Representation of the Delta function, Riesz dual of evaluation at a point $ev_0: W^{1,2}([-1,1]) \to \mathbb{R}$.

Question

Consider the Sobolev Space $W^{1,2}([-1,1])$ (so one weak derivative, $L^2$-integrable)

The linear map $ev_0:W^{1,2}([-1,1])\to \mathbb{R}$ defined by $$f\mapsto f(0)$$ is well defined and continuous because Sobolev embedding theorem tells us that $W^{1,2}([-1,1])\to C^0([-1,1])$ is continuous.

It follows by Riesz representation theorem that exists a function $F \in W^{1,2}([-1,1])$ representing $ev_0$, i.e. $$f(0) = \int_{-1}^1 f(t)F(t)dt + \int_{-1}^1 \partial_t f(t) \partial_tF(t) dt $$ for all $f\in W^{1,2}([-1,1]).$

What is $F$ explicitely?

My initial idea was to consider $\partial_tF(t)= \mathbb{1}_{[-1,0]}(t)$ characteristic function, so $F(t) = c_0 + t\mathbb{1}_{[-1,0]}(t) +\mathbb{1}_{[0,1]}(t)$; however this yields some integral terms that do not look nice: $$\langle f, F \rangle_{W^{1,2}} = \int_{-1}^0 f(t)t\ dt + \int_{0}^1f(t) \ dt \ +f(0) -f(-1) $$

Answer 1

As have said, there is solution with Fourier transform. I prefer just good old brute force method. We want to find function $F \in W^1_2$ such that for every $f \in W^1_2$ $$f(0) = \int_{-1}^1 (fF + f'F')dx = \int_{-1}^1 f(F - F'')dx + f(1)F'(1) - f(-1)F'(-1) $$ So we need to solve (in some sense) equation $F - F'' = \delta$ with boundary conditions $F'(-1) = F'(1) = 0$. Except zero, we have $ F - F'' = 0$, $F(x) = Ae^{-x} + Be^x$. So we look for $$F = \theta v + (1 - \theta) w$$ where $v$ and $w$ are of such form ($v(x) = Ae^{-x} + Be^x$, $w(x) = Ce^{-x} + De^x$), $\theta$ stands for Heaviside function, then $$F'' = \delta ' v + 2\delta v' + \theta v'' - \delta ' w - 2\delta w' + (1 - \theta) w''$$ $$F - F'' = -\delta ' v - 2\delta v' + \delta ' w + 2\delta w'$$ $$\delta = \delta '(w - v) + \delta (2w' - 2v')$$ This is the equation of generalized functions, apply it on $\phi \in C_0^{\infty}$. $$\phi(0) = (w'(0) - v'(0))\phi(0) + (w(0) - v(0))\phi'(0) + 2(w'(0) - v'(0))\phi(0)$$ So $w(0) = v(0), w'(0) - v'(0) = \frac{1}{3}$, with boundary conditions $v'(1) = w'(-1) = 0$ there are four linear equation on four variables: $A, B, C, D$ and the system can be solved.

In aftermath there should be words that found $F$ is in $W^1_2$; condition $v(0) = w(0)$ guarantees that $F'$ is the regular function, all integrabilities are obvious.