Convergence of an integral How to prove that if $f$ is continuous then
$$
F(x) =\int\limits_{-\infty}^\infty f(y)\frac{1}{x\sqrt{2\pi}}\mathrm e^{-y^2/2x^2}\,dy 
$$
is also a continuous function? I tried to make it through the definition taking $x_n\to x$ but then I can use neither Lebesgue monotone convergence theorem nor dominated convergence theorem (since the convergence of integrands is not monotone or dominated). What would you advise?
Edited: $f$ is bounded.
 A: First lets do a change of variables to get rid of the x in the exponent. Let $y=xu$. Then we have
$$F(x)=\int\limits _{-\infty}^{\infty}f(y)\frac{1}{x\sqrt{2\pi}}\mathrm{e}^{-y^{2}/2x^{2}}\, dy=\int\limits _{-\infty}^{\infty}f(xu)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2} du.$$
Next, what is the definition of continuity? Given $\epsilon>0$, we need to show that for fixed $x$ 
$$\biggr|F(x+\delta)-F(x)\biggr|=\biggr|\int\limits _{-\infty}^{\infty}\left(f((x+\delta)u)-f(xu)\right)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\, du\biggr|<\epsilon.\ \ \ \ \ \ (1)$$ 
But if $f$ is bounded, say $|f|\leq M$, then the integral on the right hand side is uniformly bounded by $2M\int\limits _{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2} du=2M$ for all $\delta$.  Hence when taking the limit as $\delta\rightarrow 0$ the dominated convergence theorem applies so we can switch the order of integration and the limit.  Consequently since $f$ is continuous the limit on the right hand side is zero, and we have $$\lim_{\delta\rightarrow 0} F(x+\delta)-F(x)=0$$ so that $F$ is continuous.
Alternative: 
We could split up the integral instead of using the dominated convergence theorem.  For the given $\epsilon$ we could choose $N$ so large that the integral
$$\int_{-\infty}^N\frac{2M}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}du+\int_N^\infty \frac{2M}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}du<\frac{\epsilon}{2}.$$  On the interval $[-Nx,Nx]$ $f$ will be uniformely continuous so we can choose $\delta$ so small that $|f((x+\delta)u)-f(x)|\leq \frac{\epsilon}{4N}$.  This implies $$\biggr|\int\limits _{-N}^{N}\left(f((x+\delta)u)-f(xu)\right)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\, du\biggr|<\frac{\epsilon}{2}.$$  Upon adding these inequalities we obtain equation $(1)$ and the proof is complete.
A: Suppose that $f$ is a continuous bounded function. For convenience, replace $x^2$ with $t\,( > 0)$. 
If $\{X_t: t \geq 0\}$ is a standard Brownian motion, then
$$
{\rm E}f(X_t) = \int_{ - \infty }^\infty  {f(y)\frac{1}{{\sqrt {2\pi t} }}e^{ - y^2 /(2t)} dy} .
$$
Now let $(t_n)$ be a sequence of positive numbers such that $t_n \to t$.
On the one hand,
$$
{\rm E}f(X_{t_n}) = \int_{ - \infty }^\infty  {f(y)\frac{1}{{\sqrt {2\pi t_n} }}e^{ - y^2 /(2t_n)} dy}.
$$
On the other hand, $X_{t_n} \to X_t$ a.s., hence also  $X_{t_n} \to X_t$ in distribution.
But the latter condition is equivalent to
$$
{\rm E}f(X_{t_n}) \to {\rm E}f(X_t)
$$
for any continuous bounded function $f$.
EDIT: Alternatively, the result can be obtained as follows. With the same notation as above, $X_{t_n} \to X_t$ a.s. as $n \to \infty$. Since $f$ is continuous, by the Continuous mapping theorem also $f(X_{t_n}) \to f(X_{t})$ a.s. Now let $\xi_n = f(X_{t_n})$ and $\xi = f(X_{t})$, so $\xi_n \to \xi$ a.s. Since $f$ is assumed bounded, $\sup_n |\xi_n| \leq M$ for some $M > 0$ fixed. Thus by the Bounded convergence theorem, ${\rm E}(\xi_n) \to {\rm E}(\xi)$, that is
$$
{\rm E}f(X_{t_n}) \to {\rm E}f(X_t).
$$
A: The function $F(x)=u(0,x)\ $ is the Poisson potential (and solution) for the Cauchy problem $u_x-u_{ss}/2=0$, $u(s,0)=f(s)$. If $f$ is bounded and continuous, then $F$ is bounded and contuinuous fo $x\ge0$ and is $C^\infty$ for $x>0$. The proof can be found in most books on parabolic equations. For example, in N.V. Krylov Lectures on elliptic and parabolic equations in Hölder spaces. It also holds if $f$ grows not too quickly at infinity, $|f(s)|\le Ce^{|s|^{2-\varepsilon}}$ for some $\varepsilon>0$ would do. 
