I have some questions about the proof of second existence theorem (Theorem 4 in 6.2) of weak solutions in Evans’ PDE:
At the last part of step 4, it says
we recall further from D.5 that the dimension of the space $N$ of the solutions of (21) is finite and equals the dimension of the space $N^*$ of solutions of
$$v-K^*v=0.$$
We readily check however that (21) holds if and only if $u$ is a weak solution of (11) and that (22) holds if and only if $v$ is weak solution of (12).
I know the weak solution set of (11) is
$$S=\{u\in H^1_0(U)|B[u,v]=\langle 0,v \rangle,\forall v\in H^1_0(U)\}=\{u\in H^1_0(U)| u-Ku=0 \}$$
That is, $S=H^1_0(U)\cap N(I-K)=H^1_0\cap N$. Then how we can get $S=N$?(Note that $I-K$ is a operator on $L^2(U)$ but not $H^1_0(U)$) and similarly question for the adjoint $N^*$.
Best Answer
How to get $S=N$? Note that $I-K$ isn't an operator on $H_0^1(U)$.
Recall that, for $g\in L^2(U)$, the notation $L^{-1}_\gamma g$ stands for the unique weak solution $u\in H_0^1(U)$ of $L_\gamma u=g$.
Therefore, the notation $Ku:=\gamma L^{-1}_\gamma u$ means that $Kf$ is defined for each $f\in L^2$ by $$Kf=\text{unique weak solution }u\in H_0^1(U)\text{ of } Lu+\gamma u=\gamma f.$$
Thus, the image of the operator $K:L^2(U)\to L^2(U)$ is indeed a subset of $H_0^1(U)$. Hence, any solution of $u-Ku=0$ is in $H_0^1(U)$ and satisfies $Lu=0$, that is, is a weak solution of $(11)$. In other words: $N\subset S$.