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It is proved that the Poincaré inequality is still true for functions with zero mean boundary traces. Motivated by this, I have the following question:

Let $\Omega$ be an open,bounded and connected subset of $\mathbb R^3$ with a $C^2-$boundary $\partial \Omega \equiv \Gamma$. If $f \in W^{1,2}(\Omega)$ then could we claim that:

${\vert \vert f - \frac{1}{\vert \Gamma \vert } \int_{\Gamma} {\vert f \vert}^{1/2} \vert \vert}_{L^2(\Omega)} \leq C {\vert \vert \nabla f \vert \vert}_{L^2(\Omega)}$

If for any $u\in \{ u\in H^1(\Omega): \frac{1}{\vert \Gamma \vert } \int_{\Gamma} u=0 \}$ we have the estimate: ${\vert \vert u \vert \vert}_{L^2(\Omega)} \leq C {\vert \vert \nabla u \vert\vert}_{L^2(\Omega)}$ then it seems logical to me that the above claim could be true. However I wasn't able to prove it (if indeed can be proved) so any help is much appreciated.

Thanks in advance!

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  • $\begingroup$ Do you really have $|f|^{1/2}$ in the boundary integral? $\endgroup$
    – daw
    Commented Mar 29, 2019 at 13:25
  • $\begingroup$ Do you really want to have $|f|^{1/2}$ as integrand? $\endgroup$
    – gerw
    Commented Mar 29, 2019 at 13:26
  • $\begingroup$ You could try to apply the inequality for boundary-mean zero functions to $f- \frac1{|\Gamma|}\int_\Gamma f d\gamma$ $\endgroup$
    – daw
    Commented Mar 29, 2019 at 13:27
  • $\begingroup$ @gerw yes, I would like to have this specific term. I try to prove it by contradiction but I 'm stuck... $\endgroup$ Commented Mar 29, 2019 at 13:29
  • $\begingroup$ Then it cannot be true. If $f \ne 1$ is a constant, the lhs is not zero while the rhs is zero. $\endgroup$
    – gerw
    Commented Mar 29, 2019 at 13:30

1 Answer 1

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When coming up with inequalities, it is important that the equations you get be 'dimensionally consistent' In particular, we can often use scaling arguments to show that only certain types of inequalities are possible.

In this particular case, let us fix a function $g$, and let $f_\lambda = \lambda g$ for all $\lambda > 0$. Then, you inequality would would have $$\lVert \lambda g - \sqrt{\lambda} \frac{1}{|\Gamma|} \int_{\Gamma} |g|^{1/2} \rVert_{L^2} \leq C \lambda \lVert \nabla g \rVert_{L^2}$$ If we cancel a factor of $\lambda$ from both sides, we have $$\lVert g - \frac{1}{|\Gamma|\lambda^{1/2}} \int_\Gamma |g|^{1/2} \rVert_{L^2} \leq C \lVert \nabla g \rVert_{L^2}$$ But, as $\lambda \to 0$, $\lambda^{-1/2} \to \infty$, so the term on the left becomes larger without bound for nonzero $g$, while the term on the right is constant, which is a contradiction.


Of course, a physicist would not even need to go through this calculation: the terms $f$ and $\frac{1}{|\Gamma|} \int_{\Gamma} |f|^{1/2}$ have different units, so there's no way we could subtract them!

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