Let $f\in {C^{3}}(\mathbb{R})$ be such that and suppose that ${f^{\prime \prime \prime }}$ is Lipschitz continuous.

Set and define $g:[0,1]\to {\mathbb{R}_{+}}$ by Then g is a continuous function and

We start with some preparations. For $n\in \mathbb{N}$ and $j=0,1,\dots ,n-1$, let and Set Then

Clearly, for all $n\in \mathbb{N}$ and $j=0,1,...,n-1$ we have By the fundamental theorem of calculus, for the first difference we have For the second difference, the one-dimensional Taylor formula with integral remainder, applied to $F(x):=u(\frac{j}{n},x)$ and $h:=\frac{{Z_{j}^{n}}}{\sqrt{n}}$, gives Since, by Lemma 1, ${u_{t}}$ and ${u_{\mathit{xx}}}$ are Lipschitz continuous on $[0,1]\times \mathbb{R}$, combining (2) and (3) we obtain that there exists a constant $\hat{L}\gt 0$ such that, for all $n\ge 2$ and $j=0,1,\dots ,n-1$, In particular, using ${u_{t}}=-{u_{\mathit{xx}}}/2$ and increasing $\hat{L}$ if necessary, for all $n\ge 2$,

Also we observe that by applying (2), (3) and the bounds $|{u_{t}}|\le b$ and ${u_{\mathit{xx}}}\le a$ we have

We finish the preparations by noting that since ${u_{\mathit{xx}}}(t,x)\le a\lt 1$ for all $(t,x)\in [0,1]\times \mathbb{R}$, and Ψ is Lipschitz continuous on $(-\mathrm{\infty },a]$, Lemma 1 implies that $\Psi ({u_{\mathit{xx}}}(t,x))$ is Lipschitz continuous on $[0,1]\times \mathbb{R}$. Hence, g is also Lipschitz continuous.

We now divide the remainder of the proof of Theorem 1 into two steps.

Step I. In this step, we show the upper bound

Choose $n\in \mathbb{N}$. Clearly, ${S_{n}}={\sum _{j=0}^{n-1}}{I_{j}^{n}}$.

Thus, to complete the proof of the upper bound, it suffices to show that there exists a constant $L\gt 0$ such that, for all $n\ge 2$ and $j=0,1,\dots ,n-1$,

Indeed, (6) implies that for every $\epsilon \gt 0$ and all sufficiently large n,

Hence, by the tower property of conditional expectation,

Since g is continuous, Since $\epsilon \gt 0$ was arbitrary, we obtain the desired upper bound.

It remains to prove (6). Fix j and define the event We split the exponential according to the disjoint events ${A_{j}^{n}}$ and ${({A_{j}^{n}})^{c}}$. Applying (4) on ${A_{j}^{n}}$ and (5) on its complement gives We now take conditional expectations and bound the first term by dropping its indicator.

Since ${Z_{j}^{n}}\sim N(0,1)$ is independent of ${\mathcal{F}_{j/n}}$, we obtain Finally, by a direct estimate for the standard normal variable ${Z_{j}^{n}}$, This yields (6) and completes the first step.

Step II. Fix $\epsilon \gt 0$. By Lemma 2, choose a continuous, piecewise-linear function $h:[0,1]\to \mathbb{R}$, with $h(0)=0$, such that Choose $\delta \gt 0$ such that, for all $t\in [0,1]$ and $x,y\in \mathbb{R}$, For $n\ge 2$, retain the notation ${I_{j}^{n}}$, ${Z_{j}^{n}}$ from the preparations and set The local remainder estimate (4) holds on ${F_{n}}$. Consequently, Define Then set By (1), $\mathbb{E}[{D_{j}^{n}}|{\mathcal{F}_{j/n}}]=1$; successive conditioning therefore shows that this density has expectation one. Changing measure and then restricting to ${E_{n}}$ gives We prove that ${\liminf _{n}}{\mathbb{Q}_{n}}({E_{n}}\cap {F_{n}})\gt 0$.

Write $m={\inf _{x\in \mathbb{R}}}{f^{\prime \prime }}(x)\gt -\mathrm{\infty }$ and retain $a={\sup _{x\in \mathbb{R}}}{f^{\prime \prime }}(x)\lt 1$. Since ${u_{\mathit{xx}}}(t,x)=\mathbb{E}[{f^{\prime \prime }}(x+{W_{1-t}})]$, we have $m\le {u_{\mathit{xx}}}(t,x)\le a$. Define Then $0\lt \underline{\sigma }\le \sigma (t,x)\le \overline{\sigma }\lt \mathrm{\infty }$. If ${L_{q}}$ is a Lipschitz constant for ${u_{\mathit{xx}}}$ with respect to the Euclidean metric, the mean value theorem gives

Successive conditioning of the factors ${D_{k}^{n}}$ with $k\gt j$ shows that the conditional density of ${Z_{j}^{n}}$ under ${\mathbb{Q}_{n}}$, given ${\mathcal{F}_{j/n}}$, is Indeed, this density is the standard normal density multiplied by $\exp ({c_{j}^{n}}({z^{2}}-1)/2-\Psi ({c_{j}^{n}}))$. Thus, conditionally on ${\mathcal{F}_{j/n}}$, ${Z_{j}^{n}}$ is centered normal with variance ${\sigma ^{2}}(j/n,{W_{j/n}})$. The Gaussian tail bound and the union bound yield

To apply Corollary 7.4.2 of [6], we include time in the state. Extend σ to ${\mathbb{R}^{2}}$ by $\tilde{\sigma }(s,x)=\sigma (\max (0,\min (s,1)),x)$. The extension is bounded and Lipschitz. For $z=(s,x)$, let ${\mu _{n}}(z,\cdot )$ be the law of $z+{H_{n}}(z)$, where Under ${\mathbb{Q}_{n}}$, the chain ${(j/n,{W_{j/n}})_{j=0}^{n}}$ has this time-homogeneous transition kernel and starts at $(0,0)$. Extend it past time 1 with the same kernel if necessary.

The limiting drift and covariance are $\boldsymbol{b}={(1,0)^{\top }}$ and $\boldsymbol{a}(z)=diag(0,{\tilde{\sigma }^{2}}(z))$. Expectations in the following estimates are over Z. Uniformly in z, where $|\cdot |$ is the Euclidean norm, ⊤ denotes transpose, and ${C_{0}}=1+2\overline{{\sigma ^{2}}}+3\overline{{\sigma ^{4}}}$. For the truncated coefficients with their dependence on z suppressed, the fourth-moment bound gives where $\|\cdot \|$ is the matrix operator norm. Moreover, for every $r\gt 0$, Thus the coefficient limits and large-jump condition of the corollary hold. For smooth, compactly supported test functions $\varphi :{\mathbb{R}^{2}}\to \mathbb{R}$, the limiting generator is Its martingale problem is well-posed for every initial distribution: the stochastic differential equation (SDE) $d{T_{t}}=\mathit{dt}$, $d{Y_{t}}=\tilde{\sigma }({T_{t}},{Y_{t}})\hspace{1.67pt}d{B_{t}}$, where B is a standard one-dimensional Brownian motion, has bounded Lipschitz coefficients and a unique nonexplosive strong solution for every initial point. Also, a is continuous, symmetric and nonnegative definite, as required; nondegeneracy is not assumed in the corollary.

Write $D([0,1])$ for the space of real-valued right-continuous paths with left limits and $C([0,1])$ for its continuous-path subspace; ${\|v\|_{\mathrm{\infty }}}={\sup _{0\le t\le 1}}|v(t)|$. Corollary 7.4.2, applied with initial law the point mass at $(0,0)$, and projection onto the second coordinate therefore yield where $d{X_{t}}=\sigma (t,{X_{t}})\hspace{1.67pt}d{B_{t}}$, ${X_{0}}=0$. Here ${T_{t}}=t$ for the limiting process, and restriction to $[0,1]$ is valid because its paths are continuous. Convergence on the left refers to the laws under ${\mathbb{Q}_{n}}$.

Define the linear interpolation by for $j=0,\dots ,n-1$. Then ${\|{W^{n}}-{V^{n}}\|_{\mathrm{\infty }}}\le {\max _{j\lt n}}|{W_{(j+1)/n}}-{W_{j/n}}|$, and, for every $r\gt 0$, The perturbation theorem and continuity of X now give ${W^{n}}\Longrightarrow X$ in $C([0,1])$ with the uniform topology (see [2, Sections 3 and 12]).

Denote the probability governing X by P.

We also verify directly that Since h is continuous and piecewise linear, it is absolutely continuous with bounded derivative, defined almost everywhere. Set ${\theta _{t}}={h^{\prime }}(t)/\sigma (t,{X_{t}})$. The lower bound on σ implies $|{\theta _{t}}|\le {\|{h^{\prime }}\|_{\mathrm{\infty }}}/\underline{\sigma }$, so Novikov’s condition holds. Girsanov’s theorem gives a probability R, equivalent to P on the time interval $[0,1]$, with a Brownian motion under R. Hence By the Dambis–Dubins–Schwarz theorem, on an extension if necessary, ${M_{t}}={\beta _{{\langle M\rangle _{t}}}}$ for a Brownian motion β. One may extend M after time 1 by an independent Brownian motion so that its quadratic variation tends to infinity. Therefore Here the final inequality is the standard positive probability that a Brownian motion remains in an open interval for a finite time. Equivalence of R and P proves (8). For these standard stochastic-calculus results, see [17, 18].

Finally the set $G=\{x\in C([0,1]):{\|x-h\|_{\mathrm{\infty }}}\lt \delta \}$ is open, and $\{{W^{n}}\in G\}\subseteq {E_{n}}$. The Portmanteau theorem gives It follows that Returning to the lower bound for $\mathbb{E}[{e^{n{S_{n}}}}]$, taking logarithms, dividing by n and using the Riemann-sum convergence, we obtain Since $\epsilon \gt 0$ was arbitrary, this completes the proof. □

It is natural to ask whether Theorem 1 can be extended to a multidimensional Brownian motion or to more general diffusion processes. Since the aim of this note is to focus on the one-dimensional case, we do not study these extensions here. A full treatment would require further analysis and is left for future research.

Let f satisfy the assumptions of Theorem 1. Then ${u_{\mathit{xx}}}$ and ${u_{t}}=-{u_{\mathit{xx}}}/2$ are Lipschitz continuous on $[0,1]\times \mathbb{R}$ with respect to the Euclidean metric.

Set $A={\|{f^{\prime \prime }}\|_{\mathrm{\infty }}}\lt \mathrm{\infty }$ and let $K\ge 0$ be a Lipschitz constant for ${f^{\prime \prime \prime }}$. Taylor’s formula gives Consequently, This argument also covers $A=0$ and $K=0$, without division by either constant. The representations show that $|q(t,x)-q(t,y)|\le {B_{0}}|x-y|$ and that ${q_{x}}(t,\cdot )$ is K-Lipschitz. Hence, for every $z\in \mathbb{R}$, If $0\le s\lt t\le 1$, the heat-semigroup identity and a standard normal variable Z give Using the remainder bound and $\mathbb{E}Z=0$, $\mathbb{E}{Z^{2}}=1$, we obtain Thus, for arbitrary $(t,x)$, $(s,y)$, The assertion for ${u_{t}}$ follows from the heat equation. □

Let f satisfy the assumptions of Theorem 1. For every $\epsilon \gt 0$, there exists a continuous, piecewise-linear function $h:[0,1]\to \mathbb{R}$, which depends on ϵ, such that $h(0)=0$ and

Again, we notice that for the constant $a:={\sup _{x\in \mathbb{R}}}{f^{\prime \prime }}(x)\lt 1$, we have ${u_{\mathit{xx}}}(t,x)\le a$ for all $(t,x)\in [0,1]\times \mathbb{R}$. Since the function Ψ is Lipschitz continuous on $(-\mathrm{\infty },a]$, Lemma 1 implies that $\Psi ({u_{\mathit{xx}}}(t,x))$ is Lipschitz continuous on $[0,1]\times \mathbb{R}$. Hence, g is also Lipschitz continuous.

Fix $\epsilon \gt 0$. By the uniform continuity of $\Psi ({u_{\mathit{xx}}}(t,x))$ and g, we may choose a partition $0={t_{0}}\lt {t_{1}}\lt \cdots \lt {t_{N}}=1$ and points ${x_{0}},\dots ,{x_{N-1}}\in \mathbb{R}$ such that the piecewise-constant function satisfies

Notice that for the constant $b:=\frac{1}{2}{\sup _{x\in \mathbb{R}}}|{f^{\prime \prime }}(x)|$, we have $|{u_{\mathit{xx}}}(t,x)|\le 2b$ for all $(t,x)\in [0,1]\times \mathbb{R}$. Together with the relation ${u_{\mathit{xx}}}(t,x)\le a$ for all $(t,x)\in [0,1]\times \mathbb{R}$, this yields $M:={\sup _{(t,x)\in [0,1]\times \mathbb{R}}}\Psi ({u_{\mathit{xx}}}(t,x))\lt \mathrm{\infty }$. Next, replace each jump of $\tilde{h}$ by a linear interpolation on sufficiently short disjoint intervals whose total length is less than $\frac{\epsilon }{8(1+M)}$ and, on an initial interval of length less than $\frac{\epsilon }{8(1+M)}$, interpolate linearly from 0 to ${x_{0}}$. In this way, we obtain a continuous, piecewise-linear function h satisfying $h(0)=0$, which differs from $\tilde{h}$ only on a set $A\subset [0,1]$ with $|A|\lt \frac{\epsilon }{4(1+M)}$. Set $h(1)={x_{N-1}}$. Therefore, This together with (9) gives □