Riesz representation theorem in a Sobolev space on an interval

Question

I already proved that there exists a unique function $u\in H_0^1((-1;1))$ such that $\int_{-1}^{1} u' \phi'dx=\phi(0)$ $\forall \phi \in C_0^\infty((-1;1))$.

I want to find this $u$. I have found this function that satisfies the condition $$ u(x)= \begin{cases} 0.5(1+x) &\text{ if } x\leq0,\\ 0.5(1-x) &\text{ if } x>0 \end{cases} $$ We have that $$ u'(x)= \begin{cases} 0.5 &\text{ if } x\leq0,\\ -0.5 &\text{ if } x>0 \end{cases} $$ And using the fundamental theorem of calculus we obtain that $\int_{-1}^{1} u' \phi'dx=\phi(0)$.

However $u$ is not compactly supported. Do you have any hint?

Answer 1

Your function $u$ is the correct one, and it is in $H^1_0((-1,1))$. The notation $H^1_0$ does not mean the function is compactly supported. The space $H^1_0$ is the closure of compactly supported (smooth) functions with respect to the $H^1$ norm.

For each $n$, the function $u_n(x)=\max(0, u(x)-1/n)$ is compactly supported in $(-1,1)$. One can mollify it, obtaining smooth compactly supported functions; hence, $u_n\in H_0^1((-1,1))$. As $n\to\infty$, we have $u_n\to u$ in the $H^1$ norm, hence $u$ is also in $H^1_0$.