First, my question is :
$U$ : bounded, open w/ $\partial U$ : $C^{1}$.
Let $Lu=-\sum_{i,j=1}^{n}\left(a^{ij}u_{x_{i}}\right)_{x_{j}}$ where $a^{ij}\in L^{\infty}\left(U\right)$ : symmetric and uniformly elliptic.
For each $f\in L^{2}\left(U\right)$, consider $(*)\begin{cases} Lu-\lambda u=f & U\\ u=0 & \partial U \end{cases}$. Then I want to find a solution or solutions of this elliptic equation with drichlet boundary.
My approach is : Since $L^{2}(U)$ is separable, there are eigenvalues $\left\{ \lambda_{k}\right\} $ organizing $0<\lambda_{1}\leq\lambda_{2}\leq\cdots$ and corresponding countable orthonormal basis(eigenfunctions) $\left\{ \phi_{k}\right\} _{k=1}^{\infty}$ which is in turn to be eigenfunctions of $L$. Then $L\phi_k=\lambda_{k}\phi_k$.
First, consider $\lambda$ is not an eigenvalue of $L$. Then $N\left(L-\lambda I\right)$ is a trivial set.
Then $u$ and $f$ can be written as following : $f=\sum\left(f,\phi_{k}\right)_{L^{2}}\phi_{k}$, and $u=\sum a_{k}\phi_{k}$.
If we product $\phi_{k}$ on $\left(*\right)$, then $\left(f,\phi_{k}\right)=((L-\lambda I)a_{k}\phi_{k},\phi_{k})=a_{k}\left((L-\lambda I\right)\phi_{k},\phi_{k})$. (I think this part is wrong) So $a_{k}={\displaystyle \frac{(f,\phi_{k})}{\left((L-\lambda I)\phi_{k},\phi_{k}\right)}}$.
So finally we can write $u={\displaystyle \sum_{k=1}^{\infty}\frac{(f,\phi_{k})}{\left((L-\lambda I)\phi_{k},\phi_{k}\right)}\phi_{k}}$.
Next, suppose $\lambda$ is an eigenvalue of $L$, and WLOG, let $\lambda=\lambda_{1}=\lambda_{2}\cdots=\lambda_{l}$.
Then $\phi_{1},\phi_{2},\cdots,\phi_{l}$ spans the $N\left(L-\lambda I\right)$.
Then like above, $u={\displaystyle \sum_{k=l+1}^{\infty}\frac{(f,\phi_{k})}{\left((L-\lambda I)\phi_{k},\phi_{k}\right)}\phi_{k}+\sum_{k=1}^{l}a_{k}\phi_{k}}$for any $a_{1},a_{2}\cdots,a_{l}\in\mathbb{R}$.
From alternative, If $f\in$Image$\left(L-\lambda I\right)$, above form is the solutions.
If not, there is no solution.
I don't make sure whether my proof is correct or not. Tell me some comments