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The delta function (well, delta is not really a function; it is a distribution) can be defined as a limit of (among many other approximations) the following approximates of the unity:

the heat kernel:

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

or the Poisson kernel $$\delta(x)=\lim_{\epsilon\rightarrow 0^{+}} \frac{1}{\pi} \frac{\epsilon}{\epsilon^2+x^2}\qquad (2)$$

Check out the link

https://en.wikipedia.org/wiki/Dirac_delta_function#Representations_of_the_delta_function

Now, let $f$ be a tempered distribution. Can we interchange the limit with the integral so that

$$\int f(x)\delta(x)dx =\lim_{\epsilon\rightarrow 0^{+}}\int f(x) \frac{e^{-\frac{x^2}{2\epsilon}}}{\sqrt{2\pi\epsilon}}dx \qquad (*)$$

If $(*)$ is true, what is the justification ?

Notice that $(*)$ is true when $f$ is a smooth function with compact support, in which case both sides of $(*)$ give $f(0)$. With tempered distributions, we can not make sense of the pointwise value $f(0)$.

An easier question is when $f$ is a continuous oscillatory function, say $f(x)=e^{\dot{\imath}\alpha x^2}$, $\alpha \in \mathbb{C}$. What can we say then?

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I think you're thinking about this the wrong way round. There isn't a function $\delta$ that is obtainable from kernels by pointwise convergence, and satisfies $\int_{\Bbb R}f(x)\delta (x) dx=f(0)$. Indeed, if we define $\delta$ as a pointwise limit, $\delta(x)=0$ for all $x\ne 0$ (while $\delta(0)=\infty$), so $\int_{\Bbb R}f(x)\delta (x) dx=0$. But what we can say is this: given a non-negative function $\phi(x)$ with $\phi(0)>0,\,\int_{\Bbb R}\phi(x) dx=1$, $$\lim_{\epsilon\to 0^+}\int_{\Bbb R}f(x)\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx=\lim_{\epsilon\to 0^+}\int_{\Bbb R}f(\epsilon y)\phi(y)dy.$$If we can move the limit inside the integral, we obtain$$\int_{\Bbb R}\lim_{\epsilon\to 0^+}f(\epsilon y)\phi(y)dy=\int_{\Bbb R}f(0)\phi(y)dy=f(0).$$Note that this argument never tries to move the limit so as to write $$\int_{\Bbb R}f(x)\lim_{\epsilon\to 0^+}\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx,$$which would be $0$ as aforesaid. The definition of the measure we call the Dirac delta is$$\int_a^b f(x)\delta(x)dx=\lim_{\epsilon\to 0^+}\int_a^bf(x)\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx,$$not$$\delta(x)=\lim_{\epsilon\to 0^+}\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon}).$$The Dirac delta is not a function.

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  • $\begingroup$ Also, in the pointwise sense, the limits in (1) and (2) do not exist because there is no FUNCTION to which these approximates of the identity converge to. Right ? $\endgroup$ – Medo Nov 10 '18 at 16:49
  • $\begingroup$ @Medo The functions exist, but they don't give the desired integrals; the measure we call $\delta$ is therefore not a function. In particular, no function $D(x)$ exists with the identity $\int_{\Bbb R}f(x)D(x) dx=f(0)$. $\endgroup$ – J.G. Nov 10 '18 at 16:53
  • $\begingroup$ Makes perfect sense. Thank you very much. $\endgroup$ – Medo Nov 10 '18 at 22:04

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