$f(x)=\sin(x)\in H_{0}^{1}(\Omega)$ with $\Omega=(0,\pi)$?

Question

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 $f\in H_{0}^{1}(0,\pi)$?

Answer 1

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 $$ \lim_{\epsilon\to0}\|\sin x-\phi_\epsilon(x)\|_{H^1}=0. $$