If $\operatorname P_U$ denotes the orthogonal projection from $L^2$ onto a closed subspace $U$, can we conclude that $\text P_UH_0^1\subseteq H_0^1$

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be open
• $U$ be a closed subspace of $L^2(\Lambda,\mathbb R^d)$
• $\operatorname P_U$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $U$

Can we show that $\text P_Uu\in H_0^1(\Lambda,\mathbb R^d)$ for all $u\in H_0^1(\Lambda,\mathbb R^d)$?

Answer 1

Take $\Lambda = (0,1)$, $U = \{f \in L^2(\Lambda) \mid f(x) = 0 \text{ for a.e. } x \in (1/3,2/3)\}$, and $u(x) = x\,(1-x)$.