sinc is not $L^1$

The main results used in the proof will be that $f\in L^1(A)\iff |f|\in L^1(A)$ and the dominated convergence theorem.

Proof by contradiction:

Let $f(x)=|\mathrm{sinc}(x)|$ and suppose it’s Lebesgue integrable in ${ℝ}^{+}$.

Consider the intervals $I_k=[k\pi ,(k+1)\pi ]$ and $U_k={\bigcup }_{i=0}^k{I}_k=[0,(k+1)\pi ]$.

and the succession of functions $f_n(x)=f(x){\chi }_{U_n}(x)$, 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 $f_n(x)\nearrow f(x)$ (pointwise)

in each $I_k$, $f(x)\ge \frac{|\sin (x)|}{(k+1)\pi }$.

So

${\displaystyle {\int }_{{ℝ}^{+}}}{f}_n={\displaystyle \sum _{k=0}^n}{\displaystyle {\int }_{k\pi }^{(k+1)\pi }}{\displaystyle \frac{|\sin (x)|}{x}}𝑑x\ge {\displaystyle \sum _{k=0}^n}{\displaystyle {\int }_{k\pi }^{(k+1)\pi }}{\displaystyle \frac{|sin(x)|}{(k+1)\pi }}={\displaystyle \sum _{k=0}^n}{\displaystyle \frac{2}{(k+1)\pi }}$.

Suppose $f$ is integrable in ${ℝ}^{+}$. Then by the dominated convergence theorem ${lim}_{n\to \mathrm{\infty }}{\int }_{{ℝ}^{+}}{f}_n={\int }_{{ℝ}^{+}}f$.

But ${lim}_{n\to \mathrm{\infty }}{\int }_{{ℝ}^{+}}{f}_n\ge {lim}_{n\to \mathrm{\infty }}{\sum }_{k=0}^n\frac{2}{(k+1)\pi }=+\mathrm{\infty }$ and we get the contradiction ${\int }_{{ℝ}^{+}}f\ge +\mathrm{\infty }$.

So $f$ cannot be integrable in ${ℝ}^{+}$. This implies that $f$ cannot be integrable in $ℝ$ and since a function is integrable in a set iff its absolute value is

$\mathrm{sinc}(x)\notin L^1(ℝ)$

Title	sinc is not $L^1$
Canonical name	SincIsNotL1
Date of creation	2013-03-22 15:44:32
Last modified on	2013-03-22 15:44:32
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	14
Author	cvalente (11260)
Entry type	Result
Classification	msc 26A06