# proof of Ingham Inequality

Let

 $f(x)=\sqrt{\frac{1}{2\pi}}\sum_{j=-n}^{n}c_{j}e^{it_{j}x}.$

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

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

It is easy to prove the equalities.

 $\displaystyle\int_{-\infty}^{\infty}k(x)|f(x)|^{2}dx$ $\displaystyle=$ $\displaystyle\sum_{j,h=-n}^{n}c_{j}\overline{c_{h}}K(t_{j}-t_{h}).$ $\displaystyle\int_{-\infty}^{\infty}k(x)f(x)e^{-ixt_{h}}dx$ $\displaystyle=$ $\displaystyle\sum_{j=-n}^{n}c_{j}K(t_{j}-t_{h}).$

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

 $k(x)=\left\{\begin{array}[]{cl}\cos(x/2),&|x|\leq\pi\\ 0,&|x|>\pi\end{array}\right.$

Then, after computation, we have

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

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

 $\displaystyle\sum_{j\neq h}|K(t_{j}-t_{h})|$ $\displaystyle=$ $\displaystyle\sum_{j\neq h}\left|\frac{4\cos\pi(t_{j}-t_{h})}{1-4(t_{j}-t_{h})% ^{2}}\right|\leq\sum_{j\neq h}\frac{4}{4(j-h)^{2}\gamma^{2}-1}\leq$ $\displaystyle\leq$ $\displaystyle\frac{4}{\gamma^{2}}\sum_{j\neq h}\frac{1}{4(j-h)^{2}-1}<\frac{8}% {\gamma^{2}}\sum_{r=1}^{\infty}\frac{1}{4r^{2}-1}=$ $\displaystyle=$ $\displaystyle\frac{8}{2\gamma^{2}}\sum_{r=1}^{\infty}\left(\frac{1}{2r-1}-% \frac{1}{2r+1}\right)=\frac{4}{\gamma^{2}}=\frac{K(0)}{\gamma^{2}}.$

But, since $2|c_{j}\overline{c_{h}}|\leq|c_{j}|^{2}+|c_{h}|^{2}$ and $K(t_{j}-t_{h})=K(t_{h}-t_{j})$, we find

 $\displaystyle\int_{-\infty}^{\infty}k(x)|f(x)|^{2}dx$ $\displaystyle=$ $\displaystyle\sum_{j,h=-n}^{n}c_{j}\overline{c_{h}}K(t_{j}-t_{h})=$ $\displaystyle=$ $\displaystyle\sum_{j}|c_{j}|^{2}K(0)+\sum_{j,h;\;j\neq h}\mathcal{O}(j,h)\frac% {|c_{j}|^{2}+|c_{h}|^{2}}{2}|K(t_{j}-t_{h})|$ $\displaystyle=$ $\displaystyle\sum_{j}|c_{j}|^{2}\left(K(0)+\sum_{h\neq j}\mathcal{O}(j,h)|K(t_% {j}-t_{h})|\right),$

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

 $\int_{-\pi}^{\pi}|f(x)|^{2}dx\geq\sum_{j}|c_{j}|^{2}K(0)(1-\frac{1}{\gamma^{2}% }),$

and we have obtained the conclusion.

Title proof of Ingham Inequality ProofOfInghamInequality 2013-03-22 15:55:18 2013-03-22 15:55:18 ncrom (8997) ncrom (8997) 8 ncrom (8997) Proof msc 42B05