Let $C=\Omega \times (0,\infty)$ for a bounded $C^1$ domain $\Omega$. Consider a function $u \in H^1(C)$. Write $u=u(x,y)$ where $x \in \Omega$ and $y \in (0,\infty).$

Is it true that if $u \in H^1(C)$ with $\int_\Omega u(x,y)dx =0$ for almost all $y$, then:

$$\int_0^\infty \int_\Omega |\nabla_x u|^2 + u_y^2 \geq C\lVert u \rVert_{H^1(C)}^2?$$ That is, Poincare inequality holds?

It seems to me the answer is trivially "yes". Since $u(\cdot,y)$ has mean value zero for a.a. $y$, we have $$\lVert{u(\cdot,y)}\rVert_{L^2(\Omega)} \leq C\lVert \nabla_x u(\cdot,y)\rVert_{L^2(\Omega)}$$ a.e. $y$ by the usual Poincare inequality on $\Omega$. Then we can just integrate since the integral ignores sets of measure zero.

My confusion is that I was told that we actually need $\int_\Omega u(x,y)dx = 0$ for all $y \in [0,\infty)$. Is there something I am missing with this? maybe some subtlety with the null sets?!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.