# Representation of function in $W^{1,2}([-1,1])$ by function in $W^{1,2}_0([-1,1])$

I read some notes from a lecture, where it is claimed that any function $$\varphi \in W^{1,2}([-1,1])$$ can be represented by $$\varphi(x)= \psi(x) +c_1e^{x}+c_2 e^{-x}$$ where $$c_1,c_2 \in \mathbb{C}$$ and $$\psi\in W^{1,2}_0([-1,1])$$.

Mabye someone can help me clarifying this.

You know that any element in $$\varphi \in W^{1,2}([-1,1])$$ is Hölder continuous, in particular, continuous up to the boundary. Let $$c_1,c_2 \in \mathbb{C}$$ solve $$\begin{pmatrix}e & e^{-1} \\ e^{-1} & e \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}=\begin{pmatrix} \varphi(1) \\ \varphi(-1) \end{pmatrix},$$ which we can always solve uniquely since $$\det (M)=e^2-e^{-2} \neq 0$$ .Note that, by plugging in the boundary values, you also have $$T(\varphi-c_1e^x-c_2e^{-x})=0,$$ where $$T$$ is the trace operator. Hence $$\psi=\varphi-c_1e^x-c_2e^{-x} \in W^{1,2}_0([-1,1]),$$ since $$T \psi=0$$, and rearranging yields your desired identity.
• Just to make sure If I could follow you. So we basically using $W_{0}^{1,2}([-1,1])=\lbrace u \in W^{1,2}([-1,1]) | Tu=0\rbrace$ and that $Tu= u\vert_{\partial[-1,1]}$ since $u \in W^{1,2}([-1,1])$ is a.e. continuous according to that post: math.stackexchange.com/questions/3928184/… and determine the constants accordingly such that $\psi$ disapears on the boundary. Right? But are you sure $c_1, c_2$ is calculated right? How is $\varphi(1)-\varphi(1)e-\varphi(-1)e^{-1}=0=\varphi(-1)-\varphi(1)e^{-1}-\varphi(-1)e^{1}$? Sep 19 at 23:13