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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.

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    @Eric: sure, you right. I've edited.2011-07-03

3 Answers 3

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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.

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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). $

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    For the fact that $X_n \to X$ in distribution if and only if ${\rm E}f(X_n ) \to {\rm E}f(X )$ for all bounded continuous functions $f$, see http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties. (This is sometimes the definition.) In this comment, $X$ is an arbitrary random variable.2011-07-03
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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.

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    Andrew, thank you very much - the only thing is that I need to prove it with the smallest use of the external theory, namely quite direct.2011-07-03