# Distribution-Theory: Show that a function tends to the Dirac delta function

Problem. Prove that the function $$\frac{1}{\pi x}\sin(\frac{x}{\epsilon})$$ tends to $$\delta(x)$$, when $$\epsilon \rightarrow +0$$.

I asked a similar question which was answered here. In the answer, the idea was to use Lebesgue integration and the dominant convergence theorem.

However, in Vladimirov's book Equations of mathematical physics he does something else. Namely, he defines the Dirac delta function like this:

He takes the average density $$f_{\epsilon}$$ by distributing the unit mass uniformly inside an open ball $$U_{\epsilon}$$ such that

$$\begin{equation*} f_{\epsilon} = \begin{cases} \frac{3}{4 \pi \epsilon^{3}} & |x| < \epsilon \\ 0 & |x| > \epsilon \end{cases} \end{equation*}$$

Then, he calculates the weak limit of the sequence of functions $$f_{\epsilon}$$ as $$\epsilon \rightarrow 0$$; that is, for any continuous function $$\phi$$ he shows that

$$\lim_{\epsilon \rightarrow 0} \int f_{\epsilon} \phi(x) dx = \phi(0).$$

The argument goes like this. Since $$\phi$$ is continuous we have for all $$\eta > 0$$ that there is $$\epsilon_{0} > 0$$ such that $$|x - 0| = |x| < \epsilon_{0}$$ implies $$|\phi(x) - \phi(0)| < \eta$$. Thus, we can use continuity of $$\phi$$ to show that for all $$\epsilon \leq \epsilon_{0}$$ we have

\begin{aligned} |\int f_{\epsilon} \phi(x) dx - \phi(0)| &= |\int_{|x| < \epsilon} \frac{3}{4 \pi \epsilon^{3}} \phi(x) dx - \phi(0)| \\ & = \frac{3}{4 \pi \epsilon^{3}} |\int_{|x| < \epsilon} \phi(x) - \phi(0) dx| \\ & \leq \frac{3}{4 \pi \epsilon^{3}} \int_{|x| < \epsilon}| \phi(x) - \phi(0)| dx \\ & < \frac{3}{4 \pi \epsilon^{3}} \eta \int_{|x| < \epsilon} dx \\ & = \eta \end{aligned}

First of what happens in the last line? I.e. why is $$\frac{3}{4 \pi \epsilon^{3}} \eta \int_{|x| < \epsilon} dx = \eta$$?

So according to the above, the weak limit of the sequence of functions $$f_{\epsilon}$$ as $$\epsilon \rightarrow 0$$ is the functional $$\phi(0)$$ assigning to each continuous function $$\phi$$ its value at $$x = 0$$ and this functional is taken as the definition of the density $$\delta(x)$$ - namely, the Dirac delta function.

When I solve problems as the above statement (first line of this question), do I have to show the weak limit? I.e.

$$\lim_{\epsilon \rightarrow 0} \int \frac{1}{\pi x}\sin(\frac{x}{\epsilon}) \phi(x) dx = \phi(0).$$

I tried doing this but can't seem to proceed. Or does one always use the Lebesgue dominated convergence theorem to solve such problems? (Wasn't able to do it this way either)

• $\int_{|x|<\epsilon}dx$ is the volume of the open ball of radius $\epsilon$ around the oring in 3 dimensions. So it is $\frac{4\pi\epsilon^3}{3}$. Commented Mar 4, 2020 at 22:27
• @AbdelmalekAbdesselam thanks. Any ideas how to solve the problem?
– user744734
Commented Mar 5, 2020 at 0:32
• math.stackexchange.com/a/2866459/168433 Commented Mar 6, 2020 at 23:22