Suppose $g$ is differentiable such that $g, g'\in L^p(\mathbb{R})$ for all $p\in [1,\infty)$, and assume that the Fourier transform of $g$ satisfies $\widehat{g}(\xi)=0$ for all $\xi\in [-R,R]$ for some $R>0$. I want to prove that in that case there is a constant $C>0$ (which doesn't depend on $p$) such that $||g'||_p\geq CR||g||_p$ for all $p$. My definition of a Fourier transform is $\widehat{g}(\xi)=\int_{\mathbb{R}}g(x)e^{-2\pi ix\xi}dx$. This inequality is called Bohr's inequality.
My work: My idea was to contruct a function $\psi$ which satisfies $g'*\psi=g$. By properties of the Fourier transform we know that $\widehat {g' * \psi}=\widehat{g'}\times \widehat{\psi}=2\pi i\xi\widehat {g}(\xi)\times \widehat{\psi}(\xi)$. Now suppose we found a function $\psi$ such that $\widehat{\psi}(\xi)=\frac{1}{2\pi\xi}$ for all $|\xi|>1$. Setting $h(x)=\psi(Rx)$ we have $\widehat{h}(\xi)=\frac{1}{R} \frac{1}{2\pi {\frac{\xi}{R}}}=\frac{1}{2\pi\xi}$ for all $|\xi|>R$. Combining it with the fact that $\widehat{g}=0$ in $[-R,R]$ we conclude that the equality $\widehat {g' * h}=\widehat{ig}$ holds on the whole real line. Hence $g' * h=ig$. And now from Young's inequality:
$||g||_p=||ig||_p=||g' * h||_p\leq ||g'||_p\times ||h||_1=||g'||_p\times\frac{1}{R}||\psi||_1$
So setting $C=\frac{1}{||\psi||_1}$ we are done.
So now the only question is how can such $\psi$ be defined, if at all. Using Polya's criterion I can prove there is a function in $L^1$ such that its Fourier transform equals to $\frac{1}{2\pi |\xi|}$ for $|\xi|>1$. However, I need it to be equal to $\frac{1}{2\pi\xi}$, without the absolute value. Any ideas on how to "fix" the function to get it? Or maybe my proof doesn't work from the beginning?