# On large deviations in the averaging principle for SDE’s with a “full
dependence”, correction

A. Yu. Veretennikov
(School of Mathematics, University of Leeds, UK
& Institute for Information Transmission Problems, Russia)

We establish the large deviation principle for stochastic differential
equations with averaging in the case when all coefficients of the fast
component depend on the slow one, including diffusion. 111AMS 1991 subject
classifications. 60F10, 60J60. 222Key words and phrases. Large deviations,
averaging, stochastic differential equation.

##  1 Introduction

This is a corrected version of the paper Veretennikov (1999). We consider the
SDE system

$\displaystyle dX_{t}=f(X_{t},Y_{t})dt,\quad X_{0}=x_{0},$ $\displaystyle
dY_{t}=\varepsilon^{-2}B(X_{t},Y_{t})dt+\varepsilon^{-1}C(X_{t},Y_{t})dW_{t},\quad
Y_{0}=y_{0}.$ (1)

Here $X_{t}\in E^{d},\,Y_{t}\in M$, $M$ is a compact manifold of dimension
$\ell$ (e.g. torus $T^{\ell}$), $f$ is a function with values in
$d$–dimensional Euclidean space $E^{d}$, $B$ is a function with values in
$TM$, $C$ is a function with values in $(TM)^{\ell}$ (i.e., in local
coordinates an $\ell\times\ell$ matrix), $W_{t}$ is an $\ell$–dimensional
Wiener process on some probability space $(\Omega,F,P)$, $\varepsilon>0$ is a
small parameter. Concerning SDE’s on manifolds we refer to Watanabe and Ikeda
(1989).

The large deviation principle (LDP) for such systems with a “full dependence”,
that is, $C(X_{t},Y_{t})$, was not treated earlier. Only the case $C(Y_{t})$
was considered in papers by Freidlin (1976), Freidlin (1978), Freidlin and
Wentzell (1984) for a compact state space and by Veretennikov (1994) for a
non-compact one. There are, as well, recent papers on more general systems
with small additive diffusions by Liptser and by the author which also only
concern the case $C(Y_{t})$.

The LDP for systems like (1) is important in averaging and homogenization, in
the KPP equation theory, for stochastic approximation algorithms with
averaging and so forth. The problem of an LDP for the case $C(X_{t},Y_{t})$
has arisen since Freidlin (1976), Freidlin (1978). Intuitively, the scheme
used for $C(Y_{t})$ should work; at least, almost all main steps go well.
Indeed, there was only one lacuna; the use of Girsanov’s transformation did
not allow freezing of $X_{t}$ if $C$ depended on the slow motion while it
worked well and very naturally for the drift $B(X_{t},Y_{t})$. Yet the problem
remained unresolved for years and the answer was not clear at all. Notice that
this difficulty does not appear in analogous discrete-time systems [see
Gulinsky and Veretennikov (1993), Chapter 11].

It turned out that the use of Girsanov’s transformation in some sense
prevented resolving the problem. Our approach in this paper is based on a new
technical lemma, Lemma 5 below. The main idea is to use two different scales
of partitions of the interval $[0,T]$, a “first-order partition” by points
$\Delta,\,2\Delta,\,\ldots$, which do not depend on the small parameter
$\varepsilon$ and “second-order partitions” which depend on $\varepsilon$ in a
special way, by points
$\varepsilon^{2}t(\varepsilon),\,2\varepsilon^{2}t(\varepsilon),\ldots\,$.
Then the exponential estimates needed for the proof of the result can be
established by two steps. First, the estimates for a “small” partition
interval are derived using the uniform bound of Lemma 3 (see below) and the
estimates for stochastic integrals. It is important that in the “second” scale
the fast motion is still close enough to its frozen version [the bound (13)
below]. Second, the bounds for “small” partitions and induction give one the
estimate for a “large” partition interval.

The original proof in the paper 1999 contains a gap. It relates to a
boundedness of some auxiliary constant $b$ in the proof, – in the original
version this constant may depend implicitly on the partition size $\Delta$,
hence generating a vicious circle. The main aim of this version of the paper
is to present the “patch”. The correction uses improved approximations that
keep this constant $b$ bounded in the lower and upper bounds, and it uses also
a truncated Legendre transformation in the upper bound. The author is deeply
indebted to Professor Yuri Kifer for discovering this vicious circle in the
original version of the paper. The main technical tool remains the Lemma 5.
All standing assumptions are the same as in the original version.

The main result is stated in Section 2. In Section 3 we expose auxiliary
lemmas, among them the main technical Lemma 5 with its proof and a version of
an important lemma from Freidlin and Wentzell (1984) (see Lemma 6) which
requires certain comments. Those comments along with other related remarks are
given in the Appendix, the latter has been also slightly extended. The proof
of the main theorem is presented in Section 4.

##  2 Main result

We make the following assumptions.

$(A_{f})$

The function $f$ is bounded and satisfies the Lipschitz condition.

$(A_{C})$

The function $CC^{*}$ is bounded, uniformly nondegenerate, $C$ satisfies the
Lipschitz condition.

$(A_{B})$

The function $B$ is bounded and satisfies the Lipschitz condition.

Some conditions may be relaxed; for example, $B$ can be locally bounded, $C$
locally (w.r.t. $x$) nondegenerate and so on.

The family of processes $X^{\varepsilon}$ satisfies a large deviation
principle in the space $C([0,T];R^{d})$ with a normalizing coefficient
$\varepsilon^{-2}$ and a rate function $S(\varphi)$ if three conditions are
satisfied:

$\limsup_{\varepsilon\to 0}\varepsilon^{2}\log P_{x}(X^{\varepsilon}\in
F)\leq-\inf_{F}S(\varphi),\quad\forall F\mbox{ closed },$ (2)
$\liminf_{\varepsilon\to 0}\varepsilon^{2}\log P_{x}(X^{\varepsilon}\in
G)\geq-\inf_{G}S(\varphi),\quad\forall G\mbox{ open },$ (3)

and $S$ is a “good” rate function; that is, for any $s\geq 0$, the set

$\Phi(s):=(\varphi\in C([0,T];R^{d}):\,\,S(\varphi)\leq s,\,\,\varphi(0)=x)$

is compact in $C([0,T];R^{d})$. In fact, we will establish the following
equivalent set of assertions due to Freidlin and Wentzell:

$\limsup_{\delta\to 0}\limsup_{\varepsilon\to 0}\varepsilon^{2}\log
P_{x}(\rho_{0,T}(X^{\varepsilon},\Phi(s)\geq\delta)\leq-s,\quad\forall s>0,$
(4)

where $\Phi(s):=\\{\varphi\in C[0,T;R^{d}],\,S(\varphi)\leq s\\}$, and

$\liminf_{\delta\to 0}\liminf_{\varepsilon\to 0}\varepsilon^{2}\log
P_{x}(\rho_{0,T}(X^{\varepsilon},\varphi)<\delta)\geq-S(\varphi),\quad\forall\varphi,$
(5)

where $S$ is a “good” rate function (see above).

Let $\tilde{W}_{t}=\varepsilon^{-1}W_{t\varepsilon^{2}}$,
$y_{t}=Y_{t\varepsilon^{2}}$, $x_{t}=X_{t\varepsilon^{2}}$ and let $y^{x}_{t}$
denote a solution of SDE,

$dy^{x}_{t}=B(x,y^{x}_{t})dt+C(x,y^{x}_{t})d\tilde{W}_{t},\quad
y^{x}_{0}=y_{0}.$ (6)

###### Theorem 1

Let $(A_{f})$, $(A_{B})$, $A_{C})$ be satisfied. Then the family
$(X^{\varepsilon}_{t}=X_{t},\;0\leq t\leq T)$ satisfies the LDP as
$\varepsilon\to 0$ in the space $C([0,T];R^{d})$ with an action function

$S(\varphi)=\int_{0}^{T}L(\varphi_{t},\dot{\varphi}_{t})dt,$

where

$L(x,\alpha)=\sup_{\beta}(\alpha\beta-H(x,\beta)),$
$H(x,\beta)=\lim_{t\to\infty}t^{-1}\log
E\exp\left(\int_{0}^{t}f(x,y^{x}_{s})ds\right).$

The limit $H$ exists and is finite for any $\beta$, the functions $H$ and $L$
are convex in their last arguments $\beta$ and $\alpha$ correspondingly,
$L\geq 0$ and $H$ is continuously differentiable in $\beta$.

The differentiability of $H$ at any $\beta$ is provided by the compactness of
the state space of the fast component.

##  3 Auxiliary lemmas

Let us consider the semigroup of operators $T^{\beta}_{t},t\geq 0$, on $C(M)$
defined by the formula

$T^{x^{\prime},x,\beta}_{t}g(y)=T^{\beta}_{t}g(y)=E_{y}g(y^{x}_{t})\exp\left(\int^{t}_{0}\beta
f(x^{\prime},y^{x}_{s})ds\right),$

where $\beta\in E^{d}$, $\beta f$ is a scalar product.

###### Lemma 1

Let assumptions $(A_{f})$, $(A_{B})$, $(A_{C})$ be satisfied. Then for any
$\beta$ the operator $T^{\beta}_{1}$ is compact in the space $C([0,T];R^{d})$.

###### Lemma 2

Let assumptions $(A_{f})$, $(A_{B})$, $(A_{C})$ be satisfied. Then the
spectral radius $r(T^{\beta}_{1})$ is a simple eigenvalue of $T^{\beta}_{1}$
separated from the rest of the spectrum and its eigen–function $e_{\beta}$
belongs to the cone $C^{+}(M)$. Moreover, function $r(T^{\beta}_{1})$ is
smooth (of $C^{\infty}$) in $\beta$ and the function $e_{\beta}$ is bounded
and separated away from zero uniformly in $|\beta|<b$ and any $x^{\prime},x$.

###### Lemma 3

Let $\beta\in E^{d}$, and let assumptions $(A_{f})$, $(A_{B})$, $(A_{C})$ be
satisfied. Then there exists a limit

$H(x^{\prime},x,\beta)=\lim_{t\to 0}t^{-1}\log
E_{y}\exp\left(\beta\int^{t}_{0}f(x^{\prime},y^{x}_{s})ds\right);$

moreover, $H(x^{\prime},x,\beta)=\log r(T^{x^{\prime},x,\beta}_{1})$. The
function $H(x^{\prime},x,\beta)$ is of $C^{\infty}$ in $\beta$ and convex in
$\beta$. For any $b>0$ there exists $C(b)$ such that, for any $y$, $|\beta|<b$

$|t^{-1}\log
E_{y}\exp\left(\beta\int^{t}_{0}f(x^{\prime},y^{x}_{s})ds\right)-H(x^{\prime},x,\beta)|\leq
C(b)t^{-1}.$ (7)

Notice that $|H(x^{\prime},x,\beta)|\leq\|f\|_{C}|\beta|$.

###### Lemma 4

Let assumptions $(A_{f})$, $(A_{B})$, $(A_{C})$ be satisfied. Then for any
$b>0$ the functions $H$ and $\nabla_{\beta}H$ are uniformly continuous in
$(x^{\prime},x,\beta)$, $|\beta|<b$.

Lemmas 1-4 are standard [cf. Veretennikov (1994) or (1992)]. They are based on
Frobenius-type theorems for positive compact operators [see Krasnosel’skii,
Lifshitz and Sobolev (1989)] and the theory of perturbations of linear
operators [see Kato (1976), Chapter 2]. Denote
$\tilde{F}_{t}=F_{t\varepsilon^{2}}$.

###### Lemma 5

Let assumptions $(A_{f})$, $(A_{B})$, $(A_{C})$, $b>0$,
$t(\varepsilon)\to\infty$ and $t(\varepsilon)=o(\log\varepsilon^{-1})$ as
$\varepsilon\to 0$. Then for any $\nu>0$ there exist $\delta(\nu)>0$,
$\varepsilon(\nu)>0$ such that for $\varepsilon\leq\varepsilon(\nu)$ uniformly
w.r.t. $t_{0},\,x^{\prime},\,x,\,x_{0},\,y_{0}$, $x_{t_{0}}=x_{0}$,
$|\beta|\leq b$, the inequality holds on the set
$\\{|x_{t_{0}}-x|<\delta(\nu)\\}$,

$\left|\log
E(\exp(\beta\int\limits_{t_{0}}^{t_{0}+t(\varepsilon)}f(x^{\prime},y_{s})ds)|\tilde{F}_{t_{0}})-t(\varepsilon)H(x^{\prime},x,\beta)\right|\leq\nu
t(\varepsilon).$ (8)

Moreover, if $\Delta\leq\Delta(\nu)=(1+\|f\|_{C})^{-1}\delta(\nu)/2$ and
$\varepsilon$ is small enough, then uniformly w.r.t.
$t_{0},\,x^{\prime},\,x,\,x_{0},\,y_{0}$, $\delta\leq\delta(\nu)$,
$|x_{0}-x|<\delta$, and $|\beta|\leq b$,

$\displaystyle\exp(\varepsilon^{-2}\Delta
H(x^{\prime},x,\beta)-\nu\Delta\varepsilon^{-2})$ $\displaystyle\leq
E\left(\exp(\beta\varepsilon^{-2}\int\limits_{t_{0}}^{t_{0}+\Delta}f(x^{\prime},Y_{s})ds)|F_{t_{0}}\right)$
$\displaystyle\leq\exp(\varepsilon^{-2}\Delta
H(x^{\prime},x,\beta)+\nu\Delta\varepsilon^{-2}).$ (9)

Remark. We reiterate and emphasize that any couple $(\Delta,\delta)$
satisfying only $\Delta\leq\Delta(\nu)$ and $\delta\leq\delta(\nu)$ would do.

Proof. Step 1\. It is sufficient to prove (8) and (5) for $t_{0}=0$. Moreover,
since $H$ is continuous, it suffices to check both inequalities for $x=x_{0}$.
Indeed, the bound

$\left|\log
E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y_{s})ds\right)-t(\varepsilon)H(x^{\prime},x_{0},\beta)\right|\leq\nu
t(\varepsilon)$

implies

$\displaystyle\left|\log
E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y_{s})ds\right)-t(\varepsilon)H(x^{\prime},x,\beta)\right|$
$\displaystyle\leq
t(\varepsilon)(\nu+|H(x^{\prime},x,\beta)-H(x^{\prime},x_{0},\beta)|),$

and we use the uniform continuity of the function $H$ on compact sets (remind
that $|\beta|\leq b$). The same arguments are applicable to the second
inequality of the assertion of the lemma. So, in the sequel we consider the
case $x_{0}=x$.

Let us show first that

$\sup_{x^{\prime},x_{0}}\left|t(\varepsilon)^{-1}\log
E\exp\left(\beta\int_{0}^{t(\varepsilon)}f(x^{\prime},y_{s})ds\right)-H(x^{\prime},x,\beta)\right|\leq\nu$
(10)

if $\varepsilon$ is small enough. Due to Lemma 3, it would be correct if
$y_{s}$ were replaced by $y^{x}_{s}$ and $t(\varepsilon)\geq\nu^{-1}C(b)$. We
will also use the bounds

$\sup_{0\leq s\leq t}|x_{s}-x_{0}|\leq\|f\|_{C}\varepsilon^{2}t\quad\forall
C,\;\;\;\exp(Ct(\varepsilon))t(\varepsilon)^{2}\varepsilon^{2}\to
0,\;\;\varepsilon\to 0.$ (11)

Let $|f(x^{\prime},y)-f(x^{\prime},y^{\prime})|\leq L_{f}|y-y^{\prime}|$ for
all $y,y^{\prime},x^{\prime}$, $L_{f}>0$, $C_{f}=\|f\|_{C}$. We estimate for
$t(\varepsilon)>\nu^{-1}C(b)/4$,

$\displaystyle
E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y_{s})ds\right)$
$\displaystyle\times\left\\{I\left(\sup_{0\leq t\leq
t(\varepsilon)}|y_{t}-y^{x}_{t}|\leq\nu/(4L_{f}b)\right)+I\left(\sup_{0\leq
t\leq t(\varepsilon)}|y_{t}-y^{x}_{t}|>\nu/(4L_{f}b)\right)\right\\}$
$\displaystyle\leq
E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y^{x}_{s})ds+t(\varepsilon)\nu/4\right)I\left(\sup_{0\leq
t\leq t(\varepsilon)}|y_{t}-y^{x}_{t}|\leq\nu/(4L_{f}b)\right)$
$\displaystyle+\exp(C_{f}bt(\varepsilon)\nu)EI\left(\sup_{0\leq t\leq
t(\varepsilon)}|y_{t}-y^{x}_{t}|>\nu/(4L_{f}b)\right)$ $\displaystyle\leq
E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y^{x}_{s})ds\right)\exp(t(\varepsilon)\nu/4)$
$\displaystyle+\exp\left(C_{f}bt(\varepsilon)\nu\right)\nu^{-2}E\sup_{t\leq
t(\varepsilon)}|y_{t}-y^{x}_{t}|^{2}.$ (12)

By virtue of Lemma 3 we have

$E\exp\left(\beta\int\limits_{0}^{t(\varepsilon)}f(x^{\prime},y^{x}_{s})ds\right)\leq\exp(t(\varepsilon)(H(x^{\prime},x,\beta)+\nu/4))$

if $\varepsilon$ is small enough. A similar lower bound holds true also.

Let us estimate the second term. By virtue of the inequalities for the Itô and
Lebesgue integrals, we have

$\displaystyle E\sup_{t^{\prime}\leq
t}|y_{t^{\prime}}-y^{x}_{t^{\prime}}|^{2}$ $\displaystyle\leq
CE\int\limits_{0}^{t}|C(x_{s},y_{s})-C(x,y^{x}_{s}))|^{2}ds$
$\displaystyle+CtE\int\limits_{0}^{t}|B(x_{s},y_{s})-B(x,y^{x}_{s}))|^{2}ds$
$\displaystyle\leq
C\int\limits_{0}^{t}E|x_{s}-x|^{2}ds+C\int\limits_{0}^{t}E\sup_{u\leq
s}|y_{s}-y^{x}_{s}|^{2}ds$ $\displaystyle\leq
Ct^{2}\varepsilon^{2}+C\int\limits_{0}^{t}E\sup_{u\leq
s}|y_{u}-y^{x}_{u}|^{2}ds.$

By virtue of Gronwall’s lemma, one gets

$E\sup_{t^{\prime}\leq t}|y_{t^{\prime}}-y^{x}_{t^{\prime}}|^{2}\leq
Ct^{2}\varepsilon^{2}\exp(Ct).$

In particular,

$E\sup_{t^{\prime}\leq
t(\varepsilon)}|y_{t^{\prime}}-y^{x}_{t^{\prime}}|^{2}\leq
Ct(\varepsilon)^{2}\varepsilon^{2}\exp(Ct(\varepsilon)).$ (13)

So the second term in (3) does not exceed the value
$\exp(C_{f}bt(\varepsilon)\nu)\nu^{-2}Ct(\varepsilon)^{2}\varepsilon^{2}$
which is $o(\exp(Kt(\varepsilon)))$ for any $K>0$. Indeed,
$\exp(t(\varepsilon)(C_{f}b\nu-K))\nu^{-2}Ct(\varepsilon)^{2}\varepsilon^{2}\to
0$ due to the assumption
$t(\varepsilon)=o(\log\varepsilon^{-1}),\;\varepsilon\to 0$. This proves (10).

Notice that the bound (10) is uniform w.r.t. $|\beta|\leq b$ and
$x^{\prime},\,x,\,y_{0}$. Since the function $H$ is continuous, we get on the
set $\\{|x_{t_{0}}-x|<\delta(\nu)\\}$,

$\displaystyle\sup\limits_{x^{\prime},x,y_{0},t_{0},|\beta|\leq b}\left|\log
E\left(\exp\left(\beta\int\limits_{t_{0}}^{t_{0}+t(\varepsilon)}f(x^{\prime},y^{x}_{s})ds\right)\mid\tilde{F}_{t_{0}}\right)-t(\varepsilon)H(x^{\prime},x,\beta)\right|\leq\nu
t(\varepsilon)$ (14)

if $\delta(\nu)$ is small enough.

Step 2\. Let $\Delta\leq(1+\|f\|_{C})^{-1}\delta(\nu)/2=\Delta(\nu)$ and
$N=\Delta\varepsilon^{-2}t(\varepsilon)^{-1}$. Then $\sup_{0\leq s\leq
Nt(\varepsilon)}|x_{s}-x_{0}|\leq\delta(\nu)/2$. Let
$|x-x_{0}|<\delta(\nu)/2$. So, $\sup_{0\leq s\leq
Nt(\varepsilon)}|x_{s}-x|<\delta(\nu)$. In particular,
$|x_{kt(\varepsilon)}-x|<\delta(\nu)$ for any $1\leq k\leq N$. By induction,
we get from (14) for such $k$,

$\displaystyle\exp(kt(\varepsilon)H(x^{\prime},x,\beta)-\nu kt(\varepsilon))$
$\displaystyle\leq
E\exp\left(\beta\int\limits_{0}^{kt(\varepsilon)}f(x^{\prime},y_{s})ds\right)$
$\displaystyle\leq\exp(kt(\varepsilon)H(x^{\prime},x,\beta)+\nu
kt(\varepsilon)),$

or, after the time change,

$\displaystyle\exp(kt(\varepsilon)H(x^{\prime},x,\beta)-\nu kt(\varepsilon))$
$\displaystyle\leq
E\exp\left(\beta\varepsilon^{-2}\int\limits_{0}^{kt(\varepsilon)\varepsilon^{-2}}f(x^{\prime},Y_{s})ds\right)$
$\displaystyle\leq\exp(kt(\varepsilon)H(x^{\prime},x,\beta)+\nu
kt(\varepsilon)).$

Since $H$ is continuous then we obtain for $k=N$,

$\displaystyle\exp(\varepsilon^{-2}\Delta
H(x^{\prime},x,\beta)-\nu\Delta\varepsilon^{-2})$ $\displaystyle\leq
E\exp\left(\beta\varepsilon^{-2}\int\limits_{0}^{\Delta}f(x^{\prime},Y_{s})ds\right)$
$\displaystyle\leq\exp(\varepsilon^{-2}\Delta
H(x^{\prime},x_{0},\beta)+\nu\Delta\varepsilon^{-2}).$ (15)

Lemma 5 is proved. QED

The next Lemma is an improved version of the Lemma 7.5.2 from Freidlin and
Wentzell. Although we will not use it, its technique is essential.

###### Lemma 6

[Freidlin (1978), Freidlin and Wentzell (1984)]. Let $S(\varphi)<\infty$. If
$\psi^{n}$ is a sequence of step functions tending uniformly to $\varphi$ in
$C[0,T];R^{d})$ as $n\to\infty$, then there exists a sequence of piecewise
linear functions $\chi^{n}$ (with the same partitions) which also tend
uniformly to $\varphi$ and such that

$\limsup_{n\to\infty}\int_{0}^{T}L(\psi^{n}_{s},\dot{\chi}^{n}_{s})ds\leq
S(\varphi).$

Moreover, one may assume without loss of generality that for any $s$ there
exists a value

$\beta_{s}=\mathop{\rm
argmax}\nolimits\limits_{\beta}(\beta\dot{\chi}^{n}_{s+}-H(\psi^{n}_{s},\psi^{n}_{s},\beta))$

and

$L(\psi^{n}_{s},\alpha)>L(\psi^{n}_{s},\dot{\chi}^{n}_{s+})+(\alpha-\dot{\chi}^{n}_{s+})\beta_{s}\quad\forall\alpha\not=\dot{\chi}^{n}_{s}.$

If $\hat{\psi}$ is close enough to $\psi^{n}_{s}$ then there exists a value

$\hat{\beta}_{s}=\mathop{\rm
argmax}\nolimits\limits_{\beta}(\beta\dot{\chi}^{n}_{s+}-H(\psi^{n}_{s},\hat{\psi},\beta)),$
$L(\psi^{n}_{s},\hat{\psi},\alpha)>L(\psi^{n}_{s},\hat{\psi},\dot{\chi}^{n}_{s+})+(\alpha-\dot{\chi}^{n}_{s+})\hat{\beta}_{s}\quad\forall\alpha\not=\dot{\chi}^{n}_{s}$

and

$L(\psi^{n}_{s},\hat{\psi},\dot{\chi}^{n}_{s+})\to
L(\psi^{n}_{s},\psi^{n}_{s},\dot{\chi}^{n}_{s+}),\quad\hat{\psi}\to\psi^{n}_{s}.$

We added to the original assertion the property that $\chi^{n}_{t}$ may be
chosen piecewise linear. Indeed, such functions are used in the proof; see
Freidlin and Wentzell (1984), Section 7.5. The existence of $\beta_{s}$
asserted in the lemma also follows from the proof; see Freidlin and Wentzell
(1984) or Freidlin (1978). Assertions about $\hat{\psi}$ and $\hat{\beta}_{s}$
also added to the original assertion can be deduced from the proof using
similar arguments.

In fact, there is a little gap in the original proof, namely, an additional
assumption was used which was not formulated explicitly. This is why we have
to present a precise statement and give necessary comments on it in the
Appendix.

##  4 Proof of theorem 1

  1. 1.

First part of the proof: the lower bound. Let $S(\varphi)<\infty$, and
$\nu>0$. To establish the lower bound, we will show the inequality: given any
$\nu>0$, and any $\delta>0$, we have for $\varepsilon>0$ small enough,

$\varepsilon^{2}\log
P_{x}(\rho_{0,T}(X^{\varepsilon},\varphi)<\delta)\geq-S(\varphi)-\nu.$

Denote $H(x,\beta)=H(x,x,\beta)$. The existence of the limit
$H(x,x^{\prime},\cdot)$ for any $x,x^{\prime}$, and its differentiability and
continuity are asserted in Lemmas 3 and 4. Throughout the proof, we may and
will assume that for any $s$, $L(\varphi_{s},\dot{\varphi}_{s})<\infty$.
Indeed, this may be violated only on a set of $s$ of Lebesgue measure zero.
Notice that due to the boundedness of the function $f$, this inequality
implies $\sup_{s}|\dot{\varphi}_{s}|\leq\|f\|_{C}$. Indeed, for any
$|\alpha|>\|f\|_{C}$, we have $L(x,\alpha)=+\infty$.

In the sequel, both $X_{0}=x$ and $Y_{0}=y$ are fixed, hence, the probability
symbol $P$ will be used without indices.

  2. 2.

We are going to reduce the problem of estimation from below the probability

$P(\rho(X,\varphi)<\delta)$

to that for the probability

$P(\rho(X^{\varphi},\varphi)<\delta^{\prime}),\quad\mbox{where}\quad
X^{\psi}_{t}:=x_{0}+\int_{0}^{t}f(\psi_{s},Y_{s})ds,\;\forall\psi,$

and further to

$P(\rho(X^{\psi},\chi)<\delta^{\prime}),$

where both $\psi,\chi$ approximate $\varphi$. The rough idea is eventually to
choose a step function as $\psi$ and piecewise linear one as $\chi$, however
we are going to perform these approximations gradually. A step function is
needed because we only have a technical tool – the Lemma 5 – established for
this very case. A piecewise linear $\psi$ is not necessary, but convenient.
Eventually we consider a finite-dimensional subset of the set
$\\{\rho(X,\varphi)<\delta\\}$, of the form (slightly abusing notations which
will be explained in the sequel)

$\\{\rho(X^{\psi}_{\Delta},\chi_{\Delta})<\delta^{\prime}_{1},\rho(X^{\psi}_{2\Delta},\chi_{2\Delta)}<\delta^{\prime}_{2},\ldots,\rho(X^{\psi}_{T},\chi_{T})<\delta^{\prime}_{T/\Delta}\\},$

with appropriately chosen $\Delta$, $X^{\psi}$, deterministic curves
$\psi,\chi$, and constants $\delta^{\prime}_{k}$: in particular, we will
choose
$\delta^{\prime}_{1}<<\delta^{\prime}_{2}<<\ldots<<\delta^{\prime}_{T/\Delta}<<\delta$.

While performing all these approximations, we need to establish simultaneously
a special property: at any point $s$, the Fenchel-Legendre adjoint to the
$\dot{\chi}_{s}$ variable $\beta_{s}=\beta_{s}[\psi_{s},\dot{\chi}_{s}]$ (see
below) can be chosen uniformly bounded.

There will be several, – actually, a lot of, – small constants chosen
consequently throughout this proof. In the beginning, $\delta>0$ and $\nu>0$
are fixed; due to the Lemma 5, we have also $\Delta(\nu)$ and $\delta(\nu)$.
Here we prompt in advance the order of the choice for most of them (by this
diagram we do not claim that every following constant only depends on the
previous one, just the order):

$b\,\&\,\tilde{\delta}^{\prime}\mapsto\delta^{\prime}\mapsto\Delta\,\&\,\delta^{\prime\prime}\mapsto\delta^{\prime}_{m}\mapsto\delta^{\prime}_{m-1}\mapsto
z_{m-1}\mapsto\nu\,^{\prime}_{m-1}\ldots\mapsto\delta^{\prime}_{1}\mapsto
z_{1}\mapsto\nu\,^{\prime}_{1}.$

Remark. Emphasize that the final set (above) will be a subset of the
$\\{\rho(X,\varphi)<\delta\\}$, however, it is not necessary that

$\\{\rho(\varphi_{\Delta},\chi_{\Delta})<\delta^{\prime}_{1},\rho(\varphi_{2\Delta},\chi_{2\Delta)}<\delta^{\prime}_{2},\ldots,\rho(\varphi_{T},\chi_{T})<\delta^{\prime}_{T/\Delta}\\},$

that is, the curve $\varphi$ itself does not have to belong to this subset.

  3. 3.

For any nonrandom curve $\psi$, – although we will apply this firstly to
$\varphi$, but other functions are also necessary for the analysis below, – we
have, due to the Lipschitz condition on $f$,

$\\{\rho(X,\varphi)<\delta\\}\supset\\{\rho(X^{\psi},\chi)<\delta^{\prime}\\}$
(16)

if $\delta^{\prime}$ and $\lambda:=\rho_{0,T}(\varphi,\psi)$ are small enough
with respect to $\delta$. (A small constant $\lambda>0$ is used just within
this step.) E.g.,
$\delta^{\prime}<\delta(e^{CT}CT+1)^{-1}/2,\quad\lambda<\delta(e^{CT}CT+1)^{-1}/2$
suffice, see below. Indeed,

$X_{t}=x+\int_{0}^{t}f(X_{s},Y_{s})ds,\quad
X^{\psi}_{t}=x+\int_{0}^{t}f(\psi_{s},Y_{s})ds,$

thence,

$\displaystyle|X_{t}-X^{\psi}_{t}|\leq\int_{0}^{t}|f(X_{s},Y_{s})ds-f(\psi_{s},Y_{s})|ds\leq
C\int_{0}^{t}|X_{s}-\psi_{s}|ds$ $\displaystyle\leq
C\int_{0}^{t}|X_{s}-X^{\psi}_{s}|ds+C\int_{0}^{t}|X^{\psi}_{s}-\chi_{s}|ds+C\int_{0}^{t}|\chi_{s}-\psi_{s}|ds;$

so on the set $\\{\rho(X^{\psi},\chi)<\delta^{\prime}\\}$,

$\displaystyle|X_{t}-X^{\psi}_{t}|\leq
C\int_{0}^{t}|X_{s}-X^{\psi}_{s}|ds+C\delta^{\prime}t+C\lambda t,$

and, moreover (on the same set),

$\sup_{0\leq t^{\prime}\leq t}|X_{t^{\prime}}-X^{\psi}_{t^{\prime}}|\leq
C\int_{0}^{t}\sup_{0\leq s^{\prime}\leq
s}|X_{s^{\prime}}-X^{\psi}_{s^{\prime}}|ds+C(\delta^{\prime}+\lambda)t,$

which implies by Gronwall’s inequality that on the same set,

$\rho(X,X^{\psi})\leq e^{CT}C(\delta^{\prime}+\lambda)T.$

Now,

$\displaystyle\rho(X,\varphi)\leq\rho(X,X^{\psi})+\rho(X^{\psi},\chi)+\rho(\chi,\varphi)$
$\displaystyle\leq
e^{CT}C(\delta^{\prime}+\lambda)T+\delta^{\prime}+\lambda=(\delta^{\prime}+\lambda)(e^{CT}CT+1).$

Therefore, (16) holds true. E.g.,

$\delta^{\prime}<\delta(e^{CT}CT+1)^{-1}/2,\quad\lambda<\delta(e^{CT}CT+1)^{-1}/2$

suffice. In particular, it is true that

$\\{\rho(X,\varphi)<\delta\\}\supset\\{\rho(X^{\varphi},\chi)<\delta^{\prime}\\},$

if $\delta^{\prime}$ and $\lambda$ are small enough with respect to $\delta$.
This bound will be used while establishing a lower bound.

  4. 4.

While establishing an upper bound, an opposite inclusion will be useful,

$\\{\rho(X,\varphi)<\delta\\}\subset\\{\rho(X^{\psi},\chi)<2\delta(KT+1)\\},$
(17)

if $\lambda=\max\left(\rho(\varphi,\psi),\rho(\varphi,\chi)\right)\leq\delta$.
Indeed,

$\displaystyle|X_{t}-X^{\psi}_{t}|\leq\int_{0}^{t}|f(X_{s},Y_{s})ds-f(\psi_{s},Y_{s})|ds\leq
K\int_{0}^{t}|X_{s}-\psi_{s}|ds$ $\displaystyle\leq
K\int_{0}^{t}|X_{s}-\varphi_{s}|ds+K\int_{0}^{t}|\psi_{s}-\varphi_{s}|ds;$

so on the set $\\{\rho(X,\varphi)<\delta\\}$,

$\displaystyle|X_{t}-X^{\psi}_{t}|\leq K\delta t+K\lambda t,$

and, moreover (on the same set),

$\rho(X,X^{\psi})\leq K(\delta+\lambda)T.$

Now, (17) follows from the inequalities,

$\displaystyle\rho(X^{\psi},\chi)\leq\rho(X,X^{\psi})+\rho(X,\varphi)+\rho(\chi,\varphi)$
$\displaystyle\leq K(\delta+\lambda)T+\delta+\lambda.$

  5. 5.

Our next goal is the choice of an appropriate $\chi=\varphi^{b}$; essential is
to keep the integral
$\int\limits_{0}^{T}L(\varphi_{s},\dot{\varphi}^{b}_{s})\,ds$ close to
$S(\varphi)$. Suppose for some $s\in[0,T]$, the set
$\\{\alpha:\,L(\varphi_{s},\alpha)<\infty\\}$ has a non-empty interior, – for
the latter we will use a notation ${\cal L}^{\circ}[f,\varphi_{s}]$, – with
respect to its linear hull ${\cal L}[f,\varphi_{s}]$. Since
$L(\varphi_{s},\dot{\varphi}_{s})<\infty$, this value is attained as a
$\,\liminf\,$ of the values $L(\varphi_{s},\alpha)$, $\alpha\in{\cal
L}^{\circ}[f,\varphi_{s}]$, as $\alpha\to\dot{\varphi}$, see Rockafellar
(1970). It is a property of any such $\alpha$ that there exists a finite
adjoint vector $\beta=\mathop{\rm
argmax}\nolimits_{\beta}(\alpha\beta-H(\varphi_{s},\beta))$ given $\alpha$,
although this adjoint may be not unique which we will discuss shortly. Notice
that, in particular, we have

$H(\varphi_{s},\beta)=(\alpha\beta-L(\varphi_{s},\alpha)),\;\;\mbox{and}\;\;L(\varphi_{s},\alpha)=(\alpha\beta-H(\varphi_{s},\beta)).$

Then we choose a vector $\dot{\tilde{\varphi}_{s}}:=\alpha\in{\cal
L}^{\circ}[f,\varphi_{s}]$ so that the value
$L(\varphi_{s},\dot{\tilde{\varphi}_{s}})$ is close enough to
$L(\varphi_{s},\dot{\varphi_{s}})$.

The adjoint $\beta$-value is unique iff $L$ is differentiable at $\alpha$,
which is also equivalent to strict convexity of $H$ at $\beta$ (see
Rockafellar (1970)). If this is not a case, that is, $L$ is non-differentiable
at $\alpha$, then there is a sub-gradient to the graph of $L$ at $\alpha$
which is a non-trivial cone (although its dimension may be less than $d$). In
this case one can choose various adjoint vectors $\beta$’s. Although not
unique, both $\dot{\tilde{\varphi}_{s}}$ and
$\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]$ can be chosen as Borel
functions of $s$ (due to the Measurable Choice Theorem). Denote thus chosen
adjoint by $\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]$.

If the set ${\cal L}^{\circ}[f,\varphi_{s}]$ is empty, one can choose
$\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]=0$, see Appendix A.

Now set

$\tilde{\varphi}_{t}:=x+\int_{0}^{t}\dot{\tilde{\varphi}_{s}}\,ds.$

We can choose this new curve to be as close to $\varphi$ as we like, and the
values $\int_{0}^{T}L(\varphi_{s},\dot{\tilde{\varphi}_{s}})\,ds$ and
$\int_{0}^{T}L(\varphi_{s},\dot{\varphi}_{s})\,ds$ are arbitrarily close to
each other, too, say,

$\left|\int\limits_{0}^{T}L(\varphi_{s},\dot{\tilde{\varphi}_{s}})\,ds-S(\varphi)\right|\leq\nu/3.$

  6. 6.

For any $s$, let us choose a (measurable) vector $\hat{\alpha}_{s}$ such that
$L(\varphi_{s},\hat{\alpha}_{s})=0$.

Notice that $L(\varphi_{s},\cdot)\geq 0$ since $H(\varphi_{s},0)=0$; moreover,
$H(\varphi_{s},\cdot)$ is convex at the origin, hence, there does exist
$\alpha$ such that $L(\varphi_{s},\alpha)=0$. If not unique, this vector can
be still chosen measurable due to the Measurable Choice Theorem. Moreover,
$|\hat{\alpha}_{s}|\leq\|f\|_{C}$ for any vector where $L$ is finite.

It is shown in the Appendix C that $\hat{\alpha}_{s}\in{\cal
L}^{\circ}[f,\varphi_{s}]$, if the latter set is not empty. A corresponding
adjoint vector $\beta$ (i.e.
$\beta=\mbox{argsup}\,[\langle\cdot\,,\hat{\alpha}_{s}\rangle-H(\varphi_{s},\cdot)]$)
does exist, and may be chosen as a zero-vector in $R^{d}$ that we denote by
$\bar{0}$. This vector is, indeed, one of (if not unique) adjoint vectors for
the vector $\hat{\alpha}_{s}$, since $H(\varphi_{s},\bar{0})=0$, and
$\bar{0}\hat{\alpha}_{s}-H(\varphi_{s},\bar{0})=0$. If unique (that depends on
the differentiability of the function $L$, or, equivalently, on strict
convexity of the function $H$), this adjoint is precisely $\bar{0}$.

  7. 7.

Next, given $b$, define

$\displaystyle\varphi^{b}_{t}:=x+\int_{0}^{t}\left(\dot{\tilde{\varphi}_{s}}\,1(|\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]|\leq
b)+\hat{\alpha}_{s}\,1(|\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]|>b)\right)\,ds.$

Since $|\hat{\alpha}_{s}|\leq\|f\|_{C}$, we can find such a $b$ that the curve
$\varphi^{b}$ is still as close to $\varphi$ as we like in the uniform norm,
and the values $\,\int_{0}^{T}L(\varphi_{s},\dot{\varphi^{b}_{s}})\,ds\,$ and
$\,\int_{0}^{T}L(\varphi_{s},\dot{\varphi}_{s})\,ds\,$ are arbitrarily close
to each other.

At the same time, if for some $s$ we have
$|\beta[\varphi_{s},\dot{\tilde{\varphi}_{s}}]|>b$, then the $\beta$-value is
transformed into
$\beta[\varphi_{s},\dot{\varphi_{s}^{b}}]=\beta[\varphi_{s},\hat{\alpha}_{s}]=0$,
so that for any $s\in[0,T]$ the inequality holds true,

$|\beta[\varphi_{s},\dot{\varphi^{b}_{s}}]|\leq b.$

We have,

$\\{\rho(X,\varphi)<\delta\\}\supset\\{\rho(X,\varphi^{b})<\delta/2\\}$,

if $b$ is large enough.

  8. 8.

Next, the values of rate functions $S^{\varphi}_{s}(\tilde{\varphi})$ and
$S^{\varphi}(\varphi^{b})$ are arbitrarily close, say,
$|S^{\varphi}_{s}(\tilde{\varphi})-S^{\varphi}_{s}(\varphi^{b})|<\nu/3$, so
that

$\left|S^{\varphi}(\varphi^{b})-S(\varphi)\right|\leq 2\nu/3,$

if $b$ is large enough. Moreover, in addition,

$\\{\rho(X,\varphi^{b})<\delta/2\\}\supset\\{\rho(X^{\varphi},\varphi^{b})<\tilde{\delta}^{\prime}\\}$,
(18)

if $\tilde{\delta}^{\prime}$ and $\lambda=\rho(\varphi,\varphi^{b})$ are small
enough with respect to $\delta$; hence, we can fix the values $b$ and
$\tilde{\delta}^{\prime}$ here.

  9. 9.

The next transform is the change of both $\varphi$ and $\varphi^{b}$ so that
the first becomes a step function, while the second becomes piecewise linear
on $[0,T]$. In the meantime, all adjoint $\beta$-values will remain bounded.

Consider the approximations

$\psi_{s}=\varphi_{\kappa_{m}(s+a)-a},\quad\dot{\chi}_{s}=\dot{\varphi^{b}}_{\kappa_{m}(s+a)-a},$

where $m=T/\Delta$ is a positive integer, $\kappa_{m}(s):=[s/\Delta]\Delta$,
and $a$ is ‘almost any’ value from $[0,T]$; we assume
$\varphi_{s}\equiv\varphi_{0},\,s<0$, and
$\dot{\varphi^{b}}_{s}\equiv\dot{\varphi^{b}}_{0},\,s<0$. Since
$\sup_{t}|\dot{\varphi}_{t}|<\infty$, we have $\rho(\varphi,\psi)\to
0,\,m\to\infty$. Next, it is well-known333Possibly this folklore fact can be
found in N. V. Krylov’s Lectures on Stochastic Processes (Moscow State
University, 1980s, Russian, vol. 2; the English translation of this book has
been published recently by the AMS). For the reader’s convenience we provide a
folklore proof here. Indeed, due to a standard technique, it suffices to show
for the first integral that
$\int_{0}^{T}\int_{0}^{T}|\dot{\varphi^{b}}_{\kappa_{m}(s+a)-a}-\dot{\varphi^{b}}_{s}|\,da\,ds\to
0,\quad m\to\infty.$ Let $g_{s},\,0\leq s\leq T,$ be a smooth bounded function
such that $\int|\dot{\varphi^{b}}_{s}-g_{s}|\,ds<\nu$; all functions are
extended to $-\infty<s<\infty$ so that all vanish outside $[0,T]$. Then
$\displaystyle\int_{0}^{T}\,da\int|\dot{\varphi^{b}}_{\kappa_{m}(s+a)-a}-g_{\kappa_{m}(s+a)-a}|\,ds=\int\,ds\int_{0}^{T}|\dot{\varphi^{b}}_{\kappa_{m}(s+a)-a}-g_{\kappa_{m}(s+a)-a}|\,da$
$\displaystyle=\sum_{i=1}^{m}\Delta\left(\Delta^{-1}\int_{(i-1)\Delta}^{i\Delta}|\dot{\varphi^{b}}_{s}-g_{s}|\,ds\right)\equiv\int|\dot{\varphi^{b}}_{s}-g_{s}|\,ds<\nu.$
Hence, it suffices to show that
$\int_{0}^{T}\int_{0}^{T}|g_{\kappa_{m}(s+a)-a}-g_{s}|\,da\,ds\to 0,\quad
m\to\infty,$ which for smooth functions follows from the Lebesgue dominated
convergence theorem, or even from uniform convergence under the integrals (of
course, the smooth function $g$ with a compact support is bounded). In turn,
the existence of a smooth function $g$ claimed above is due to the property
that smooth functions are dense in $L_{1}[0,T]$; this is, e.g., because so are
step functions, while the indicator of any Borel set may be approximated in
this function space by a finite set of intervals, and any indicator of an open
interval can be smoothed. In turn, the ‘standard technique’ above means that
one chooses next $m_{n}$ so that the Lebesgue measure of the set
$\\{a:\,\int|\dot{\varphi^{b}}_{\kappa_{m}(s+a)-a}-\dot{\varphi^{b}}_{s}|\,ds>2^{-n}\\}$
does not exceed $2^{-n}$, then finally the almost sure convergence is due to
the Borel-Cantelli Lemma. The reasoning for the second integral is similar: we
treat the integrand as a function of $s$ and use the same freezing. that there
exists such a subsequence $m\to\infty$ (we keep notation $m$ for this
subsequence) that almost surely w.r.t. $a$, the values of the two integrals
are close to zero,

$\int_{0}^{T}|\dot{\varphi}_{s}-\dot{\chi}_{s}|\,ds+\int_{0}^{T}|L(\varphi_{s},\dot{\varphi^{b}}_{s})-L(\psi_{s},\dot{\chi}_{s})|\,ds\approx
0.$ (19)

So, we can choose a value $m$ from this subsequence ($m\to\infty$) so that,
firstly, $\Delta\leq\Delta(\nu)$ (a value from the Lemma 5); secondly,
$\int_{0}^{T}|L(\varphi_{s},\dot{\varphi^{b}}_{s})-L(\psi_{s},\dot{\chi}_{s})|\,ds\leq\nu/3$,
so that

$\left|S^{\psi}(\chi)-S(\varphi)\right|\leq\nu;$ (20)

thirdly,
$\int_{0}^{T}|\dot{\varphi}^{b}_{s}-\dot{\chi}_{s}|\,ds\leq\tilde{\delta}^{\prime}/10$,
and finally,

$\\{\rho(X^{\varphi},\varphi^{b})<\tilde{\delta}^{\prime}\\}\supset\\{\rho(X^{\varphi},\chi)<\tilde{\delta}^{\prime}\times
9/10\\}$, (21)

and (see above (16)), for $\delta^{\prime}$ small enough (i.e. we can fix the
value $\delta^{\prime}$ here),

$\\{\rho(X^{\varphi},\chi)<\tilde{\delta}^{\prime}\times
9/10\\}\supset\\{\rho(X^{\psi},\chi)<\delta^{\prime}\\}$, (22)

for the latter we need $\rho(\varphi,\chi)+\rho(\varphi,\psi)$ to be small
enough which means, in particular, $m$ large enough.

  10. 10.

For simplicity of presentation, we assume that $a=0$; otherwise the
discretisations below should be read
$\varphi^{\Delta}=(\varphi_{\Delta-\tilde{a}},\varphi_{2\Delta-\tilde{a}},\ldots,\varphi_{m\Delta-\tilde{a}},\varphi_{T})$,
where $\tilde{a}=a-[a/\Delta]$; this general case is considered similarly. So,
we assume $a=0$, and denote
$\varphi^{\Delta}=(\varphi_{\Delta},\varphi_{2\Delta},\ldots,\varphi_{m\Delta})$,
$m\Delta=T$. Since $\|f\|_{C}<\infty$, we have (see Freidlin and Wentzell,
proof of the Lemma 7.5.1),

$\framebox{$\\{\rho(X^{\psi},\chi)<\delta^{\prime}\\}\supset\\{\rho((X^{\psi})^{\Delta},\chi^{\Delta})<\delta^{\prime\prime}\\},$}$
(23)

if $\delta^{\prime\prime}$ and $\Delta$ are small enough,

$\delta^{\prime\prime}<\delta^{\prime\prime}(\delta^{\prime})\quad\mbox{and}\quad\Delta\leq\Delta(\delta^{\prime})$
(24)

(however, $\Delta\leq\Delta(\delta^{\prime\prime})$ is not required!), and
assuming all our curves start at $x_{0}$ at time zero (hence, we do not
include the starting point into the definition of $\varphi^{\Delta}$). Here
for discretised curves we use the metric,

$\rho(\psi^{\Delta},\chi^{\Delta}):=\sup_{k}|\psi_{k\Delta}-\chi_{k\Delta}|.$

Now, we are going to estimate from below the value in the right hand side of
the inequality,

$\framebox{$P(\rho((X^{\psi})^{\Delta},\chi^{\Delta})<\delta^{\prime\prime})\geq
E\prod_{k=1}^{m}I(|X^{\psi}_{k\Delta}-\chi_{k\Delta}|<\delta^{\prime}_{i}),$}$
(25)

where
$\delta^{\prime}_{1}<\delta^{\prime}_{2}<\ldots<\delta^{\prime}_{m}=\min(\delta(\nu),\delta^{\prime\prime})$,
$i=1,\ldots,m$, and $\delta(\nu)$ is from the Lemma 5; here all values
$\delta^{\prime}_{i}$ and certain auxiliary values $z_{i}$ will be chosen in
the next two steps as follows:

$m_{\nabla
H}(\delta^{\prime}_{k-1}+z_{k-1})+\frac{\kappa}{2}\,\delta^{\prime}_{k-1}\leq\frac{\kappa}{2}\,\delta^{\prime}_{k},\quad\&\quad\delta^{\prime}_{k-1}\leq\frac{\delta^{\prime}_{k}}{2},\quad\&\quad
m_{H}(\delta^{\prime}_{k})\leq\nu,$

where $0<\kappa\leq 1$, and $m_{g}$ stands for the modulus of continuity of
any function $g$ with respect to all its variables restricted to $|\beta|\leq
b+1$. Emphasize that $\delta^{\prime\prime}$ and $\Delta$ can be chosen
arbitrarily small at this stage, in particular, in addition to they should
satisfy the conditions of the Lemma 5 that will be used in the sequel, that
is, we require also $\delta^{\prime\prime}\leq\delta(\nu)$ and
$\Delta\leq\Delta(\nu)$. Hence, both $\delta^{\prime\prime}$ and $\Delta$ are
fixed at this stage.

  11. 11.

Now everything is prepared for the lower estimate. We start with the
estimation of the conditional expectation
$E(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\mid
F_{(m-1)\Delta})$ on the set
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$, or,
in the other words, on the set
$\\{X^{\psi}_{(m-1)\Delta}=\hat{\psi}_{(m-1)\Delta}\\}$ with
$|\hat{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|~{}<~{}\delta^{\prime}_{m-1}$.
Let us apply the Cramér transformation of measure. Let $|\beta|\leq b$, we
will choose this vector a bit later (as
$\beta[\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+}]$). We get,

$\displaystyle
E\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)=E^{\beta}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\times\right.$
$\displaystyle\left.\times\exp\left(-\varepsilon^{-2}\beta(X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta})+\varepsilon^{-2}\Delta
H^{\varepsilon,\psi}_{m}(\hat{\psi}_{m-1},\beta)\right)|F_{(m-1)\Delta}\right),$

where $E^{\beta}$ is the (conditional) expectation with respect to the measure
$P^{\beta}$ defined on the sigma-field $F_{m\Delta}$ given $F_{(m-1)\Delta}$,
by its density

$\frac{dP^{\beta}}{dP}(\omega)=\exp\left(\varepsilon^{-2}\beta(X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta})-\varepsilon^{-2}\Delta
H^{\varepsilon,\psi}_{m}(\hat{\psi}_{(m-1)\Delta},\beta)\right),$

where

$\varepsilon^{-2}\Delta H^{\varepsilon,\psi}_{m}(\hat{\psi}_{m-1},\beta)=\log
E\left(\exp\left(\varepsilon^{-2}\beta(X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta})\right)|F_{(m-1)\Delta}\right).$

(We remind that $\hat{\psi}_{(m-1)\Delta}=X^{\psi}_{(m-1)\Delta}$.) By virtue
of the Lemma 5, on the set
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$ we
estimate,

$\displaystyle
E\left[I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\mid
F_{(m-1)\Delta}\right]$ (26)
$\displaystyle=E^{\beta}\left[I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\right.$
$\displaystyle\left.\times\exp\left(\varepsilon^{-2}\beta(X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta})-\varepsilon^{-2}\Delta
H^{\varepsilon,\psi}_{m}(\hat{\psi}_{(m-1)\Delta},\beta)\right)\mid
F_{(m-1)\Delta}\right]$ $\displaystyle\geq
E^{\beta}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\exp\left(-\varepsilon^{-2}\Delta\beta\left((\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta\right)\right.\right.$
$\displaystyle\left.\left.-\frac{\Delta}{\varepsilon^{2}}(H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta)+\nu)-\frac{(\delta^{\prime}_{m}+\delta^{\prime}_{m-1})}{\varepsilon^{2}}\right)|F_{(m-1)\Delta}\right).$

Let us choose
$\beta=\beta(m)=\beta[\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+}]\;\;\\{=\mathop{\rm
argmax}\nolimits_{\beta}(\beta\dot{\chi}_{(m-1)\Delta+}-H(\psi_{(m-1)\Delta},\beta))\\}$.
As was explained above, $|\beta(m)|\leq b$, moreover,

$\displaystyle\beta(m)\dot{\chi}_{(m-1)\Delta+}-H(\psi_{(m-1)\Delta},\beta(m))=L(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+}),$

and

$\dot{\chi}_{(m-1)\Delta+}=\nabla_{\beta}H(\psi_{(m-1)\Delta},\beta(m)).$

So (26) implies (with $\beta=\beta(m)$),

$\displaystyle
E\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)$
$\displaystyle\geq\exp\left(-\varepsilon^{-2}\Delta(L(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+})+\nu)-\varepsilon^{-2}(\delta^{\prime}_{m}+\delta^{\prime}_{m-1})\right)\times$
$\displaystyle\times\exp\left(-\varepsilon^{-2}\Delta(H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta)-H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta))\right)$
$\displaystyle\times
E^{\beta(m)}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)$
$\displaystyle\geq\exp\left(-\varepsilon^{-2}\Delta\left(L(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+})+2\nu\right)-\varepsilon^{-2}(\delta^{\prime}_{m}+\delta^{\prime}_{m-1})\right)\times$
$\displaystyle\times
E^{\beta(m)}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right).$
(27)

We have used the uniform continuity of $H(x,\cdot,\beta)$ over $|\beta|\leq b$
and $x~{}\in~{}R^{d}$:

$\displaystyle|H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta)-H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta)|$
$\displaystyle\leq m_{H}(|\hat{\psi}_{(m-1)\Delta}-\psi_{(m-1)\Delta}|)\leq
m_{H}(\delta^{\prime}_{m-1})\leq\nu$

(here $m_{H}$ stands for the modulus of continuity of $H$ for $|\beta|\leq b$
with $b$ fixed), as $\delta^{\prime}_{m-1}$ is small enough.

  12. 12.

Let us show the bound

$E^{\beta(m)}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)\geq
1-\exp(-C_{m}\Delta\varepsilon^{-2})$ (28)

with some $C_{m}>0$ which may depend on $\nu$, on the set
$\\{|\hat{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$,
if $\varepsilon$ is small enough. There exists a finite number of vectors
$v_{1},\,v_{2},\,\ldots,v_{N}$ such that $\|v_{k}\|=1\;\forall k$, $N=2d$ (any
orthonormal basis would do accomplished by its “symmetric” transformation,
i.e. with each coordinate vector $v$ we consider $-v$ as well), and

$\displaystyle
E^{\beta(m)}(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|>\delta^{\prime}_{m})\mid
F_{(m-1)\Delta})$
$\displaystyle\leq\sum^{N}_{k=1}E^{\beta(m)}(I((X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta}-\chi_{m\Delta}+\chi_{(m-1)\Delta})v_{k}$
$\displaystyle>\kappa(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\mid
F_{(m-1)\Delta}),$

given
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$,
where $\kappa=(2/N)^{1/2}$ (notice that $\kappa\leq 1$). Given any
$\nu\,^{\prime}_{m-1}>0$ (a new constant which has nothing to do with $\nu$
and will be fixed shortly; we need it so to say only locally, while
establishing the inequality (28)), we estimate, for any $v:=v_{k}$, $0\leq
z\leq 1$,

$\displaystyle
E^{\beta(m)}\left(I((X^{\psi}_{m\Delta}-X^{\psi}_{(m-1)\Delta}-\chi_{m\Delta}+\chi_{(m-1)\Delta})v>\kappa(\delta^{\prime}_{m}-\delta^{\prime}_{m-1}))|F_{(m-1)\Delta}\right)$
$\displaystyle\leq\exp(-(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\Delta
z\kappa\varepsilon^{-2})\exp\left(\Delta\varepsilon^{-2}[-zv\dot{\chi}_{(m-1)\Delta+}\right.$
$\displaystyle\left.+H^{\varepsilon,\psi}(\hat{\psi}_{(m-1)\Delta},\beta(m)+vz)-H^{\varepsilon,\psi}(\hat{\psi}_{(m-1)\Delta},\beta(m))+2\nu\,^{\prime}_{m-1}]\right)$
$\displaystyle\leq\exp(-(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\Delta
z\kappa\varepsilon^{-2})\exp\left(\Delta\varepsilon^{-2}[-zv\dot{\chi}_{(m-1)\Delta+}\right.$
$\displaystyle\left.+H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m)+vz)\right.$
$\displaystyle\left.-H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m))+2\nu\,^{\prime}_{m-1}]\right),$
(29)

if $\varepsilon$ is small enough. Denote

$\displaystyle h(z):=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa
z+\dot{\chi}_{(m-1)\Delta+}vz$
$\displaystyle-[H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m)+vz)-H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m))].$

We have, $h(0)=0$. Moreover,

$\displaystyle
h^{\prime}(0)=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa+\dot{\chi}_{(m-1)\Delta+}v-\nabla_{\beta}H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m))v$
$\displaystyle=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa+\nabla_{\beta}H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta(m))v$
$\displaystyle-\nabla_{\beta}H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m))v$
$\displaystyle\geq(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa-m_{\nabla
H}(\delta^{\prime}_{m-1})=:C_{m-1}>0$

(where $m_{\nabla H}$ stands for the modulus of continuity of the function
$\nabla_{\beta}H$ given $|\beta(m)|\leq b+1$ ($b+1$ is needed for the sequel,
although here $b$ would be enough)), because
$\dot{\chi}_{(m-1)\Delta+}=\nabla_{\beta}H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta(m))$.
The latter inequality,
$C_{m-1}=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa-m_{\nabla
H}(\delta^{\prime}_{m-1})>0$, holds true provided $\delta^{\prime}_{m-1}$ is
small enough in compare to $(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})$,
e.g.,

$m_{\nabla
H}(\delta^{\prime}_{m-1})\leq\frac{\kappa}{2}(\delta^{\prime}_{m}-\delta^{\prime}_{m-1}),$

or, equivalently,

$m_{\nabla
H}(\delta^{\prime}_{m-1})+\frac{\kappa}{2}\,\delta^{\prime}_{m-1}\leq\frac{\kappa}{2}\,\delta^{\prime}_{m}.$
(30)

Moreover, since $\nabla_{\beta}H$ is bounded and continuous due to the Lemma
4, then $h^{\prime}(z)\geq C_{m-1}/2$ for small $z$, say, for $0\leq z\leq
z_{m-1}$ (thus, $z_{m-1}$ has been chosen here). Indeed,

$\displaystyle
h^{\prime}(z)=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa+\dot{\chi}_{(m-1)\Delta+}v$
$\displaystyle-\nabla_{\beta}H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m)+vz)v$
$\displaystyle=(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa+\nabla_{\beta}H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta(m))v$
$\displaystyle-\nabla_{\beta}H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta(m)+vz)v$
$\displaystyle\geq(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa-m_{\nabla
H}(\delta^{\prime}_{m-1}+z).$

So, $h(z_{m-1})\geq C_{m-1}z_{m-1}/2$, provided $z_{m-1}$ along with
$\delta^{\prime}_{m-1}$ are small in compare to
$(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})$, e.g.,

$m_{\nabla
H}(\delta^{\prime}_{m-1}+z_{m-1})\leq(\delta^{\prime}_{m}-\delta^{\prime}_{m-1})\kappa/2,$
(31)

rather than (30). Hence, under assumption of (31), the r.h.s. in (12) with
$z=z_{m}$ does not exceed the value

$\exp(\Delta\varepsilon^{-2}(2\nu\,^{\prime}_{m-1}-h(z)))\leq\exp(-C_{m-1}\Delta
z_{m-1}\varepsilon^{-2}/4])$

if we choose

$\nu\,^{\prime}_{m-1}<C_{m-1}z_{m-1}/8.$ (32)

Remind that the constant $\nu\,^{\prime}_{m-1}$ should be fixed in the
beginning of this step of the proof; hence, we can do it now, once we have
chosen $z_{m-1}$, since the latter does not require any knowledge of
$\nu\,^{\prime}_{m-1}$. This gives the bound, given
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$,

$E^{\beta(m)}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|\geq\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)\leq\exp(-C_{m-1}\Delta\varepsilon^{-2}/4),$

which is equivalent to (28). In turn, (28) implies the estimate

$\displaystyle
P(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m}|F_{(m-1)\Delta})$
$\displaystyle\geq\exp\left(-\varepsilon^{-2}\Delta(L(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+})+3\nu)-\varepsilon^{-2}(\delta^{\prime}_{m}+\delta^{\prime}_{m-1})\right),$

if $\varepsilon$ is small enough. Indeed, $\nu$ being fixed, one can choose
$\varepsilon$ so that

$1-\exp(-C_{m-1}\Delta\varepsilon^{-2})\geq\exp(-1)\geq\exp(-\nu(\Delta\varepsilon^{-2})).$

  13. 13.

By “backward” induction from $k=m$ to $k=1$, choosing at each step
$\delta^{\prime}_{k-1}$ and $z_{k-1}$ small enough in compare to
$\delta^{\prime}_{k}-\delta^{\prime}_{k-1}$,

$m_{\nabla
H}(\delta^{\prime}_{k-1}+z_{k-1})+\frac{\kappa}{2}\delta^{\prime}_{k-1}\leq\frac{\kappa}{2}\delta^{\prime}_{k},\;\;\&\;\;\delta^{\prime}_{k-1}\leq\delta^{\prime}_{k}/2,\;\;\&\;\;m_{H}(\delta^{\prime}_{k-1})<\nu$
(33)

(cf. (31)), as well as all auxiliary values $C_{k-1}$, we get the for
$\varepsilon$ small enough the desired lower bound:

$\displaystyle P(|X^{\psi}_{\Delta m}-\varphi_{\Delta
m}|<\delta^{\prime}_{m},\ldots,|X^{\psi}_{\Delta}-\varphi_{\Delta}|<\delta^{\prime}_{1})$
$\displaystyle\geq\exp\left(-\varepsilon^{-2}\Delta\sum^{m}_{i=1}(L(\psi_{(m-i)\Delta},\dot{\chi}_{(m-i)\Delta+})+3\nu)-2\varepsilon^{-2}\sum_{k=1}^{m}\delta^{\prime}_{k}\right)$
$\displaystyle=\exp\left(-\varepsilon^{-2}(\int_{0}^{T}L(\psi_{s},\dot{\chi}_{s})\,ds+3\nu
T)-4\varepsilon^{-2}\delta^{\prime}_{m}\right)$
$\displaystyle\geq\exp\left(-\varepsilon^{-2}(S_{0T}(\varphi)+\nu(3T+2))\right),\qquad\varepsilon\to
0,$

provided $4\delta^{\prime}_{m}<\nu$, and due to (20). This is equivalent to
(5). This bound is uniform in $x\in E^{d},\,|y|\leq r$, and
$\varphi\in\Phi_{x}(s)$ for any $r,s>0$ (similar to the Lemma 7.4.1 from
Freidlin and Wentzell (1984)).

For the reader’s convenience we remind the order of all choices by repeating
the diagram in the third step of the proof. The (any) constants $\delta>0$ and
$\nu>0$ are fixed; due to the Lemma 5, we have also $\Delta(\nu)$ and
$\delta(\nu)$; all other constants are chosen consequently as follows:

$b\,\&\,\tilde{\delta}^{\prime}\mapsto\delta^{\prime}\mapsto\Delta\,\&\,\delta^{\prime\prime}\mapsto\delta^{\prime}_{m}\mapsto\delta^{\prime}_{m-1}\mapsto
z_{m-1}\mapsto\nu\,^{\prime}_{m-1}\ldots\mapsto\delta^{\prime}_{1}\mapsto
z_{1}\mapsto\nu\,^{\prime}_{1};$

one may say that all these values have been constructed via $\nu$. Some other
constants (namely, $C_{k}$’s) are constructed via $\delta^{\prime}_{k}$’s; the
value $m$ is defined as soon as $\Delta$ is chosen, as $\Delta=T/m$; and
eventually $\varepsilon$ should be small enough to ensure all our asymptotic
inequalities.

  14. 14.

The property of the rate function $S$ to be a “good rate function” can be
shown as in Freidlin and Wentzell (1984), using the semi-continuity of the
function$L(x,y)$ w.r.t. $y$ and continuity w.r.t. $x$ variable (see Lemma
7.4.2 from Freidlin and Wentzell).

  15. 15.

Second part of the proof: the upper bound. Assume that the assertion is not
true, that is, there exist $s$ and $\nu>0$ with the following properties:

$\forall\bar{\delta}>0,\;\mbox{there
exists}\;\delta_{0}<\bar{\delta},\;\forall\bar{\varepsilon},\;\mbox{there
exists}\;\varepsilon<\bar{\varepsilon}:$
$P(\rho(X,\Phi_{x}(s))>\delta_{0})>\exp(-\varepsilon^{-2}(s-\nu)).$

In the other words, for some (hence, actually, for any) $\delta_{0}>0$
arbitrarily close to zero, there exists a sequence $\varepsilon_{n}\to 0$ such
that

$P(\rho(X,\Phi_{x}(s))>\delta_{0})>\exp(-\varepsilon_{n}^{-2}(s-\nu)).$ (34)

We fix any such $\delta_{0}>0$.

  16. 16.

Since $f$ is bounded, all possible paths $X^{\psi}$ for any $\psi$ belong to
some compact $F\subset C[0,T;R^{d}]$. Due to semi-continuity of the functional
$S^{\psi}(\varphi)$ w.r.t. $\psi$, for any $\nu>0$ there exists a value
$\delta_{\nu}(\varphi)>0$ such that $\rho(\varphi,\psi)<\delta_{\nu}(\varphi)$
and $S(\varphi)>s$ imply $S^{\psi}(\varphi)>s-\nu/2$.

Since $S^{\psi}(\varphi)$ is lower semi-continuous w.r.t. $\varphi\,$, too, it
follows that $\delta_{\nu}(\varphi)$ is also lower semi-continuous w.r.t.
$\varphi$. Hence, it attains its minimum on any compact.

Consider $F_{1}$ = the compact obtained from $F$ by dropping the
$\delta_{0}/2$-neighbourhood of the set $\Phi_{x}(s)=\\{\varphi\in
C[0,T;R^{d}]:\,\varphi_{0}=x,\,S(\varphi)\leq s\\}$. Denote
$\bar{\delta}_{\nu}=\inf_{\varphi\in F_{1}}\delta_{\nu}(\varphi)$, and take
any
$\delta^{\prime}\leq\min\left(\bar{\delta}_{\nu}/(4KT+2),\delta_{0}/2\right)$
where $K$ is a Lipschitz constant of $f$.

Choose a $\delta^{\prime}$-net in $F_{1}$, let
$\varphi^{1},\ldots,\varphi^{N}$ be its elements. All of them do not belong to
$\Phi_{x}(s)$, hence, $S(\varphi^{i})\geq s^{\prime}>s$.

  17. 17.

Note that

$\\{\rho(X,\Phi_{x}(s))>\delta_{0}\\}\subset\bigcup_{i=1}^{N}\\{\rho(X,\varphi^{i})<\delta^{\prime}\\}.$

Then, by the Dirichlet principle, for any $n$ there exists an index444Indeed,
otherwise for any $i$,  $P(\rho(X,\varphi^{i})\leq\delta^{\prime})\leq
N^{-1}\exp(-\varepsilon_{n}^{-2}(s-\nu)),$ and
$P(\rho(X,\Phi_{x}(s))>\delta_{0})\leq
P\left(\bigcup_{i=1}^{N}\\{\rho(X,\varphi^{i})<\delta^{\prime}\\}\right)\leq\exp(-\varepsilon_{n}^{-2}(s-\nu)),$
which does contradict (34). $i$ such that

$P(\rho(X,\varphi^{i})\leq\delta^{\prime})>N^{-1}\exp(-\varepsilon_{n}^{-2}(s-\nu)).$
(35)

There is a finite number of $i=1,\ldots,N$. Thus, there exists at least one
$i$ such that (35) holds true for this $i$ for some subsequence
$n^{\prime}\to\infty$ and correspondingly $\varepsilon_{n^{\prime}}\to 0$;
however, we will keep a notation $n$ for simplicity.

We can rewrite (35) as

$P(\rho(X,\varphi^{i})\leq\delta^{\prime})>\exp(-\varepsilon_{n}^{-2}(s-\nu)),$
(36)

since $N$ does not depend on $\varepsilon_{n}$, strictly speaking with a new
$\nu>0$; however, it is convenient to keep the same notation. Denote
$\varphi^{i}=:\varphi(\delta^{\prime})$.

  18. 18.

Consider a sequence $\delta^{\prime}\to 0$ such that a corresponding function
$\varphi(\delta^{\prime})$ does exist for any $\delta^{\prime}$ from this
sequence. Recall that $\delta_{0}$ is fixed. All these functions satisfy the
inequality

$S(\varphi(\delta^{\prime}))\geq s^{\prime}>s,$

since $\rho(\varphi^{i},\Phi_{x}(s))\geq\delta_{0}/2$, and also

$S(\varphi(\delta^{\prime}))<\infty,$

which implies

$\sup_{t}|\dot{\varphi}_{t}(\delta^{\prime})|\leq C.$

Due to the Arcela-Ascoli Theorem, it is possible to extract from this set of
functions a subsequence which converges in $C[0,T;R^{d}]$ to some limit,
$\bar{\varphi}$. Since
$\rho(\varphi(\delta^{\prime}),\Phi_{x}(s))\geq\delta_{0}/2$, we have,
$\rho(\bar{\varphi},\Phi_{x}(s))\geq\delta_{0}/2$, hence,

$S(\bar{\varphi})>s,$

and, in particular, the lower bound (5) can be applied. However, due to the
construction, the function $\bar{\varphi}$ satisfies one more lower bound,

$\liminf_{\delta^{\prime}\to 0}\limsup_{\varepsilon\to 0}\varepsilon^{2}\ln
P(\rho(X,\bar{\varphi})<\delta^{\prime})\geq-s+\nu.$ (37)

Indeed, the latter follows from (36) because, e.g.,

$P\left(\rho(X,\bar{\varphi})\leq\delta^{\prime}+\rho(\bar{\varphi},\varphi(\delta^{\prime}))\right)\geq
P(\rho(X,\varphi(\delta^{\prime}))\leq\delta^{\prime})>\exp(-\varepsilon_{n}^{-2}(s-\nu)).$

Due to (37), there exists $\hat{\delta}^{\prime}>0$ such that for smaller
$\delta^{\prime}$’s (a sequence)

$\limsup_{\varepsilon\to 0}\varepsilon^{2}\ln
P(\rho(X,\bar{\varphi})<\delta^{\prime})\geq-s+\nu/2.$

In fact, this implies the same inequality for any $\delta^{\prime}>0$, because
with any $\delta^{\prime}$ for which the inequality holds true, each greater
value would do as well. Therefore, for any $\delta^{\prime}$, there exists
$\varepsilon>0$ (arbitrarily small) such that

$\varepsilon^{2}\ln
P(\rho(X,\bar{\varphi})<\delta^{\prime})\geq-s+\nu/3=-(s-\nu/3).$ (38)

We are going to show that this leads to a contradiction.

  19. 19.

Consider the case $S(\bar{\varphi})<\infty$. Remind that $S(\bar{\varphi})>s.$
Denote

$\displaystyle L^{b}(x,y)=\sup_{|\beta|\leq b}(\beta y-H(x,\beta)),$
$\displaystyle\ell^{b}(x,y):=L(x,y)-L^{b}(x,y)$
$\displaystyle\equiv\sup_{\beta}(\beta y-H(x,\beta))-\sup_{|\beta|\leq
b}(\beta y-H(x,\beta)).$

Consider the function $\ell^{b}(\bar{\varphi}_{t},\dot{\bar{\varphi}}_{t})$.
We have,

$0\leq\ell^{b}(\bar{\varphi}_{t},\dot{\bar{\varphi}}_{t})\leq
L(\bar{\varphi}_{t},\dot{\bar{\varphi}}_{t}).$

Moreover,

$\ell^{b}(\bar{\varphi}_{t},\dot{\bar{\varphi}}_{t})\to 0,\quad b\to\infty,$

and the function $\ell$ is decreasing with $b\to\infty$. Hence, given $\nu>0$,
one can choose a $b>0$ such that

$\int_{0}^{T}\ell^{b}(\bar{\varphi}_{t},\dot{\bar{\varphi}}_{t})\,dt<\nu/20.$

Notice that we have chosen $b$. Moreover, one can also choose a discretisation
step $\Delta$ (see above, item 9 of the proof) such that

$\int_{0}^{T}\ell^{b}(\bar{\varphi}_{\kappa_{m}(t+a)-a},\dot{\bar{\varphi}}_{\kappa_{m}(t+a)-a})\,dt<\nu/10,$

and, correspondingly,

$\int_{0}^{T}L^{b}(\bar{\varphi}_{\kappa_{m}(t+a)-a},\dot{\bar{\varphi}}_{\kappa_{m}(t+a)-a})\,dt>s-\nu/10;$
(39)

again assume for simplicity of presentation that $a=0$. In addition, we
require $\Delta\leq\Delta(\nu/20)$ (this notation is from the Lemma 5). Hence,
we have chosen $\Delta$ and $m=T/\Delta$.

  20. 20.

So, with $a=0$, let

$\psi_{t}:=\bar{\varphi}_{\kappa_{m}(t)},\quad\dot{\chi}_{t}:=\dot{\bar{\varphi}}_{\kappa_{m}(t)},\quad\chi_{0}=x.$

We have, with a unique $C=2(KT+1)$ (see (17)) and for any $\delta^{\prime}$,

$P(\rho(X,\bar{\varphi})<\delta^{\prime})\leq
P(\rho(X^{\psi},\chi)<C\delta^{\prime})\leq
P(\rho(X^{\psi,\Delta},\chi^{\Delta})<C\delta^{\prime}).$

Denote $\delta^{\prime\prime}=C\delta^{\prime}$. Let us choose
$\delta^{\prime\prime}\leq\delta(\nu/20)$ (the notation from the Lemma 5 is
used), and consider the following inequality, with the sequence
$(\delta^{\prime}_{i},\ 1\leq i\leq m)$,
$\delta^{\prime}_{m}=\delta^{\prime\prime}$, constructed via the value
$\nu/20$ instead of $\nu$ (compare to (33), where we can now drop the
requirement related to $m_{\nabla H}$),

$P(\rho(X,\bar{\varphi})<\delta^{\prime}_{1})\leq
E\prod_{i=1}^{m}1(|X^{\psi,\Delta}-\chi^{\Delta}|<\delta^{\prime}_{i}).$

In particular, we require
$4\delta^{\prime\prime}=4\delta^{\prime}_{m}\leq\nu/20$, and
$\sum_{i=1}^{m}\delta^{\prime}_{i}\leq 2\delta^{\prime\prime}$. Then, due to
the Lemma 5 and using the calculus as in the steps (11-12), – the only change
is that now we need an upper bound, and an estimation of the indicator
function can be skipped, – we get on the set
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$ and
for any $|\beta|\leq b$,

$\displaystyle
E\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)$
$\displaystyle\leq
E^{\beta}\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\exp\left(-\varepsilon^{-2}\Delta\beta\left((\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta\right)\right.\right.$
$\displaystyle\left.\left.-\varepsilon^{-2}\Delta(H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta)-\nu/20)+\frac{\delta^{\prime}_{m}+\delta^{\prime}_{m-1}}{\varepsilon^{2}}\right)|F_{(m-1)\Delta}\right)$
(40)

(compare to (26)). Estimate here
$I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})\leq 1$ and drop
the expectation sign, then on the set
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$, for
any $|\beta|\leq b$,

$\displaystyle
E\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)$
$\displaystyle\leq\exp\left(-\varepsilon^{-2}\Delta\beta\left((\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta\right)\right.$
$\displaystyle\left.-\varepsilon^{-2}\Delta(H(\psi_{(m-1)\Delta},\hat{\psi}_{(m-1)\Delta},\beta))+\frac{\delta^{\prime}_{m}+\delta^{\prime}_{m-1}}{\varepsilon^{2}}\right),$
$\displaystyle\leq\exp\left(-\varepsilon^{-2}\Delta\beta\left((\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta\right)\right.$
$\displaystyle\left.-\varepsilon^{-2}\Delta(H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta)-\nu/20)+\frac{\delta^{\prime}_{m}+\delta^{\prime}_{m-1}}{\varepsilon^{2}}\right),$
(41)

once we have chosen $m_{H}(\delta^{\prime\prime})\leq\nu/20$ (remind that
$m_{H}$ is the modulus of continuity of the function $H$ on the set
$|\beta|\leq b$), because of the inequality
$|\hat{\psi}_{(m-1)\Delta}-\psi_{(m-1)\Delta}|\leq\delta^{\prime}_{m-1}\leq\delta^{\prime\prime}$.

Let $\beta$ satisfy a condition,

$\displaystyle\beta(\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta-H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta)$
$\displaystyle=\sup_{|\beta|\leq
b}\left(\beta(\chi_{m\Delta}-\chi_{(m-1)\Delta})/\Delta-H(\psi_{(m-1)\Delta},\psi_{(m-1)\Delta},\beta)\right)$
$\displaystyle=L^{b}(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+}).$

Then, on the set
$\\{|X^{\psi}_{(m-1)\Delta}-\chi_{(m-1)\Delta}|<\delta^{\prime}_{m-1}\\}$,

$\displaystyle
E\left(I(|X^{\psi}_{m\Delta}-\chi_{m\Delta}|<\delta^{\prime}_{m})|F_{(m-1)\Delta}\right)$
$\displaystyle\leq\exp\left(-\varepsilon^{-2}\Delta(L^{b}(\psi_{(m-1)\Delta},\dot{\chi}_{(m-1)\Delta+})+\varepsilon^{-2}\Delta\frac{\nu}{20}+\frac{\delta^{\prime}_{m}+\delta^{\prime}_{m-1}}{\varepsilon^{2}}\right).$
(42)

Similarly, using induction and due to (39), we get

$\displaystyle P(\rho(X,\chi)<\delta^{\prime}_{1})$
$\displaystyle\leq\exp\left(-\varepsilon^{-2}\int\limits_{0}^{T}L^{b}(\bar{\varphi}_{\kappa_{m}(t)},\dot{\bar{\varphi}}_{\kappa_{m}(t)})\,dt+\varepsilon^{-2}\nu/20+4\varepsilon^{-2}\delta^{\prime}_{m}\right)$
$\displaystyle\leq\exp\left(-\varepsilon^{-2}(s-\nu/5)\right).$ (43)

This evidently contradicts (38) where $\delta^{\prime}$ may be arbitrarily
small.

  21. 21.

Consider the case $\bar{\varphi}$ absolute continuous, and
$S(\bar{\varphi})=\infty$. In this case, due to monotone convergence $L^{b}\to
L$, there exist $b>0$, $m$ and $a\in[0,T]$ such that

$\int_{0}^{T}L^{b}(\bar{\varphi}_{t},\dot{\bar{\varphi}_{t}})\,dt\geq
s-\nu/20,\;\int_{0}^{T}L^{b}(\bar{\varphi}_{\kappa_{m}(t+a)-a},\dot{\bar{\varphi}}_{\kappa_{m}(t+a)-a})\,dt\geq
s-\nu/10.$

The rest is similar to the main case, $S(\bar{\varphi})<\infty$, and leads
again to

$P(\rho(X,\bar{\varphi})<\delta^{\prime}_{1})\leq\exp\left(-\varepsilon^{-2}(s-\nu/5)\right).$

This contradicts (38).

  22. 22.

Consider the last possible case, $\bar{\varphi}$ not absolute continuous. In
this case, for any constant $c$, in particular, for $c=\|f\|_{C}+1$, there
exist two values $0\leq t_{1}<t_{2}\leq T$, such that
$|\bar{\varphi}_{t_{2}}-\bar{\varphi}_{t_{1}}|>c(t_{2}-t_{1})$; indeed,
otherwise $\bar{\varphi}$ must be Lipschitz with $|\dot{\bar{\varphi}}|\leq
c$. Therefore, for $\delta<(t_{2}-t_{1})/2$, the probability
$P(\rho(X,\bar{\varphi})<\delta)$ necessarily equals zero, because the event
$\\{\rho(X,\bar{\varphi})<\delta\\}$ is empty. This evidently contradicts
(38). In all possible cases, we got to contradictions. Hence, the assumption
is wrong, that is, the upper bound (4) holds true. The Theorem is proved.

APPENDIX

A. Comments on the Lemma 6. To explain that the Lemma 6 is valid without
additional assumptions, we have to review very briefly its proof and show
those assumptions.

Let $0=t_{0}<t_{0}<\ldots<t_{m}=T$ be a partition,
$\gamma_{k}(\beta):=\int_{t_{k-1}}^{t_{k}}H(\varphi_{s},\beta)ds$,
$\ell_{k}(\alpha)=\sup_{\beta}(\alpha\beta-\gamma_{k}(\beta))$,
$A_{k}=\\{\alpha:\;\ell_{k}(\alpha)<\infty\\}$, $A_{k}^{\circ}$ its interior
w.r.t. the linear hull $L_{A_{k}}$.

The inequality
$S(\varphi)=\int_{0}^{T}L(\varphi_{t},\dot{\varphi}_{t})dt<\infty$ implies

$\sum\limits_{k=1}^{m}\sup\limits_{\beta}\left((\varphi_{t_{k}}-\varphi_{t_{k-1}})-\gamma_{k}(\beta)\right)=\sum\limits_{k=1}^{m}\ell_{k}(\varphi_{t_{k}}-\varphi_{t_{k-1}})\leq
S(\varphi).$

Under additional assumption $A^{\circ}_{k}\not=\emptyset$ it is proved in
Freidlin and Wentzell (1984) using the arguments from Rockafellar (1970) that
for any $\nu>0$, there exists a function $\tilde{\varphi}$ such that
$\rho(\varphi,\tilde{\varphi})<\nu$ and there exist $\beta_{k}$ such that

$\ell_{k}(\tilde{\varphi}_{t_{k}}-\tilde{\varphi}_{t_{k-1}})=(\tilde{\varphi}_{t_{k}}-\tilde{\varphi}_{t_{k-1}})\beta_{k}-\gamma_{k}(\beta_{k})$
(44)

and

$\tilde{\varphi}_{t_{k}}-\tilde{\varphi}_{t_{k-1}}=\nabla\gamma_{k}(\beta_{k}).$
(45)

The proof goes well if $A^{\circ}_{k}\not=\emptyset\;\forall k$.

Let us show that the same is true if $A^{\circ}_{k}=\emptyset$ for some $k$’s.
The property $A^{\circ}_{k}=\emptyset$ is equivalent to $\dim L_{A_{k}}=0$. In
this case, $\gamma_{k}(\beta)=c_{k}\beta$ with some $c_{k}\in R^{d}$. Hence,
$\ell_{k}(\alpha_{k})<\infty$ means that $\ell_{k}(\alpha_{k})=0$ and for any
other $\alpha$, $\ell_{k}(\alpha)=+\infty$ and
$\gamma_{k}(\beta)=\alpha_{k}\beta$. So, we have

$\ell_{k}(\varphi_{t_{k}}-\varphi_{t_{k-1}})=0=(\varphi_{t_{k}}-\varphi_{t_{k-1}})\beta-\gamma_{k}(\beta)$

for any $\beta$. Let $\beta_{k}=0$. Evidently,

$\varphi_{t_{k}}-\varphi_{t_{k-1}}=\nabla\gamma_{k}(\beta_{k}).$

Hence, in the case $A^{\circ}_{k}=\emptyset$, one should not just change the
curve $\varphi_{s}$ on the interval $(t_{k-1},t_{k})$; that is, (44)and (45)
are valid in this case also.

The rest of the proof is not changed. For any step function $\zeta$, one
defines a piecewise linear $\chi$ by the formula

$\chi_{0}=\varphi_{0},\quad\dot{\chi}_{s}=\nabla_{\beta}H(\zeta_{s},\beta_{k})),\;t_{k-1}<s<t_{k},\;k=1,2,\ldots,m.$

Then it is shown that $\zeta^{n}\to\varphi$ implies $\chi^{n}\to\varphi$ due
to the property that the convergence of smooth convex functions to the limit
implies the convergence of their gradients. Then there exists a partition such
that this construction gives one

$\int_{0}^{T}L(\zeta_{t},\dot{\chi}_{t})\,dt\leq S(\varphi)+\nu.$

So, the lemma holds true without additional assumptions. The assertions about
$\hat{\zeta}$ and $\hat{\beta}_{s}$ can be shown similarly.

B. Comments on the property $A^{\circ}_{k}\not=\emptyset$, and
characterization of the set ${\cal L}^{\circ}[f,x]$. Denote the interior of
$A(x)=\\{\alpha:\,L(x,\alpha)<\infty\\}$ with respect to its linear hull
$L_{A(x)}$ by $A^{\circ}(x)$. Then $A^{\circ}_{k}=\emptyset\Longleftrightarrow
A^{\circ}(\varphi_{t_{k-1}})=\emptyset$. In this section we show the following
equivalence:

$card(f\in R^{d}:\;f=f(x,y),y\in M)=1\Longleftrightarrow\dim
L_{A(x)}=0\Longleftrightarrow A^{\circ}(x)=\emptyset.$

Since $A(x)$ is convex, clearly the first two conditions are equivalent.

If $\\{f(x,\cdot)\\}$ contains only one point then $H(x,\beta)$ is linear
w.r.t. $\beta$; hence, $A(x)$ consists of a unique point and
$A^{\circ}(x)=\emptyset$.

Now, let $\\{f(x,\cdot)\\}$ contain at least two different points, say,
$f(x,y_{1})\not=f(x,y_{2})$. Then there exists $1\leq k\leq d$ such that
$(f(x,y_{1})-f(x,y_{2}))_{k}\not=0$. Denote
$M_{k}=\sup_{y}f^{k}(x,y),\;m_{k}=\inf_{y}f^{k}(x,y)$. Let
$0<\nu<(f(x,y_{1})-f(x,y_{2}))_{k}/2$. Take two points $y^{\prime}$ and
$y^{\prime\prime}$ such that $f^{k}(x,y^{\prime})<m_{k}+\nu/5$ and
$f^{k}(x,y^{\prime\prime})>M_{k}-\nu/5$. There exist two open sets
$B^{\prime}\subset M$ and $B^{\prime\prime}\subset M$ such that $\sup_{y\in
B^{\prime}}f^{k}(x,y)<m_{k}+\nu/4$ and $\inf_{y\in
B^{\prime\prime}}f^{k}(x,y)>M_{k}-\nu/4$.

Since the process $y^{x}_{t}$ is a nondegenerate ergodic diffusion, there
exists $\lambda>0$ such that

$P(y^{x}_{s}\in B^{\prime},\;1\leq s\leq t)\geq\lambda^{t-1},\quad
P(y^{x}_{s}\in B^{\prime\prime},\;1\leq s\leq t)\geq\lambda^{t-1},\quad
t\to\infty.$

Let $\beta=z\beta_{k}$ where $\beta_{k}\in E^{d}$ is a $k$th unit coordinate
vector and $z\in R$. Then for $z>0$ we have,

$\displaystyle z^{-1}t^{-1}\log
E\exp(z\beta_{k}\int_{0}^{t}f(x,y_{s}^{x})\,ds)$ $\displaystyle\geq
z^{-1}t^{-1}\log E\exp(z\beta_{k}\int_{0}^{t}f(x,y_{s}^{x})\,ds)I(y^{x}_{s}\in
B^{\prime\prime},\;1\leq s\leq t)$ $\displaystyle\geq
z^{-1}t^{-1}\log\\{\exp(z(M_{k}-\nu/2)t)\lambda^{t-1}\\}$
$\displaystyle=M_{k}-\nu/4+\frac{t-1}{t}z^{-1}\log\lambda\geq\frac{t-1}{t}M_{k}-\nu/2,$

if $z$ is large enough. In the other words, for large positive $z$ one has
$H(x,z\beta_{k})\geq z(M_{k}-2\nu)$. Similarly, for large negative $z$

$\displaystyle|z|^{-1}t^{-1}\log
E\exp(z\beta_{k}\int_{0}^{t}f(x,y_{s}^{x})\,ds)$
$\displaystyle\geq|z|^{-1}t^{-1}\log
E\exp(z\beta_{k}\int_{0}^{t}f(x,y_{s}^{x})\,ds)I(y^{x}_{s}\in
B^{\prime\prime},\;1\leq s\leq t)$
$\displaystyle\geq|z|^{-1}t^{-1}\log\\{\exp(z(m_{k}+\nu/4)t)\lambda^{t-1}\\}$
$\displaystyle=-(m_{k}+\nu/4)+\frac{t-1}{t}|z|^{-1}\log\lambda\geq-\frac{t-1}{t}m_{k}-\nu/2,$

if $|z|$ is large enough. In other words, for negative $z$ with large absolute
values one has $H(x,z\beta_{k})\geq z(m_{k}+\nu)$. Therefore,
$\\{\alpha:\;\alpha=\beta_{k}\theta,\,m_{k}+\nu<\theta<M_{k}-\nu\\}\,\subset\,A(x)$.

On the other hand, it is obvious that if $\alpha=\beta_{k}\theta$, $\theta\in
R^{1}$, with $\theta>M_{k}$ or $\theta<m_{k}$, then $L(x,\alpha)=\infty$,
because $m_{k}z\leq H(x,\beta_{k}z)\leq M_{k}z$, and, hence (say, if
$\theta>M_{k}$), for $z>>1$,

$\beta_{k}\theta\beta_{k}z-H(x,\beta_{k}z)\geq(\theta-M_{k})z\to+\infty,\quad
z\to+\infty.$

A similar calculus and inequalities are valid for any unit vector $\beta_{0}$.
This shows, in particular, that $\dim L_{A}(x)=\dim L_{f}(x)$, and, moreover,
that $L_{A}(x)=L_{f}(x)$. Since $A(x)$ is convex, it shows also that the
interior $A^{\circ}(x)$ w.r.t. $L_{A(x)}$ is not empty., except only the case
dim$(L_{A(x)})=1$. Hence, the third condition is equivalent to the second one
and the first.

So, the condition $A^{\circ}_{k}\not=\emptyset$ is always satisfied if the set
$\\{f(x,\cdot)\\}$ for any $x$ consists of more than one point. In fact, if
$card\\{f(x,\cdot)\\}=1$ for any $\,x$ then $f$ does not depend on $y$. In
this case, one has nothing to average.

Notice that our considerations above provide the following description of the
set ${\cal L}^{\circ}[f,x]$:

$\displaystyle{\cal L}^{\circ}[f,x]=\\{\alpha\in
R^{d}:\,m_{\beta}(x)<\langle\alpha,\beta\rangle<M_{\beta}(x),\;\forall|\beta|=1,\;\mbox{with}\;m_{\beta}(x)<M_{\beta}(x),$
$\displaystyle\mbox{and}\;\langle\alpha,\beta\rangle=M_{\beta}(x),\;\forall|\beta|=1,\;\mbox{with}\;m_{\beta}(x)=M_{\beta}(x)\\},$

where
$m_{\beta}(x):=\inf\limits_{y}\langle\frac{\beta}{|\beta|},f(x,y)\rangle,\,M_{\beta}(x):=\sup\limits_{y}\langle\frac{\beta}{|\beta|},f(x,y)\rangle$.
Moreover, it can be shown similarly that for any $x,\tilde{x}$ (although we do
not need it here),

${\cal L}^{\circ}[f,x,\tilde{x}]={\cal L}^{\circ}[f,x].$

C. About $\hat{\alpha}_{s}\in{\cal L}^{\circ}[f,\varphi_{s}]$. Let
$x=\varphi_{s}$, $\hat{\alpha}=\hat{\alpha}[x,\dot{\chi}]$ as described in the
proof of the theorem 1. If we show that for any direction $v$ (a unit vector)
satisfying the property $m_{v}<M_{v}$, the strict double inequality holds true

$m_{v}<\partial H(x,zv)/\partial z|_{z=0}<M_{v},$

$z\in R^{1}$, then it would follow $\hat{\alpha}_{s}\in{\cal
L}^{\circ}[f,\varphi_{s}]$. Let $\nu>0$ and again two open sets $B^{\prime}$
and $B^{\prime\prime}$ be chosen such that $\sup_{y\in
B^{\prime}}vf(x,y)<m_{v}+\nu/2$, and $\inf_{y\in
B^{\prime\prime}}vf(x,y)>M_{v}-\nu/2$. Let $\mu_{inv}(B^{\prime\prime})$ be
invariant measure for the event $\\{y^{x}_{t}\in B^{\prime\prime}\\}$. We can
choose $\nu$ and correspondingly $B^{\prime\prime}$ so that
$\mu_{inv}(B^{\prime\prime})<1$. Then, due to large deviation asymptotics for
the process $y^{x}_{t}$, for any $\mu_{inv}(B^{\prime\prime})<\zeta<1$ there
exists $\lambda>0$ such that

$P\left(t^{-1}\int_{0}^{t}1(y^{x}_{s}\in
B^{\prime\prime})\,ds\geq\zeta\right)\leq\exp(-\lambda t),\quad t\geq
t_{\zeta}.$

Denote $A_{\zeta}=\left\\{t^{-1}\int_{0}^{t}1(y^{x}_{s}\in
B^{\prime\prime})\,ds<\zeta\right\\}$,
$A^{c}_{\zeta}=\left\\{t^{-1}\int_{0}^{t}1(y^{x}_{s}\in
B^{\prime\prime})\,ds\geq\zeta\right\\}$, then for $z>0$,

$\displaystyle E\exp(zv\int_{0}^{t}f(x,y^{x})\,ds)$ $\displaystyle\leq
E\exp(z\int_{0}^{t}\left(M_{v}1(y^{x}_{s}\in
B^{\prime\prime})+(M_{v}-\nu)1(y^{x}_{s}\not\in
B^{\prime\prime})\right)\,ds)\,1(A^{c}_{\zeta})$
$\displaystyle+E\exp(z\int_{0}^{t}\left(M_{v}1(y^{x}_{s}\in
B^{\prime\prime})+(M_{v}-\nu)1(y^{x}_{s}\not\in
B^{\prime\prime})\right)\,ds)\,1(A_{\zeta})$ $\displaystyle\leq
E\exp(ztM_{v}+zt(M_{v}-\nu))\,1(A^{c}_{\zeta})+E\exp(ztM_{v}\zeta+zt(M_{v}-\nu))\,1(A_{\zeta})$
$\displaystyle\leq\exp(ztM_{v}+zt(M_{v}-\nu)-z\lambda
t/z)+\exp(ztM_{v}\zeta+zt(M_{v}-\nu)),$

hence,

$\limsup_{z\to
0}\,\limsup_{t\to\infty}\,(tz)^{-1}E\exp(zv\int_{0}^{t}f(x,y^{x})\,ds)<M_{v}.$

Similarly, using $B^{\prime}$ one can get

$\liminf_{z\to
0}\,\liminf_{t\to\infty}\,(tz)^{-1}E\exp(zv\int_{0}^{t}f(x,y^{x})\,ds)>m_{v}.$

Thus,

$m_{v}<\partial H(x,zv)/\partial z|_{z=0}<M_{v}.$

Therefore, $\hat{\alpha}\in{\cal L}^{\circ}[x,f]$.

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