $\mathcal{J}(\dot{k})=\langle c,R(\dot{k})\rangle$ is clearly
continuous as a map from $(V_{\ker},\|\cdot\|)_{L^{2}([0,1]^{2})}$ to
$\mathbb{R}$. Suppose that $P$ is a closed, bounded, convex subset of
$V_{\ker}$ containing the zero perturbation, and that $\mathcal{J}$ is not
uniformly vanishing on $P$. We wish to solve the following problem:

######  General Problem 1.

Find $\dot{k}\in P$ such that

$\big{\langle}c,R(\dot{k})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}=\max_{\dot{h}\in
P}\big{\langle}c,R(\dot{h})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}.$ (17)

We may immediately apply Proposition 4.1 to obtain that there exists a
solution to (17). If, in addition, $P$ is strictly convex, then by Proposition
4.3 the solution to (17) _is unique_.

To end this subsection we note that without loss of generality, we may assume
that $c\in$ span$\\{f_{0}\\}^{\perp}$. This is because for $c\in L^{2}$, we
have

$\langle c,R(\dot{k})\rangle_{L^{2}([0,1],{\mathbb{R}})}=\langle c-\langle
c,f_{0}\rangle_{L^{2}([0,1],{\mathbb{R}})}\mathbf{1},R(\dot{k})\rangle_{L^{2}([0,1],{\mathbb{R}})},$

since $R(\dot{k})\in V$. From $\int f_{0}(x)dx=1,$ we have that
$f\mapsto\langle f,f_{0}\rangle_{L^{2}([0,1],{\mathbb{R}})}\mathbf{1}$ is a
projection onto span$\\{\mathbf{1}\\}$ and so $f\mapsto f-\langle
f,f_{0}\rangle_{L^{2}([0,1],{\mathbb{R}})}\mathbf{1}$ is a projection onto
span$\\{f_{0}\\}^{\perp}$.

###  4.3 Optimising the response of the rate of mixing

We now consider the linear response problem of optimising the rate of mixing.
Let $\lambda_{0}\in{\mathbb{C}}$ denote an eigenvalue of $L_{0}$ strictly
inside the unit circle with largest magnitude. From now on, whenever
discussing the linear response of eigenvalues to kernel perturbations we
assume the conditions of Corollary 3.6. We recall that $e$ and $\hat{e}$ are
the eigenfunctions of $L_{0}$ and $L_{0}^{*}$, respectively, corresponding to
the eigenvalue $\lambda_{0}$.

To find the kernel perturbations that enhance mixing, we follow the approach
taken in [1] (see also [18, 16] in the continuous time setting), namely
perturbing our original dynamics $L_{0}$ in such a way that the modulus of the
second eigenvalue of the perturbed dynamics decreases. Equivalently, we want
to decrease the real part of the logarithm of the perturbed second eigenvalue.
The following result provides an explicit formula for this instantaneous rate
of change. Define

$E(x,y):=\left(\Re(\hat{e})(x)\Re(e)(y)+\Im(\hat{e})(x)\Im(e)(y)\right)\Re(\lambda_{0})+\left(\Im(\hat{e})(x)\Re(e)(y)-\Re(\hat{e})(x)\Im(e)(y)\right)\Im(\lambda_{0}).$
(18)

######  Lemma 4.4.

One has

$\frac{d}{d\delta}\Re\left(\log\lambda_{\delta}\right)\bigg{|}_{\delta=0}=\frac{\big{\langle}\dot{k},E\big{\rangle}_{L^{2}([0,1]^{2},{\mathbb{R}})}}{|\lambda_{0}|^{2}}.$

###### Proof.

From (15), we have that

$\Re(\dot{\lambda}_{0})=\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Re(\hat{e})(x)\Re(e)(y)+\Im(\hat{e})(x)\Im(e)(y)\right)dydx$
(19)

and

$\Im(\dot{\lambda}_{0})=\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)\left(\Im(\hat{e})(x)\Re(e)(y)-\Re(\hat{e})(x)\Im(e)(y)\right)dydx.$
(20)

Next, we note that

$\frac{d}{d\delta}\Re(\log\lambda_{\delta})=\Re\left(\frac{d}{d\delta}\log\lambda_{\delta}\right)=\Re\left(\frac{d\lambda_{\delta}}{d\delta}\frac{1}{\lambda_{\delta}}\right).$
(21)

From (19)-(21), we obtain

$\frac{d}{d\delta}\Re\left(\log\lambda_{\delta}\right)\bigg{|}_{\delta=0}=\Re\left(\frac{\dot{\lambda}_{0}}{\lambda_{0}}\right)=\Re\left(\frac{\dot{\lambda}_{0}}{\lambda_{0}}\frac{\overline{\lambda_{0}}}{\overline{\lambda_{0}}}\right)=\frac{\Re(\dot{\lambda}_{0})\Re(\lambda_{0})+\Im(\dot{\lambda}_{0})\Im(\lambda_{0})}{|\lambda_{0}|^{2}}=\frac{\big{\langle}\dot{k},E\big{\rangle}_{L^{2}([0,1]^{2},{\mathbb{R}})}}{|\lambda_{0}|^{2}}.$

The function $\mathcal{J}(\dot{k})=\langle\dot{k},E\rangle$ is clearly
continuous as a map from $(V_{\ker},\|\cdot\|_{L^{2}([0,1]^{2})})$ to
$\mathbb{R}$. As in subsection 4.2, suppose that $P$ is a closed, bounded,
strictly convex subset of $V_{\ker}$ containing the zero element, and that
$\mathcal{J}$ is not uniformly vanishing on $P$. We wish to solve the
following problem:

######  General Problem 2.

Find $\dot{k}\in P$ such that

$\langle\dot{k},E\rangle_{L^{2}([0,1]^{2},{\mathbb{R}})}=\min_{\dot{h}\in
P}\langle\dot{k},E\rangle_{L^{2}([0,1]^{2},{\mathbb{R}})}.$ (22)

We may immediately apply Proposition 4.1 to obtain that there exists a
solution to (17). If, in addition, $P$ is strictly convex, then by Proposition
4.3 the solution to (22) _is unique_.

##  5 Explicit formulae for the optimal perturbations

Thus far we have not been specific about the feasible set $P$; we take up this
issue in this and the succeeding subsections to provide explicit formulae for
the optimal responses in both problems (17) and (22). First, we have not
required that the perturbed kernel $k_{\delta}$ in (9) be nonnegative for
$\delta>0$, however, this is a natural assumption. To facilitate this, for
$0<l<1$, define

$F_{l}:=\\{(x,y)\in[0,1]^{2}:k_{0}(x,y)\geq l\\}\quad\mbox{ and }\quad
S_{k_{0},l}:=\\{k\in L^{2}([0,1]^{2}):\text{supp}(k)\subseteq F_{l}\\}.$ (23)

The set of allowable perturbations that we will consider in the sequel is

$P_{l}:=V_{\ker}\cap S_{k_{0},l}\cap B_{1},$ (24)

where $B_{1}$ is the closed unit ball in $L^{2}([0,1]^{2})$.

We now begin verifying the conditions on $P_{l}$ and $\mathcal{J}$ required by
Proposition 4.3. First, $P_{l}$ is clearly bounded in $L^{2}([0,1]^{2})$.
Second, we note that as long as $F_{l}$ has positive Lebesgue measure, the
zero kernel is in the relative interior of $P_{l}$. Third, the following lemma
handles closedness of $P_{l}$. Fourth, from this, since $V_{\ker}$ and
$S_{k_{0},l}$ are closed subspaces, $V_{\ker}\cap S_{k_{0},l}$ is itself a
Hilbert space, and hence, $P_{l}$ is strictly convex. Finally, sufficient
conditions for the objective function to not uniformly vanish are given in
Lemma 5.2.

######  Lemma 5.1.

The set $S_{k_{0},l}$ is a closed subspace of $L^{2}([0,1]^{2})$.

###### Proof.

The fact that $S_{k_{0},l}$ is a subspace is trivial. Let $\\{k_{n}\\}\subset
S_{k_{0},l}$ and suppose $k_{n}\rightarrow_{L^{2}}k\in L^{2}([0,1]^{2})$.
Further suppose $\\{(x,y)\in[0,1]^{2}:k_{0}(x,y)<l\\}$ is not a null set;
otherwise $S_{k_{0},l}=L^{2}([0,1]^{2})$ and the result immediately follows.
Then, we have

$\int_{\\{k_{0}\geq
l\\}}(k_{n}(x,y)-k(x,y))^{2}dydx+\int_{\\{k_{0}<l\\}}k(x,y)^{2}dxdy\rightarrow
0.$

Since $\int_{\\{k_{0}\geq l\\}}(k_{n}(x,y)-k(x,y))^{2}dydx\geq 0$, if
$\int_{\\{k_{0}<l\\}}k(x,y)^{2}dxdy>0$ then we obtain a contradiction; thus,
$\int_{\\{k_{0}<l\\}}k(x,y)^{2}dxdy=0$ and therefore $k=0$ a.e. on
$\\{(x,y)\in[0,1]^{2}:k_{0}(x,y)<l\\}$. Hence, $S_{k_{0},l}$ is closed.

Let

$F_{l}^{y}:=\\{x\in[0,1]:(x,y)\in F_{l}\\},$ (25)

and for $F_{l}\subset[0,1]^{2}$, define

$\Xi(F_{l})=\\{y\in[0,1]:m(F_{l}^{y})>0\\}.$

The following lemma provides sufficient conditions for a functional of the
general form we wish to optimise to not uniformly vanish. The general
objective has the form
$\mathcal{J}(\dot{k})=\int\int\dot{k}(x,y)\mathcal{E}(x,y)\ dy\ dx$; in our
first specific objective (optimising response of expectations) we put
$\mathcal{E}(x,y)=((\text{Id}-L_{0}^{*})^{-1}c)(x)\cdot f_{0}(y)$ and in our
second specific objective (optimising mixing) we put $\mathcal{E}(x,y)=E(x,y)$
from (18). Let $\mathcal{E}^{+}$ and $\mathcal{E}^{-}$ denote the positive and
negative parts of $\mathcal{E}$. For $y\in\Xi(F_{l})$, let
$A(y)=\int_{F_{l}^{y}}\mathcal{E}^{+}(x,y)\ dx$ and
$a(y)=\int_{F_{l}^{y}}\mathcal{E}^{-}(x,y)\ dx$.

######  Lemma 5.2.

Assume that there is $\Xi^{\prime}\subset\Xi(F_{l})$ such that
$m(\Xi^{\prime})>0$ and $A(y),a(y)>0$ for $y\in\Xi^{\prime}$. Then there is a
$\dot{k}\in P_{l}$ such that $\mathcal{J}(\dot{k})>0$.

###### Proof.

For $y\in\Xi(F_{l})$, set
$\dot{k}(x,y)=\mathbf{1}_{F_{l}^{y}}(x)\left(a(y)\mathcal{E}^{+}(x,y)-A(y)\mathcal{E}^{-}(x,y)\right)$.
To show $\dot{k}\in P_{l}$ we need to check that (i) the support of $\dot{k}$
is contained in $F_{l}$ and (ii) $\int_{F_{l}^{y}}\dot{k}(x,y)\ dx=0$ for a.e.
$y\in\Xi(F_{l})$; these points show $\dot{k}\in S_{k_{0},l}\cap V_{\ker}$ and
by trivial scaling we may obtain $\dot{k}\in B_{1}$. Item (i) is obvious from
the definition of $\dot{k}$. For item (ii) we compute

$\int_{F_{l}^{y}}\dot{k}(x,y)\
dx=\int_{F_{l}^{y}}(a(y)\mathcal{E}^{+}(x,y)-A(y)\mathcal{E}^{-}(x,y))\
dx=a(y)A(y)-A(y)a(y)=0.$

Finally, we check that $\mathcal{J}(\dot{k})>0$. One has

$\displaystyle\int_{F_{l}}\dot{k}(x,y)\mathcal{E}(x,y)\ dx\ dy$
$\displaystyle=$
$\displaystyle\int_{F_{l}}\left(a(y)\mathcal{E}^{+}(x,y)-A(y)\mathcal{E}^{-}(x,y)\right)\cdot\mathcal{E}(x,y)\
dx\ dy$ $\displaystyle=$
$\displaystyle\int_{F_{l}}a(y)(\mathcal{E}^{+}(x,y))^{2}+A(y)(\mathcal{E}^{-}(x,y))^{2}\
dx\ dy$ $\displaystyle=$
$\displaystyle\int_{\Xi(F_{l})}\left[\left(\int_{F_{l}^{y}}\mathcal{E}^{-}(x,y)\
dx\right)\cdot\left(\int_{F_{l}^{y}}(\mathcal{E}^{+}(x,y))^{2}\
dx\right)+\left(\int_{F_{l}^{y}}\mathcal{E}^{+}(x,y)\
dx\right)\cdot\left(\int_{F_{l}^{y}}(\mathcal{E}^{-}(x,y))^{2}\
dx\right)\right]\ dy.$

This final expression is positive due by the hypotheses of the Lemma.

######  Remark 5.3.

We note that in the situation where $\mathcal{E}(x,y)$ is in separable form
$\mathcal{E}(x,y)=h_{1}(x)h_{2}(y)$—as in the case of optimising the
derivative of the expectation of an observable $c$ , and in the case of
optimising the derivative of a real eigenvalue—then
$A(y)=h_{2}(y)\int_{F_{l}^{y}}h_{1}^{+}(x)\ dx$ and
$a(y)=h_{2}(y)\int_{F_{l}^{y}}h_{1}^{-}(x)\ dx$. Because $h_{2}=f_{0}$ and
$h_{2}=e$ are not the zero function, and $h_{1}=(\text{Id}-L_{0}^{*})^{-1}c$
and $h_{1}=\hat{e}$ are both nontrivial signed functions, the conditions of
Lemma 5.2 are relatively easy to satisfy.

###  5.1 Maximising the expectation of an observable

In this section we provide an explicit formula for the optimal kernel
perturbation to increase the expectation of an observation function $c$ by the
greatest amount. Since the objective function in (17) is linear in $\dot{k}$,
a maximum will occur on $\partial B_{1}\cap V_{\ker}\cap S_{k_{0},l}$ (i.e. we
only need to consider the optimization over the unit sphere and not the unit
ball). Thus, we consider the following reformulation of the general problem 1:

######  Problem A.

Given $l>0$ and $c\in$ span$\\{f_{0}\\}^{\perp}$, solve

$\displaystyle\min_{\dot{k}\in V_{\ker}\cap S_{k_{0},l}}$
$\displaystyle-\big{\langle}c,R(\dot{k})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}$
(26) subject to $\displaystyle\|\dot{k}\|_{L^{2}([0,1]^{2})}^{2}-1=0.$ (27)

Our first main result is:

######  Theorem 5.4.

Let $L_{0}:L^{2}\rightarrow L^{2}$ be an integral operator with the stochastic
kernel $k_{0}\in L^{2}([0,1]^{2})$. Suppose that $L_{0}$ satisfies $(A1)$ of
Theorem 2.2 and that there is a $\Xi^{\prime}\subset\Xi(F_{l})$ with
$m(\Xi^{\prime})>0$ and
$f_{0}(y)>0,\int_{F_{l}^{y}}((\text{Id}-L_{0}^{*})^{-1}c)^{+}(x)\ dx>0$, and
$\int_{F_{l}^{y}}((\text{Id}-L_{0}^{*})^{-1}c)^{-}(x)\ dx>0$ for
$y\in\Xi^{\prime}$. Then the unique solution to Problem A is

$\dot{k}(x,y)=\begin{cases}\frac{f_{0}(y)}{\alpha}\left(((\text{Id}-L_{0}^{\ast})^{-1}c)(x)-\frac{\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz}{m(F_{l}^{y})}\right)&(x,y)\in
F_{l},\\\ 0&\text{otherwise},\end{cases}$ (28)

where $\alpha>0$ is selected so that $\|\dot{k}\|_{L^{2}([0,1]^{2})}=1$.
Furthermore, if $c\in W:=$ span$\\{f_{0}\\}^{\perp}\cap L^{\infty}$, $k_{0}\in
L^{\infty}([0,1]^{2})$, and $k_{0}$ is such that $L_{0}:L^{1}\rightarrow
L^{1}$ is compact, then $\dot{k}\in L^{\infty}([0,1]^{2})$.

###### Proof.

See Appendix A.

Note that the expression for the optimal perturbation $\dot{k}$ in (28)
depends only on $k_{0}$ and $c$. This is in part a consequence of the fact
that the linear response formula (12) depends only on the first order term
$\dot{k}$ (the “direction” of the perturbation) in the expansion of
$k_{\delta}$. Thus, in order to find the unique perturbation that optimises
our linear response, we seek the best “direction” for the perturbation.
Similar comments hold for our other three optimal linear perturbation results
in later sections.

######  Remark 5.5.

In certain situations we may desire to make non-infinitesimal perturbations
$k_{\delta}:=k_{0}+\delta\cdot\dot{k}$ that remain stochastic for small
$\delta>0$. If $\dot{k}\in L^{\infty}([0,1]^{2})\cap V_{\ker}\cap
S_{k_{0},l}$, clearly $k_{\delta}=k_{0}+\delta\cdot\dot{k}$ satisfies $\int
k_{\delta}(x,y)dx=1$ for a.e. $y$. Also, as we are only perturbing at values
where $k_{0}\geq l>0$, and since $\dot{k}$ is essentially bounded, there
exists a $\bar{\delta}>0$ such that $k_{\delta}\geq 0$ a.e. for all
$\delta\in(0,\bar{\delta})$. In summary, for $\delta\in(0,\bar{\delta})$,
$k_{\delta}$ is a stochastic kernel.

The compactness condition on $L_{0}:L^{1}\to L^{1}$ required for essential
boundedness of $\dot{k}$ can be addressed as follows. A criterion for $L_{0}$
to be compact on $L^{1}([0,1])$ is the following (see [12]): Given
$\varepsilon>0$ there exists $\beta>0$ such that for a.e. $y\in[0,1]$ and
$\gamma\in{\mathbb{R}}$ with $|\gamma|<\beta$,

$\int_{{\mathbb{R}}}\big{|}\tilde{k}(x+\gamma,y)-\tilde{k}(x,y)\big{|}dx<\varepsilon,$

where $\tilde{k}:\mathbb{R}\times[0,1]\to\mathbb{R}$ is defined by

$\tilde{k}(x,y)=\begin{cases}k_{0}(x,y)&x\in[0,1],\\\
0&\text{otherwise}.\end{cases}$

A class of kernels that satisfy this are essentially bounded kernels
$k_{0}:[0,1]\times[0,1]\rightarrow{\mathbb{R}}$ that are uniformly continuous
in the first coordinate. Such a class naturally arises in our dynamical
systems settings.

###  5.2 Maximally increasing the mixing rate

Let $\lambda_{0}\in{\mathbb{C}}$ denote a geometrically simple eigenvalue of
$L_{0}$ strictly inside the unit circle and $e$ and $\hat{e}$ denote the
corresponding eigenvectors of $L_{0}$ and $L_{0}^{*}$, respectively. Our
results concerning optimal rate of movement of $\lambda_{0}$ under system
perturbation work for any $\lambda_{0}$ as above, but eigenvalues of largest
magnitude inside the unit circle have the additional significance of
controlling the exponential rate of mixing. We therefore primarily focus on
these eigenvalues and in this section we consider again the linear response
problem for enhancing the rate of mixing, now providing explicit formulae for
optimal perturbations and the response.

Since we are again interested in kernel perturbations that will ensure that
the perturbed kernel $k_{\delta}$ is nonnegative, we consider the constraint
set $P_{l}$, as in Section 4.1, where $0<l<1$. The objective function of (22)
is linear and therefore, we only need to consider the optimization problem on
$V_{\ker}\cap S_{k_{0},l}\cap\partial B_{1}$. Thus, to obtain the perturbation
$\dot{k}$ that will enhance the mixing rate, we solve the following
optimization problem:

######  Problem B.

Given $l>0$, solve

$\displaystyle\min_{\dot{k}\in V_{\ker}\cap S_{k_{0},l}}$
$\displaystyle\big{\langle}\dot{k},E\big{\rangle}_{L^{2}([0,1]^{2},{\mathbb{R}})}$
(29) such that
$\displaystyle\|\dot{k}\|_{L^{2}([0,1]^{2},{\mathbb{R}})}^{2}-1=0,$ (30)

where $E$ is defined in (18).

######  Theorem 5.6.

Let $L_{0}:L^{2}([0,1],{\mathbb{C}})\rightarrow L^{2}([0,1],{\mathbb{C}})$ be
an integral operator with the stochastic kernel $k_{0}\in
L^{2}([0,1]^{2},{\mathbb{R}})$. Suppose that $L_{0}$ satisfies $(A1)$ of
Theorem 2.2 and that there is a $\Xi^{\prime}\subset\Xi(F_{l})$ with
$m(\Xi^{\prime})>0$, and $\int_{F_{l}^{y}}E(x,y)^{+}\ dx>0$ and
$\int_{F_{l}^{y}}E(x,y)^{-}\ dx>0$ for $y\in\Xi^{\prime}$. Then, the unique
solution to Problem B is

$\dot{k}(x,y)=\begin{cases}\frac{1}{\alpha}\left(\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}E(x,y)dx-E(x,y)\right)&(x,y)\in
F_{l}\\\ 0&\text{otherwise},\end{cases}$ (31)

where $E$ is given in (18) and $\alpha>0$ is selected so that
$\|\dot{k}\|_{L^{2}([0,1]^{2},{\mathbb{R}})}=1$. Furthermore, if $k_{0}\in
L^{\infty}([0,1]^{2},{\mathbb{R}})$ then $\dot{k}\in
L^{\infty}([0,1]^{2},{\mathbb{R}})$.

###### Proof.

See Appendix B.

If $\lambda_{0}$ is real, the optimal kernel has a simpler form:

######  Corollary 5.7.

If $\lambda_{0}$ is real and $k_{0}\geq l$, then the solution to Problem B is

$\dot{k}(x,y)=sgn(\lambda_{0})\frac{e(y)}{\|e\|_{2}}\left(\frac{\langle\hat{e},\mathbf{1}\rangle_{L^{2}([0,1],{\mathbb{R}})}\mathbf{1}-\hat{e}(x)}{\|\langle\hat{e},\mathbf{1}\rangle_{L^{2}([0,1],{\mathbb{R}})}\mathbf{1}-\hat{e}\|_{2}}\right).$
(32)

###### Proof.

We have $E(x,y)=\lambda_{0}\hat{e}(x)e(y)$; thus, the solution to the
optimization problem (29)-(30) is

$\dot{k}(x,y)=(\lambda_{0}/\alpha)\left(\int_{0}^{1}\hat{e}(x)dx-\hat{e}(x)\right)e(y),$

where $\alpha>0$ is the normalization constant such that
$\|\dot{k}\|_{L^{2}([0,1]^{2},{\mathbb{R}})}^{2}=1$.

##  6 Linear response for map perturbations

In this section we consider random dynamics governed by the composition of a
deterministic map $T_{\delta}$, $\delta\in[0,\bar{\delta})$, and additive
i.i.d. perturbations, or “additive noise”. We will assume that the noise is
distributed according to a certain Lipschitz kernel $\rho$ and impose a
reflecting boundary condition that ensures that the dynamics remain in the
interval $[0,1]$. More precisely, we consider a random dynamical system whose
trajectories are given by

$x_{n+1}=T_{\delta}(x_{n})\ \hat{+}\ \omega_{n},$ (33)

where $\hat{+}$ is the “boundary reflecting” sum, defined by
$a\hat{+}b:=\pi(a+b)$, and $\pi:\mathbb{R}\rightarrow[0,1]$ is the piecewise
linear map $\pi(x)=\min_{i\in\mathbb{Z}}|x-2i|$. We assume throughout that

  * (T1)

$T_{\delta}:[0,1]\rightarrow[0,1]$ is a Borel-measurable map for each
$\delta\in[0,\bar{\delta})$,

  * (T2)

$\omega_{n}$ is an i.i.d. process distributed according to a probability
density $\rho\in Lip(\mathbb{R})$, supported on $[-1,1]$ with Lipschitz
constant $K$.

###  6.1 Expressing the map perturbation as a kernel perturbation

In this subsection we describe precisely the kernel of the transfer operator
of the system (33). Associated with the process (33) is an integral-type
transfer operator $L_{\delta}$, which we will derive (following the method of
§10.5 in [35]). Noting that $|\pi^{\prime}(z)|=1$ for all $z\in\mathbb{R}$,
the Perron-Frobenius operator $P_{\pi}:L^{1}(\mathbb{R})\rightarrow
L^{1}([0,1])$ associated to the map $\pi$ is given by

$P_{\pi}f(x)=\sum_{z\in\pi^{-1}(x)}f(z)=\sum_{i\in
2\mathbb{Z}}(f(i+x)+f(i-x)).$ (34)

For $b\in{\mathbb{R}}$ consider the shift operator $\tau_{b}$ defined by
$(\tau_{b}g)(y):=g(y+b)$ for $g\in Lip(\mathbb{R})$. For the process (33),
suppose that $x_{n}$ has the distribution $f_{n}:[0,1]\to\mathbb{R}^{+}$ (i.e.
$f_{n}\in L^{1},\ f_{n}\geq 0$ and $\int f_{n}\ dm=1$). We note that
$T_{\delta}(x_{n})$ and $\omega_{n}$ are independent and thus the joint
density of $(x_{n},\omega_{n})\in[0,1]\times[-1,1]$ is $f_{n}\cdot\rho$. Let
$h:[0,1]\rightarrow{\mathbb{R}}$ be a bounded, measurable function and let
$\mathbb{E}$ denote expectation with respect to Lebesgue measure; we then
compute

$\displaystyle\mathbb{E}(h(x_{n+1}))$
$\displaystyle=\int_{-\infty}^{\infty}\int_{0}^{1}h(\pi(T_{\delta}(y)+z))f_{n}(y)\rho(z)dydz$
$\displaystyle=\int_{0}^{1}\int_{-\infty}^{\infty}h(\pi(z^{\prime}))f_{n}(y)\rho(z^{\prime}-T_{\delta}(y))dz^{\prime}dy$
$\displaystyle=\int_{0}^{1}f_{n}(y)\int_{-\infty}^{\infty}h(\pi(z^{\prime}))(\tau_{-T_{\delta}(y)}\rho)(z^{\prime})dz^{\prime}dy$
$\displaystyle=\int_{0}^{1}f_{n}(y)\int_{0}^{1}h(z^{\prime})(P_{\pi}\tau_{-T_{\delta}(y)}\rho)(z^{\prime})dz^{\prime}dy,$

where the last equality follows from the duality of the Perron-Frobenius and
the Koopman operators for $\pi$. Since
$\mathbb{E}(h(x_{n+1}))=\int_{0}^{1}h(x)f_{n+1}(x)dx$, and $h$ is arbitrary,
the map $f_{n}\mapsto f_{n+1}$ is given by

$f_{n+1}(z^{\prime})=\int_{0}^{1}(P_{\pi}\tau_{-T_{\delta}(y)}\rho)(z^{\prime})f_{n}(y)dy$

for all $z^{\prime}\in[0,1]$. Thus, for $\delta\in[0,\bar{\delta})$ the
integral operator $L_{\delta}:L^{2}([0,1])\rightarrow L^{2}([0,1])$ associated
to the process (33) is given by

$L_{\delta}f(x)=\int k_{\delta}(x,y)f(y)dy,$ (35)

where

$k_{\delta}(x,y)=(P_{\pi}\tau_{-T_{\delta}(y)}\rho)(x)$ (36)

and $x,y\in[0,1]$.

######  Lemma 6.1.

The kernel (36) is a stochastic kernel in $L^{\infty}([0,1]^{2})$.

###### Proof.

Stochasticity and nonnegativity of $k_{\delta}$ follow from stochasticity and
nonnegativity of $\rho$ and the fact that Perron-Frobenius operators preserve
these properties. Essential boundedness of $k_{\delta}$ follows from the facts
that $\rho$ is Lipschitz (thus essentially bounded), $\tau$ is a shift, and
$P_{\pi}$ is constructed from a finite sum because $\rho$ has compact support.

######  Proposition 6.2.

Assume that $k_{\delta}$ arising from the system $(T_{\delta},\rho)$ is given
by (36). Suppose that the family of interval maps
$\\{T_{\delta}\\}_{\delta\in[0,\bar{\delta})}$ satisfies

$T_{\delta}=T_{0}+\delta\cdot\dot{T}+t_{\delta},$

where $\dot{T},t_{\delta}\in L^{2}$ and $\|t_{\delta}\|_{2}=o(\delta)$. Then

$k_{\delta}=k_{0}+\delta\cdot\dot{k}+r_{\delta}$

where $\dot{k}\in L^{2}([0,1]^{2})$ is given by

$\dot{k}(x,y)=-\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot\dot{T}(y)$
(37)

and $r_{\delta}\in L^{2}([0,1]^{2})$ satisfies
$\|r_{\delta}\|_{L^{2}([0,1]^{2})}=o(\delta)$.

If additionally, $d\rho/dx$ is Lipschitz and the derivative of the map
$\delta\mapsto T_{\delta}$ with respect to $\delta$ varies continuously in
$L^{2}$ in a neighborhood of $\delta=0$, then $\delta\mapsto k_{\delta}$ has a
continuous derivative with respect to $\delta$ in a neighborhood of
$\delta=0$.

###### Proof.

We show that
$\|k_{\delta}(x,y)-k_{0}(x,y)-\delta\cdot\dot{k}(x,y)\|_{L^{2}([0,1]^{2})}=o(\delta)$,
where $\dot{k}$ is as in (37). We have

$\displaystyle\left\|k_{\delta}(x,y)-k_{0}(x,y)-\delta\cdot\dot{k}(x,y)\right\|_{L^{2}([0,1]^{2})}$
$\displaystyle\leq$
$\displaystyle\left\|(P_{\pi}\tau_{-T_{\delta}(y)}\rho)(x)-(P_{\pi}\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho)(x)\right\|_{L^{2}([0,1]^{2})}$
$\displaystyle+\left\|(P_{\pi}\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho)(x)-(P_{\pi}\tau_{-T_{0}(y)}\rho)(x)-\delta\left(-\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot\dot{T}(y)\right)\right\|_{L^{2}([0,1]^{2})}.$

We begin by showing that the first term on the right hand side of (6.1) is
$o(\delta)$. Since $\rho$ is Lipschitz with constant $K$, one has

$\big{|}(\tau_{-(T_{\delta}(y))}\rho)(x)-(\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho)(x)\big{|}=\big{|}\rho(x-T_{\delta}(y))-\rho(x-T_{0}(y)-\delta\cdot\dot{T}(y))\big{|}\leq
K|t_{\delta}(y)|.$ (39)

Because the support of
$\tau_{-(T_{\delta}(y))}\rho-\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho$ is
contained in 2 intervals, each of length 2, by (39) and Lemma C.1, we
therefore see that

$\left\|(P_{\pi}\tau_{-T_{\delta}(y)}\rho)(x)-(P_{\pi}\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho)(x)\right\|_{L^{2}([0,1]^{2})}\leq
6K\|t_{\delta}\|_{L^{2}}=o(\delta).$

Next we show that the second term on the right hand side of (6.1) is
$o(\delta)$. Using the definition of the derivative and the fact that $\rho$
is differentiable a.e. we see that

$\lim_{\delta\rightarrow 0}D(\delta):=\lim_{\delta\rightarrow
0}\left[\frac{\rho(x-T_{0}(y)-\delta\cdot\dot{T}(y))-\rho(x-T_{0}(y))}{\delta}-\left(-\frac{d\rho}{dx}(x-T_{0}(y))\dot{T}(y)\right)\right]=0$
(40)

for a.e. $x,y$. Since
$\bigg{|}\frac{\rho(x-T_{0}(y)-\delta\cdot\dot{T}(y))-\rho(x-T_{0}(y))}{\delta}\bigg{|}\leq
K\dot{T}(y)$, by dominated convergence the limit (40) also converges in
$L^{2}.$ Hence, applying Lemma C.1 to the second term on the right hand side
of (6.1), noting that $D(\delta)$ in (40) is square-integrable and supported
in at most 3 intervals of length at most 2, we obtain

$\displaystyle{\left\|(P_{\pi}\tau_{-(T_{0}(y)+\delta\cdot\dot{T}(y))}\rho)(x)-(P_{\pi}\tau_{-T_{0}(y)}\rho)(x)-\delta\left(-\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot\dot{T}(y)\right)\right\|_{L^{2}([0,1]^{2})}}$
$\displaystyle\leq$ $\displaystyle 9\delta D(\delta)=o(\delta).$

Regarding the final statement, suppose that $\delta\mapsto T_{\delta}$ has a
continuous derivative with respect to $\delta$ at a neighborhood of
$\delta=0$. This implies that $\dot{T}$ exists and varies continuously on a
small interval $[0,\delta^{*}]$, with $0<\delta^{*}\leq\bar{\delta}$. Denote
the derivative $dT_{\delta}/d\delta$ at $\delta$ by $\dot{T}_{\delta}$, and
similarly for $\dot{k}$. One has

$\displaystyle\|\dot{k}_{\delta}-\dot{k}_{0}\|_{L^{2}([0,1]^{2})}$
$\displaystyle=$
$\displaystyle\left\|\left(P_{\pi}\left(\tau_{-T_{\delta}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot\dot{T}_{\delta}(y)-\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot\dot{T}_{0}(y)\right\|_{L^{2}([0,1]^{2})}$
$\displaystyle\leq$
$\displaystyle\left\|\left(P_{\pi}\left(\tau_{-T_{\delta}(y)}\frac{d\rho}{dx}\right)\right)(x)\cdot(\dot{T}_{\delta}(y)-\dot{T}_{0}(y))\right\|_{L^{2}([0,1]^{2})}$
$\displaystyle\quad+\left\|\left[\left(P_{\pi}\left(\tau_{-T_{\delta}(y)}\frac{d\rho}{dx}\right)\right)(x)-\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\right]\cdot\dot{T}_{0}(y)\right\|_{L^{2}([0,1]^{2})}$
$\displaystyle\leq$ $\displaystyle
3\|d\rho/dx\|_{2}\|\dot{T}_{\delta}-\dot{T}_{0}\|_{2}+6{\mathrm{L}ip}(d\rho/dx)\|\delta\cdot\dot{T}_{0}+r_{\delta}\|_{2}\|\dot{T}_{0}\|_{2},$

where the final inequality follows from Lemma C.1 applied to each term in the
previous line, noting that $\rho$ is supported in a single interval of length
2. The first term in the final inequality goes to zero as $\delta\to 0$ by
continuity of $\dot{T}$, and the second term goes to zero as $\delta\to 0$
since $\|r_{\delta}\|_{2}\to 0$.

###  6.2 A formula for the linear response of the invariant measure and
continuity with respect to map perturbations

By considering the kernel form of map perturbations, we can apply Corollary
3.5 to obtain the following.

######  Proposition 6.3.

Let $L_{\delta}:L^{2}\rightarrow L^{2}$, $\delta\in[0,\bar{\delta})$, be the
integral operators in (35) with the kernels $k_{\delta}$ as in (36). Suppose
that $L_{0}$ satisfies $(A1)$ of Theorem 2.2. Then the kernel $\dot{k}$ in
(37) is in $V_{\ker}$ and

$\lim_{\delta\rightarrow
0}\frac{f_{\delta}-f_{0}}{\delta}=-(\text{Id}-L_{0})^{-1}\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\dot{T}(y)f_{0}(y)dy,$

with convergence in $L^{2}.$

###### Proof.

The result is a direct application of Corollary 3.5; we verify its
assumptions. From Lemma 6.1, $k_{\delta}\in L^{2}([0,1]^{2})$ is a stochastic
kernel and so $L_{\delta}$ is an integral-preserving compact operator. From
Proposition 6.2, $k_{\delta}$ has the form (9). Thus, we can apply Corollary
3.5 to obtain the result.

######  Remark 6.4.

If $T$ is covering444We say $T$ is covering if for each small open interval
$I\subseteq[0,1]$ there is $n=n(I)$ such that $T^{n}(I)=[0,1]$. and $\rho$ is
strictly positive in a neighbourhood of zero one can show the corresponding
transfer operator $L_{0}$ satisfies assumption $(A1)$ of Theorem 2.2, using
arguments similar to e.g. [49] Proposition 8.1, [14] Lemmas 3 and 10, or [20],
Lemma 41. Let $f\in L^{1}$ have zero average: $\int_{[0,1]}f=0$. If $f$ is $0$
almost everywhere, $L_{0}^{n}(f)=0$ and we are done. Otherwise, given
$\epsilon<0$, we can find an $f_{1}$ such that $\|f-f_{1}\|_{1}<\epsilon$ and
$f_{1}$ is positive in some small interval $I\subset[0,1]$. Since $\rho$ is
positive in a neighbourhood of zero, $\mathrm{s}upp(L_{0}(f_{1}^{+}))\supset
T(I)$. By the covering condition there is some $n^{\prime}\in{\mathbb{N}}$
such that $\mathrm{s}upp(L_{0}^{n^{\prime}}(f_{1}^{+}))=[0,1]$. It is then
standard to deduce that there is an $n_{0}\geq n^{\prime}$ such that
$\|L_{0}^{n}(f_{1})\|_{1}<\epsilon$ for $n\geq n_{0}$. Since the transfer
operator contracts the $L^{1}$ norm, then $||L_{0}^{n}f||_{1}\leq 2\epsilon$
for $n\geq n_{0}$ and since $\epsilon$ was arbitrary, this implies that
$L_{0}$ satisfies $(A1)$.

Let the linear response $\widehat{R}:L^{2}\rightarrow L^{2}$ of the invariant
density be defined as

$\widehat{R}(\dot{T}):=-(\text{Id}-L_{0})^{-1}\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\dot{T}(y)f_{0}(y)dy.$
(41)

######  Lemma 6.5.

The function $\widehat{R}:L^{2}\rightarrow L^{2}$ is continuous.

###### Proof.

We have

$\widehat{R}(\dot{T}_{1})-\widehat{R}(\dot{T}_{2})=-(\text{Id}-L_{0})^{-1}\int_{0}^{1}\tilde{k}(x,y)\left(\dot{T}_{1}(y)-\dot{T}_{2}(y)\right)dy,$

where
$\tilde{k}(x,y):=\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)f_{0}(y)$.
Since $\frac{d\rho}{dx}\in L^{\infty}$, we have
$\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\in
L^{\infty}([0,1]^{2})$. From inequality (7), we then have $f_{0}\in
L^{\infty}$ and so $\tilde{k}\in L^{\infty}([0,1]^{2})$. We finally have

$\|\widehat{R}(\dot{T}_{1})-\widehat{R}(\dot{T}_{2})\|_{2}\leq
l\|(\text{Id}-L_{0})^{-1}\|_{V\rightarrow
V}\|\tilde{k}\|_{L^{2}([0,1]^{2})}\cdot\|\dot{T}_{1}-\dot{T}_{2}\|_{2}.$

###  6.3 A formula for the linear response of the dominant eigenvalues and
continuity with respect to map perturbations

We are also able to express the linear response of the dominant eigenvalues as
a function of the perturbing map $\dot{T}$. Define

$H(y)=-\bar{e}(y)\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\hat{e}(x)dx.$

######  Proposition 6.6.

Let $L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow
L^{2}([0,1],{\mathbb{C}})$, $\delta\in[0,\bar{\delta})$, be integral operators
generated by the kernels $k_{\delta}$ as in (36), assume that $d\rho/dx$ is
Lipschitz and $\delta\mapsto T_{\delta}$ is $C^{1}$. Let $\lambda_{\delta}$ be
an eigenvalue of $L_{\delta}$ with second largest magnitude strictly inside
the unit disk. Suppose that $L_{0}$ satisfies $(A1)$ of Theorem 2.2 and
$\lambda_{0}$ is geometrically simple. Then

$\frac{d\lambda_{\delta}}{d\delta}\bigg{|}_{\delta=0}=\langle
H,\dot{T}\rangle_{L^{2}([0,1],\mathbb{C})},$ (42)

where $e$ is the eigenvector of $L_{0}$ associated to the eigenvalue
$\lambda_{0}$ and $\hat{e}$ is the eigenvector of $L_{0}^{*}$ associated to
the eigenvalue $\lambda_{0}$.

###### Proof.

Since $k_{\delta}\in L^{2}([0,1]^{2},{\mathbb{R}})$,
$L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow L^{2}([0,1],{\mathbb{C}})$ is
compact. From Lemma 6.1 we have that $k_{\delta}$ is a stochastic kernel and
so $L_{\delta}$ preserves the integral (i.e. it satisfies (3)). By Proposition
6.2 the kernel $k_{\delta}$ is in the form (9) and the map $\delta\mapsto
k_{\delta}$ is $C^{1}$. By Lemma 3.4 we see that $\delta\mapsto L_{\delta}$ is
$C^{1}$, where the derivative operator $\dot{L}$ is the integral operator with
the kernel $\dot{k}$. Using the assumption that $L_{0}$ is mixing and
$\lambda_{0}$ is geometrically simple, we apply Proposition 2.6 to obtain
$\frac{d\lambda_{\delta}}{d\delta}\big{|}_{\delta=0}=\langle\hat{e},\dot{L}e\rangle_{L^{2}([0,1],\mathbb{C})}$.
Finally, we compute

$\displaystyle\langle\hat{e},\dot{L}e\rangle_{L^{2}([0,1],\mathbb{C})}$
$\displaystyle=\int_{0}^{1}\hat{e}(x)\overline{\int_{0}^{1}\dot{k}(x,y)e(y)dy}dx$
$\displaystyle=\int_{0}^{1}\int_{0}^{1}\hat{e}(x)\dot{k}(x,y)\bar{e}(y)dxdy$
$\displaystyle=-\int_{0}^{1}\bar{e}(y)\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\hat{e}(x)dx\
\dot{T}(y)dy$ $\displaystyle=\langle
H,\dot{T}\rangle_{L^{2}([0,1],\mathbb{C})}.$

From (42), the linear response of the dominant eigenvalues is continuous with
respect to map perturbations.

######  Lemma 6.7.

The eigenvalue response function $\check{R}:L^{2}\to\mathbb{C}$ given by
$\check{R}(\dot{T})=\langle H,\dot{T}\rangle$ is continuous.

###### Proof.

This follows from Cauchy-Schwarz and the fact that $H\in
L^{2}([0,1],{\mathbb{C}})$; the latter claim follows from the fact that
$\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\in
L^{\infty}([0,1]^{2},{\mathbb{R}})$ (see proof of Lemma 6.5) and that
$e,\hat{e}\in L^{\infty}([0,1],{\mathbb{C}})$ (which follows from (7) and the
fact that $k_{0}\in L^{\infty}([0,1]^{2},{\mathbb{R}})$, see Lemma 6.1).

##  7 Optimal linear response for map perturbations

In this section we derive formulae for the map perturbations that maximise our
two types of linear response. We begin by formalising the set of allowable map
perturbations then state the formulae.

###  7.1 The feasible set of perturbations

Before we formulate the optimization problem, we note that in this setting, we
require some restriction on the space of allowable perturbations to $T_{0}$ if
we are to interpret $T_{0}+\delta\dot{T}$ as a map of the unit interval for
some $\delta$ strictly greater than 0 (a non-infinitesimal perturbation). With
this in mind, let $\ell>0$ and $\widetilde{F}_{\ell}:=\\{x\in[0,1]:\ell\leq
T_{0}(x)\leq 1-\ell\\}$; it will turn out that we obtain for free that
$\dot{T}\in L^{\infty}$. Note that in principle, $\ell>0$ can be taken as
small as one likes, and indeed if one wishes to consider only infinitesimal
perturbations $\dot{T}$ then one may set
$\widetilde{F}_{\ell}=\widetilde{F}_{0}=[0,1]$. Of course if $T:S^{1}\to
S^{1}$ then may may use $\widetilde{F}_{\ell}=\widetilde{F}_{0}=[0,1]$ even
for non-infinitesimal perturbations. Recalling that in Proposition 6.2 we are
considering $L^{2}$ perturbations $\dot{T}$ of the map $T_{0}$, we define

$S_{T_{0},\ell}:=\\{T\in
L^{2}:\text{supp}(T)\subseteq\widetilde{F}_{\ell}\\}.$ (43)

######  Lemma 7.1.

$S_{T_{0},\ell}$ is a closed subspace of $L^{2}$.

###### Proof.

It is clear that $S_{T_{0},\ell}$ is a subspace. To show it is closed, let
$\\{f_{n}\\}\subset S_{T_{0},\ell}$ and suppose that
$f_{n}\rightarrow_{L^{2}}f\in L^{2}$. Further, suppose that
$\widetilde{F}_{\ell}$ is not $[0,1]$ up to measure zero; otherwise
$S_{T_{0},\ell}=L^{2}$, which is closed. Then, we have

$\|f_{n}-f\|_{2}^{2}=\int_{\widetilde{F}_{\ell}}(f_{n}(x)-f(x))^{2}dx+\int_{\widetilde{F}_{\ell}^{c}}f(x)^{2}\
dx\rightarrow 0.$

If $\int_{\widetilde{F}_{\ell}^{c}}f(x)^{2}dx>0$, we obtain a contradiction
since $\int_{\widetilde{F}_{\ell}}(f_{n}(x)-f(x))^{2}dx\geq 0$; thus,
$\int_{\widetilde{F}_{\ell}^{c}}f(x)^{2}dx=0$ and so $f=0$ a.e. on
$\widetilde{F}_{\ell}^{c}$. Hence, $S_{T_{0},\ell}$ is closed.

For the remainder of this section, the set of allowable perturbations that we
consider is

$P_{\ell}:=S_{T_{0},\ell}\cap B_{1},$ (44)

where $B_{1}$ is the unit ball in $L^{2}$. Since $S_{T_{0},\ell}$ is a closed
subspace of $L^{2}$, it is itself a Hilbert space and so $P_{\ell}$ is
strictly convex. The following lemma concerns the existence of a perturbation
$\dot{T}$ for which our objectives will be nonzero; that is, our objective
$\mathcal{J}$ is not uniformly vanishing. Denote
$\mathcal{P}(x,y):=P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)(x)$
and let

$\mathcal{J}(\dot{T}):=\int_{\Xi(\widetilde{F}_{\ell})}\int_{0}^{1}\mathcal{P}(x,y)\dot{T}(y)\mathcal{E}(x,y)\
dx\ dy$

be our objective. In our first specific objective (optimising response of
expectations) we will insert
$\mathcal{E}(x,y)=((\text{Id}-L_{0}^{*})^{-1}c)(x)f_{0}(y)$ and in our second
specific objective (optimising mixing) we will insert
$\mathcal{E}(x,y)=E(x,y)$ from (18).

######  Lemma 7.2.

Assume that there is $F^{\prime}\subset\widetilde{F}_{\ell}$ such that
$m(F^{\prime})>0$ and
$\mathcal{E}(\cdot,y)\notin{\mathrm{s}pan}\\{\mathcal{P}(\cdot,y)\\}^{\perp}$
for all $y\in F^{\prime}$. Then there is a $\dot{T}\in P_{\ell}$ such that
$\mathcal{J}(\dot{T})>0$.

###### Proof.

Because

$\mathcal{J}(\dot{T})=\int_{\Xi(\widetilde{F}_{\ell})}\dot{T}(y)\left(\int_{0}^{1}\mathcal{P}(x,y)\mathcal{E}(x,y)\
dx\right)\ dy,$

we may set $\dot{T}(y)=\int_{0}^{1}\mathcal{P}(x,y)\mathcal{E}(x,y)\ dx$ for
$y\in F^{\prime}$ and $\dot{T}(y)=0$ otherwise to obtain
$\mathcal{J}(\dot{T})>0$. Trivial scaling yields $\dot{T}\in B_{1}$.

We expect the hypotheses of Lemma 7.2 to be satisfied “generically”.

###  7.2 Explicit formula for the optimal map perturbation that maximally
increases the expectation of an observable

In this section we consider the problem of finding the optimal map
perturbation that maximizes the expectation of some observable $c\in L^{2}$.
We first present a result that ensures a unique solution exists and then
derive an explicit expression for the optimal perturbation.

We begin by noting that $\widehat{R}(\dot{T})\in V$; this follows from the
fact that
$\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)f_{0}(y)\in
V_{\ker}$ (since $\dot{k}\in V_{\ker}$, see Proposition 6.3) and therefore
$\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)f_{0}(y)g(y)dy\in
V$ for $g\in L^{2}$ (see Lemma 3.2). Hence, we only need to consider $c\in$
span$\\{f_{0}\\}^{\perp}$ (see the discussion at the end of Section 4.2).

######  Proposition 7.3.

Let $c\in$ span$\\{f_{0}\\}^{\perp}$ and $P_{\ell}$ be the set in (44). Assume
that the function
$\mathcal{J}(\dot{T}):=\big{\langle}c,\widehat{R}(\dot{T})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}$
is not uniformly vanishing on $P_{\ell}$. Then the optimisation problem

$\big{\langle}c,\widehat{R}(\dot{T})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}=\max_{\dot{h}\in
P_{\ell}}\big{\langle}c,\widehat{R}(\dot{h})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})},$
(45)

where $\widehat{R}$ is as in (41), has a unique solution $\dot{T}\in L^{2}$.

###### Proof.

Let $\mathcal{H}=L^{2}$, $P=P_{\ell}$ and $\mathcal{J}(\dot{h})=\langle
c,\widehat{R}(\dot{h})\rangle_{L^{2}([0,1],{\mathbb{R}})}$. Using Lemma 7.1 we
note that $P_{\ell}$ is closed, as well as bounded, strictly convex and that
it contains the zero element of $\mathcal{H}$. From Lemma 6.5, it follows that
$\langle c,\widehat{R}(\dot{h})\rangle_{L^{2}([0,1],{\mathbb{R}})}$ is
continuous as a function of $\dot{h}$; note that it is also linear in
$\dot{h}$. By hypothesis, $\mathcal{J}$ is not uniformly vanishing on
$P_{\ell}$. We can therefore apply Propositions 4.1 and 4.3 to conclude that
(45) has a unique solution.

Before we present the explicit formula for the optimal solution, we will
reformulate the optimization problem (45) to simplify the analysis. We first
note that since the objective function in (45) is linear in $\dot{T}$, the
maximum will occur on $S_{T_{0},\ell}\cap\partial B_{1}$. Combining this with
the fact that we only need $c\in$ span$\\{f_{0}\\}^{\perp}$, we consider the
following reformulation of (45):

######  Problem C.

Given $\ell\geq 0$ and $c\in$ span$\\{f_{0}\\}^{\perp}$ solve

$\displaystyle\min_{\dot{T}\in S_{T_{0},\ell}}$
$\displaystyle-\big{\langle}c,\widehat{R}(\dot{T})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}$
(46) subject to $\displaystyle\|\dot{T}\|^{2}_{2}-1=0.$ (47)

######  Theorem 7.4.

Suppose the transfer operator $L_{0}$ associated with the system
$(T_{0},\rho)$ has a kernel $k_{0}$ as in (36), which satisfies $(A1)$ of
Theorem 2.2, and there is a $F^{\prime}\subset\widetilde{F}_{\ell}$ such that
$m(F^{\prime})>0$, and $f_{0}(y)>0$ and
$(\text{Id}-L_{0}^{*})^{-1}c\notin{\mathrm{s}pan}\\{\mathcal{P}(\cdot,y)\\}^{\perp}$
for all $y\in F^{\prime}$. Let $\mathcal{G}:L^{2}\rightarrow L^{2}$ be defined
as

$\mathcal{G}f(y):=\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)f(x)dx.$
(48)

Then, the unique solution to Problem C is

$\dot{T}(y)=\begin{cases}-f_{0}(y)\mathcal{G}((\text{Id}-L_{0}^{*})^{-1}c)(y)/\|f_{0}\mathcal{G}((\text{Id}-L_{0}^{*})^{-1}c)\mathbf{1}_{\widetilde{F}_{\ell}}\|_{2}&y\in\widetilde{F}_{\ell},\\\
0&\text{otherwise}.\end{cases}$ (49)

Furthermore, $\dot{T}\in L^{\infty}$.

###### Proof.

See Appendix D.

###  7.3 Explicit formula for the optimal map perturbation that maximally
increases the mixing rate

In this section we set up the optimisation problem for mixing enhancement and
derive a formula for the optimal map perturbation. We remark that related
spectral approaches to mixing enhancement for continuous-time flows were
developed in [18, 16].

Recall that to enhance mixing in Section 5.2, we perturbed $k_{0}$ so that the
logarithm of the real part of the second eigenvalue decreases. From Lemma 4.4,
we have

$\frac{d}{d\delta}\Re(\log\lambda_{\delta})\bigg{|}_{\delta=0}=\frac{\langle\dot{k},E\rangle_{L^{2}([0,1]^{2},{\mathbb{R}})}}{|\lambda_{0}|^{2}},$
(50)

where $\lambda_{\delta}$ denotes the second largest eigenvalue in magnitude
(assumed to be simple) of the integral operator $L_{\delta}$ with the kernel
$k_{\delta}=k_{0}+\delta\cdot\dot{k}+o(\delta)$, where $\delta\mapsto
k_{\delta}$ is $C^{1}$ at $\delta=0$. Since we want to perturb $T_{0}$ by
$\dot{T}$, we reformulate the above inner product. Define

$\widehat{E}(y)=-\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)E(x,y)dx,$
(51)

where $E(x,y)$ is as in (18).

######  Proposition 7.5.

Let $L_{\delta}:L^{2}([0,1],{\mathbb{C}})\rightarrow
L^{2}([0,1],{\mathbb{C}})$, $\delta\in[0,\bar{\delta})$, be integral operators
generated by the kernels $k_{\delta}$ as in (36), assume that $d\rho/dx$ is
Lipschitz and $\delta\mapsto T_{\delta}$ is $C^{1}$. Let $\lambda_{\delta}$ be
an eigenvalue of $L_{\delta}$ with second largest magnitude strictly inside
the unit disk. Suppose that $L_{0}$ satisfies $(A1)$ of Theorem 2.2 and
$\lambda_{0}$ is geometrically simple. Let $e$ and $\hat{e}$ be the
eigenvectors of $L_{0}$ and $L_{0}^{*}$, respectively, corresponding to the
eigenvalue $\lambda_{0}$. Then $\widehat{E}\in L^{\infty}([0,1],{\mathbb{R}})$
and

$\big{\langle}\dot{k},E\big{\rangle}_{L^{2}([0,1]^{2},{\mathbb{R}})}=\big{\langle}\dot{T},\widehat{E}\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}.$

###### Proof.

We first show that $\widehat{E}\in L^{\infty}([0,1],{\mathbb{R}})$. We can
write

$\displaystyle-\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)E(x,y)dx$
$\displaystyle=-\sum_{i=1}^{4}\beta_{i}h_{i}(y)\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)g_{i}(x)dx$
$\displaystyle=-\sum_{i=1}^{4}\beta_{i}h_{i}(y)(\mathcal{G}g_{i})(y),$

where $\beta_{1}=\beta_{2}=\Re(\lambda_{0})$,
$\beta_{3}=-\beta_{4}=\Im(\lambda_{0}),g_{1}=g_{4}=\Re(\hat{e}),g_{2}=g_{3}=\Im(\hat{e})$,
$h_{1}=h_{3}=\Re(e),h_{2}=h_{4}=\Im(e)$. From the proof of Theorem 7.4, we
have $\mathcal{G}g_{i}\in L^{\infty}([0,1],{\mathbb{R}})$. Also, from Lemma
6.1, we have that $k_{0}\in L^{\infty}([0,1]^{2})$ and therefore $h_{i}\in
L^{\infty}([0,1],{\mathbb{R}})$; thus, $\widehat{E}\in
L^{\infty}([0,1],{\mathbb{R}})$.

Finally, we compute

$\displaystyle\langle\dot{k},E\rangle_{L^{2}([0,1]^{2},{\mathbb{R}})}$
$\displaystyle=$ $\displaystyle\int_{0}^{1}\int_{0}^{1}\dot{k}(x,y)E(x,y)dxdy$
$\displaystyle=$
$\displaystyle-\int_{0}^{1}\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)\dot{T}(y)E(x,y)dxdy$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\dot{T}(y)\widehat{E}(y)dy=\big{\langle}\dot{T},\widehat{E}\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}.$

From equation (50), in order to maximally increase the spectral gap, by
Proposition 7.5, we should choose the map perturbation $\dot{T}$ to minimise
$\langle\dot{T},\widehat{E}\rangle$. We first show this optimisation problem
has a unique solution.

######  Proposition 7.6.

Let $P_{\ell}$ be the set in (44) and assume that
$\mathcal{J}(\dot{T})=\langle\dot{T},\widehat{E}\rangle$ does not uniformly
vanish on $P_{\ell}$. Then, the problem of finding $\dot{T}\in P_{\ell}$ such
that

$\big{\langle}\dot{T},\widehat{E}\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}=\min_{\dot{h}\in
P_{\ell}}\big{\langle}\dot{h},\widehat{E}\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}$
(52)

has a unique solution.

###### Proof.

Note that $P_{\ell}$ is closed (by Lemma 7.1), bounded, strictly convex and
contains the zero element of $L^{2}$. Now, since
$\mathcal{J}(\dot{h}):=\langle\dot{h},\widehat{E}\rangle_{L^{2}([0,1],{\mathbb{R}})}$
is linear and continuous and by hypothesis does not vanish everywhere on
$P_{\ell}$, we may apply Propositions 4.1 and 4.3 to obtain the result.

Since the objective function in (52) is linear, all optima will lie in
$S_{T_{0},\ell}\cap\partial B_{1}$. Hence, we equivalently consider the
following optimization problem:

######  Problem D.

Given $\ell\geq 0$, solve

$\displaystyle\min_{\dot{T}\in S_{T_{0},\ell}}$
$\displaystyle\big{\langle}\dot{T},\widehat{E}\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})}$
(53) such that $\displaystyle\|\dot{T}\|_{2}^{2}-1=0.$ (54)

We now state a formula for the unique optimum.

######  Theorem 7.7.

Let $(T_{0},\rho)$ be a deterministic system with additive noise satisfying
$(T1)$ and $(T2)$. Suppose the associated transfer operator
$L_{0}:L^{2}([0,1],{\mathbb{C}})\rightarrow L^{2}([0,1],{\mathbb{C}})$, with
the kernel $k_{0}$ as in (36), satisfies $(A1)$ of Theorem 2.2, and that there
is a $F^{\prime}\subset\tilde{F}_{\ell}$ with $m(F^{\prime})>0$ and
$E(\cdot,y)\notin{\mathrm{s}pan}\\{\mathcal{P}(\cdot,y)\\}^{\perp}$ for all
$y\in F^{\prime}$. Suppose $\lambda_{0}$ is geometrically simple. Then, the
unique solution to the optimization problem D is

$\dot{T}(y)=\begin{cases}\frac{1}{\alpha}\int_{0}^{1}\left(P_{\pi}\left(\tau_{-T_{0}(y)}\frac{d\rho}{dx}\right)\right)(x)E(x,y)dx&y\in\widetilde{F}_{\ell},\\\
0&\text{otherwise},\end{cases}$ (55)

where $E(x,y)$ is as in (18) and $\alpha>0$ is selected so that
$\|\dot{T}\|_{2}=1$. Furthermore, $\dot{T}\in L^{\infty}$.

###### Proof.

See Appendix E.

######  Corollary 7.8.

If $\lambda_{0}$ is real, then

$\dot{T}(y)=\begin{cases}\text{sgn}(\lambda_{0})\frac{e(y)(\mathcal{G}\hat{e})(y)}{\|e\mathcal{G}\hat{e}\mathbf{1}_{\widetilde{F}_{\ell}}\|_{2}}&y\in\widetilde{F}_{\ell},\\\
0&\text{otherwise},\end{cases}$

where $\mathcal{G}$ is the operator in (48). Furthermore, if there exists an
$\ell>0$ such that $\ell\leq T_{0}(x)\leq 1-\ell$ for $x\in[0,1]$, then

$\dot{T}=\text{sgn}(\lambda_{0})\frac{e\cdot\mathcal{G}\hat{e}}{\|e\cdot\mathcal{G}\hat{e}\|_{2}}.$
(56)

###### Proof.

Since $e,\hat{e}$ and $\lambda_{0}$ are real, we have
$E(x,y)=\hat{e}(x)e(y)\lambda_{0}$ and the expression for $\dot{T}$ follows
from (55). Finally, if $\ell\leq T_{0}(x)\leq 1-\ell$, then
$\widetilde{F}_{\ell}=[0,1]$ and we have (56).

##  8 Applications and numerical experiments

In this section we will consider two stochastically perturbed deterministic
systems, namely the Pomeau-Manneville map and a weakly mixing interval
exchange map. For each of these maps we numerically estimate:

  1. 1.

The unique kernel perturbation that maximises the change in expectation of a
prescribed observation function (see Problem A). An expression for this
optimal kernel is given by (28).

  2. 2.

The unique kernel perturbation that maximally increases the mixing rate (see
Problem B). An expression for this optimal kernel is given by (31) and (32).

  3. 3.

The unique map perturbation that maximises the change in expectation of a
prescribed observation function (see Problem C). An expression for this
optimal map perturbation is given by (49).

  4. 4.

The unique map perturbation that maximally increases the mixing rate (see
Problem D). An expression for this optimal map perturbation is given by (55)
and (56).

The numerics will be explained as we proceed through these four optimisation
problems. We refer the reader to [1] for additional details on the
implementation and related experiments.

###  8.1 Pomeau-Manneville map

We consider the Pomeau-Manneville map [36]

$T_{0}(x)=\left\\{\begin{array}[]{ll}x(1+(2x)^{\alpha}),&\hbox{$x\in[0,1/2)$;}\\\
2x-1,&\hbox{$x\in[1/2,1]$}\end{array}\right.,$ (57)

with parameter value $\alpha=1/2$. For this parameter choice it is known that
the map $T_{0}$ admits a unique absolutely continuous invariant probability
measure, but only algebraic decay of correlations [36]. With the addition of
noise as per (33), the transfer operator defined by (35) and (36) for
$\delta=0$ becomes compact as an operator on $L^{2}$. In our numerical
experiments we will use the smooth noise kernel
$\rho_{\epsilon}:[-\epsilon,\epsilon]\to\mathbb{R}$, defined by
$\rho_{\epsilon}(x)=N(\epsilon)\exp(-\epsilon^{2}/(\epsilon^{2}-x^{2}))$,
where $N(\epsilon)$ is a normalisation factor ensuring
$\int\rho_{\epsilon}(x)\ dx=1$.

We now begin to set up our numerical procedure for estimating $L_{0}$, which
is a standard application of Ulam’s method [47]. Let
$B_{n}=\\{I_{1},\dots,I_{n}\\}$ denote an equipartition of $[0,1]$ into $n$
subintervals, and set $\mathcal{B}_{n}=$
span$\\{\mathbf{1}_{I_{1}},\dots,\mathbf{1}_{I_{n}}\\}$. We define the (Ulam)
projection $\pi_{n}:L^{2}([0,1])\rightarrow\mathcal{B}_{n}$ by
$\pi_{n}(g)=\sum_{i=1}^{n}\left(\frac{1}{m(I_{i})}\int_{I_{i}}g(x)dx\right)\mathbf{1}_{I_{i}}$.
The finite-rank transfer operator
$L_{n}:=\pi_{n}L_{0}:L^{2}([0,1])\rightarrow\mathcal{B}_{n}$ can be computed
numerically. We use MATLAB’s built-in functions `integral.m` and `integral2.m`
to perform the $\rho$-convolution (using an explicit form of
$\rho_{\epsilon}$) and the Ulam projections, respectively. Figure 1 displays
the nonzero entries in the column-stochastic matrix corresponding to $L_{n}$
for $\epsilon=0.1$.

Figure 1: Transition matrix $L_{n}$ for the system (33) generated by the
Pomeau-Manneville map $T_{0}$ (57) using $n=500$ subintervals of equal length.
The matrix entries are located according to the subinterval positions in the
domain $[0,1]$, so that the image appears as a “blurred” version of the graph
of $T_{0}$. The additive noise in (33) is drawn according to $\rho_{\epsilon}$
with $\epsilon=1/10$.

Approximations to the invariant probability densities for our stochastic
dynamics are displayed in Figure 2 (left) for large and small noise supports.

Figure 2: Approximate invariant densities (left) and eigenfunctions
corresponding to the 2nd largest eigenvalue of $L_{0}$ (right) for the system
(33) with $T_{0}$ given by the Pomeau-Manneville map (57). The additive noise
in (33) is drawn according to $\rho_{\epsilon}$ with $\epsilon$ taking the
values 1/10 (blue) and $\sqrt{6}/100$ (red). The Ulam matrix $L_{n}$ is
constructed with 500 subintervals.

A lower level of noise permits greater concentration of invariant probability
mass near the fixed point $x=0$ of the map $T_{0}$. Also shown in Figure 2
(right) are the estimated eigenfunctions corresponding to the second-largest
eigenvalue of $L_{n}$. The signs of these second eigenfunctions split the
interval $[0,1]$ into left and right hand portions, broadly indicating that
the slow mixing is due to positive mass near $x=0$ and negative mass away from
$x=0$ [9]; see [17] for further discussion of this point in the Pomeau-
Manneville setting.

####  8.1.1 Kernel perturbations

In the framework of Problems A and B we use the (arbitrarily chosen)
monotonically increasing observation function $c(x)=-\cos(x)$. In order to
estimate $\dot{k}$ as in (28) we use the code from Algorithm 3 [1]; the inputs
are the Ulam matrix $L_{n}$ and $c_{n}$ (obtained as $\pi_{n}(c)$).
Equivalently, directly using (28) one may substitute $f_{n}$ (obtained as the
leading eigenvector of $L_{n}$) for $f$, $L_{n}$ for $L$, $c_{n}$ as above for
$c$, and solve $(Id-L_{n}^{*})^{-1}c_{n}$ (obtained as a vector
$y\in\mathbb{R}^{n}$ by numerically solving the linear system $(Id-
L_{n}^{*})y=c_{n},f_{n}^{\top}y=0$). Figure 3 shows the optimal kernel
perturbations $\dot{k}_{n}$ for $n=500$.

Figure 3: Optimal kernel perturbations for the Pomeau-Manneville map to
maximise the change in expectation of $c(x)=-\cos(x)$, based on an Ulam
approximation of (28) with $n=500$ subintervals. Left: $\epsilon=1/10$, Right:
$\epsilon=\sqrt{6}/100$.

Because $c$ is an increasing function, intuitively one might expect the kernel
perturbation to try to shift mass in the invariant density from left to right.
Broadly speaking, this is what one sees in the high-noise case in Figure 3
(left): vertical strips typically have red above blue, corresponding to a
shift of mass to the right in $[0,1]$. The main exception to this is around
the $y$-axis value of 1/2, where red is strongly below blue along vertical
strips. This is because at the next iteration, these red regions will be
mapped near $x=1$ and achieve the highest value of $c$, while the blue regions
will be mapped near to $x=0$ with the least value of $c$. In the low-noise
case of Figure 3 (right), we see a similar solution with higher spatial
frequencies, and strong perturbations near the critical values of $x=0$ and
$T_{0}(x)=1/2$.

To investigate the optimal kernel perturbation to maximally increase the rate
of mixing in the stochastic system, we use the expression $\dot{k}$ in (31). A
natural approximate version (31) requires estimates of the left and right
eigenfunctions of $L_{0}$ corresponding to the second largest eigenvalue
$\lambda_{2}$; these are obtained directly as eigenvectors of $L_{n}$. Figure
4 shows the resulting optimal kernel perturbations, computed using the code
from Algorithm 4 [1] with input $L_{n}$.

Figure 4: Optimal kernel perturbation for the Pomeau-Manneville map to
maximally increase the mixing rate, computed with $n=500$ subintervals. Left:
$\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

Because the fixed point at $x=0$ is responsible for the slow algebraic decay
of correlations for the deterministic dynamics of $T_{0}$, the fixed point
will also play a dominant role in the mixing rate of the stochastic system for
low to moderate levels of noise. Indeed, Figure 4 shows that the optimal
perturbation concentrates its effort in a neighbourhood of the fixed point,
and pushes mass away from the fixed point as much as possible. This is
particularly extreme in the low noise case of Figure 4 (right) with the
perturbation almost exclusively concentrated in a small neighbourhood of
$x=0$.

####  8.1.2 Map perturbations

We now turn to the problem of finding the unique map perturbation $\dot{T}$
that maximises the change in expectation of the observation $c(x)=-\cos(x)$
(see Problem C for a precise formulation) and maximises the speed of mixing
(see Problem D). We use the natural Ulam discretisation of the
expression555Note that since $T_{0}^{-1}(\\{0,1\\})$ is a finite set, we may
take $\ell>0$ as small as we like. In the computations we set $\ell=0$, so
that $\widetilde{F}_{\ell}=[0,1]$ mod $m$. (49). The objects $f_{n}$ and $(Id-
L_{n}^{*})^{-1}c_{n}$ are computed exactly as before in Section 8.1.1. The
action of the operator $\mathcal{G}$ in (49) is computed using MATLAB’s built-
in function `integral.m` using an explicit form of $d\rho_{\epsilon}/dx$ for
$d\rho/dx$ in (49).

Figure 5 (left) shows the optimal $\dot{T}$ for the two noise amplitudes
$\epsilon=1/10$ and $\epsilon=\sqrt{6}/100$.

Figure 5: Left: Optimal map perturbation $\dot{T}$ for the Pomeau-Manneville
map to maximise the change in expectation of $c(x)=-\cos(x)$, computed using
(49) with $n=500$. Right: Illustration of $T_{0}+\dot{T}/100$.

Note that for the noise amplitude $\epsilon=0.1$ (blue curve in Figure 5) the
map perturbation $\dot{T}$ is mostly positive, corresponding to moving
probability mass to the right, as expected because we are maximising the
change in expectation of an increasing observation function $c$. The blue
curve is most negative in neighbourhoods of the two preimages of $x=1/2$,
corresponding to moving probability mass to the left. The reason for this is
identical to the discussion of the “blue above red” effect in Figure 3, namely
moving mass to the left creates a very large increase in the objective
function value at the next iterate. This “look ahead” effect is even more
pronounced in the low noise case (red curve of Figure 5), where $\dot{T}$ is
mostly positive, but has deep negative perturbations at multiple preimages of
$x=1/2$ reaching further into the past.

Figure 5 (right) illustrates the Pomeau-Manneville map (black) with perturbed
maps $T_{0}+\dot{T}/100$. We have chosen a scale factor of 1/100 for
visualisation purposes; one should keep in mind we have optimised for an
infinitesimal change in the map. Figure 6 shows the kernel derivatives
$\dot{k}$ corresponding to the optimal map derivatives $\dot{T}$ for the two
noise levels.

Figure 6: Kernel perturbations corresponding to the optimal map perturbations
in Figure 5. Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

These kernel derivatives have a restricted form because they arise purely from
a derivative in the map. One may compare Figure 6 with Figure 3 and note that
the kernel derivative in Figure 6 (left) attempts to follow the general
structure of the kernel derivative in Figure 3 (left), while obeying its
structural restrictions arising from the less flexible map perturbation.
Broadly speaking, in Figure 6 (left), red lies above blue (mass is shifted to
the right). Exceptions are near $y=1/2$ because at the next iteration these
red points will land near $x=1$, achieving very high objective value, while
the blue region will get mapped to near $x=0$, encountering the lowest value
of $c$. Note that the perturbation decreases from a peak to very close to zero
near $x=0$. This is because in a small neighbourhood of $x=0$ there is already
some stochastic perturbation away from $x=0$ “for free” due to the reflecting
boundary conditions. Thus, the map perturbation $\dot{T}$ does not need to
invest energy in large perturbations very close to $x=0$.

The map perturbation that maximally increases the rate of mixing is a
particularly interesting question. Our computations use the natural Ulam
discretisation of (56). The computations follow as in Section 8.1.1 with the
action of $\mathcal{G}$ computed as above. Figure 7 (left) shows the optimal
$\dot{T}$ for the two noise amplitudes $\epsilon=1/10$ and
$\epsilon=\sqrt{6}/100$.

Figure 7: Left: Optimal map perturbation $\dot{T}$ for the Pomeau-Manneville
map to maximise the change in the mixing rate, computed using (56) with
$n=500$. Right: Illustration of $T_{0}+\dot{T}/100$.

A sharp map perturbation away from $x=0$ is seen for both noise levels, with
the perturbation sharper for the lower noise case. In both cases, the
perturbations far from $x=0$ are weak (low magnitude values of $\dot{T}$).
This result corresponds well with the results seen for the optimal kernel
perturbations in Figure 4, where mass was primarily moved away from $x=0$. As
in the optimal solution shown in Figure 5 (left), the optimal perturbation in
Figure 7 decreases from a sharp peak down to zero near $x=0$. This is again
because in a small neighbourhood of $x=0$ the system experiences “free”
stochastic perturbations away from $x=0$ due to the reflecting boundary
conditions, and thus the map perturbation $\dot{T}$ need not need invest
energy in large perturbations very close to $x=0$. Figure 7 (right)
illustrates the Pomeau-Manneville map (black) with perturbed maps
$T_{0}+\dot{T}/100$, where again the factor $1/100$ is just for illustrative
purposes. When inspecting the kernel derivatives $\dot{k}$ corresponding to
the optimal map perturbations $\dot{T}$ in Figure 8, we see similar behaviour
to those in Figure 7.

Figure 8: Kernel perturbations corresponding to the optimal map perturbations
in Figure 7. Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

###  8.2 Interval exchange map

In our second example, we consider a weak-mixing interval exchange map. This
is because of an existing literature in mixing optimisation for these classes
of maps with the addition of noise. Avila and Forni [3] prove that a typical
interval exchange is either weak mixing or an irrational rotation. We use a
specific weak-mixing [45] interval exchange map $T_{0}$ with interval
permutation $(1234)\mapsto(4321)$ and interval lengths given by the normalised
entries of the leading eigenvector of the matrix
$\left(\begin{array}[]{cccc}13&37&77&47\\\ 10&30&60&37\\\ 3&10&24&14\\\
4&10&19&12\end{array}\right)$; see equation (51) in [45]. We again form a
stochastic system using the same noise kernels as for the Pomeau-Manneville
map in Section 8.1. The mixing properties of this map have been studied in
[15]. Figure 9 shows the column-stochastic matrix corresponding to $L_{n}$ for
$n=500$ and $\epsilon=0.1$.

Figure 9: Transition matrix for the system (33) for $\delta=0$ and $T_{0}$
given by the interval exchange map above using $n=500$ subintervals. The
additive noise is drawn from the density $\rho_{\epsilon}$ with
$\epsilon=1/10$.

####  8.2.1 Kernel perturbations

In the framework of Problem A, we use the same observation function
$c(x)=-\cos(x)$ as in the Pomeau-Manneville case study, and estimate the
optimal kernel perturbation $\dot{k}$ that maximally increases the expectation
of $c$ in an identical fashion. In broad terms, one again sees that $\dot{k}$
attempts to shift invariant probability mass to the right in $[0,1]$. In
Figure 10 (left), in each smooth part of the support of $\dot{k}$, red is
“above” blue, meaning mass is pushed to the right.

Figure 10: Optimal kernel perturbation for the interval exchange map to
maximise the change in expectation of $c(x)=-\cos(x)$, computed with $n=500$
Ulam subintervals. Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

Clear exceptions to the “red above blue” scheme are seen as three sharp
horizontal lines. The $y$-coordinates of these three sharp horizontal lines
coincide with the three points of discontinuity in the domain of the interval
exchange at approximately $x=0.43,0.77,0.89$. Consider the sharp horizontal
“blue above red” line at $y\approx 0.43$. According to Figure 9, under the
action of the kernel $k_{0}$, mass in the vicinity of $x=0.6$ will be
transported near to $x=0.43$. The perturbation $\dot{k}$ shown in Figure 10
will then tend to push this mass to the left of $x=0.43$. Thus, on the next
iteration there will be a bias for mass to be mapped near to $x=1$ rather than
near $x=0.25$, achieving a much larger objective value at this iterate. A
similar reasoning applies to the “blue above red” horizontal lines at
$y\approx 0.77$ and $0.89$; the contrast is a little weaker because the
potential gain at the next iterate is also weaker. In the low noise case,
Figure 10 (right), displays similar behaviour to the higher noise case of
Figure 10 (left). With lower noise, the deterministic dynamics plays a greater
role and additional preimages are taken into account, leading to a more
oscillatory optimal $\dot{k}$.

To investigate the optimal kernel perturbation to maximally increase the rate
of mixing in the stochastic system (in the framework of Problem B) we use the
expression $\dot{k}$ in (31). The method of numerical approximation is
identical to that used for the Pomeau-Manneville map. Figure 11 shows the
signed distribution of mass that is responsible for the slowest real666In our
numerical experiments the largest magnitude real eigenvalue appears as the
sixth (resp. fourth) eigenvector of $L_{500}$ for $\epsilon=1/10$ (resp.
$\epsilon=\sqrt{6}/100$). Slightly larger complex eigenvalues are present, but
we do not investigate these in order to make the dynamic interpretation more
straightforward. exponential rate of decay in the stochastic system.

Figure 11: Approximate second eigenfunctions of the transfer operator $L_{0}$
of the system (33) with $T_{0}$ given by the interval exchange map above. The
additive noise in (33) is drawn from the density $\rho_{\epsilon}$ with
$\epsilon$ taking the values 1/10 (blue) and $\sqrt{6}/100$ (red).

This eigenfunction becomes more oscillatory as the level of noise decreases,
and as must be the case, the magnitude of the corresponding eigenvalue
increases from $\lambda\approx-0.7476$ ($\epsilon=1/10$) to
$\lambda\approx-0.9574$ ($\epsilon=\sqrt{6}/100$). Because the sign of these
eigenvalues is negative, one expects a pair of almost-2-cyclic sets [10],
consisting of three subintervals each, given by the positive and negative
supports of the eigenfunctions.

Figure 12 shows the approximate optimal kernel perturbations.

Figure 12: Optimal kernel perturbation for the interval exchange map to
maximally increase the mixing rate, computed with $n=500$ Ulam subintervals.
Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

In the high-noise situation of Figure 12 (left), the sharp horizontal changes
are present at preimages of the deterministic dynamics, as they were in to
Figure 10 (left). The importance of the break points to the overall mixing
rate is thus clearly borne out in the optimal $\dot{k}$; a precise
interpretation of the optimal $\dot{k}$ is not very straightforward. For the
low noise case (Figure 12 (right)) it appears that there is an alternating
shifting of mass left and right with alternating “red above blue” and “blue
above red”. This leads to greater mixing at smaller spatial scales than is
possible in a single iteration of the deterministic interval exchange. We
anticipate that decreasing the noise amplitude further will result in more
rapid alternation of “red above blue” and “blue above red”. As the diffusion
amplitude decreases, the efficient large-scale diffusive mixing is no longer
possible and so a transition is made to small-scale mixing, accessed by
increasing oscillation in the kernel.

####  8.2.2 Map perturbations

The computations in this section follow those of Section 8.1.2. Figure 13
(left) shows the optimal map perturbations $\dot{T}$ at two different noise
levels.

Figure 13: Left: Optimal map perturbation $\dot{T}$ for the interval exchange
map to maximise the change in expectation of $c(x)=-\cos(x)$, computed using
(49) with $n=500$. Right: Illustration of $T_{0}+\dot{T}/100$.

Figure 13 (right) illustrates $T_{0}+\dot{T}/100$ for the two different levels
of noise. The kernel perturbations generated by these optimal map
perturbations are displayed in Figure 14.

Figure 14: Kernel perturbations corresponding to the optimal map perturbations
in Figure 13. Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

If one compares the kernel perturbations in Figure 14 with those more flexible
kernel perturbations in Figure 10, one sees that the two sets of kernel
perturbations are broadly equivalent with one another in terms of the relative
positions of the positive and negative (red and blue) perturbations. Note that
the more restrictive kernel derivative in Figure 14 by construction cannot
replicate the sharp horizontal red-blue switches in Figure 10. It turns out
that the strongest of these red-blue switches, namely the one at $y\approx
0.43$ in Figure 10 (left) is approximated as best as is allowed by a map
perturbation, see Figure 14 (left), while the other two (weaker) horizontal
red/blue switches seen in 10 are ignored.

We now turn to optimal map perturbations for the mixing rate. The combined
effect of the “cutting and shuffling” of interval exchanges with diffusion on
mixing rates has been widely studied, e.g. [2, 46, 15, 34, 48], including
investigations of the impact of changing the diffusion or the interval
exchange on mixing. The very general type of formal map optimisation we
consider here has not been attempted before, and we hope that our novel
techniques will stimulate interesting new research questions and motivate more
sophisticated experiments in the field of mixing optimisation.

Under repeated iteration, the original interval exchange $T_{0}$ cuts and
shuffles the unit interval into an increasing number of smaller pieces,
assisting the small scale mixing of diffusion. Our results in Figure 15 (left)
show an oscillatory $\dot{T}$, with increasing oscillations as the noise
amplitude decreases.

Figure 15: Left: Optimal map perturbation $\dot{T}$ for the interval exchange
map to maximise the change in the mixing rate, computed using (56) with
$n=500$. Right: Illustration of $T_{0}+\dot{T}/100$.

This increased oscillation effect is also seen when comparing the left and
right panes of Figure 16. Thus, the optimisation attempts to include some
additional mixing by rapid local warping of the phase space. It is plausible
that this additional warping effect enhances mixing beyond the rigid shuffling
of the interval exchange. An illustration of $T_{0}+\dot{T}/100$ is given in
Figure 15.

Figure 16: Kernel perturbations corresponding to the optimal map perturbations
in Figure 15. Left: $\epsilon=1/10$, Right: $\epsilon=\sqrt{6}/100$.

We emphasise that the factor $1/100$ is only for visualisation purposes and
for smaller factors, the perturbed map would remain a piecewise homeomorphism
(modulo small overshoots at the boundaries, which are taken care of by the
reflecting boundary conditions on the noise).

##  9 Acknowledgments

FA is supported by a UNSW University Postgraduate Award. GF is partially
supported by an ARC Discovery Project. FA and GF thank the Department of
Mathematics at the University of Pisa for generous support and hospitality. SG
is partially supported by the research project `PRIN 2017S35EHN_004` “Regular
and stochastic behaviour in dynamical systems” of the Italian Ministry of
Education and Research.

##  Appendix A Proof of Theorem 5.4

First we need a technical lemma. We note that the statement of the lemma is
analogous to the continuity of $(\text{Id}-L_{0})^{-1}$, which was treated in
the proof of Theorem 2.2.

######  Lemma A.1.

Consider the closed subspace span$\\{f_{0}\\}^{\perp}\subset L^{2}$ equipped
with the $L^{2}$ norm. Then, the operator $(\text{Id}-L_{0}^{*})^{-1}:$
span$\\{f_{0}\\}^{\perp}\rightarrow$ span$\\{f_{0}\\}^{\perp}$ is bounded.

###### Proof.

We begin by finding the kernel and range of the operator
$\text{Id}-L_{0}^{*}$. Recall that $L_{0}(V)\subset V$ and that $L_{0}$
preserves a one-dimensional eigenspace span$\\{f_{0}\\}$, with eigenvalue $1$.
Thus, we have $\ker(\text{Id}-L_{0})=$ span$\\{f_{0}\\}$ and
ran$(\text{Id}-L_{0})\subset V$. Recalling that $L_{0}:V\rightarrow V$ is
compact and $f_{0}\not\in V$, we have by the Fredholm alternative (see [11],
VII.11) that for any $g\in V$, there exists a unique $h\in V$ such that
$g=(\text{Id}-L_{0})h$. Hence, ran($\text{Id}-L_{0})=V$. Since $V$ is closed,
the range of $\text{Id}-L_{0}$ is closed and so, by the Closed Range Theorem
(Theorem 5.13, IV-§5.2,[29]), we have
$\text{ran}((\text{Id}-L_{0})^{*})=\ker(\text{Id}-L_{0})^{\perp}=$
span$\\{f_{0}\\}^{\perp}$, which is a co-dimension $1$ space, and
$\ker((\text{Id}-L_{0})^{*})=$ ran($\text{Id}-L_{0})^{\perp}=V^{\perp}=$
span$\\{\mathbf{1}\\}^{\perp\perp}=$ span$\\{\mathbf{1}\\}$, where the last
equality follows from Corollary 1.41 in III-§1.8 [29] and the fact that
span$\\{\mathbf{1}\\}$ is a finite-dimensional closed subspace of $L^{2}$.

To prove that $(\text{Id}-L_{0}^{*})^{-1}:$
span$\\{f_{0}\\}^{\perp}\rightarrow$ span$\\{f_{0}\\}^{\perp}$ is bounded, we
will use the Inverse Mapping Theorem (Theorem III.11, [41]). Since the
integral operator $L_{0}^{*}$ has an $L^{2}$ kernel, by (6) and the triangle
inequality it follows that $\text{Id}-L_{0}^{*}$ is bounded. Also, from the
Fredholm alternative argument above,
$\text{Id}-L_{0}^{*}:{\mathrm{s}pan}\\{f_{0}\\}^{\perp}\to{\mathrm{s}pan}\\{f_{0}\\}^{\perp}$
is surjective. Thus, to apply the Inverse Mapping Theorem, we just need to
show that $\text{Id}-L_{0}^{*}$ is injective on span$\\{f_{0}\\}^{\perp}$. Let
$f_{1},f_{2}\in$ span$\\{f_{0}\\}^{\perp}$ be such that
$(\text{Id}-L_{0}^{*})f_{1}=(\text{Id}-L_{0}^{*})f_{2}$. Thus,
$f_{1}-f_{2}\in\ker(\text{Id}-L_{0}^{*})=$ span$\\{\mathbf{1}\\}$ and so
$f_{1}-f_{2}=\gamma\mathbf{1}$ for some $\gamma\in{\mathbb{R}}$. Since
$f_{1}-f_{2}\in$ span$\\{f_{0}\\}^{\perp}$, we have that
$0=\int(f_{1}(x)-f_{2}(x))f_{0}(x)dx=\gamma\int f_{0}(x)dx$ and so $\gamma=0$
(since $\int f_{0}(x)dx=1$), i.e. $f_{1}=f_{2}$; thus, $(\text{Id}-L_{0}^{*})$
is injective and the result follows.

###### Proof of Theorem 5.4.

We will use the method of Lagrange multipliers to derive the expression (28)
from the first-order necessary conditions for optimality and then show that
such a $\dot{k}$ satisfies the second-order sufficient conditions. To this
end, we consider the following Lagrangian function

$\mathcal{L}(\dot{k},\mu):=f(\dot{k})+\mu g(\dot{k}),$

where
$f(\dot{k}):=-\big{\langle}c,R(\dot{k})\big{\rangle}_{L^{2}([0,1],{\mathbb{R}})},$
$g(\dot{k}):=\|\dot{k}\|_{L^{2}([0,1]^{2})}^{2}-1$ and $\dot{k}\in
V_{\ker}\cap S_{k_{0},l}$.

Necessary conditions: We verify the conditions in Theorem 2, §7.7, [37]. We
want to find $\dot{k}$ and $\mu$ that satisfy the first-order necessary
conditions:

$\displaystyle g(\dot{k})=0$ $\displaystyle
D_{\dot{k}}\mathcal{L}(\dot{k},\mu)$ $\displaystyle\tilde{k}=0\text{ for all
}\tilde{k}\in V_{\ker}\cap S_{k_{0},l},$

where
$D_{\dot{k}}\mathcal{L}(\dot{k},\mu)\in\mathcal{B}(L^{2}([0,1]^{2}),{\mathbb{R}})$
is the Frechet derivative with respect to the variable $\dot{k}$. Since $f$ is
linear, we have $(D_{\dot{k}}f)\tilde{k}=f(\tilde{k})$. Also,
$(D_{\dot{k}}g)\tilde{k}=2\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}$
since

$\displaystyle\frac{|g(\dot{k}+\tilde{k})-g(\dot{k})-2\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}|}{\|\tilde{k}\|_{L^{2}([0,1]^{2})}}$
$\displaystyle=\frac{|\|\dot{k}+\tilde{k}\|_{L^{2}([0,1]^{2})}^{2}-\|\dot{k}\|_{L^{2}([0,1]^{2})}^{2}-2\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}|}{\|\tilde{k}\|_{L^{2}([0,1]^{2})}}$
$\displaystyle=\frac{|\langle\dot{k}+\tilde{k},\dot{k}+\tilde{k}\rangle_{L^{2}([0,1]^{2})}-\langle\dot{k},\dot{k}\rangle_{L^{2}([0,1]^{2})}-2\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}|}{\|\tilde{k}\|_{L^{2}([0,1]^{2})}}$
$\displaystyle=\frac{|\langle\tilde{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}|}{\|\tilde{k}\|_{L^{2}([0,1]^{2})}}=\|\tilde{k}\|_{L^{2}([0,1]^{2})}.$

Thus, for the necessary conditions of the Lagrange multiplier method to be
satisfied, we need that

$D_{\dot{k}}\mathcal{L}(\dot{k},\mu)\tilde{k}=(D_{\dot{k}}f)\tilde{k}+\mu(D_{\dot{k}}g)\tilde{k}=f(\tilde{k})+2\mu\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}=0$
(58)

for all $\tilde{k}\in V_{\ker}\cap S_{k_{0},l}$ and

$g(\dot{k})=0.$ (59)

Noting Lemma A.1 and the fact that $c\in$ span$\\{f_{0}\\}^{\perp}$, we have

$\displaystyle f(\tilde{k})$
$\displaystyle+2\mu\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}$ (60)
$\displaystyle=-\langle
c,R(\tilde{k})\rangle_{L^{2}([0,1],{\mathbb{R}})}+2\mu\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}$
$\displaystyle=-\bigg{\langle}c,(\text{Id}-L_{0})^{-1}\int\tilde{k}(x,y)f_{0}(y)dy\bigg{\rangle}_{L^{2}([0,1],{\mathbb{R}})}+2\mu\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}$
$\displaystyle=\int\int-((\text{Id}-L_{0}^{*})^{-1}c)(x)\tilde{k}(x,y)f_{0}(y)dydx+\int\int
2\mu\dot{k}(x,y)\tilde{k}(x,y)dydx$
$\displaystyle=\int\int\left[-((\text{Id}-L_{0}^{*})^{-1}c)(x)f_{0}(y)+2\mu\dot{k}(x,y)\right]\tilde{k}(x,y)dydx.$

We claim that

$\dot{k}(x,y)=\frac{1}{2\mu}\mathbf{1}_{F_{l}}(x,y)f_{0}(y)\left(((\text{Id}-L_{0}^{\ast})^{-1}c)(x)-\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\right)$

satisfies the necessary condition (58) and lies in $V_{\ker}\cap S_{k_{0},l}$.
Before we verify this, we show that

$M(x,y):=\mathbf{1}_{F_{l}}(x,y)f_{0}(y)\left(((\text{Id}-L_{0}^{\ast})^{-1}c)(x)-\hat{g}(y)\right),$

where
$\hat{g}(y):=\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz$,
is in $L^{2}([0,1]^{2})$. Since $f_{0},(\text{Id}-L_{0}^{\ast})^{-1}c\in
L^{2}$, we just need to show that $\mathbf{1}_{F_{l}}(x,y)f_{0}(y)\hat{g}(y)$
is in $L^{2}([0,1]^{2})$. First, we note that

$\displaystyle|\hat{g}(y)|$
$\displaystyle\leq\frac{1}{m(F_{l}^{y})}\int\big{|}\mathbf{1}_{F_{l}^{y}}(z)((\text{Id}-L_{0}^{\ast})^{-1}c)(z)\big{|}dz$
$\displaystyle\leq\frac{1}{m(F_{l}^{y})}\|(\text{Id}-L_{0}^{\ast})^{-1}c\|_{2}\|\mathbf{1}_{F_{l}^{y}}\|_{2}$
$\displaystyle=\frac{1}{m(F_{l}^{y})}\|(\text{Id}-L_{0}^{\ast})^{-1}c\|_{2}\sqrt{m(F_{l}^{y})}$
$\displaystyle=\frac{\|(\text{Id}-L_{0}^{\ast})^{-1}c\|_{2}}{\sqrt{m(F_{l}^{y})}}$

and therefore

$\displaystyle\hat{g}(y)^{2}\leq\frac{\|(\text{Id}-L_{0}^{\ast})^{-1}c\|^{2}_{2}}{m(F_{l}^{y})}.$

We then have

$\displaystyle\int\int\mathbf{1}_{F_{l}}(x,y)\hat{g}(y)^{2}f_{0}(y)^{2}dxdy$
$\displaystyle=\int_{\Xi(F_{l})}\int_{F_{l}^{y}}\hat{g}(y)^{2}f_{0}(y)^{2}dxdy$
$\displaystyle=\int_{\Xi(F_{l})}m(F_{l}^{y})\hat{g}(y)^{2}f_{0}(y)^{2}dy$
$\displaystyle\leq\int_{\Xi(F_{l})}m(F_{l}^{y})\frac{\|(\text{Id}-L_{0}^{\ast})^{-1}c\|^{2}_{2}}{m(F_{l}^{y})}f_{0}(y)^{2}dy$
$\displaystyle\leq\|(\text{Id}-L_{0}^{\ast})^{-1}c\|^{2}_{2}\|f_{0}\|_{2}^{2}.$

Thus, $\mathbf{1}_{F_{l}}(x,y)f_{0}(y)\hat{g}(y)$ is in $L^{2}([0,1]^{2})$ and
therefore $M\in L^{2}([0,1]^{2})$.

Now, to verify $\dot{k}$ satisfies (58), we compute, for $\tilde{k}\in
V_{\ker}\cap S_{k_{0},l}$,

$\displaystyle f$
$\displaystyle(\tilde{k})+2\mu\langle\dot{k},\tilde{k}\rangle_{L^{2}([0,1]^{2})}$
$\displaystyle=\int_{F_{l}}\left[-((\text{Id}-L_{0}^{*})^{-1}c)(x)f_{0}(y)+2\mu\dot{k}(x,y)\right]\tilde{k}(x,y)dxdy$
$\displaystyle=\int_{F_{l}}\bigg{[}-((\text{Id}-L_{0}^{*})^{-1}c)(x)f_{0}(y)$
$\displaystyle\qquad\quad+f_{0}(y)\left(((\text{Id}-L_{0}^{\ast})^{-1}c)(x)-\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\right)\bigg{]}\tilde{k}(x,y)dxdy$
$\displaystyle=-\int_{\Xi(F_{l})}\left[\int_{F^{y}_{l}}\left(f_{0}(y)\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\right)\tilde{k}(x,y)dx\right]dy$
$\displaystyle=-\int_{\Xi(F_{l})}\left(f_{0}(y)\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\right)\left[\int_{F^{y}_{l}}\tilde{k}(x,y)dx\right]dy$
$\displaystyle=0,$

where the last equality follows from $\tilde{k}\in V_{\ker}\cap S_{k_{0},l}$.
To conclude checking that $\dot{k}$ satisfies the necessary condition (58), we
need to check that $\mu\neq 0$. Since $M\in L^{2}([0,1]^{2})$, note that the
necessary condition (59) yields $\mu=\pm\frac{1}{2}\|M\|_{L^{2}([0,1]^{2})}$;
thus, to finish the proof that $\dot{k}$ satisfies both necessary conditions
(58)-(59), we will show that $\|M\|_{L^{2}([0,1]^{2})}\neq 0$. From the
hypotheses on $f_{0}$ and $(\text{Id}-L_{0}^{\ast})^{-1}c$ we conclude that

$\|M\|_{L^{2}([0,1]^{2})}^{2}=\int_{F_{l}}f_{0}(y)^{2}\left(((\text{Id}-L_{0}^{\ast})^{-1}c)(x)-\frac{1}{m(F_{l}^{y})}\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\right)^{2}dxdy\neq
0.$

Hence, $\mu=\pm\frac{1}{2}\|M\|_{L^{2}([0,1]^{2})}\neq 0$. The sign of $\mu$
is determined by checking the sufficient conditions.

We can now verify that $\dot{k}\in V_{\ker}\cap S_{k_{0},l}$. We note from
$M\in L^{2}([0,1]^{2})$ and $\mu\neq 0$ that $\dot{k}\in L^{2}([0,1]^{2})$. By
construction supp$(\dot{k})\subseteq F_{l}$. Finally, we have

$\displaystyle\int\dot{k}(x,y)dx$
$\displaystyle=\frac{1}{2\mu}f_{0}(y)\left(\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(x)dx-\int_{F_{l}^{y}}((\text{Id}-L_{0}^{\ast})^{-1}c)(z)dz\frac{\int_{F_{l}^{y}}\mathbf{1}_{F_{l}}(x,y)dx}{m(F_{l}^{y})}\right)$
$\displaystyle=0.$

Sufficient conditions: We want to show that $\dot{k}$ in (28) is a solution to
the optimization problem (26)-(27) by checking that it satisfies the second-
order sufficient conditions. We first demonstrate the set of Lagrange
multipliers $\Lambda(\dot{k})$ (in Definition 3.8, §3.1 [6]) is not empty in
our setting; this will enable us to use the second-order sufficient conditions
of Lemma 3.65 [6]. Note that in terms of the notation used in [6] versus our
notation, $Q=X=V_{\ker}\cap S_{k_{0},l}$, $x_{0}=\dot{k}$,
$Y^{\ast}={\mathbb{R}}$, $G(x_{0})=g(\dot{k})$, $K=\\{0\\}$,
$N_{K}(G(x_{0}))={\mathbb{R}}$, $T_{K}(G(x_{0}))=\\{0\\}$ and
$N_{Q}(x_{0})=\\{0\\}$ (since $Q=X$, see discussion in §3.1 following
Definition 3.8). Thus, to show that $\Lambda(\dot{k})$ is not empty, we need
to show that $\dot{k}$ and $\mu$ satisfy

$D_{\dot{k}}\mathcal{L}(\dot{k},\mu)\dot{k}=0,\;g(\dot{k})=0,\;\mu\in\\{0\\}^{-},\;\mu
g(\dot{k})=0,$ (61)

where $\\{0\\}^{-}:=\\{a\in\mathbb{R}:ax\leq 0\ \forall
x\in\\{0\\}\\}=\mathbb{R}$ (this simplification of conditions (3.16) in [6]