# Is this Poincaré-type inequality valid?

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.

• Do you really have $|f|^{1/2}$ in the boundary integral?
– daw
Mar 29 '19 at 13:25
• Do you really want to have $|f|^{1/2}$ as integrand?
– gerw
Mar 29 '19 at 13:26
• You could try to apply the inequality for boundary-mean zero functions to $f- \frac1{|\Gamma|}\int_\Gamma f d\gamma$
– daw
Mar 29 '19 at 13:27
• @gerw yes, I would like to have this specific term. I try to prove it by contradiction but I 'm stuck... Mar 29 '19 at 13:29
• Then it cannot be true. If $f \ne 1$ is a constant, the lhs is not zero while the rhs is zero.
– gerw
Mar 29 '19 at 13:30

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!