# Poincaré's inequality in Fourier space

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 $s>0$.)

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.

• Thank you very much! This is exactly what I was looking for. I had a vague idea that (1) could be a consequence of some form of the uncertainty principle but I couldn't go any further than that. Mar 18, 2013 at 4:18

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 $C_{R,s} = (2R)^s$.

• The point is that $s$ is typically non integer. Mar 13, 2013 at 20:08
• If $s \in (k,k+1)$, you might be able to get this by comparing $|\xi|^s |\widehat{f}|$ to $|\xi|^{k+1} |\widehat{f}|$ and $|\xi|^k |\widehat{f}|$ for $|\xi| \leq 1$ and $|\xi| > 1$. Moreover, what does such an inequality demonstrate? As $s$ grows, so do $(2R)^s$ and $|\xi|^s |\widehat{f}|$ for $|\xi| > 1$. So, unless $\widehat f$ is concentrated around $0$, both quantities on the right hand side get larger as $s$ grows. I don't think this is the case though - compact support of $f$ will mean $\widehat{f}$ is entire, and has sub-exponential decay. Mar 14, 2013 at 1:40
• (if $\widehat{f}$ has exponential decay, this implies $f$ is analytic, which it can't be, since it's compactly supported) Mar 14, 2013 at 1:42