# sinc is $L^{2}$

Our objective will be to prove the integral $\int_{\mathbb{R}}f^{2}(x)dx$ exists in the Lebesgue sense when $f(x)=\operatorname{sinc}(x)$.

The integrand is an even function and so we can restrict our proof to the set $\mathbb{R}^{+}$.

Since $f$ is a continuous function   , so will $f^{2}$ be and thus for every $a>0$, $f\in L^{2}([0,a])$.

Thus, if we prove $f\in L^{2}([\pi,\infty[)$, the result will be proved.

Consider the intervals $I_{k}=[k\pi,(k+1)\pi]$ and $U_{k}=\bigcup_{i=1}^{k}I_{k}=[\pi,(k+1)\pi]$.

Each $f_{n}$ is a continuous function of compact support and will thus be integrable in $\mathbb{R}^{+}$. Furthermore $f_{n}(x)\nearrow f^{2}(x)$ (pointwise) in this set.

In each $I_{k}$,$0\leq f^{2}(x)\leq\frac{\sin^{2}(x)}{(k\pi)^{2}}$, for $k>0$.

So:

$\displaystyle\int_{x\geq\pi}f_{n}(x)dx=\sum_{k=1}^{n}\int_{k\pi}^{(k+1)\pi}% \frac{\sin(x)^{2}}{x^{2}}dx\leq\sum_{k=1}^{n}\int_{k\pi}^{(k+1)\pi}\frac{\sin(% x)^{2}}{(k\pi)^{2}}=\sum_{k=1}^{n}\frac{1}{2k^{2}\pi}$ 11we have used the well known result $\int_{0}^{\pi}\sin^{2}(x)dx=\frac{\pi}{2}$

So: $\lim_{n\to\infty}\int_{x\geq\pi}f_{n}(x)dx\leq\lim_{n\to\infty}\sum_{k=1}^{n}% \frac{1}{2k^{2}\pi}$ and since the series on the right side converges22asymptotic behaviour as $k^{-2}$ and $f_{n}\nearrow f^{2}$ we can use the monotone convergence theorem  to state that $f^{2}\in L([\pi,\infty[)$.

So we get the result that $\operatorname{sinc}\in L^{2}(\mathbb{R})$

Title sinc is $L^{2}$ SincIsL2 2013-03-22 15:44:44 2013-03-22 15:44:44 cvalente (11260) cvalente (11260) 9 cvalente (11260) Result msc 26A06