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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.

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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 $1$.

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    Nice answer, it is much shorter then showing the claim directly with the definition. – 2017-02-24
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    Thanks for your very nice answer. But I want to write down a non-zero function in $(H^1(\Omega))^{\bot}$ explicitly, without knowing the P-F inequality. Can we prove it just from the definition that $H^1_0(\Omega)$ is the closure of $C^{\infty}_0$ with respect to the $H^1$ norm? – 2017-02-25
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    Then you can do the following: let $v \in H_0^1(\Omega)$ be the orthogonal projection of $1$ onto $H_0^1(\Omega)$. Then, $v - 1 \in H_0^1(\Omega)^\perp$. – 2017-02-25
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    You are right. But this still relies on that $f(x)=1$ is not in $H^1_0(\Omega)$. – 2017-02-25
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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.

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    Thanks a lot for your nice answer. Roughly speaking, if constant function $f(x)=1$ can be approximated by sequence of $C^{\infty}_0$ functions, then near the boundary the approximating functions have to jump the gap between 1 an 0, so the gradient of the approximating sequence will change rapidly and this will destroy the continuity of gradient of the approximating functions. Is this what you actually want to tell us? – 2017-02-25
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    @fengxiao: almost. what I'm saying is that you destroy the integrability of the gradients, not the continuity, although these concepts are actually closely related. – 2017-02-25
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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 $\Delta u-u=0$.