Poincaré's inequality in Fourier space

Question

If $f\in C^{\infty}_c(\mathbb{R})$ is supported in the interval $[-R, R]$, then by means of the fundamental theorem of calculus one can show that $$ \lVert f\rVert_{L^2(\mathbb{R})}\le 2R\left\lVert\frac{df}{dx}\right\rVert_{L^2(\mathbb{R})}.$$ Plancherel's theorem converts this inequality into the following: $$ \tag{1}\left\lVert \widehat{f}\right\rVert_{L^2(\mathbb{R})} \le 2R\left\lVert \xi \widehat{f}(\xi)\right\rVert_{L^2_\xi(\mathbb{R})}.$$ Is there a way of proving (1) by working purely on the frequency space, avoiding the fundamental theorem of calculus? (I think that such a proof could be easily generalized to prove inqualities such as $$\lVert \hat{f}\rVert_{L^2(\mathbb{R})}\le C_{R,s}\left\lVert \lvert \xi \rvert^s\widehat{f}(\xi)\right\rVert_{L^2_\xi(\mathbb{R})},$$ where $s>0$.)

Answer 1

As BaronVT notes, in order to do something in the frequency space, one has to translate the condition $\operatorname{supp}f\subseteq [-R,R]$ there. This is what the various uncertainty inequalities do. The classical Heisenberg-Pauli-Weyl uncertainty inequality $$ \|x f(x)\|_{L^2}\, \| \xi \widehat f(\xi)\|_{L^2} \ge \frac{1}{4\pi}\|f\|_{L^2}^2\tag{HPW}$$ immediately gives (1) because $\|xf(x)\|_{L^2}\le R\|f\|_{L^2}$ under your assumption. In order to handle fractional $s$ one needs a generalization of (HPW), $$ \||x|^s f(x)\|_{L^2}\, \| |\xi|^s \widehat f(\xi)\|_{L^2} \ge C(s)\,\|f\|_{L^2}^2\tag{H}$$ Inequality (H) is a consequence of Hirschman's entropy inequality (proved here). Since $\||x|^sf(x)\|_{L^2}\le R^s\|f\|_{L^2}$, it follows that $$ \| |\xi|^s \widehat f(\xi)\|_{L^2} \ge C(s) R^{-s}\|f\|_{L^2}$$ as you wanted.

Answer 2

Why only work in frequency space? The key fact is that, in physical space, the function is compactly supported, so you either have to translate that into a condition on $\widehat f$ you can work with, or just start from $f$ to begin with.

But, you can just iterate your first statement to get $$ \lVert f \rVert \leq (2R)^s \left\lVert \frac{d^s f}{dx^s} \right\rVert $$ so that $C_{R,s} = (2R)^s$.