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I'm trying to show that that sequence $\sqrt{\frac{n}{\pi}}e^{-nx^{2}}$ converges to dirac delta function. Proving it, I need to show that for any $r>0$ $$ , \lim_{n\rightarrow\infty}\int_{\mathbb{R}\backslash[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}=0$$

So, I have

$$\lim_{n\rightarrow\infty}\int{}_{\mathbb{R}\backslash[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}=\lim_{n\rightarrow\infty}\left(\int_{\mathbb{R}}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}-\int_{[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)\\=\lim_{n\rightarrow\infty}\left(1-\int_{-r}^{r}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)\\=\lim_{n\rightarrow\infty}\left(1-\int_{-r\sqrt{n}}^{r\sqrt{n}}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt\right)$$

Intuitively, It is zero. But How can I show it rigorously?

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  • $\begingroup$ So: first the dirac delta is a distribution, not a function. Secondly: your last limit clearly converges to $1$, as one can easily apply the inequality $|\int_a^b f\ dx|\leq (b-a)\sup_{x\in(a,b)}|f(x)|$, and see that the latter converges to $0$, as the supremum of $f$ is bounded, and the interval length goes to $0$. Are you sure you need to prove what you claim? $\endgroup$ – b00n heT Jun 6 '16 at 6:43
  • $\begingroup$ @b00nheT In this case, $(b-a)\sup _{x\in(a,b)} |f(x)|=2r\sqrt n |f(0)|$ and seems not to be convergent as $n\rightarrow \infty$, isn't it? $\endgroup$ – Darae-Uri Jun 6 '16 at 6:49
  • $\begingroup$ you are most definitely right... I thought the boundary terms read $r/\sqrt{n}$, but I was clearly mistaken. So yes, you are correct, and to prove that just use the fact that $e^{-t^2}$ is an integrable function, so that the limit for $n\rightarrow \infty$ converges to the well defined integral $\int_\mathbb{R}\frac{1}{\sqrt{\pi}}e^{-t^2}dt$, as $r\sqrt{n}\rightarrow \infty$. for any $r>0$. $\endgroup$ – b00n heT Jun 6 '16 at 7:00
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In most cases, you can explicitly bound the integrand by integrable functions so that you can integrate and then take the limit, especially if you want to prove the limit is zero. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$ $\def\less{\smallsetminus}$

Given $r > 0$ and as $n \in \nn \to \infty$:

  Thus $\int_{\rr\less[-r,r]} \sqrt{\frac{n}{π}} e^{-nx^2}\ dx \le 2 \int_r^\infty \sqrt{\frac{n}{π}} \frac{x}{r} e^{-nx^{2}}\ dx = \left[ - \sqrt{\frac{n}{π}} \frac{1}{rn} e^{-nx^2} \right]_r^\infty \to 0$.

I'll leave the last limit for you to check, but it's kind of obvious. Note that it is crucial that $r > 0$.

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  • $\begingroup$ Look! No hard-to-evaluate integrals! $\endgroup$ – user21820 Jun 6 '16 at 7:24
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Hint. Since $$ \int_{-\infty}^{\infty}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt=1 $$ then we have $$ \lim_{n\rightarrow\infty}\left(1-\int_{-r}^{r}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)=\lim_{n\rightarrow\infty}\left(1-\int_{-r\sqrt{n}}^{r\sqrt{n}}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt\right)=0. $$

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  • $\begingroup$ Hint you gave me is already in my question. See the 6th line of question. $\endgroup$ – Darae-Uri Jun 6 '16 at 7:07
  • $\begingroup$ @Darae-Uri Please, unless you would like to prove that, as $z \to \infty$, we have something like $\text{erf}(z)=1-\frac{1}{\sqrt{\pi }}e^{-z^2} \left(\frac{1}{ z}-\frac{1}{2 z^3}+O\left(\frac{1}{z^4}\right)\right)$, the hint I gave is OK. $\endgroup$ – Olivier Oloa Jun 6 '16 at 7:24
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A probabilistic approach. $$ f_n(x) = \sqrt{\frac{n}{\pi}} e^{-nx^2} $$ is the density function of a random variable $X_n$ that is normally distributed, with mean zero and variance $\sigma_n^2 = \frac{1}{2n}$. By Chebyshev's inequality, $$ \int_{\mathbb{R}\setminus[-r,r]}f_n(x)\,dx = \mathbb{P}[|X_n|>r\sqrt{2n}\sigma_{n}]\leq \frac{1}{2nr^2} $$ and the RHS tends to zero as $n\to +\infty$ for any $r\in\mathbb{R}^+$.

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