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Prove that

$$\lim_{k \to \infty} \int_{-\infty}^{\infty} \frac{ke^{-k^2x^2}}{\pi}h(x) \, dx= h(0) \tag{A}$$

Proof

From the left side of (A)

$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx$$

Integration by parts?

$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx = \frac{k}{\pi}\lim_{k \to \infty} \left[ \left( \int e^{-k^2x^2} \right)h(x)\bigg|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \left( \int e^{-k^2x^2} \right)h'(x)\, dx \right]$$

But here there's no antiderivative for the Gaussian-type function, unless I can apply those limits of integration to it, but I don't see how. The definite integral of that piece would be $\sqrt{\pi}/{k}. $

Then we'd have

$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx = \frac{k}{\pi}\lim_{k \to \infty} \left[ \frac{\sqrt{\pi}}{{k}}h(x)\bigg|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \left( \int e^{-k^2x^2} \right)h'(x)\, dx \right]$$

Don't know where to go from here. The actual question is to prove that the kernel in the integrand in (A) is Dirac Delta sequence.

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  • $\begingroup$ What are the assumptions on $h$? Just integrable? Or continuous as well? $\endgroup$ – M10687 Mar 8 '18 at 18:51
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    $\begingroup$ You cannot factor the dummy parameter $k$ from the limit. $\endgroup$ – Mark Viola Mar 8 '18 at 18:53
  • $\begingroup$ Oh yep, mistake. $\endgroup$ – Zduff Mar 8 '18 at 18:54
  • $\begingroup$ This is not correct as written. Take $h(x)= c$ then the limit is $\frac{c}{\sqrt{\pi}} = \frac{h(0)}{\sqrt{\pi}}$. Change the denominator to $\sqrt{\pi}$ and you are good $\endgroup$ – Winther Mar 8 '18 at 19:03
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If $h\in C^{0}(-\infty,\infty)\cap L^{\infty}(-\infty,\infty)$, then \begin{align*} \dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}(h(u/k)-h(0))du\rightarrow 0,~~~~k\rightarrow\infty \end{align*} because $h(u/k)\rightarrow h(0)$ for every $u$ as $k\rightarrow\infty$ and $|h(u/k)-h(0)|\leq 2\|u\|_{L^{\infty}(-\infty,\infty)}$ and $e^{-u^{2}}\in L^{1}(-\infty,\infty)$, so Lebesgue Dominated Convergence Theorem.

Now use the fact that $\displaystyle\int_{-\infty}^{\infty}e^{-u^{2}}du=\sqrt{\pi}$, we have \begin{align*} \dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}h(u/k)du\rightarrow h(0),~~~~k\rightarrow\infty. \end{align*} For the question, let $u=kx$ for the substitution. If the denominator in question is $\pi$, then it may not go through.

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  • $\begingroup$ You beat me to it. (+1) Of course, the tacit assumption is that $h$ is continuous. $\endgroup$ – Mark Viola Mar 8 '18 at 18:54

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