proof of Ingham Inequality

Let

$$f(x)=\sqrt{\frac{1}{2\pi }}\sum _{j=-n}^n{c}_j{e}^{i{t}_{j}x}.$$

If $k(x)$ is a integrable function in $(-\mathrm{\infty },\mathrm{\infty })$, let

$$K(u)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}k(x)e^{ixu}𝑑x.$$

It is easy to prove the equalities.

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}k(x){|f(x)|}^2𝑑x$ $=$ ${\displaystyle \sum _{j,h=-n}^n}{c}_j\overline{c_h}K(t_j-t_h).$

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}k(x)f(x)e^{-ix{t}_h}𝑑x$ $=$ ${\displaystyle \sum _{j=-n}^n}{c}_{j}K(t_j-t_h).$

For the rest of the proof we make the following choices:

Then, after computation, we have

$$K(u)=\frac{4\cos (\pi u)}{1-4u^2}.$$

Let $T=\pi$ and $\gamma =1+\delta$ ($\delta =\epsilon /\pi >0$). Since $|t_{j+1}-t_j|\ge \gamma$, we have that $|t_j-t_h|\ge |j-h|\gamma$. If we suppose that $h$ is fixed, we have

${\displaystyle \sum _{j\ne h}}|K(t_j-t_h)|$ $=$ ${\displaystyle \sum _{j\ne h}}\left|{\displaystyle \frac{4\cos \pi (t_j-t_h)}{1-4{(t_j-t_h)}^2}}\right|\le {\displaystyle \sum _{j\ne h}}{\displaystyle \frac{4}{4{(j-h)}^2{\gamma }^2-1}}\le$

$\le$

$=$ ${\displaystyle \frac{8}{2{\gamma }^2}}{\displaystyle \sum _{r=1}^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{2r-1}}-{\displaystyle \frac{1}{2r+1}}\right)={\displaystyle \frac{4}{{\gamma }^2}}={\displaystyle \frac{K(0)}{{\gamma }^2}}.$

But, since $2|c_j\overline{c_h}|\le {|c_j|}^2+{|c_h|}^2$ and $K(t_j-t_h)=K(t_h-t_j)$, we find

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}k(x){|f(x)|}^2𝑑x$ $=$ ${\displaystyle \sum _{j,h=-n}^n}{c}_j\overline{c_h}K(t_j-t_h)=$

$=$ ${\displaystyle \sum _j}{|c_j|}^{2}K(0)+{\displaystyle \sum _{j,h;j\ne h}}𝒪(j,h){\displaystyle \frac{{|c_j|}^2+{|c_h|}^2}{2}}|K(t_j-t_h)|$

$=$ ${\displaystyle \sum _j}{|c_j|}^2\left(K(0)+{\displaystyle \sum _{h\ne j}}𝒪(j,h)|K(t_j-t_h)|\right),$

where $|𝒪(j,h)|\le 1$. Using the definition of $k$ and the previous inequality, we have

$${\int }_{-\pi }^{\pi }{|f(x)|}^2𝑑x\ge \sum _j{|c_j|}^{2}K(0)(1-\frac{1}{{\gamma }^2}),$$

and we have obtained the conclusion.

Title	proof of Ingham Inequality
Canonical name	ProofOfInghamInequality
Date of creation	2013-03-22 15:55:18
Last modified on	2013-03-22 15:55:18
Owner	ncrom (8997)
Last modified by	ncrom (8997)
Numerical id	8
Author	ncrom (8997)
Entry type	Proof
Classification	msc 42B05