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$\newcommand\R{\mathbb R} \newcommand\E{\mathbf E}$ Consider the family of operator $(F_T:T>0)$ on $L^2(\R)$ defined as $$F_Tf(x)=\E\left[f(x+B_T)\exp\left(-\int_0^T|x+B_t|~dt\right)\right],$$ where $B$ is a standard Brownian motion on $[0,T]$ and the expected value $\E$ is taken with respect to $B$.

I'm trying to understand why $F_0$ can be considered the identity operator, more precisely, I want to show that $$\lim_{T\downarrow0}\|F_Tf-f\|_2^2=0.$$

My first instinct is to use the dominated convergence theorem. This works well for half of the solution: if we take $f$ continuous (should be no problem since we're in $L^2$), then $$f(x+B_T)\to f(x)$$ and $$\exp\left(-\int_0^T|x+B_t|~dt\right)\to 1$$ almost surely, so $$\E\left[f(x+B_T)\exp\left(-\int_0^T|x+B_t|~dt\right)\right]-f(x)\to0$$ pointwise (that the dominated convergence theorem applies here is trivial).

However, the problem is to then apply dominated convergence to the integral $$\int_0^\infty\left(\E\left[f(x+B_T)\exp\left(-\int_0^T|x+B_t|~dt\right)\right]-f(x)\right)^2d x.$$ Try as I might, I have absolutely no idea how to dominate the inner function as an integrable function in $x$. Any idea would be appreciated.

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You have the inequality $(a-b)^2\leq 2(a^2+b^2)$ where $a$ is your expectation and $b$ is $f(x)$. This means you only need to show that the expectation is square integrable, since you already know $f$ is square integrable. Furthermore the inequality $E(X)^2\leq E(X^2)$ means you only need to show that $E(\dots^2)$ is integrable. Since the random variable inside the expectation is now non-negative, Fubini's theorem can be applied to switch the expectation and the integral. The Cauchy-Schwarz inequality applied to the integral (which now lies inside the expectation) will tell you that you are taking the expectation of a uniformly bounded random variable.