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Let $X=C(\mathbb{R})$ the set of all continuous functions with compact support such that there exist $r>0$ such that $f(x)=0$ when $|x|\geq r$.

For any $f\in X$ and $n\in\mathbb{N}$ define $\Lambda_n(f)(x)=\frac{n}{\sqrt\pi}\int_\mathbb{R}f(x-y)e^{-(ny)^2}dy$.

I've been trying to show that $\lim_{n\rightarrow\infty}\Lambda_n(f)(x)=f(x)$.

I've tried to use the product rule for integration, but no luck there. Tried to use the definition of uniform convergence, but nothing there as well. Also tried to use the $\sup$ method.

I would really appreciate clues!

BTW: I may use that $\frac{n}{\sqrt\pi}\int_{-\infty}^{\infty}e^{-(ny)^2}dy=1$ and $\lim_{n\rightarrow\infty}\frac{n}{\sqrt\pi}\int_{-R}^{R}e^{-(ny)^2}dy=1$.

stressed out
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Mathiaspilot123
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    This is the usual notion of an approximate identity, which you should be able to find in textbooks. In this particular case you could also apply the Laplace method, since the approximate identity decays exponentially. – Ian Feb 20 '19 at 21:10
  • Which notion, those on the bottom? – Mathiaspilot123 Feb 20 '19 at 21:11
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    The entire question is demonstrating that the sequence $\phi_n(x)=\frac{n}{\sqrt{\pi}} e^{-nx^2}$ is an approximate identity. This is the notion I'm alluding to. – Ian Feb 20 '19 at 21:12
  • The main idea of the proof is to get a small enough $\delta$ so that $f(x-y)$ is "almost constant" for $y \in (-\delta,\delta)$ and then select $n$ large enough that $\int_{|y|>\delta} ne^{-ny^2} dy$ is "almost zero". Then the problem is just one of normalization (showing that the limit is actually $f(x)$ and not some multiple of $f(x)$). – Ian Feb 20 '19 at 21:15
  • I think this is related: https://math.stackexchange.com/a/3115617/281471 – Botond Feb 20 '19 at 21:18

2 Answers2

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Hint:

\begin{align*} |\Lambda_n(f)(x)-f(x)| &= \Bigg|\int_{\mathbb{R}}{(f(x-y)-f(x))\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy\,}\Bigg| \\ &\leq \int_{\mathbb{R}}{\left(\sup_{|r-x| \leq n^{-1/2}}\,|f(r)-f(x)|\right)\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy} + \int_{|y| \geq n^{-1/2}}{2\|f\|_{\infty}\frac{ne^{-n^2y^2}}{\pi^{1/2}}\,dy} \\ &\leq \left(\sup_{|r-x| \leq n^{-1/2}}\,|f(r)-f(x)|\right) + 2\|f\|_{\infty}\int_{|y| \geq n^{1/2}}{e^{-y^2}\,dy}. \end{align*}

stressed out
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Aphelli
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  • I took the liberty of changing | with \Bigg| to make your answer more beautiful. I hope that you don't mind that. (+1) – stressed out Feb 20 '19 at 21:48
  • Could you explain the first inequality a bit more? Why do you use $|r-x|\leq n^{-1/2}$ under the $sup$? And what does $||f||_\infty$ mean? – Mathiaspilot123 Feb 20 '19 at 21:53
  • The first inequality is about splitting the integral: when $y$ is “large” (greater than $n%{-1/2}$) the exponential gets ludicrously small and everything is negligible. So the assessment to make is when $y$ is “small”, and then it is the variation of $f$ which is quite small (uniformly in $x$) while the integral of the exponential is at most $1$. – Aphelli Feb 20 '19 at 21:56
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    @Mathiaspilot123 The actual form of the approximate identity gets in the way here. You are just splitting the integral up into two pieces and controlling one with the uniform continuity of $f$ and the other with a property of the approximate identity. – Matematleta Feb 20 '19 at 22:10
  • Sorry, but I still don't get why you use $|r-x|\leq n^{-1/2}$ under the first sup in the first inequality, and what $||f||_\infty$ means. – Mathiaspilot123 Feb 20 '19 at 22:11
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    @Mathiaspilot123 if your comment is addressed to me, I'll point out that the answer, which is correct, is not mine. But the idea is as I indicated. You choose $\delta$ to respond to the $\epsilon$ challenge for the uniform continuity of $f$. Then, since for large $n$, the exponential is small if $|y|\ge \delta$, you can split the integral up and control each piece separately. $|f|_{\infty}$ is the sup norm of $f$. – Matematleta Feb 20 '19 at 22:17
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    @Mathiaspilot123 The answer isn't mine, but if I understood correctly, $|f|_{\infty}$ denotes the norm of $f$ in the supremum norm. Isn't $X=C(\mathbb{R})$ with the supremum norm? Anyway, if you're asking why the radius is $n^{-1/2}$, it's because $|n^2y^2| \leq n$ gives that $|y| \leq n^{-1/2}$. Now set $x-y=r$. – stressed out Feb 20 '19 at 22:18
  • It doesn't specify any norms in the problem. – Mathiaspilot123 Feb 20 '19 at 22:22
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    @Mathiaspilot123 Uniform convergence is equivalent to convergence in the supremum norm. What kind of space $C(X)$ is then? It's analysis, you need to define a metric on it to talk about taking limit. What metric do you put on it? – stressed out Feb 20 '19 at 22:23
  • The info doesn't specifically inform which metric to use (part of why this confused me), but it does say that it is supported with compactness. – Mathiaspilot123 Feb 20 '19 at 22:25
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    @Mathiaspilot123 but the info specifically says that you want to show that the convergence is uniform on $\mathbb{R}$. Uniform convergence on $\mathbb{R}$ is the same as convergence in $(C(\mathbb{R}), |\cdot|{\infty})$ where $|\cdot|{\infty}$ denotes the supremum norm. – stressed out Feb 20 '19 at 22:26
  • Okay, thanks didn't know that. – Mathiaspilot123 Feb 20 '19 at 22:27
  • @Mathiaspilot123 You're welcome. – stressed out Feb 20 '19 at 22:27
  • @Mindlack One more question: In the last inequality, the second term changed from $|y|\geq n^{-1/2}$ to $|y|\geq n^{1/2}$ in the integral. Is that correct? – Mathiaspilot123 Feb 20 '19 at 23:00
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    Yes, I did a variable change. – Aphelli Feb 20 '19 at 23:13
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The other answer provides the solution, but I will add another, in hopes of clearing up the questions posed by the OP in the comments.

$1).\ $ Uniform convergence/convergence in sup norm: a sequence $(f_n)$ converges to $f$ uniformly on a set $S\subseteq \mathbb R$ if and only if to each $\epsilon>0$, there is an integer $N$ such that $n>N\Rightarrow |f_n(x)-f(x)|<\epsilon$ $\textit{for all}\ x\in S.$ This means that if $n>N,$ we must have $\sup_{x\in S}|f_n(x)-f(x)|\le \epsilon.$ Now, by definition, the sup norm on $S$ is $\|f\|_{\infty}=\sup_{x\in S}|f_n(x)-f(x)|$. The argument is reversible: if we start with the definition of the sup norm, we arrive back at the definition of uniform convergence of $(f_n).$ So the two notions of convergence are the same.

$2).$ The sequence functions $(K_n(x))=\left(\frac{ne^{-n^2x^2}}{\pi^{1/2}}\right )$ is an $\textit{approximate identity};\ $ i.e., it satisfies the following conditions:

$i).\ \int K_n(x) dx = 1$ for every $n$

$ii).\ $ there exists a constant $C ≥ 0$ such that $\int |K_n(x)| dx ≤ C$ for every $n$;

$iii).\ \lim_{n\to \infty} \int_{|x|≥\delta}|K_n(x)| dx = 0$ for every $\delta >0.$

Now, if we use these observations about our sequence, it is easy to prove the claim. Let $\epsilon>0$ and suppose $f$ is supported on the compact set $K$. Since $f$ is continuous on $K$ and $K$ is compact, $f$ is also uniformly continuous there. This means that there exists a number $\delta>0$ such that $|f(x) − f(x − y)| < \epsilon$ for every $x \in K$ and every $y\in \mathbb R$ that satisfies $|y| < \delta$. For this $\delta$, now choose $N$ so large that $\int_{|y|≥\delta}|Kn(y)|dy < \epsilon$ for $n>N.$ Then, if $x\in K$, and $n>N,$ we have

$\Lambda_n(f)(x)-f(x)\le$

$ \int_{|y|<\delta} |f(x) − f(x − y)||K_n(y)| dy + \int_{|y|≥\delta}|f(x) − f(x − y)||K_n(y)| dy \le$

$C\epsilon + 2\epsilon \|f\|_{\infty}.$

Matematleta
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