As a part of a question about $W^{1,2}((0,1))$, I want to get a boundary ODE on $g$ and don't quite know how to integrate (?) in order to get the equation.
let $g\in C^2[0,1]$ be our variable, $f\in W^{1,2}((0,1))$ a general function (my wish is that the following would be right for every $f$, for a (single) $g$ that would solve the equ. the equ is $f(0)=\int f\overline g + \int f'\overline g'$ (=$\langle f,g\rangle_{1,2}$)
If it makes it easier for you to explain, don't think of that as $f(0)=$ but as $\langle f,g\rangle_{1,2}=$
How can I get a differential equ from here? I believe I'd be able to solve and continue once I get it. I didn't quite know what to do, but I mainly tried playing with integration by parts
Thanks
Choose any $g \in \mathcal{C}^{\infty}[0,1]$ that is identically $1$ near $0$ and identically $0$ near $1$. Then, for any continuously differentiable $f$ on $[0,1]$, you have $$ f(0)=-fg|_{0}^{1} = -\int_{0}^{1}f(t)g'(t)+f'(t)g(t)\,dt. $$ That's enough to give you constants $C$ and $D$ such that $$ |f(0)| \le C(\|f\|_{L^{2}}+\|f'\|_{L^{2}})\le D\|f\|_{W^{1,2}}. $$ This functional $\Phi(f) = -\int_{0}^{1}f(t)g'(t)+f'(t)g(t)\,dt$ has a continuous extension to $W^{1,2}$. By the Riesz Representation Theorem, there exists $g \in W^{1,2}$ such that $$ f(0) = (f,g)_{W^{1,2}}, \;\;\; f \in C^{1}[0,1]. $$ I'll let you fill in the details of what this means for a general $f \in W^{1,2}$. All the point evaluations in $[0,1]$ are continuous in the $W^{1,2}$ norm. $\mathbb{R}^{1}$ is a very nice and classical setting.
Added: To find the unique $g$ you're looking for, $$ \int_{0}^{1}fg+f'g'\,dt = \int_{0}^{1}fg-fg''\,dt+fg'|_{0}^{1} \\ = \int_{0}^{1}f(g-g'')\,dt + f(1)g'(1)-f(0)g'(0). $$ That means solving for $g$ such that $$ g-g''=0,\;\; g'(0)=-1,\;g'(1)=0. $$ The obvious solution is $$ g(t)=\frac{1}{e-1/e}(e^{1-t}+e^{t-1}) $$