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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.

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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.

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  • $\begingroup$ Thank you. I guess I have some homework to do :D $\endgroup$ Commented Sep 19, 2023 at 20:35
  • $\begingroup$ You are welcome. This is just a really fancy way of "subtracting boundary values". $\endgroup$
    – F. Conrad
    Commented Sep 19, 2023 at 20:37
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    $\begingroup$ 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 $\varphi(1)-\varphi(1)e-\varphi(-1)e^{-1}=0=\varphi(-1)-\varphi(1)e^{-1}-\varphi(-1)e^{1}$? $\endgroup$ Commented Sep 19, 2023 at 23:13
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    $\begingroup$ You are right on all accounts and thanks for pointing out my mistake. I think I salvaged the proof. Please double check again, if something is still wrong, feel free to unaccept this answer. $\endgroup$
    – F. Conrad
    Commented Sep 20, 2023 at 10:28

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