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
\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.
 A: 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...
