# Show that $\bigg|\int_{0}^{T}\overline {f(t)}g(t)~dt\bigg|^{2}\le\frac{T^{2}}{4\pi^{2}}\int_{0}^{T}|f(t)|^{2}dt\int_{0}^{T}|g'(t)|^{2}dt$

If $f$ is T-periodic , continuous , and piecewise $C^{1}([0,T])$ with $\displaystyle\int_{0}^{T}f(t)~dt=0$ and $g$ is $C^{1}([0,T])$ and T-periodic then we have $$\bigg|\int_{0}^{T}\overline {f(t)}g(t)~dt\bigg|^{2}\le\frac{T^{2}}{4\pi^{2}}\int_{0}^{T}|f(t)|^{2}dt\int_{0}^{T}|g'(t)|^{2}dt$$

Here's a my attempt:

Let $S$ denote the set of complex-valued Riemann integrable functions on $[0,T]$ and define the inner product for $f,g\in S$ by $$\langle f,g\rangle=\frac{1}{T}\int_{0}^{T}f(x)\overline {g(x)}~dx~.$$

Denote the $m^{th}$ Fourier coeffcient of $f$ by $\widehat{f}(m)=\displaystyle\frac{1}{T}\int_{0}^{T}f(t)e^{-\frac{2\pi i mt}{T}}dt$ and for all $0<p<\infty$ , $f\in L^{p}([0,T])$ provided that $\|f\|_{L^{p}([0,T])}=\bigg(\displaystyle\int_{0}^{T}|f(x)|^{p}dx\bigg)^{1/p}<\infty~.$

Firstly we note that $f,g\in L^{2}([0,T]) .$ Indeed , as $g\in C^{1}([0,T])$ then $g\in L^{1}([0,T])\cap L^{\infty}([0,T])$ and hence that $g\in L^{2}([0,T])$ . Similar reason for $f\in L^{2}([0,T])$ as $f$ is continuous on $[0,T]$ .Then we see on account of the Parseval's relation to yield that

\begin{align} \bigg|\int_{0}^{T}\overline {f(t)}g(t)~dt\bigg|&=T\bigg|\sum_{m\in{\bf Z} }\overline {\widehat {f}(m)}{\widehat g(m)}\bigg|\\ &\color{blue}=T\bigg|\sum_{m\ne 0}\overline{\widehat{f}(m)}{\widehat g(m)}\bigg|\\ &\color{red}\le T\bigg(\sum_{m\ne 0}\bigg|\overline{\widehat{f}(m)}~\bigg|^{2}\bigg)^{1/2}\bigg(\sum_{m\ne 0}|\widehat{g}(m)|^{2}\bigg)^{1/2}\\ &=T\bigg(\sum_{m\ne 0}|{\widehat{f}(m)}|^{2}\bigg)^{1/2}\bigg(\sum_{m\ne 0}|\widehat{g}(m)|^{2}\bigg)^{1/2}\\ &=\bigg(T\sum_{m\ne 0}|{\widehat{f}(m)}|^{2}\bigg)^{1/2}\bigg(T\sum_{m\ne 0}|\widehat{g}(m)|^{2}\bigg)^{1/2}\\ &\color{red}=\bigg(T\sum_{m\ne 0}|\widehat {f}(m)|^{2}\bigg)^{1/2}\bigg(\frac{T^{2}}{4\pi^{2} }T\sum_{m\ne 0}\bigg|\frac{\widehat{g'}(m)}{m}\bigg|^{2}\bigg)^{1/2}\\ &\le \bigg(T\sum_{m\ne 0}|\widehat {f}(m)|^{2}\bigg)^{1/2}\bigg(\frac{T^{3}}{4\pi^{2} }\sum_{m\ne 0}|\widehat{g'}(m)|^{2}\bigg)^{1/2}\\ &\color{blue}=\bigg(T\sum_{m\in{\bf Z}}|\widehat {f}(m)|^{2}\bigg)^{1/2}\bigg(\frac{T^{3}}{4\pi^{2} }\sum_{m\in{\bf Z}}|\widehat{g'}(m)|^{2}\bigg)^{1/2}\\ &=\bigg(\int_{0}^{T}|f(t)|^{2}dt\bigg)^{1/2}\bigg(\frac{T^{2}}{4\pi^{2}}\int_{0}^{T}|g'(t)|^{2}dt\bigg)^{1/2} \end{align} , where the blue equalities hold by the assumption $\widehat {f}(0)=0$ and the fact that $\widehat {g'}(0)=0$ as $g$ is T-periodic on $[0,T]$ and red inequality holds by the Hölder's inequality for series by applying the counting measure and the red equality is according to the following :

for all $m\in {\bf Z}-\{0\}$ , we have $${\widehat{g'}(m)}=\frac{1}{T}\int_{0}^{T}g'(t)e^{-\frac{2\pi i mt}{T}}dt=\frac{2\pi im}{T^{2}}\int_{0}^{T}g(t)e^{-\frac{2\pi imt}{T}}dt=\frac{2\pi im}{T}\widehat{g}(m)$$ , where the second equality holds by the integration by parts and $g$ is T-periodic with $g(0)=g(T)$ . So for any $m\in {\bf Z}-\{0\},$ one has $$|\widehat{g}(m)|^{2}=\frac{T^{2}}{4\pi^{2}}\bigg|\frac{\widehat{g'}(m)}{m}\bigg|^{2}$$

Can anyone check my working for validity ? Any comment or advice will be appreciated . Thanks for considering my request .