Let $\Omega\subset \mathbb{R}^d$, with $d\in \{1,2,3\}$ be an open bounded, simply connected domain. Define $H_0^1$ as the subspace of $H^1$ whose member functions have vanishing trace on the boundary. Does it make sense to define $H^1_\perp$ as $\{u\in H^1\colon \forall v\in H^1_0\, \text{we have} \int_{\Omega} v u\, \mathrm{d} x = 0\}?$
My thinking is that we can define $H^1_\perp$ this way but we can't write $H^1$ as $H_0^1\oplus H^1_\perp$, since $H_0^1$ is not closed (I don't think), with respect to the $L^2$ norm. Is my thinking correct?
Actually, you have $H_\perp^1 = \{0\}$, the set containing only the zero function. Indeed, suppose that $u \in H^1(\Omega)$ satisfies $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in H_0^1(\Omega)$. Then, using the density of $H_0^1(\Omega)$ in $L^2(\Omega)$ you get $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in L^2(\Omega)$ and this shows $u = 0$.