# 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.

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

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 ${\mathbb{R}}^{+}$. Furthermore ${f}_{n}(x)\nearrow f(x)$ (pointwise)

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

So

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

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

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

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

$\mathrm{sinc}(x)\notin {L}^{1}(\mathbb{R})$

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 |