Gaussian smoothing density and forward equation

For $a>0$, direct differentiation gives ${\mathrm{\partial }}_a{\varphi }_a(z)=\frac{1}{2}{\mathrm{\Delta }}_z{\varphi }_a(z)$. Every finite spatial derivative of ${\varphi }_a(z)$ is a polynomial in $z$ and $a^{-1}$ times ${\varphi }_a(z)$. Fix $ϵ>0$ and compact sets $K\subset {\mathbb{R}}^d$ and $J\subset [ϵ,T]$. Since $a=v(t)$ is bounded above and away from zero on $J$, and $z$ lies in the compact support of $\mu$, each such derivative is uniformly bounded for $x\in K$, $t\in J$, and $z\in \mathrm{supp}\mu$. Difference quotients are bounded by the same type of constant on a slightly larger compact set, by the one-variable fundamental theorem of calculus. Dominated convergence [F4] therefore permits differentiation under the finite measure $\mu$ to every spatial order and once in time. The chain rule and $v^{\mathrm{\prime }}(t)=g(t{)}^2$ give ${\mathrm{\partial }}_t{p}_t(x)=\int \frac{g(t{)}^2}{2}{\mathrm{\Delta }}_x{\varphi }_{v(t)}(x-z)\mu (dz)=\frac{g(t{)}^2}{2}{\mathrm{\Delta }}_x{p}_t(x)\mathrm{.}$ Positivity follows from the positive kernel and the nonzero probability measure.

The SDE setup in [F2] and well-posedness theorem [F3] give a common continuous-path event. For bounded continuous $f$, $X_t\to X_s$ almost surely as $t\to s$ on that event, and [F4] applied to the bounded variables $f(X_t)$ gives $\mathbb{E}f(X_t)\to \mathbb{E}f(X_s)$. By [F9], these expectations are $\int fd{\mu }_t$; also ${\mu }_0=\mu$.

Fix $t\in (0,T]$ and $f\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$. For $0\le s<t$ set $h(s,x)={\int }_{{\mathbb{R}}^d}f(x+z){\varphi }_{v(t)-v(s)}(z)dz,h(t,x)=f(x)\mathrm{.}$ For $a=v(t)-v(s)>0$, write $P_{a}f(x)=\int f(x+z){\varphi }_a(z)dz$. Differentiating the explicit Gaussian and integrating by parts twice, with boundary terms zero because $f$ is compactly supported, gives ${\mathrm{\partial }}_a{P}_{a}f=\frac{1}{2}{P}_a\mathrm{\Delta }f$. Thus ${\mathrm{\partial }}_{s}h+\frac{1}{2}g(s{)}^2{\mathrm{\Delta }}_{x}h=0$ for $s<t$. Spatial derivatives may be moved onto $f$, so $h$, $\mathrm{\nabla }h$, and $\mathrm{\Delta }h$ are bounded by the corresponding sup norms of $f$, $\mathrm{\nabla }f$, and $\mathrm{\Delta }f$. For any such derivative $q$ and $\eta >0$, ${\sup }_x\mathrm{\mid }{P}_{a}q(x)-q(x)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_{\mathrm{\infty }}\eta +2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }z\mathrm{\mid }>\eta }{\varphi }_a(z)dz\mathrm{.}$ The tail equals ${\int }_{\mathrm{\mid }u\mathrm{\mid }>\eta \mathrm{/}\sqrt{a}}{\varphi }_1(u)du$ and tends to zero as $a\downarrow 0$ because ${\varphi }_1$ is integrable. Letting first $a\downarrow 0$ and then $\eta \downarrow 0$ proves uniform convergence of $P_{a}q$ to $q$, including $q=\mathrm{\Delta }f$; hence the backward equation and bounds extend continuously to $s=t$.

At $t=0$ the weak identity has equal endpoint terms and a zero right-hand integral. For $t>0$, fix an admissible test $\psi$ and $0<\delta <t$. On the common compact spatial support, $p_s$ and ${\mathrm{\partial }}_s{p}_s$ are continuous and bounded for $s\in [\delta ,t]$ by step 1.1. Differentiation under the space integral is therefore justified by [F4]. Using step 1.1 and integrating by parts twice in each coordinate, with no boundary terms because $\psi$ has compact spatial support, gives $\frac{d}{ds}\int \psi (s,x)p_s(x)dx=\int ({\mathrm{\partial }}_s\psi (s,x)+\frac{g(s{)}^2}{2}{\mathrm{\Delta }}_x\psi (s,x))p_s(x)dx\mathrm{.}$ Integrating from $\delta$ to $t$ gives the weak identity with lower endpoint $\int \psi (\delta ,x){\mu }_{\delta }(dx)$. Its difference from $\int \psi (0,x)\mu (dx)$ tends to zero: the term ${\sup }_x\mathrm{\mid }\psi (\delta ,x)-\psi (0,x)\mathrm{\mid }$ tends to zero by the bounded continuous time derivative on the common compact support, and $\int \psi (0,\cdot )d{\mu }_{\delta }\to \int \psi (0,\cdot )d\mu$ by step 1.2. The omitted right-hand integral over $[0,\delta ]$ tends to zero because each ${\mu }_s$ has mass one and the derivatives of $\psi$ and $g$ are bounded on the common compact support. Letting $\delta \downarrow 0$ proves the weak identity, so ${\mu }_t$ is a weak forward solution.

Let $({\nu }_s{)}_{0\le s\le T}$ be any weak forward solution with initial law $\mu$. Choose a smooth cutoff ${\chi }_L$ equal to one on $\mathrm{\mid }x\mathrm{\mid }\le L$ and zero on $\mathrm{\mid }x\mathrm{\mid }\ge 2L$, with $\mathrm{\mid }\mathrm{\nabla }{\chi }_L\mathrm{\mid }\le C\mathrm{/}L$ and $\mathrm{\mid }\mathrm{\Delta }{\chi }_L\mathrm{\mid }\le C\mathrm{/}{L}^2$. For completeness, take ${\chi }_L(x)=\eta (\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{L}^2)$, where $\rho (u)=0$ for $u\le 0$, $\rho (u)=e^{-1\mathrm{/}u}$ for $u>0$, and $\eta (u)=\rho (4-u)\mathrm{/}(\rho (4-u)+\rho (u-1))$. When $t<T$, extend $h$ to $t<s\le T$ by $h(s,x)=f(x)-\frac{1}{2}(s-t)g(t{)}^2\mathrm{\Delta }f(x)$. Its value and spatial derivatives through order two match at $t$, and both one-sided time derivatives equal $-\frac{1}{2}g(t{)}^2\mathrm{\Delta }f$ by step 1.3. Thus ${\psi }_L(s,x)={\chi }_L(x)h(s,x)$ is an admissible test on all of $[0,T]$. On $[0,t]$, step 1.3 gives ${\mathrm{\partial }}_s{\psi }_L+\frac{g(s{)}^2}{2}\mathrm{\Delta }{\psi }_L=\frac{g(s{)}^2}{2}(2\mathrm{\nabla }{\chi }_L\cdot \mathrm{\nabla }h+h\mathrm{\Delta }{\chi }_L),$ whose absolute value is at most $C_t(L^{-1}\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{\mathrm{\infty }}+L^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }})$. Since each ${\nu }_s$ is a probability, the time-integrated error tends to zero as $L\to \mathrm{\infty }$. For all sufficiently large $L$, ${\chi }_L=1$ on the support of $f$ and on the support of $\mu$. Letting $L\to \mathrm{\infty }$ in the weak identity therefore gives $\int fd{\nu }_t=\int h(0,x)\mu (dx)\mathrm{.}$ The same cutoff argument applied to the weak solution ${\mu }_t$ from step 2.1 gives $\int fd{\mu }_t=\int h(0,x)\mu (dx)$. Thus ${\nu }_t$ and ${\mu }_t$ agree on every compactly supported smooth real function.

For any fixed $t>0$, any $\theta \in {\mathbb{R}}^d$, and integer $n\ge 1$, apply step 3.1 to ${\chi }_n(x)\cos (\theta \cdot x)$ and ${\chi }_n(x)\sin (\theta \cdot x)$ and combine the resulting identities into one complex identity. Let $n\to \mathrm{\infty }$. The left side converges to the characteristic function of ${\nu }_t$ by dominated convergence [F4]. For each fixed $x$, the backward averages on the right converge by [F4] and [F10] to $e^{i\theta \cdot x}{e}^{-v(t)\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}2}$; they are bounded by one, so another application of [F4] gives the limit $e^{-v(t)\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}2}\int e^{i\theta \cdot x}\mu (dx)$. By [F1], ${\mu }_t$ is the convolution of $\mu$ with the centered Gaussian of covariance $v(t)I_d$; Fubini [F5] and [F10] show that its characteristic function is exactly this product. Hence the vector characteristic functions of ${\nu }_t$ and ${\mu }_t$ agree for every $\theta$. For each $u\in {\mathbb{R}}^d$, take $\theta =ru$ for arbitrary $r\in \mathbb{R}$; scalar uniqueness [F7] gives equality of the laws of $u\cdot x$ under the two measures, and [F8] then gives ${\nu }_t={\mu }_t$. This holds for every $t>0$; at $t=0$ both laws equal $\mu$ by the initial condition, proving uniqueness. AC is used only through the inherited Brownian/Itô interfaces and the cited Fourier-uniqueness results.