Exist unique $g_0 \in H^1(0, 1)$ such that $f(0) = \int_0^1 (f'g_0' + fg_0) \text{ for all }f \in H^1(0, 1)$?

Question

The mapping $f \mapsto f(0)$ from $H^1(0, 1)$ into $\mathbb{R}$ is a continuous linear functional on $H^1(0, 1)$. Does there exist a unique $g_0 \in H^1(0, 1)$ such that$$f(0) = \int_0^1 (f'g_0' + fg_0) \text{ for all }f \in H^1(0, 1)?$$

Answer 1

Hint:

$H^1(0,1)$ is a Hilbert space. If you can show that $$L : H^1(0,1) \to \mathbb R, \ \ \ L(f) = f(0)$$ is a bounded linear functional, the Riesz representation theorem tells you that

$$L(f) = \langle f, g_0\rangle\ \ \ \forall f\in H^1(0,1),$$

which is exactly what you want.

Answer 2

Existence is demonstrated using $g(x)=-\cos(x-1)/\sin(1)$, which is the unique solution of $$ -g''+g = 0 \\ g'(0)=-1,\;g'(1)=0. $$ Then $$ \int_{0}^{1}\{f'(x)g'(x)+f(x)g(x)\}dx \\ = \left.f(x)g'(x)\right|_{0}^{1}+\int_{0}^{1}f(x)\{-g''(x)+g(x)\}dx =f(0). $$ Uniquess follows from the Riesz Representation Theorem.