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Suppose $f$ is a Lipschitz continuous real-valued function over a bounded domain $\Omega \subset \mathbb{R}^d$ with smooth boundary, and let $\overline{f} := \frac{1}{|\Omega|}\int_\Omega f(x) dx$. Is it possible to upper bound $\|f-\overline{f}\|_{L^\infty(\Omega)}$ in terms of $\|\nabla f\|_{L^2(\Omega)}$ using a generalized Poincare inequality or a related inequality? Of course, standard Poincare gives: $$\|f-\overline{f}\|_{L^2(\Omega)} \leq C\|\nabla f \|_{L^2(\Omega)}$$ but can this be "upgraded" to an $L^\infty$-norm bound in terms of $\|\nabla f\|_{L^2(\Omega)}$ using the fact that $f$ is Lipschitz?

This question asks something similar: Bounding supremum norm of Lipschitz function by L1 norm and I think it might be possible to string together an answer from the replies there. But I wonder if there is a more direct approach, using, for example, the Gagliardo-Nirenberg interpolation inequality. However, all versions of Gagliardo-Nirenberg for bounded domains that I have found do not allow for $p=\infty$ norm control (see, e.g., https://arxiv.org/abs/2110.12967).

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  • $\begingroup$ Could you precise a little bit what type of dependence you allow for $C$ ? I mean, if $\nabla f \neq 0$, obviously $C:= \text{diam}(\Omega) \frac{\|\nabla f\|_\infty}{\|\nabla f\|_2}$ does the job .... $\endgroup$
    – Ayman Moussa
    Commented Apr 3 at 20:11
  • $\begingroup$ Yes, I am hoping for a bound of the form: $\|f-\overline{f}\|_\infty \leq C L^p \|\nabla f\|_2^q$ where L is the Lipschitz constant of $f$, $p,q>0$, and $C$ is a constant independent of $f$ but could depend on $\Omega$ and $d$. Also, it doesn't have to be a power of $L$ and $\|\nabla f\|_2$. Any increasing functions of these quantities would suffice, too. $\endgroup$
    – Greg O.
    Commented Apr 3 at 21:12

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For $p>\max(2,d)$ you have by Sobolev embedding $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_p + \|\nabla f\|_p.$$ The interpolation $L^p(\Omega) = [L^2(\Omega),L^\infty(\Omega)]_\theta$ with $\theta\in(0,1)$ entails $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_2^\theta \|f-\overline{f}\|_\infty^{1-\theta} + \|\nabla f\|_2^\theta\|\nabla f\|_\infty^{1-\theta}.$$ Now, you can absorb part of the first term of r.h.s. in the l.h.s. with Young's inequality $a b \leq \theta a^{1/\theta} + (1-\theta) b^{1/(1-\theta)}$ : $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_2 + \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta}.$$ The usual Poincaré-Wirtinger inequality allows to write $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|\nabla f\|_2 + \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta},$$ and this implies what you hope for, because you can write (but this is rough ...) $$\|\nabla f\|_2 = \|\nabla f\|_2^\theta \|\nabla f\|_2^{1-\theta}\leq \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta}.$$

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  • $\begingroup$ I wrote it using the hilbertian Poincaré-Wirtinger inequality as you referred to it, but you could directly use it on $W^{1,p}(\Omega)$ for $p>\max(2,d)$ right after the Sobolev embedding, and then directly interpolate $\nabla f$ between $L^2(\Omega)$ and $L^\infty(\Omega)$ avoiding the absorption argument with Young's inequality and probably getting a better constant than this rough estimate ... $\endgroup$
    – Ayman Moussa
    Commented Apr 4 at 14:55
  • $\begingroup$ Looks good, thanks! Not too worried about constants, but yes, I see what you mean about using P-W inequality to simplify the argument. $\endgroup$
    – Greg O.
    Commented Apr 4 at 17:52

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