sinc is $L^2$

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

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

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

Consider the intervals $I_k=[k\pi ,(k+1)\pi ]$ and $U_k={\bigcup }_{i=1}^k{I}_k=[\pi ,(k+1)\pi ]$.

and the succession of functions $f_n(x)=f^2(x){\chi }_{U_n}(x)$, where ${\chi }_{U_n}$ is the characteristic function of the set $U_n$.

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

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

So:

${\displaystyle {\int }_{x\ge \pi }}{f}_n(x)𝑑x={\displaystyle \sum _{k=1}^n}{\displaystyle {\int }_{k\pi }^{(k+1)\pi }}{\displaystyle \frac{\sin {(x)}^2}{x^2}}𝑑x\le {\displaystyle \sum _{k=1}^n}{\displaystyle {\int }_{k\pi }^{(k+1)\pi }}{\displaystyle \frac{\sin {(x)}^2}{{(k\pi )}^2}}={\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{2k^2\pi }}$

So: ${lim}_{n\to \mathrm{\infty }}{\int }_{x\ge \pi }{f}_n(x)𝑑x\le {lim}_{n\to \mathrm{\infty }}{\sum }_{k=1}^n\frac{1}{2k^2\pi }$ and since the series on the right side converges and $f_n\nearrow f^2$ we can use the monotone convergence theorem to state that $f^2\in L([\pi ,\mathrm{\infty }[)$.

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

Title	sinc is $L^2$
Canonical name	SincIsL2
Date of creation	2013-03-22 15:44:44
Last modified on	2013-03-22 15:44:44
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	9
Author	cvalente (11260)
Entry type	Result
Classification	msc 26A06