# Feynman-Kac formula on $L^2(\mathbb{R}^d)$ (the right hand side)

Let $$B_t$$ be a $$d$$ dimensional Brownian motion and $$V\in C^\infty_0(\mathbb{R}^d)$$. Define the operator $$T_t$$ on $$L^2(\mathbb{R}^d)$$ by $$$$(T_tf)(x)=\mathrm{E}_x[ e^{-\int_0^t V(B_s) ds} f(B_t) ].$$$$

I want to show that $$T_t$$ is bounded on $$L^2(\mathbb{R}^d)$$ for each $$t$$: $$\| T_t f\|^2 \leq e^{-2\inf V}\| f \|^2$$. But I cannot show this inequality:

\begin{align} \| T_t f \|^2 &= \int_{\mathbb{R}^d} \mathrm{E}_x[ e^{-\int_0^t V(B_s) ds} f(B_t) ]^* \mathrm{E}_x[ e^{-\int_0^t V(B_s) ds} f(B_t) ] dx \\ &= ? \end{align}

It is clear that $$|(T_t f)(x)| \leq e^{-t \inf V} \mathrm{E}_x[f(B_t)]$$, but this is dependent on $$t$$.

This question is not correct. Example: t=0.

• Precisely, what do you mean by the expectation notation $\mathrm{E}_x[...]$ and why does the stuff within the [...] does seem to depend --at least not explicitly-- on $x$ ? What's the relation between $B_t$ and $x$ ? The question as written doesn't make much sense mathematically. It's probably just a problem of notation though... Mar 29, 2020 at 11:52
• @dohmatob $\mathrm{E}_x[\cdot]$ is expectation with respect to measure $\mathcal{W}_x$, such that $\mathcal{W}_x(B_0 = x) = 1$.
– sate
Mar 29, 2020 at 12:07
• OK, makes sense. See answer below. Mar 29, 2020 at 12:41

By the Cauchy-Schwarz inequality for expectations, for every $$x \in \mathbb R^d$$, we have $$\begin{split} |(T_tf)(x)|^2 &= |\mathbb E_x[e^{-\int_0^t V(B_s)ds}f(B_t)]|^2 \le \mathbb E_x[e^{-2\int_0^t V(B_s)ds}]\mathbb E_x[f(B_t)^2]\\ &\le G_{V,t}(x)\mathbb E_x[f(B_t)^2], \end{split}$$ where $$G_{V,t}(x) := \mathbb E_x[e^{-2\int_0^t V(B_s)ds}] \ge 0$$.

Note that if the function $$V$$ is such that $$\inf_{x \in \mathbb R^d} V(x) > -\infty$$, then we have the crude bound

$$0 \le G_{V,t}(x) \le e^{-2t\inf V}< \infty,$$ for all $$t \ge 0$$ and $$x \in \mathbb R^d$$.

Thus, one computes

$$\begin{split} \|T_tf\|^2 &:= \int_{\mathbb R^d}|(T_f)(x)|^2dx \le e^{-2t\inf V}\int_{\mathbb R^d}\mathbb E_x[f(B_t)^2]dx \le e^{-2t\inf V}\int_{\mathbb R^d}\mathbb E_x[f(B_t)^2]dx \\ &\le e^{-2t\inf V}\int_{\mathbb R^d}\mathbb E_x[\|f\|^2]dx = e^{-2t\inf V}\|f\|^2\underbrace{\int_{\mathbb R^d}\mathcal W_x(B_0=x)dx}_{1}\\ &= e^{-2t\inf V}\|f\|^2, \end{split}$$ where

• the 2nd inequality is an application of Jensens's inequality, and
• the last equality is because $$\mathbb E_x$$ is epectation w.r.t to a measure $$\mathcal W_x$$ such that $$\mathcal W_x(B_0=x) = 1$$.

We have thus proved that $$\|T_tf\| \le e^{-t\inf V}\|f\|$$, for all $$f \in L^2(\mathbb R^d)$$.

Thus,

For every $$t \ge 0$$, the operator $$T_t$$ is an endomorphism on the vector space $$L^2(\mathbb R^d)$$.

N.B.: The OP claims one sould be able to do away with the $$t$$ dependence in the exponential factor in the above bound, but I doubt this...

• Thank you for your answer. I don't get it. Why $\left|\mathbb E_x[e^{-\int_0^t V(B_s)ds}]\right| \le e^{-\inf V}$? $\left|\mathbb E_x[e^{-\int_0^t V(B_s)ds}]\right| \le e^{-t\inf V}$ is clear.
– sate
Mar 29, 2020 at 13:06
• Can you be precise concerning which part(s) you don't get ? Mar 29, 2020 at 13:41
• First, I think the R.H.S of $|\mathbb E_x[e^{-\int_0^t V(B_s)ds}f(B_t)]| \le \left|\mathbb E_x[e^{-\int_0^t V(B_s)ds}]\right||\mathbb E_x[f(B_t)]|$ is not valid because this is not Cauchy-Schwartz. $\left|\mathbb E_x[e^{-2\int_0^t V(B_s)ds}]\right|^{1/2}\mathbb E_x[|f(B_t)|^2]^{1/2}$ is correct.
– sate
Mar 29, 2020 at 13:54
• Second, I don't know the inequality $E[e^{-\int_0^t V(B_s)ds}]\leq e^{-\inf V}$. I think $-\int_0^t V(B_s) ds \leq -t\inf V$ not $-\int_0^t V(B_s) ds \leq -\inf V$
– sate
Mar 29, 2020 at 13:57
• Actually, I can't have confidence in that. I'll think a little more time. Thank you.
– sate
Mar 29, 2020 at 14:43