Dynkin formula for bounded Brownian stopping

The functions $f$, its first partial derivatives and its second partial derivatives are continuous and vanish outside a compact set: outside the support of $f$ it vanishes on a neighborhood, so all these derivatives are zero. They are bounded: continuity provides a neighborhood with a finite bound at each point of the compact support, and a finite subcover gives a common bound. The Hessian $H_f$ is uniformly continuous on all of ${\mathbb{R}}^d$. To see the latter, enclose the support in a ball of radius $R$ and apply [F3] on the ball of radius $R+1$. For points at distance less than $1$, either both are in that larger ball or both Hessians vanish; this gives global uniform continuity. Fix $C$ bounding the Hessian operator norm and put $\omega (r)={\sup }_{\mathrm{\mid }y-z\mathrm{\mid }\le r}\mathrm{\parallel }{H}_f(y)-H_f(z)\mathrm{\parallel }$. Then $0\le \omega (r)\le 2C$ and $\omega (r)\to 0$ as $r\downarrow 0$. Applying the degree-one formula in [F3] and subtracting the base Hessian gives $f(y+z)-f(y)=\mathrm{\nabla }f(y)\cdot z+\frac{1}{2}{z}^{\mathsf{T}}{H}_f(y)z+R(y,z),\mathrm{\mid }R(y,z)\mathrm{\mid }\le \frac{1}{2}\omega (\mathrm{\mid }z\mathrm{\mid })\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{.}$ The remainder is defined by the displayed difference, so no measurable selection of the Lagrange point is used.

Fix a positive integer $n$, put $m=2^n$, $h=K\mathrm{/}m$ and $t_j=jh$ for $0\le j\le m$. Let ${\tau }_n=h\lceil \tau \mathrm{/}h\rceil$; then $\tau \le {\tau }_n\le K$ and $0\le {\tau }_n-\tau <h$ unless equality already holds. Each grid event $\{{\tau }_n>t_j\}=\{\tau >t_j\}$ belongs to ${\mathcal{F}}_{t_j}$ by [F2]. Pathwise telescoping for the original process $Y_t=x+B_t$ gives $f(Y_{{\tau }_n})-f(Y_0)={\sum }_{j=0}^{m-1}{1}_{\{\tau >t_j\}}(f(Y_{t_{j+1}})-f(Y_{t_j}))\mathrm{.}$ This includes $\tau =0$ (every summand vanishes) and $\tau =K$ (every grid increment is included).

For every random vector $Y$ and Gaussian increment $Z$ of variance $h{I}_d$, regardless of their dependence, the uniform bound of step 1.1 and [F4] imply $E\mathrm{\mid }R(Y,Z)\mathrm{\mid }\le \frac{1}{2}\omega (\delta )dh+C{\delta }^{-2}3d^2{h}^2$ for every $\delta >0$: split at $\mathrm{\mid }Z\mathrm{\mid }\le \delta$, and use $\mathrm{\mid }Z{\mathrm{\mid }}^2{1}_{\mathrm{\mid }Z\mathrm{\mid }>\delta }\le {\delta }^{-2}\mathrm{\mid }Z{\mathrm{\mid }}^4$ on the complement. Thus there is a deterministic function $\epsilon (h)\to 0$ as $h\downarrow 0$ such that $E\mathrm{\mid }R(Y,Z)\mathrm{\mid }\le h\epsilon (h)$ uniformly in $Y$. Indeed divide the displayed bound by $h$, first send $h$ to zero for fixed $\delta$, and then send $\delta$ to zero.

Every grid evaluation is unchanged almost surely when $Y$ is replaced by $B^x=x+\widehat{B}$, because the processes agree on the common full event in [F1]. Joint measurability of $B^x$ makes $B_{\tau }^x$ measurable: the map $\omega \mapsto (\tau (\omega ),\omega )$ is measurable into the product sigma-algebra, as is seen on rectangles. Alternatively its coordinates are the limits of the measurable finite grid evaluations $B_{{\tau }_n}^x$, since every path is continuous. Consequently $f(B_{{\tau }_n}^x)\to f(B_{\tau }^x)$ everywhere and the variables are bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Their expectations converge by [F6].

In each summand apply step 1.1 with $y=Y_{t_j}$ and $z=B_{t_{j+1}}-B_{t_j}$. The indicator, gradient and Hessian at $Y_{t_j}$ are bounded ${\mathcal{F}}_{t_j}$-measurable factors. The entire increment vector is independent of that sigma-algebra by the explicit hypothesis. Hence its coordinate means are zero and its conditional coordinate products have means $h{\delta }_{ik}$ by [F4] and [F5]. The linear term therefore has expectation zero, and the quadratic term has expectation $hE[1_{\{\tau >t_j\}}Lf(Y_{t_j})]$. All terms are integrable by bounded derivatives and Gaussian moments. The sum of the absolute remainder expectations is at most $mh\epsilon (h)=K\epsilon (h)$ by step 2.1. Since $Y_0=x$ almost surely, $\mid Ef(Y_{{\tau }_n})-f(x)-E{\sum }_{j=0}^{m-1}h{1}_{\{\tau >t_j\}}Lf(Y_{t_j})\mid \le K\epsilon (h)⟶0.$

For each normalized path put $g(s)=Lf(B_s^x)$. This is continuous on $[0,K]$ and bounded by $\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}$. The sum in step 3.1 with $B^x$ is the left Riemann sum on $[0,{\tau }_n]$. Its difference from ${\int }_0^{{\tau }_n}g(s)ds$ is bounded by $K{\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le h}\mathrm{\mid }g(s)-g(t)\mathrm{\mid }$, which tends to zero by [F3]. The extra interval between $\tau$ and ${\tau }_n$ contributes at most $h\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus these measurable sums converge everywhere to the stated pathwise Lebesgue integral, using [F8] (and the zero integral if ${\tau }_n=0$), which is therefore measurable, and each sum and the limit are bounded by $K\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F6] their expectations converge.

Passing to the limit in step 3.1 using steps 2.2 and 4.1 proves the asserted identity. Changing the normalization event changes neither random expression on the intersection of the two measurable full events, so the expectations are unchanged. The argument uses the original adapted process only on finite deterministic grids and never claims that the normalized process is adapted to the original filtration.

For $\tau =0$ the integral is zero and $B_0^x=x$ everywhere, and for $f=0$ both sides vanish. Deterministic stopping times are included; $d=1$ gives the scalar statement, while $d=0$ is excluded. If $Lf=0$ the displayed identity directly reduces to $Ef(B_{\tau }^x)=f(x)$; no maximum principle or non-compact affine test is invoked. Compact support supplies uniform boundedness and Hessian continuity, and the deterministic bound $K$ controls the summed remainders and both dominated limits. Full AC is declared for the conditional-expectation interfaces identified in [F7] and supplies the Countable Choice used in [F8]; the Brownian and Gaussian data remain hypotheses. There is no additional path selection and no assertion for unbounded $\tau$.