I was trying to prove that function $\frac{\sin{nx}}{\pi x}$ sutisfies these conditions:

1) $ \forall D \; \exists c : \forall a,b \;\; |a|<D, \; |b|<D \quad \left|\int_a^b \frac{\sin{nx}}{\pi x} dx\right| \leq c$

2) $ \underset{n \to \infty}{\lim} \int_a^b \frac{\sin{nx}}{\pi x} dx = \begin{cases} 1 & 0\in(a,b) \\ 0 & 0\notin(a,b)\\ \end{cases} $

Then it converges to $\delta$-distribution. I do not know how I can show that the function sutisfies these conditions. Can someone please explain how I should proceed or at least give some hints?

  • $\begingroup$ I think this is false. The Dirichlet kernel is not a summability kernel. IN particular 1) fails. $\endgroup$ – Dunham Nov 27 '16 at 1:58
  • $\begingroup$ @Dunham did you read $\int_a^b | \frac{\sin n x}{\pi x}| dx$ ? $\endgroup$ – reuns Nov 27 '16 at 2:01
  • $\begingroup$ yes, i did. And that is an important difference $\endgroup$ – Dunham Nov 27 '16 at 2:13
  • $\begingroup$ For 2) it might reduce to showing the Fourier series of $f(x) = 1_{x \in [a,b]}$ converges, using the Riemann-Lebesgue lemma or the power series for $-\log(1-z)$ $\endgroup$ – reuns Nov 27 '16 at 2:16
  • $\begingroup$ Since $\left|\frac {\sin nx}{\pi x}\right| \le \frac n\pi$ everywhere, for (1), you can just take $c = \frac {2nD}\pi$. Even if the inequality has been $\int_a^b | \frac{\sin n x}{\pi x}| dx \le c$ $\endgroup$ – Paul Sinclair Nov 27 '16 at 5:22

First, we enforce the substitution $x\to x/n$ to obtain

$$\int_a^b \frac{\sin(nx)}{\pi x} \,dx=\frac1\pi \int_{na}^{nb} \frac{\sin(x)}{ x}\,dx \tag 1$$

PART $(1):$

Using $\frac{\sin(x)}{x}\le 1$ for $x\in [-\pi ,\pi]$, it is easy to see that for any $na$ and $nb$ we have

$$\left| \frac1\pi \int_{na}^{nb}\frac{\sin(x)}{x}\,dx\right| \le \frac2\pi\int_0^\pi (1)\,dx=2$$

PART $(2)$:

Using $(1)$, we have

$$\lim_{n\to \infty}\int_a^b \frac{\sin(nx)}{\pi x}\,dx= \begin{cases} \frac1\pi\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx=1 &,a<0<b\\\\ \frac1\pi\int_{\infty}^\infty \frac{\sin(x)}{x}\,dx=0 &,0<a<b\\\\ \frac1\pi\int_{-\infty}^{-\infty} \frac{\sin(x)}{x}\,dx=0 &,a<b<0 \end{cases}$$


To show that $\lim_{n\to \infty}\int_{na}^{nb}\frac{\sin(x)}{x}\,dx=0$ for $0<a<b$ or $a<b<0$, we integrate by parts with $u=1/x$ and $v=-\cos(x)$ to write

$$\lim_{n\to \infty}\int_{na}^{nb}\frac{\sin(x)}{x}\,dx=\lim_{n\to \infty}\left(\frac{\cos(na)}{na}-\frac{\cos(nb)}{nb}-\int_{na}^{nb}\frac{\cos(x)}{x^2}\,dx\right)=0$$

To show that $\int_0^\infty \frac{\sin(x)}{x}\,dx=\pi/2$, we write

$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\int_0^\infty \sin(x)\int_0^\infty e^{-kx}\,dk\,dx$$

Now, since we have for any $L>0$

$$\int_0^L\int_0^\infty \left|e^{-kx}\sin(x)\right|\,dk\,dx\le L$$

the Fubini-Tonelli Theorem guarantees that

$$\begin{align} \int_0^L\int_0^\infty e^{-kx}\sin(x)\,dk\,dx&=\int_0^\infty \int_0^L e^{-kx}\sin(x)\,dx\,dk\\\\ &=\int_0^\infty \frac{1-e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\,dk\\\\ &=\frac\pi2-\int_0^\infty \frac{e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\,dk \end{align}$$

Finally, since $\left|\frac{e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\right|\le g(k)$ where $g(k)$ is given by

$$g(k)= \begin{cases}e^{-k}&,k\ge 1\\\\\frac{1+\sqrt2}{2}&,k\le 1\end{cases}$$

and $\int_0^\infty g(k)\,dk <\infty$, we can apply the Dominated Convergence Theorem to obtain

$$\begin{align}\lim_{L\to \infty}\int_0^L\int_0^\infty e^{-kx}\sin(x)\,dk\,dx&=\lim_{L\to \infty}\int_0^\infty\int_0^L e^{-kx}\sin(x)\,dk\,dx\\\\ &=\int_0^\infty \lim_{L\to \infty}\left(\frac{1-e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\right)\,dk\\\\ &= \pi/2 \end{align}$$


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