Modern Stochastics: Theory and Applications

Recall that a family $\{{X}^{\varepsilon }\}$ is called exponentially tight with the speed function $r(\varepsilon )$ if for each $Q>0$, there exists a compact set $K\subset \mathbb{X}$ such that For an exponentially tight family, the LDP is equivalent to the weak LDP; the latter by definition means that the upper bound (2) holds for all compact sets F, whereas the lower bound (3) still holds for all open sets G. An equivalent formulation of the weak LDP is the following: for each $x\in \mathbb{X}$, where $B_{\delta }(x)$ denotes the open ball with center x and radius δ.

To prove (5), we will use a certain extension of the contraction principle, which in its classical form (e.g., [8], Section 3.1, and [7], Section 4.2.1) states the LDP for a family ${X}^{\varepsilon }=F({Y}^{\varepsilon })$, where ${Y}^{\varepsilon }$ is a family of random elements in a Polish space $\mathbb{Y}$ that satisfies an LDP with a good rate function J, and $F:\mathbb{Y}\to \mathbb{X}$ is a continuous mapping. The rate function for ${X}^{\varepsilon }$ in this case has the form In the sequel, we use two different representations of our particular family ${X}^{\varepsilon }$ as an image of certain family whose LDP is well understood; however, the functions F in these representations fail to be continuous. Within such a framework, the following general lemma appears quite useful. We denote by $\rho _{\mathbb{X}},\rho _{\mathbb{Y}}$ the metrics in $\mathbb{X},\mathbb{Y}$ and by $\varLambda _{F}$ the set of continuity points of a mapping $F:\mathbb{Y}\to \mathbb{X}$. Note that $\varLambda _{F}$ is Borel measurable; see Appendix II in [3].

Lemma 1 is a simplified and more precise version of Lemma 4 in [12]. The functions $I_{\mathit{upper}}$, $I_{\mathit{lower}}$ are lower semicontinuous: we can show this easily using that, for any sequence $x_{n}\to x$, the sets $\varTheta _{\delta /2}(x_{n})$ are embedded into $\varTheta _{\delta }(x)$ for n large enough (see, e.g., Proposition 3 in [12]). In fact, Lemma 1 says that for an arbitrary image of a family $\{{Y}^{\varepsilon }\}$, one part of an LDP (the upper bound) holds with one rate function, whereas the other part (the lower bound) holds with another rate function. This is our reason to call (6) and (7) the upper and the lower semicontraction principles. To prove (5), it suffices to verify the inequalities We refer to [12] for a more discussion and an example where the pair of semicontraction principles do not provide an LDP.

In this section, we prove directly that the functional I specified in Theorem 1 is lower semicontinuous, that is, it is indeed a rate functional. This will explain the particular choice of the modified functions $\bar{a},\bar{\sigma }$. In addition, this will simplify the proofs, where we will use the representation for $I(x)$ presented further.

Define Then $I(f)$, if it is finite, can be represented as The function S is continuous; hence, the functional $I_{2}$ is just continuous. The function ${\bar{a}}^{2}/{\bar{\sigma }}^{2}$ is lower semicontinuous by the choice of $\bar{a},\bar{\sigma }$; thus, the functional $I_{3}$ is lower semicontinuous. Finally, we can represent $I_{1}$ in the form $I_{1}(f)=I_{0}(\varSigma (f_{\cdot }))$, where the function is continuous, and the functional is known to be lower semicontinuous (this is just the rate functional for the family $\{\varepsilon W\}$). Hence, $I_{1}$ is lower semicontinuous, which completes the proof of the statement.

In this section, we prove that the family $\{{X}^{\varepsilon }\}$ is exponentially tight with the speed function $r(\varepsilon )={\varepsilon }^{2}$. Note that is a continuous martingale with the quadratic characteristics see (4). Recall that ${M}^{\varepsilon }$ can be represented as a Wiener process with the time change $t\mapsto \langle {M}^{\varepsilon }\rangle _{t}$; see, for example, [10], II. §7. Then, for each R, On the other hand, for each $\omega \in \varOmega $ such that $\varepsilon |{M_{t}^{\varepsilon }}(\omega )|>R$, the corresponding trajectory of ${X}^{\varepsilon }$ satisfies and therefore, by the Gronwall inequality, Therefore, for any $Q>0$, there exists R such that

Next, recall the Arzelà–Ascoli theorem: for a closed set $K\subset C(0,T)$ to be compact, it is necessary and sufficient that it is bounded and equicontinuous. The family $\varepsilon {M}^{\varepsilon }$ is represented as a time changed family $\varepsilon {W}^{\varepsilon }$, where each ${W}^{\varepsilon }$ is a Wiener process, and the derivative of $\langle {M}^{\varepsilon }\rangle _{t}$ is bounded by C. Using these observations, it is easy to deduce the exponential tightness for $\{\varepsilon {M}^{\varepsilon }\}$ using the well-known fact that the family $\{\varepsilon W\}$ is exponentially tight. On the other hand, for any ω such that the trajectory of ${X_{t}^{\varepsilon }}$ is bounded by R, the corresponding trajectory of the process ${X_{t}^{\varepsilon }}-\varepsilon {M_{t}^{\varepsilon }}$ satisfies the Lipschitz condition w.r.t. t with the constant $C(1+R)$. Combined with the previous calculation, this easily yields the required exponential tightness.

In what follows, we proceed with the proof of (5). Since now the state space $\mathbb{X}=C(0,T)$ is specified, we change the notation and denotes the points in this space by $f,g,\dots \hspace{0.1667em}$. Since the set $B_{1}(f)$ is bounded, the law of ${X}^{\varepsilon }$ restricted to any $B_{\delta }(f)$ does not change if we change the coefficients $a,\sigma $ on the intervals $(-\infty ,-R],[R,\infty )$ with $R>0$ large enough. Hence, we furthermore assume the coefficients $a,\sigma $ to be constant on such intervals for some R.

We proceed with the further proof in a step-by-step way, increasing gradually the classes of the coefficients $a,\sigma $ for which the corresponding LDP is proved. First, let $a,\sigma $ be constant on the intervals $(-\infty ,z)$ and $(z,\infty )$ with some $z\in \mathbb{R}$. Without loss of generality, we can assume that $z=0$. Then we can use Theorem 2.2 [11], where the LDP with the speed function $r(\varepsilon )={\varepsilon }^{2}$ is established for the pair $({X}^{\varepsilon },{Z}^{\varepsilon })$ with The corresponding rate function in [11] is given in the following form. Denote $a_{\pm }=a(0\pm ),\sigma _{\pm }=\sigma (0\pm )$ and define the class $H(f)$ of functions $\psi \in \mathit{AC}(0,T)$ such that Then the rate functional for $({X}^{\varepsilon },{Z}^{\varepsilon })$ equals with for all pairs $(f,\psi )$ such that $f\in \mathit{AC}(0,T),f_{0}=x_{0},\psi \in H(f)$ and, for all other pairs, $I(f,\psi )=\infty $.

From this result, using the contraction principle (see Section 2), we easily derive the LDP for ${X}^{\varepsilon }$ with the rate function for $f\in \mathit{AC}(0,T)$, $f_{0}=x_{0}$ and $I(f)=\infty $ otherwise. Now only a minor analysis is required to show that this rate function actually coincides with that specified in Theorem 1. First, we observe that This is obvious if either $a_{-}/{\sigma _{-}^{2}}\le a_{+}/{\sigma _{+}^{2}}$ or $a_{-}>0$, $a_{+}<0$. In the case where $a_{-}/{\sigma _{-}^{2}}>a_{+}/{\sigma _{+}^{2}}$ and $a_{-}$, $a_{+}$ have the same sign, we can verify directly that ${L^{\prime }_{z}}(0,y,z)$ have the same sign for $z\in [0,1]$, which completes the proof of the required identity.

We will use repeatedly the following fact, which follows easily from the change-of-variables formula: for any $f\in \mathit{AC}(0,T)$ and any set $A\subset \mathbb{R}$ with zero Lebesgue measure, the Lebesgue measure of the set see, for example, Lemma 1 in [13]. Applying (10) with $A=\{0\}$, we conclude that in the above expression for $I(f)$, the function L can be changed to which completes the proof of Theorem 1 in this case.

Let, for some $z_{1}<\cdots <z_{m}$, the functions $a,\sigma $ be constant on the intervals $(-\infty ,z_{1})$, $(z_{1},z_{2}),$ $\dots ,(z_{m},\infty )$. Assume that $x_{0}\notin \{z_{k},k=1,\dots ,m\}$, which does not restrict the generality of the construction given further, and define the functions $a_{k},\sigma _{k},k=0,\dots ,m$ by Consider a family of independent processes ${Y}^{0,\varepsilon },{Y}^{n,k,\varepsilon }$, $k=1,\dots ,m$, $n\ge 1$, such that ${Y}^{0,\varepsilon }$ solves SDE (1) with the coefficients $a_{0},\sigma _{0}$ and each ${Y}^{n,k,\varepsilon }$ solves a similar SDE with the coefficients $a_{k},\sigma _{k}$ and the initial value $z_{k}$. Define iteratively the process ${\tilde{X}}^{\varepsilon }$ in the following way: put ${\tilde{X}}^{\varepsilon }$ equal ${Y}^{0,\varepsilon }$ until the time moment Define the random index $\kappa _{1}\in \{1,\dots ,m\}$ such that ${Y_{\tau _{1}}^{0,\varepsilon }}=z_{\kappa _{1}}$. Then put ${\tilde{X}_{t}^{\varepsilon }}={Y_{t-\tau _{1}}^{1,\kappa _{1},\varepsilon }}$ until the first time moment $\tau _{2}$ when this process hits $\{z_{k},k=1,\dots ,m\}\setminus \{z_{\kappa _{1}}\}$. Iterating this procedure, we get a process ${\tilde{X}_{t}^{\varepsilon }}$ with It follows from the strong Markov property of ${X}^{\varepsilon }$ that ${\tilde{X}}^{\varepsilon }$ has the same law with ${X}^{\varepsilon }$. Hence, the given construction in fact represents the law of ${X}^{\varepsilon }$ as the image of the joint law family of independent processes ${Y}^{0,\varepsilon },{Y}^{n,k,\varepsilon }$, $k=1,\dots ,m$, $n\ge 1$. Each of these processes is a solution to (1) with corresponding coefficients having at most one discontinuity point; hence, the LDP for them is provided in the previous section. Our idea is to deduce the LDP ${X}^{\varepsilon }$ via a version of the contraction principle. With this idea in mind, we first perform a simplification of the above representation. For some N (the choice of N will be discussed below), we consider the space $\mathbb{Y}=C{(0,T)}^{1+mN}$ and construct a function $F:\mathbb{Y}\to \mathbb{X}=C(0,T)$ in the following way. For $y=({y}^{0},{y}^{n,k},k=1,\dots ,m,n=1,\dots ,N)$, we first define $\tau _{1}(y)=\inf \{t:{y_{t}^{0}}\in \{z_{k},k=1,\dots ,m\}\}$ with the usual convention that $\inf \varnothing =T.$ The function $[F(y)]_{t}$, $t\in [0,T]$, is defined to be equal to ${y_{t}^{0}}$ for $t\le \tau _{1}(y)$. If $\tau _{1}(y)<T$, then the construction is iterated: we define $\kappa _{1}(y)$ by ${y_{\tau _{1}(y)}^{0}}=x_{\kappa _{1}(y)}$ and put, for $t\ge \tau _{1}(y)$, $[F(y)]_{t}$ equal to ${y_{t-\tau _{1}(y)}^{1,\kappa _{1}(y)}}$ up to the first moment when this function hits $\{z_{k},k=1,\dots ,m\}\setminus \{x_{\kappa _{1}(y)}\}$. We iterate this procedure at most N times; that is, if $\tau _{N}(y)<T$, then we put Denote For any fixed $f\in C(0,T)$, we can choose $\delta _{f}>0$ small enough and $N_{f}$ large enough so that each $g\in B_{\delta }(f)$ has less than N Δ-oscillations on $[0,T]$. Hence, if in the above construction, N is taken equal to $N_{f}$, then the restriction of the law of ${X}^{\varepsilon }$ to any ball $B_{\delta }(f),\delta \le \delta _{f}$, equals to the same restriction of the image of the joint law of the finite family ${Y}^{0,\varepsilon }$, ${Y}^{n,k,\varepsilon }$, $k=1,\dots ,m$, $n=1,\dots ,N$, under the mapping F specified before.

We aim to verify (5), and we argue in the following way. We fix f and choose $N=N_{f}$ as before, so that the laws of ${X}^{\varepsilon }$, restricted to $B_{\delta }(f)$ for δ small, can be obtained as the image under F specified before. Then we prove (8) at this particular point $x=f$, with $I_{\mathit{lower}}$, $I_{\mathit{upper}}$ being constructed by this particular F. This yields the required weak LDP (5).

Within such an argument, we have to treat for any N the image under the corresponding F of the family of laws in $\mathbb{Y}=C{(0,T)}^{1+mN}$, which, according to the result proved in the previous section, satisfies the LDP with the rate function for such that ${y}^{0},{y}^{n,k}\in \mathit{AC}(0,T)$, ${y_{0}^{0}}=x_{0}$, ${y_{0}^{n,k}}=z_{k}$ and $J(y)=\infty $ otherwise. To apply Lemma 1 in this setting, we first analyze the structure and the properties of the corresponding F.

Each trajectory $f=F(y)\in C(0,T)$ is actually a patchwork, which consists of pieces of trajectories ${y}^{0},{y}^{n,k}$, $k=1,\dots ,m$, $n=1,\dots ,N$: the pasting points are $\tau _{1}(y),\dots ,\tau _{r}(y),$ $r=r(y)\le N$, and after $\tau _{n}(y)$, the (part of the) new trajectory is used with the number $\kappa _{n}(y)$. For a $y_{l}\to y$ in $\mathbb{Y}$, the corresponding sequence of trajectories $f_{l}=F(y_{l})$ may fail to converge to f because the functionals $\tau _{n}(\cdot ),\kappa _{n}(\cdot )$ are not continuous. However, the above “patchwork representation” easily yields the following two facts:

Using the first fact, now it is easy to prove the second inequality in (8). If it fails, then for a given f, there exists a sequence $\{{y}^{l}\}$ such that $F(y_{l})\to f$ and $J(y_{l})\le c<I(f)$. Since the level set $\{y:J(y)\le c\}$ is compact, we can assume without loss of generality that $y_{l}$ converge to some y; recall that J is lower semicontinuous and thus $J(y)\le c$. The function f possesses the above patchwork representation with the trajectories taken from y, some pasting points ${\tau _{1}^{\ast }},\dots ,{\tau _{r}^{\ast }}$, and some numbers ${\kappa _{1}^{\ast }},\dots ,{\kappa _{r}^{\ast }}$. From this representation it is clear that $f\in \mathit{AC}(0,T)$ and $f_{0}=x_{0}$: if this fails, then the same properties fail at least for one trajectory from the family y and thus $J(y)=\infty $, which contradicts to $J(y)\le c$. Hence, we have where we put ${\tau _{0}^{\ast }}=0,{\tau _{r+1}^{\ast }}=T$. Let $x_{0}$ be located on some interval $(z_{k-1},z_{k})$, $k=2,\dots ,m$, say, $x_{0}\in (z_{1},z_{2})$. Then, on the interval $(0,{\tau _{1}^{\ast }})$, the trajectory f is contained in the segment $[z_{1},z_{2}]$. The functions $a_{0},\sigma _{0}$ are constant and coincide with $\bar{a},\bar{\sigma }$ on $(z_{1},z_{2})$. In addition, $a_{0}=a(z_{1}+)=a(z_{2}-)$, $\sigma _{0}=\sigma (z_{1}+)=\sigma (z_{2}-)$; hence, by the choice of $\bar{a},\bar{\sigma }$ we have Then by (10) with $A=\{z_{1},z_{2}\}$ we have Analogous inequalities hold on each of the time intervals $({\tau _{n}^{\ast }},{\tau _{n+1}^{\ast }}),$ $n=1,\dots ,r$, with $a_{0},\sigma _{0}$ changed to $\bar{a}_{{\kappa _{n}^{\ast }}},\bar{\sigma }_{{\kappa _{n}^{\ast }}}$ (the proof is similar and omitted). Thus, This gives a contradiction with inequalities $J(y)\le c$ and $I(f)>c$, which completes the proof of the second inequality in (8).

The first inequality in (8) holds immediately for f such that $I(f)=\infty $. We fix f with $I(f)<\infty $ and $\gamma >0$ and construct $y_{\gamma }$ such that $F(y_{\gamma })=f$, the functions $\tau _{1}(\cdot ),\tau _{2}(\cdot ),\dots \hspace{0.1667em}$ are continuous at $y_{\gamma }$, and $J(y_{\gamma })\le I(f)+\gamma $. This completes the proof of (8).

The construction explained gives a cue for the choice of $y=y_{\gamma }$ (we omit the index γ to simplify the notation). We put ${y}^{0}$ equal to f until its first time moment ${\tau _{1}^{\ast }}$ of hitting the set $\{z_{1},z_{2}\}$ (we still assume that $x_{0}\in (z_{1},z_{2})$). Then we extend ${y}^{0}$ to the entire time interval $[0,T]$, and we aim to make the integral small enough; that is, to make small the error in the second inequality in (11), which arises because of the integral of ${y}^{0}$. If we put ${y_{t}^{0}}=y_{{\tau _{1}^{\ast }}}+a_{0}(t-{\tau _{1}^{\ast }})$, then we obtain the trajectory at which the integral (12) equals zero; we call such a trajectory a zero-energy one. However, under such a choice, we may fail with the other our requirement that $\tau _{1}(\cdot )$ should be continuous at the point y. It is easy to verify that for such a continuity, it suffices that ${y}^{0}$, if hitting $\{z_{1},z_{2}\}$ at a point, say, $z_{1}$ at every interval $({\tau _{1}^{\ast }},{\tau _{1}^{\ast }}+\delta ),\hspace{2.5pt}\delta >0$, takes values both from $(-\infty ,z_{1})$ and $(z_{1},\infty )$. We can perturb the zero-energy trajectory introduced above on a small time interval near ${\tau _{1}^{\ast }}$ in such a way that this new trajectory possesses the continuity property explained before, and the integral (12) is $\le \gamma /N$.

Then we iterate this procedure. Observe that, for any k, by the construction of the function $\bar{a}_{k}$ there exists at least one corresponding zero-energy trajectory with the initial value $z_{k}$, which now is defined as a solution to the ODE We have ${\kappa _{1}^{\ast }}$ uniquely determined by the trajectory f (in fact, by the part of this trajectory up to time ${\tau _{1}^{\ast }}$). For $k\ne {\kappa _{1}^{\ast }}$, we define ${y}^{k,1}$ as the zero-energy trajectory on $[0,T]$ that starts from $x_{k}$ and corresponds to the coefficient $\bar{a}_{k}$. All these trajectories are “phantom” in the sense that they neither are involved into the representation of f through y nor give an impact into $J(y)$. For $k={\kappa _{1}^{\ast }}$, we define ${y}^{k,1}$ similarly as before: it equals $f_{t+{\tau _{1}^{\ast }}}$ for $t\le {\tau _{2}^{\ast }}-{\tau _{1}^{\ast }}$, and afterwards it is defined as a perturbation of a zero-energy trajectory that makes $\tau _{2}(\cdot )$ continuous in y and Repeating this construction $\le N$ times, we finally get the required function $y=y_{\gamma }$. This completes the proof of (8) and thus of (5). Together with the exponential tightness proved in Section 3.1, this completes the proof of the LDP in this case.

In this section, we remove the assumption on $a,\sigma $ to be piecewise constant, still keeping this assumption for $a/{\sigma }^{2}$; we also assume that $a,\sigma $ are constant on $(-\infty ,R]$ and $[R,\infty )$ for some R. Our basic idea is to represent $\{{X}^{\varepsilon }\}$ as the image under a time changing transformation of a family $\{{Y}^{\varepsilon }\}$ and then to use the semicontraction principles. The same approach was used in [13], where the LDP was established for a solution of (1) with $a\equiv 0$; in this case, ${Y}^{\varepsilon }$ was taken in the form ${Y_{t}^{\varepsilon }}=x_{0}+\varepsilon W_{t}$. In our current setting, the choice of the coefficients for the SDE that defines ${Y}^{\varepsilon }$ should take into account the common discontinuity points for $a/\sigma $ and σ. This becomes visible both from an analysis of the proof of Theorem 1 in [13] and from the definition of the functions $\bar{a},\bar{\sigma }$, which combines the left- and right-hand values of both a and σ at the discontinuity points. The proper choice of the family is explained below. Some parts of the arguments are similar to those in [13]. We omit detailed proofs whenever it is possible to give a reference to [13] and focus on the particularly new points.

We assume $a/{\sigma }^{2}$ to be piecewise constant with discontinuity points $z_{1}<\cdots <z_{m}$ and put (with the convention $\prod _{\varnothing }=1$) Under such a choice, $\tilde{\sigma }=\sigma \upsilon $, and thus the function $\tilde{a}/{\tilde{\sigma }}^{2}$ equals $a/{\sigma }^{2}$ and is constant on each of the intervals $(-\infty ,z_{1}),\dots ,(z_{m},\infty )$. By construction, $\tilde{\sigma }$ is constant on these intervals as well; hence, $\tilde{a},\tilde{\sigma }$ fit the case studied in the previous section, and the required LDP holds for the family ${Y}^{\varepsilon }$ of the solutions to (1) with these coefficients and ${Y_{0}^{\varepsilon }}=x_{0}$. This construction yields also the following property, which will be important below: the function $a=(a/{\sigma }^{2}){\sigma }^{2}$ does not change its sign on each of the intervals $(-\infty ,z_{1}),\dots ,(z_{m},\infty )$. Hence, denoting $B={a}^{2}/{\sigma }^{2}$ and $\bar{B}={\bar{a}}^{2}/{\bar{\sigma }}^{2}$, we get

Fix $\varepsilon >0$ and define $\tau _{t}={[\eta ]_{t}^{-1}}$ (the inverse function w.r.t. t), and ${X_{t}^{\varepsilon }}={Y_{\tau _{t}}^{\varepsilon }}$. Then ${X}^{\varepsilon }$ is a weak solution to (1) with ${X_{0}^{\varepsilon }}=x_{0}$; see [10], IV §7.

In the above construction, $\eta _{t}\ge {c}^{2}t$ and thus $\tau _{t}\le {c}^{-2}t$; see (4). We put $\tilde{T}={c}^{-2}T$, $\mathbb{Y}=C(0,\tilde{T})$, and define ${Y}^{\varepsilon }$ as a family of solutions to (1) with the coefficients $\tilde{a},\tilde{\sigma }$ and the time horizon $\tilde{T}$. Then the family ${X}^{\varepsilon }$ possesses a representation ${X}^{\varepsilon }=F({Y}^{\varepsilon })$ with the mapping $F:\mathbb{Y}\to \mathbb{X}$ defined by Observe that for F to be continuous at a point $y\in \mathbb{Y}$, it suffices that y spends zero time in the set $\varDelta _{\upsilon }$ of the discontinuity points of the function υ; see [13], Lemma 1 and Corollary 1. Now $\varDelta _{\upsilon }\subset \varDelta _{a}\cup \varDelta _{\sigma }$ is at most countable, and it is easy to see that the continuity set $\varLambda _{F}$ has probability 1 w.r.t. the distribution of each ${Y}^{\varepsilon }$, that is, we can apply Lemma 1.

Our further aim is to prove (8) in the above setting, which then would imply (5) and thus prove the LDP. The general idea of the proof is similar to that of Theorem 1 in [13], though particular technicalities differ substantially.

First, for a given $f\in \mathbb{X}$, we describe explicitly the set ${F}^{-1}(\{f\})$. We put If $f=F(y)$, then here we changed the variables $r=\tau _{s}(y)$ and used that Therefore, Observe that $\zeta _{T}(f)\le {c}^{-2}T=\tilde{T}$ and define Then we conclude that that is, for any $y\in {F}^{-1}(\{f\})$, the part of its trajectory with $t\le \zeta _{T}(f)$ is uniquely defined. On the other hand, it is easy to show that any $y\in \mathbb{Y}$ satisfying (14) belongs to ${F}^{-1}(\{f\})$.

Next, we denote by $\hat{a},\hat{\sigma }$ the modified coefficients, which correspond to the coefficients $\tilde{a},\tilde{\sigma }$ in the sense explained in Section 1. Since $\tilde{a}=a{\upsilon }^{2}$ and $\tilde{\sigma }=\sigma \upsilon $, we easily see that at every continuity point x for υ. Then, for any $y\in \mathit{AC}(0,\tilde{T})$ with $y_{0}=x_{0}$ that spends zero time in the set $\varDelta _{\upsilon }$, we have On the other hand, using (14) and making the time change $s=\pi _{t}(f)$ with $f=F(y)$, we get because now $t=\zeta _{s}(f)=\tau _{s}(y)$ and thus Thus, with

Now we are ready to proceed with the proof of the first inequality in (8).

Recall that B and $\bar{B}$ satisfy (13). For the similar pair of functions $\tilde{B}={\tilde{a}}^{2}/{\tilde{\sigma }}^{2}$ and $\hat{B}={\hat{a}}^{2}/{\hat{\sigma }}^{2}$, we have even more: the functions $\hat{a},\hat{\sigma }$ are constant on each of the intervals $(-\infty ,z_{1}),\dots ,(z_{m},\infty )$; hence, On the other hand, since $\tilde{a}=a{\upsilon }^{2}$ and $\tilde{\sigma }=\sigma \upsilon $, we have $B=\tilde{B}{\upsilon }^{-2}$, and thus This yields, for $z\notin \{z_{k}\}$, Recall that υ is continuous at each point $z_{k}$; hence, by (15) identity (18) holds for all $z\in \mathbb{R}$.

Now we are ready to proceed with the proof of the second inequality in (8).

In this last part of the proof, we remove the assumption $a/{\sigma }^{2}$ to be piecewise constant and prove the required statement in the full generality. According to Section 2.1, it suffices to prove that, for fixed $f\in C(0,T)$ and $\varkappa >0$, for some $\delta >0$ and for each $\delta >0$. While doing that, we can and will assume that $a,\sigma $ are constant on $(-\infty ,R]$ and $[R,\infty )$ for some R.

Consider, together with the original SDE (1), the SDE where α is a bounded function to be specified later. Then by the Girsanov theorem see [10], Chapter IV, Theorem 4.2. Then, for arbitrary $p,q>1:1/p+1/q=1$, we have Let $|\alpha (x)|\le \gamma $. Then we have Since α is bounded, ${\mathcal{E}}^{q\varepsilon }$ is a martingale. Thus, we can summarize the above calculation as follows:

In what follows, we will choose α such that the function $(a+\alpha \sigma )/{\sigma }^{2}$ is piecewise constant. Then the result proved in the previous section will provide that, for given $f\in C(0,T)$ and $\kappa >0$, there exists $\delta >0$ such that where $\tilde{I}$ is the rate functional that corresponds to the new drift coefficient $a+\alpha \sigma $ and the same diffusion coefficient. It is easy to verify using the representation (9) and its analogue for $\tilde{I}$ that we can choose γ small enough so that the above construction with arbitrary α such that $\| \alpha \| \le \gamma $ yields In that case, we get Now we are ready to summarize the entire argument. For given $f\in C(0,T)$ and $\kappa >0$, we take $p>1$ close enough to 1 such that Then we take $\gamma >0$ small enough such that (27) holds and

Observe that the function $a/{\sigma }^{2}$ has left and right limits at every point and is constant on $(-\infty ,R]$ and $[R,\infty )$. Then it can be approximated by piecewise constant functions in the uniform norm. This means that we can find α with $\| \alpha \| \le \gamma $ such that the function $(a+\alpha \sigma )/{\sigma }^{2}$ is piecewise constant. Then there exists $\delta >0$ such that (26) holds, and we finally deduce which completes the proof of (23).

Exactly the same argument provides the proof of (24) as well, with the minor change that now the law of ${Y}^{\varepsilon }$ should be expressed in the terms of ${X}^{\varepsilon }$; that is, we should use the Girsanov theorem in the following form: The rest of the proof remains literally the same; we omit the detailed exposition.