How to prove that $\lim_{y \rightarrow \infty} \frac{\sin^2(xy)}{yx^2}=\pi \delta(x)$

Question

I want to understand how to prove that

$$\lim_{y \rightarrow \infty} \frac{\sin^2(xy)}{yx^2}=\pi \delta(x)$$

The proof I am studying relies on doing the following Fourier Transform

$$\int dx \frac{\sin^2(xy)}{yx^2} \exp(-ipx) = \frac{\pi}{2} \Theta(2y - |p|) \Big(2-\frac{|p|}{y}\Big)$$

Which is shown by using the inverse of this formula. Then for $y \rightarrow \infty$ the right hand side goes to $\pi$ which gets you the limit proven.

However, I do not understand this proof. Could you please shed more details on it? If you have another kind of proof in mind please feel free to share it.

Thank you.

Answer 1

First we multiply the expression with a test function $\varphi.$ Then we integrate doing a variable substitution: $$ \int_{-\infty}^{\infty} \frac{\sin^2(xy)}{yx^2} \varphi(x) \, dx = \int_{-\infty}^{\infty} \left( \frac{\sin(xy)}{xy} \right)^2 \varphi(x) \, y \, dx = \{ z = xy \} = \int_{-\infty}^{\infty} \left( \frac{\sin(z)}{z} \right)^2 \varphi(\frac{z}{y}) \, dz \\ \to \int_{-\infty}^{\infty} \left( \frac{\sin(z)}{z} \right)^2 \varphi(0) \, dz = \int_{-\infty}^{\infty} \left( \frac{\sin(z)}{z} \right)^2 \, dz \, \varphi(0) . $$

Here we can use the Plancherel theorem for the left integral: $$ \int_{-\infty}^{\infty} \left( \frac{\sin(z)}{z} \right)^2 \, dz = \left< \frac{\sin(z)}{z}, \frac{\sin(z)}{z} \right> = \frac{1}{2\pi} \left< \mathcal{F}\{\frac{\sin(z)}{z}\}, \mathcal{F}\{\frac{\sin(z)}{z}\} \right> \\ = \frac{1}{2\pi} \left< \pi\chi_{[-1,1]}, \pi\chi_{[-1,1]} \right> = \frac{\pi^2}{2\pi} \int_{-1}^{1} dx = \pi . $$

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Why is $\mathcal{F}\{\frac{\sin(z)}{z}\} = \pi\chi_{[-1,1]}$?

We have $$ \mathcal{F}\{\chi_{[-1,1]}(z)\}(\zeta) = \int \chi_{[-1,1]}(z) \, e^{-i\zeta z} dz = \left[ \frac{e^{-i\zeta z}}{-i\zeta} \right]_{-1}^{1} = \frac{e^{-i\zeta}}{-i\zeta} - \frac{e^{i\zeta}}{-i\zeta} = 2\frac{\sin(\zeta)}{\zeta} , $$ so by the Fourier inversion theorem, $$ \mathcal{F}\{2\frac{\sin(z)}{z}\}(\zeta) = 2\pi \chi_{[-1,1]}(\zeta), $$ i.e. $$ \mathcal{F}\{\frac{\sin(z)}{z}\}(\zeta) = \pi \chi_{[-1,1]}(\zeta), $$

Answer 2

If $f\in L^1(\Bbb{R})$ with $ c=\int_{-\infty}^\infty f(x)dx$ then $\lim_{n\to \infty}n f(n.)=c\delta$ in the sense of distributions.

Proof : with $\phi \in C^0_c(\Bbb{R})$ then $$\int_{-\infty}^\infty n f(nx)\phi(x)dx=\int_{-\infty}^\infty f(x)\phi(x/n)dx=\int_{-A}^A f(x)(\phi(0)+o(1))dx+O(\int_{|x|>A} |f(x)|dx)$$

Answer 3

This can be seen in terms of approximations to the identity. The kernel $\mathcal{K}_R(y)=\frac{\sin^2(R\pi y)}{R(\pi y)^2}$, called the Fejér kernel, is integrable. In fact, $\mathcal{K}_R(y)=R\mathcal{K}_1(Ry)$, and $\int \mathcal{K}_1(y)\,dy=1$. This can be seen by noticing that $\mathcal{K}_1(y)$ is the Fourier transform of the tent distribution $\phi(t)=(1-|t|)_+$, which is the convolution of the uniform distribution in $(-1/2,1/2)$ with itself.

Then, for any $f\in\mathcal{L}_p$, $\|\mathcal{K}_R*f-f\|_p\xrightarrow{R\rightarrow\infty}0$, and convergene pointwise also holds at every Lebesgue point of $f$. If you use test functions $f\in\mathcal{S}$ (Schwartz functions), then $\mathcal{K}_R*f\rightarrow f$ point wise, that is $K_R$ behaves like a delta function. A brief explanation of approximations to the identity is given here.