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This question already has an answer here:

prove that for any integrable function $f$ defined on $(-\infty,\infty)$ (or maybe square integrable function.. my professor didn't gave exact imformation about $f$.) $$\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)$$

I think I have to get some $\delta>0$ and divide integral into three terms.

$$\int_{-\infty}^{-\delta} f*g_n +\int_{-\delta}^{\delta}f*g_n + \int_\delta ^\infty f*g_n$$ ($g_n(x)=\frac n {\sqrt {\pi}}e^{-(nx)^2}$)

since $f*g_n$ converges uniformly to $0$ if $x\neq0$, first and third term vanish. so what I have to do is to take $N$ and $\delta$ (from the given value $\epsilon>0$) so that if $n>N$ $$\left|\int_{-\delta}^{\delta}\frac n {\sqrt {\pi}}e^{-(nx)^2} f(x)dx - f(0) \right|<\epsilon$$

I can't go any further.

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marked as duplicate by Winther, mickep, Shailesh, C. Falcon, Daniel W. Farlow May 1 '17 at 1:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Maybe expand exp with a Taylor series around x=0 could help. $\endgroup$ – mathreadler Apr 30 '17 at 17:35
  • $\begingroup$ Would it help to represent $g_n(x)$ as the PDF of a normal random variable? You are ultimately taking an expectation of $f(x)$ as your $\text{Var}\ x \to 0$ $\endgroup$ – jameselmore Apr 30 '17 at 17:36
  • $\begingroup$ If $f$ is integrable, then LDCT applies. Do you want $f$ to be integrable? $\endgroup$ – user384138 Apr 30 '17 at 17:36
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    $\begingroup$ @Dr.MV we may write the integral as $$\frac1{\sqrt \pi} \int_{-\infty}^{\infty} e^{-u^2} f(u/n)du$$ $\endgroup$ – user384138 Apr 30 '17 at 17:39
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    $\begingroup$ Possible duplicate of How to show this sequence is a delta sequence?; See also math.stackexchange.com/questions/55137/… $\endgroup$ – Winther Apr 30 '17 at 18:01
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Hints: 1. Since we could redefine $f$ at the one point $0$ and not change any of the integrals, we need further conditions on $f$ for this to be true. For example, continuity of $f$ at $0.$ 2. This is not a DCT problem. 3.This problem is but one example of the following situation: We have a sequence integrable functions $g_n$ on $\mathbb R$ such that i)$\int_{\mathbb R}g_n = 1$ for all $n;$ ii)for all $r>0,$ $g_n(x)\to 0$ uniformly on $\{|x|>r\}.$

If we have that, then given any integrable $f$ on $\mathbb R$ that is continuous at $0,$

$$\lim_{n\to \infty} \int_{-\infty}^\infty g_n\,f = f(0).$$

Generalizing like this actually makes the proof easier I think.

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  • $\begingroup$ (+1) for the only post so far that addresses the issue of continuity and the inapplicability of the DCT $\endgroup$ – Mark Viola Apr 30 '17 at 22:46
  • $\begingroup$ I've posted a solution HERE that uses the approach you outlined herein. The only solution posted applied the DCT; the OP requested (in a comment) to see a way forward that did not use the DCT. $\endgroup$ – Mark Viola May 3 '17 at 18:40
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Not an answer, but too long for a comment.

Using a substitution, the left-hand side can be rewritten as $\lim_{n\to\infty}\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(\frac{u}{n}\right) du$. The first hard part is proving the integration commutes with the limit-taking, so that we can rewrite again to give $\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}\lim_{n\to\infty}f\left(\frac{u}{n}\right) du$. For continuous $f$, this is $$\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(\lim_{n\to\infty}\frac{u}{n}\right) du=\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(0\right) du=f\left(0\right).$$As integrable functions are not in general continuous, what can be said about integrable $f$ is the second hard part.

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This is my try. The continuity at $0$ and the boundedness of $f$ should be enough.

Rewrite the LHS as: $$\int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\frac{1}{2n^2}}} e^{-\frac{x^2}{2\frac{1}{2n^2}}} f(x)dx$$

This is the expectation of function $f(X_n)$, where $X_n \sim N(0, \frac{1}{2n^2})$. We need to show that: $E(f(X_n)) \to f(0)$.

We can show that $X_n \overset{a.s.}{\longrightarrow} 0$. As $f$ is continuous at $0$, we have: $f(X_n) \overset{a.s.}{\longrightarrow} f(0)$. $f(X_n)$ is bounded, so we can use DCT to conclude.

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