Is the sequence of function $\frac{\sin nx}{\pi x}$ convergent to $\delta(x)$ I have no doubt that the integral of the function from negative infinity to positive infinity is 1. But I don't think $f_n(x) = \frac{\sin nx}{\pi x} \rightarrow 0 $ as $ n \rightarrow \infty $ given $x\ne 0$, which doesn't fit the definition of $\delta(x)$. Do you agree with that?
 A: That depends on your definition of the Dirac delta. The mathematically sound definition is that it is a distribution defined by its application to test functions $φ\in C_0^\infty(\Bbb R)$ as
$$
δ(φ)=\langle δ,φ\rangle=φ(0)
$$
Now the $f_n$ are locally integrable functions and thus related to a regular distribution 
\begin{align}
T_{f_n}(φ)&=\int_{\Bbb R}f_n(x) φ(x)\,dx=\langle f_n,φ\rangle_{L^2}
\\
&=\langle \hat f_n,\hat φ\rangle_{L^2}=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}\hat φ(ω)\cdot 1\,dω
\end{align}
By the dominated convergence theorem, $T_{f_n}(φ)$ converges to
$$
\frac{1}{\sqrt{2\pi}}\int_{-∞}^{∞}\hat φ(ω)\cdot 1\,dω=φ(0)
$$
that is, $T_{f_n}\to δ$ as a distribution.

What you had in mind by trying to prove pointwise convergence is the idea of an approximate identity (of convolution) or a delta-approximating sequence. For such a sequence $f_n:\Bbb R\to\Bbb R$ one demands, for the reasons obvious in the sequence of Dirichlet kernels, that


*

*$f_n(x)\ge 0$, the functions are non-negative,

*$\int_{\Bbb R}f_n(x)\,dx=1$ and

*$\lim_{n\to\infty} f_n(x)\to 0$ for $x\ne 0$


Then
$$
\begin{align}
\left|\int_{\Bbb R}f_n(x)φ(x)\,dx-φ(0)\right|
&\le\int_{\Bbb R}|f_n(x)|\,|φ(x)-φ(0)|\,dx
\\
&\le \int_{\Bbb R}|f_n(x)|\,dx\cdot\sup_{x\in[-a,a]}|φ(x)-φ(0)|+\int_{|x|\ge a} |f_n(x)|\,|φ(x)-φ(0)|\,dx
\end{align}
$$
The first term is controlled by the continuity of $φ$, the second by dominated convergence and the pointwise convergence of the $f_n$.

Examples of delta-approximating sequences are


*

*$\frac{n}2\,e^{-n\,|x|}$

*$\sqrt{\frac{n}{2\pi}}\,e^{-n\,x^2}$

*$\frac{1}{\pi}\frac{n}{1+n^2x^2}$

*$C\cdot \frac{\sin^2(nx)}{nx^2}$ for some constant $C$,...

A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#00f}{\large\lim_{n \to \infty}{\sin\pars{nx} \over \pi x}}
&=\lim_{n \to \infty}\braces{%
{n \over \pi}\bracks{\sin\pars{nx} \over nx}}
=\lim_{n \to \infty}\bracks{%
{n \over \pi}\pars{\half\int_{-1}^{1}\expo{\ic knx}\,\dd k}}
\\[5mm] & =\lim_{n \to \infty}
\int_{-n}^{n}\expo{\ic kx}\,{\dd k \over 2\pi} =
\int_{-\infty}^{\infty}\expo{\ic kx}\,{\dd k \over 2\pi}
=\color{#00f}{\large\delta\pars{x}}
\end{align}
