Is $X= \{ u \in H^1(\Omega \times I) \mid \int_\Omega u(x,y)\;\mathrm{d}x = 0 \text{ for a.a. $y \in I$}\}$ a Hilbert space?

Question

Let $\Omega$ be a bounded domain and let $I$ be an unbounded interval.

Let $$X= \{ u \in H^1(\Omega \times I) \mid \int_\Omega u(x,y)\;\mathrm{d}x = 0 \text{ for a.a. $y \in I$}\}$$

Is this well-defined as a Hilbert space? I think it is possible to show that it is a closed subspace of $H^1(\Omega \times I)$, but I feel like the "a.a. $y \in I$" might be problematic. Maybe it should be replaced with "for all $y \in I$".

Answer 1

You are looking at the kernel of a map $f:H^1(\Omega\times I)\to?$ with codomain some vector space of functions on $I$ modulo measure zero such that $f(y)=\int_\Omega u(x,y)\,\mathrm dx$ for all $y\in I$. Can you pick such a space $?$ so that this linear map is continuous?