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 \begin{equation} (T_tf)(x)=\mathrm{E}_x[ e^{-\int_0^t V(B_s) ds} f(B_t) ]. \end{equation}

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 $t$.

Thank you for your cooperation.

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

  • $\begingroup$ 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... $\endgroup$
    – dohmatob
    Mar 29, 2020 at 11:52
  • 1
    $\begingroup$ @dohmatob $\mathrm{E}_x[\cdot]$ is expectation with respect to measure $\mathcal{W}_x$, such that $\mathcal{W}_x(B_0 = x) = 1$. $\endgroup$
    – sate
    Mar 29, 2020 at 12:07
  • $\begingroup$ OK, makes sense. See answer below. $\endgroup$
    – dohmatob
    Mar 29, 2020 at 12:41

1 Answer 1


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 $f \in L^2(\mathbb R^d)$.


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...

  • $\begingroup$ 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. $\endgroup$
    – sate
    Mar 29, 2020 at 13:06
  • $\begingroup$ Can you be precise concerning which part(s) you don't get ? $\endgroup$
    – dohmatob
    Mar 29, 2020 at 13:41
  • $\begingroup$ 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. $\endgroup$
    – sate
    Mar 29, 2020 at 13:54
  • $\begingroup$ 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 $-\int_0^t V(B_s) ds \leq -\inf V$ $\endgroup$
    – sate
    Mar 29, 2020 at 13:57
  • 1
    $\begingroup$ Actually, I can't have confidence in that. I'll think a little more time. Thank you. $\endgroup$
    – sate
    Mar 29, 2020 at 14:43

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