Proving norms inequality for fourier transforms

Question

Let $F\in PW_\pi$. Prove that $\Vert F^\prime\Vert_{L^2(\mathbb R)}\le\pi\Vert F\Vert_{L^2(\mathbb R)}$

I know that the derivative of $F=\hat{f}$ is given by the formula $F^\prime(\omega)=\widehat{-itf(t)}(\omega)$ so $$\Vert F^\prime\Vert^2=\int_{\mathbb{R}}|F^\prime(t)|^2dt=\int_{\mathbb{R}}\int_{[-\pi,\pi]}|t(f(t)\exp(-i\omega t)|^2dt(\omega)d\omega\le4\pi^2\int_{\mathbb{R}}\int_{[-\pi,\pi]}|f(t)\exp(-i\omega t)|^2dt(\omega)d\omega=(2\pi\Vert F\Vert^2)$$ so I got a bound of $2\pi$ instead of $\pi$. What is my mistake in calculating the integral above?

Answer 1

It starts on the first line. You should have $$ \int_{[-\pi,\pi]} |tf(t)|^2\, dt \le \pi^2 \int_{[-\pi,\pi]}|f(t)|^2 dt = \pi^2\|F\|^2.$$

This will result in $\|F'\| \le \pi\|F\|$, as desired.

Answer 2

Short one-liner $$ \|F'\|_2=\|\widehat{-it\,f(t)}\|_2=C⋅\|t\,f(t)\|_2\le\sup_{t\in[-π,π]}|t|⋅C⋅\|f\|_2=π⋅\|F\|_2 $$ where the constant $C$ depends on the chosen form of the Fourier transform.