How to show that $f(x) = \int_{-\infty}^{\infty}f(y) \frac {\sin[\pi(x-y)]}{\pi(x-y)}dy$, where $f(x)=0$ for $|x|>1/2$ and $f\in L^2(\mathbb{R})$?

Question

How to show that $f(x) = \int_{-\infty}^{\infty}f(y) \frac {\sin[\pi(x-y)]}{\pi(x-y)}dy$, where $f(x)=0$ for $|x|>1/2$ and $f\in L^2(\mathbb{R})$?

Attempt: For the identity function $g(x)=1_{[-1/2,1/2]}$ we get the Fourier transform

$\hat{g}(\gamma)=\int^{1/2}_{-1/2}e^{-2\pi i x \gamma}dx = \frac{\sin[\pi\gamma]}{\pi\gamma}$

Thus

$f(x) = f(x)g(x) = \check{[\hat{f}\hat{g}]}(x) = \int_{-\infty}^{\infty}\hat{f}(\gamma)\hat{g}(\gamma)e^{2\pi i x \gamma}d\gamma = \int_{-\infty}^{\infty}\hat{f}(\gamma)\frac{\sin[\pi\gamma]}{\pi\gamma}e^{2\pi i x \gamma}d\gamma$ $= \int_{-\infty}^{\infty}\int_{-1/2}^{1/2}f(y)e^{-2\pi i y \gamma}dy\frac{\sin[\pi\gamma]}{\pi\gamma}e^{2\pi i x \gamma}d\gamma$.

But I tend to get stuck. A was given the hint to use the Plancherel theorem but cannot see how that would apply.

Answer 1

$$\begin{split} \int_{-\infty}^{\infty}f(y) \frac {\sin[\pi(x-y)]}{\pi(x-y)}dy&=\int_{-\infty}^{\infty}f(y) \int_{-\frac 1 2}^{\frac 1 2}e^{2i\pi(x-y)\xi}d\xi dy\\ &=\int_{-\frac 1 2}^{\frac 1 2}\int_{-\infty}^{\infty}f(y)e^{2i\pi(x-y)\xi}dyd\xi\\ &=\int_{-\frac 1 2}^{\frac 1 2}e^{2i\pi x\xi}\hat f(\xi)d\xi\\ &=f(x) \end{split}$$

Answer 2

Let $g(x)=(\pi x)^{-1}\sin(\pi x)$. You want to find that $f=f*g.$ To that end, consider $h=f-f*g.$ Taking the Fourier transform gives that $\hat{h}=\hat{f}-\hat{f}\hat{g}.$ If you proceed similar as in your post, you can find $\hat{g}$ and hence $\hat{h}.$