Background
Evans PDE book presents the following result (p. 305).
Theorem 5 (Third Existence Theorem for weak solutions). There exists an at most countable set $\Sigma\subset\mathbb R$ such that the problem $$\left\{\begin{align} Lu=\lambda u+f\quad &\text{in}\quad U\\ u=0\quad &\text{on}\quad\partial U \end{align}\right.\tag{1}$$ has a unique weak solotion $u\in H_0^1$ for each $f\in L^2$ if and only if $\lambda\notin \Sigma$.
Notation. Here $L$ stands for a second-order uniformly elliptic operator in the divergence form with associated bilinear for $B[\cdot,\cdot]$.
Remark. From the proof of Theorem 5 we know that $\Sigma=\{\lambda>-\gamma \mid\tfrac{\gamma}{\lambda+\gamma}\in\sigma_p(K)\}$, where
- $\gamma> 0$ is a number such that $Lu+\mu u=f$ has a unique weak solution in $H_0^1(U)$ for each $f\in L^2(U)$ and each $\mu\geq\gamma$ (i.e., such that the bilinear form $B_\mu[u,v]:=B[u,v]+\mu(u,v)$ is coercive for all $\mu\geq \gamma$; the existence of such a number is proved in Theorem 3, p. 301).
- $K:L^2(U)\to L^2(U)$ is the linear compact operator defined by $Kf=\gamma u$, where $u\in H_0^1(U)$ is the unique weak solution of the problem $Lu+\gamma u=f$. (From the previous item, the said problem has indeed a unique weak solution and thus $K$ is well defined; the compactness is proved in Theorem 4, p. 303.)
Set $\lambda \notin \Sigma$. From the above theorem we can define a linear map $T: L^2\to L^2$ by $Tf=u$ (where $u$ is the unique weak solution of $(1)$ for the given $f$). Theorem 6 (p. 306) proves that $T$ is bounded by using a contradiction argument, without invoking $K$.
Question
I present below a different proof which uses the properties of $K$.
I'd like to know if this proof is correct.
Claim: $T$ is bounded.
Proof: Let $f\in L^2(U)$. The function $u=Tf$ satisfies $$Lu=\lambda u+f$$ in the weak sense and thus $$Lu+\gamma u=(\lambda+\gamma) u+f$$ in the weak sense. From the definition of $K$ (see Remark above), it follows that $$K((\lambda+\gamma)u+f)=\gamma u$$ which implies $$(I-\tfrac{\lambda+\gamma}{\gamma}K)u=\frac{1}{\gamma}Kf.\tag{2}$$
If $\lambda=-\gamma$, then $(2)$ implies $\|Tf\|=\|u\|=\|\tfrac{1}{\gamma}Kf\|\leq C\|f\|$ and we are done. So, let us assume $\lambda\neq -\gamma$. Note that
$$\begin{align} (I-\tfrac{\lambda+\gamma}{\gamma}K)v=0\quad &\Longrightarrow \quad K((\lambda+\gamma)v)=\gamma v\\&\Longrightarrow\quad Lv+\gamma v=(\lambda+\gamma)v\quad \text{in the weak sense} \\&\Longrightarrow\quad Lv=\lambda v\quad \text{in the weak sense}. \end{align}$$ As $\lambda\notin \Sigma$, the last equation has a unique weak solution (see Theorem above) and thus $v=0$. This shows that $\text{Ker}(I-\tfrac{\lambda+\gamma}{\gamma}K)=\{0\}$. So, from the compactness of $\tfrac{\lambda+\gamma}{\gamma}K$ (which comes from the compactness of $K$) we conclude that $R(I-\tfrac{\lambda+\gamma}{\gamma}K)=L^2(U)$.
Due to the last paragraph and the open mapping theorem, the operator $I-\tfrac{\lambda+\gamma}{\gamma}K$ has a bounded inverse $(I-\tfrac{\lambda+\gamma}{\gamma}K)^{-1}$. Hence, from $(2)$, $$\|Tf\|=\|u\|=\|(I-\tfrac{\lambda+\gamma}{\gamma}K)^{-1}\tfrac{1}{\gamma}Kf\|\leq C\|f\|.$$ As $f$ is arbitrary and $C$ does not depend on $f$, we conclude that $T$ is bounded.