How to prove $H^1_0(\Omega)$ is strictly contained in $H^1(\Omega)$ when $\Omega$ is open and bounded?

Question

One why to do this is to show the orthogonal complement of $H^1_0(\Omega)$ in $H^1(\Omega)$ is not $\{0\}$. Since $H^1(\Omega)$ is the direct sum of $H^1_0(\Omega)$ and $(H^1_0(\Omega))^{\bot}$. But how to find a non-zero function in $(H^1_0(\Omega))^{\bot}$?

If $u\in (H^1_0(\Omega))^{\bot}$ then $u$ must satisfy $\Delta u=u$ in $\Omega$, but how to prove this P.D.E has non-zero solution in $H^1(\Omega)$ ?

Thanks.

Answer 1

If $\Omega$ is open and bounded then any constant function is trivially in $H^1(\Omega)$. By definition, $H^1_0(\Omega)$ is the closure of $C_c^{\infty}(\Omega)$ with respect to the $H^1$-norm, so let's prove that $f(x) = 1$ cannot be approximated by compactly supported function.

The idea is that in every compact subset of $\Omega$ we want our approximating sequence to be close to $1,$ but to enforce the fact the sequence has to be in $C_c^{\infty}$ we'll end up paying a lot on the gradient to drop quickly from $1$ to $0$ near the boundary of $\Omega$.

For simplicity, I'll work out the details for the case $N = 1$, $\Omega = (0,1)$, but the same technique can be applied for the general setting in which you phrased the question.

Let $\{\phi_n\} \subset C^{\infty}_c(0,1)$, let $\epsilon > 0$ be given and assume by contradiction that $\|\phi_n - 1\|_{H^1} \le \epsilon$. Then in particular $\|\phi_n - 1\|_{L^2} \le \epsilon$, and if we let $A_n = \{x : \phi_n(x) \ge \frac 12\}$ we have that $$1 - \mathcal{L}^1(A_n) = \mathcal{L}^1(A_n^c)= \int_{A_n^c}2\cdot \frac 12dx \le \int_{A_n^c} 2 |1 - \phi_n|\le 2\epsilon.$$ This shows that $$\mathcal{L}^1(A_n) \ge 1 - 2\epsilon.$$ Let $y \in (0,3\epsilon)$ be such that $\phi_n(y) \ge \frac 12$ (the existence of such a $y$ is guaranteed by the estimate above). Then, by the Fundamental Theorem of Calculus and the Cauchy-Schwartz inequality, we have \begin{align} \frac 12 \le &\ \phi_n(y) - \phi_n(0) \le \int_0^{3\epsilon}|\phi_n'(x)|\,dx \le \sqrt{3\epsilon}\int_0^1|\phi'(x)|^2\,dx \\ = &\ \sqrt{3\epsilon}\|\nabla(\phi_n - 1)\|_{L^2} \le \sqrt{3\epsilon}\|\phi_n - 1\|_{H^1} \\ \le &\ \sqrt{3}\epsilon^{3/2}. \end{align}

This gives a contradiction for $\epsilon$ small enough.

Answer 2

If you already know the Poincaré-Friedrichs-inequality, this is easy to establish. This inequality gives you a constant $C > 0$, such that $$\|u\|_{L^2(\Omega)} \le C \, \|\nabla u\|_{L^2(\Omega)}$$ for all $u \in H_0^1(\Omega)$. This inequality blatantly fails for the $H^1(\Omega)$ function which is constantly $1$.

Answer 3

Actually, $\{e^{x_1}, e^{x_2}...\}$ and $\{e^{x_1+x_2}, e^{x_1+x_2+x_3}...\}$ are non-zero solutions to $\Delta u-u=0$.