so basically I want a proof for

$\lim_{\epsilon\rightarrow 0}\frac{1}{\sqrt{2\pi\epsilon}}e^{\frac{-x^2}{2\epsilon}}=\delta(x)$

I don't yet care about proving


I just want to prove that

$\delta(x)=\lim_{\epsilon\rightarrow 0}\frac{1}{\sqrt{2\pi\epsilon}}e^{\frac{-x^2}{2\epsilon}}=0$ $\forall x\neq 0$

I don't have a clue how to solve the limit. I tried using L'Hospitals rule but it doesn't work and I'm completely clueless how I should be going about this. I mean, it makes sense that the limit is zero when I look at a graph of the normal distribution, but how can I prove that with equations?

Thanks in advance!

  • 3
    $\begingroup$ Convergence in which sense? As measures (i.e., the total variation of the difference of the two measures tends to zero) or weak$^*$ convergence, against bounded continuous functions in $\mathbb R$...or in some other sense? $\endgroup$ – Yiorgos S. Smyrlis Dec 26 '16 at 12:42
  • $\begingroup$ It converges to the Dirac delta in the sense that $\lim_{\epsilon\rightarrow 0}\int_a^b \frac{1}{\sqrt{2\pi\epsilon}}e^{\frac{-x^2}{2\epsilon}}dx = \lim_{\epsilon\rightarrow 0}\int_{a/\sqrt{\epsilon}}^{b/\sqrt{\epsilon}} \frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}dy = \begin{cases} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}dy =1 \text{ if } 0 \in (a,b) \\ 0 \text{ if } 0 \not \in [a,b]\end{cases}$ $\endgroup$ – reuns Dec 26 '16 at 12:47
  • $\begingroup$ @user1952009, A little bit of modification is required when at least one of endpoints of $(a, b)$ are zero. $\endgroup$ – Sangchul Lee Dec 26 '16 at 13:09
  • $\begingroup$ @SangchulLee Yes but what happens when $a=0$ is obvious. And I think the OP wants to understand the convergence to $\delta(x)$ in physics or in the context of the Fourier transform/series, without any measures and functional analysis. $\endgroup$ – reuns Dec 26 '16 at 13:56

You know that $e^x\ge 1+x$. Then $$ e^{-\frac{x^2}{2ϵ}}\le\frac1{1+\frac{x^2}{2ϵ}}=\frac{2ϵ}{2ϵ+x^2} $$ gives a sufficiently small upper bound for the considered point-wise limit.

In general you have for any positive function $\phi$ of integral $1$ that $\frac1ϵ\phi(\frac xϵ)$ converges to $δ$. This is part of the idea of approximations of unity ($δ$ is the unit element under convolution).

  • $\begingroup$ Thank you, that's the kind of answer I was looking for. I'm studying physics so I don't really understand much of the other answers... But thanks to everyone! $\endgroup$ – Keno Goertz Dec 26 '16 at 13:56

You have to be a bit careful about what this notation you are using actually means. Rigorously, the dirac delta is only defined when acting on functions in the Schwartz space. There is no rigorous justification for handling it as a function $x \mapsto \delta(x)$ for real $x$. I'll discuss the case where we are talking about convergence in the distributional sense. Let $\varphi \in C_{c}^{\infty}(\mathbb{R})$. I'll note $ f_{\epsilon} := \frac{1}{\sqrt{2\pi\epsilon}}e^{\frac{-x^2}{2\epsilon}}$. Using the substitution $y = x/\sqrt{\epsilon}$ we obtain $$\langle f_{\epsilon},\varphi \rangle = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\epsilon}}e^{\frac{-x^2}{2\epsilon}} \varphi(x) \mathrm{d}x =\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-y^2 /2} \varphi(\sqrt{\epsilon}y) \mathrm{d}y $$ As $\varphi$ is continous and compactly supported, this integrand is dominated by $\frac{1}{\sqrt{2\pi}}e^{-y^2 /2} \|\varphi\|_{\infty}$ which integrates to $\|\varphi\|_{\infty}$. Moreover, because $\varphi$ is continous the integrand converges pointwise to $\frac{1}{\sqrt{2\pi}}e^{-y^2 /2} \varphi(0)$ as $\epsilon \to 0$. Applying the dominated convergence theorem yields

$$\lim_{\epsilon\to 0} \langle f_{\epsilon},\varphi \rangle = \int_{-\infty}^{\infty} \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}}e^{-y^2 /2} \varphi(\sqrt{\epsilon}y) \mathrm{d}y = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-y^2 /2} \varphi(0) \mathrm{d}y = \varphi(0) $$ which finally shows $$\lim_{\epsilon\to 0} \langle f_{\epsilon},\varphi \rangle = \langle \delta_0, \varphi \rangle$$

  • $\begingroup$ Precisely the way forward I would have taken. (+1) and Happy Holidays! -Mark $\endgroup$ – Mark Viola Dec 26 '16 at 16:09
  • $\begingroup$ @Dr.MV Thanks and Happy Holidays to you too! $\endgroup$ – user159517 Dec 27 '16 at 1:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.