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We know that $f\in C^{\infty}(0,\pi)$ and $f(0)=f(\pi)=0$. But $supp\ f=[0,\pi]$. From the definition, $H_{0}^{1}(0,\pi)$ is the clusear of $C_{0}^{\infty}(0,\pi)$ in $H^{1}(0,\pi)$. Since, $f$ is not defined outside $(0,\pi)$ there is not any point outside the support on which $f$ vanish. So how we can say $f\in H_{0}^{1}(0,\pi)$?

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For $\epsilon>0$ small define $\phi_\epsilon(x)=\max(\sin x-\sin\epsilon,0)$, $0<x<\pi$. The support of $\phi_\epsilon$ is $[\epsilon,\pi-\epsilon]$. Its (weak) derivative is $$ \phi_\epsilon'(x)=\begin{cases} 0 & 0<x<\epsilon\\ \cos x & \epsilon \le x\le\pi-\epsilon\\ 0 & \pi-\epsilon<x<\pi \end{cases} $$ It is easy to see that $\phi_\epsilon\in H_0^1(0\,\pi)$ and $$ \lim_{\epsilon\to0}\|\sin x-\phi_\epsilon(x)\|_{H^1}=0. $$

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  • $\begingroup$ Thank you very much @Julián Aguirre. But for each $\epsilon$, $\phi'_{\epsilon}$ is not continuous at $x=\epsilon$. I mean that $\phi_{\epsilon}$ is not in $C_{0}^{\infty}(0,\pi)$ or even in $C_{0}^{1}(0,\pi)$. $\endgroup$ – Albert May 8 at 12:05
  • $\begingroup$ But it is in $L^2$. $\endgroup$ – Julián Aguirre May 8 at 13:45
  • $\begingroup$ Yes, of course. But what happens for the definition? We have to find $C^{\infty}$ molifiers. $\endgroup$ – Albert May 8 at 13:56
  • $\begingroup$ Well, there are other equivalent definitions. $\endgroup$ – Julián Aguirre May 8 at 17:34
  • $\begingroup$ $H_0^1$ is a closed subspace of $H^1$. If a sequence in $H_0^1$ converges in the $H^1$ norm, then its limit is in $H_0^1$. All is needed is to show that each $\phi_\epsilon$ is in fact in $H^1_0$. This can be done by convolution with a compactly supported approximation of the identity. $\endgroup$ – Julián Aguirre May 9 at 11:31

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