Let $\Omega=(-1,1)$ and consider the functional $\Phi :C_0^\infty(\Omega) \rightarrow \mathbb{R}, f \mapsto f(0)$.
Can this functional be extended to a continous linear functional on $H_0^0(\Omega)$ or $H_0^1(\Omega)$?
Show that there is a unique function $u \in H_0^1(\Omega)$, such that $$\int_{-1}^1 u'f' dx = f(0) \; \operatorname{for all} \; f \in C_0^\infty(\Omega)$$ and find the function $u$.
For the first part I need your help. I know that the solution has to be that $\Phi$ can be extended to $H_0^1(\Omega)$, otherwise the second part of the task wouldn't make sense, but how can I show that?
For the second part I have the following approach:
The term $\int_{-1}^1 u'f' dx$ is one part of the usual scalarproduct on the Sobolev space $H_0^1(\Omega)$ given by
$$ (u,v)_{H^1_0(\Omega)=} = \int_\Omega u(x)v(x) dx + \int_\Omega u'(x)v'(x) dx$$
Assume that we have extended $\Phi: H_0^1(\Omega) \rightarrow \mathbb{R}$. Since $H_0^1(\Omega)$ is a Hilbert space and $\Phi$ is a bounded linear functional, the Riesz Representation Theorem can be applied and yields that there exists an unique $u \in H_0^1(\Omega)$ such that
$$ (u,f)_{H_0^1(\Omega)} = \Phi(f) = f(0) \; \operatorname{for all} f \in H_0^1(\Omega)$$
To get the desired result, $\int_\Omega u(x)f(x) dx$ has to vanish in the scalarproduct $(u,f)_{H_0^1(\Omega)}$. Do you have an argument for that?
Concerning the construction of $u$, set
$u(x)=
\begin{cases}
0.5(1-x) &\text{ for } x>0\\
0.5(1+x) &\text{ for } x\leq 0
\end{cases}$ , $\;\;$so
$
u'(x)=
\begin{cases}
-0.5 &\text{ for } x>0 \\
0.5 &\text{ for } x\leq 0
\end{cases}
$
and
$$\int_{-1}^1 u'(x)f'(x) dx = 0.5 \int_{-1}^0 f'(x) dx + -0.5\int_{0}^1 f'(x) dx \\
= 0.5[f(0)-f(-1)] – 0.5[f(1)-f(0)] = f(0) – 0.5[f(-1)+f(1)] = f(0),$$
where $f(-1) = f(1) = 0$, because $-1$ and $1$ are the boundary points of the support and f is continuous.
Best Answer
The functional can't be extended continuously to $H_0^0$ (i.e., $L^2$): Pick a sequence of functions $f_n \in C^{\infty}_0$ which converge to $0$ in $L^2$ but which concentrate near $0$ as $n \to \infty$. (E.g., $f_n=n\cdot g(n^3x), $ where $g$ is a fixed function with compact support and is non-zero at $0$.*) Then, $\Psi(f_n)$ diverges.
For the second part, you can proceed as suggested in the comments or just first construct $u$ explicitly as you did and solve everything at once. (However, note that it is not true that $1$ and $-1$ are the boundary points of the support; they are just outside of the support.) Then, define \begin{align*} \Psi:H^1_0 &\to \mathbb{R} \\ v&\mapsto \int u'v'. \end{align*} It is a continuous linear functional by Hölder's inequality** and your computations show that it is $v(0)$ when $v \in C^{\infty}_0$.
*We have that $\lVert f_n\rVert_2^2=\int n^2g^2(n^3x)dx$, and thus $$\lVert f_n\rVert_2^2=\int n^2g^2(u)\frac{1}{n^3}du=\frac{1}{n}\lVert g\rVert_2^2. $$ It follows that $\lVert f_n \rVert_2 \to 0$.
**If this is not clear, then: \begin{align*} \lvert\int u'v'\rvert &\leq \int \lvert u'v'\rvert \\ &\leq \lVert u'\rVert_2\lVert v' \rVert_2 \\ &\leq \lVert u'\rVert_2\lVert v \rVert_2+\lVert u'\rVert_2\lVert v' \rVert_2 \\ & = \lVert u'\rVert_2(\lVert v\rVert_2+\lVert v'\rVert_2) \\ & \leq \lVert u'\rVert_2\cdot\sqrt{2}(\lVert v\rVert_{H_0^1}) = C \cdot \lVert v\rVert_{H_0^1}. \end{align*}