I am struggling to understand an inequality in a proof (pg. 325 Evans PDE) of the Second Existsnce Theorem for weak solutions (Theorem 4).
I’m fine with everything up until the statement $$\frac{1}{\gamma}(Kf,v)=\frac{1}{\gamma}(f,K^*v)$$ where $v$ solves the ellpiptic problem $$\begin{cases}Lu=f &\text{in} \ \Omega \\ u=0 &\text{on} \ \partial\Omega\end{cases}.$$ and $(\cdot,\cdot)$ is the $L^2$ inner-product. The operator $Ku$ is defined as $$Ku=:\gamma L^{-1}_{\gamma}f,$$ and $$h=:L^{-1}_{\gamma}f;$$ The operator $L_{\gamma}$ is defined as $$L_{\gamma}=:Lu+\gamma u.$$ As it follows from Theorem 5, appendix §D.5, $$v-K^*v=0,$$ where $K^*$ is the adjoint of $K$.
Now, here is my attempt to explain it: expanding the equality out, we have $$\frac{1}{\gamma}\int_{\Omega}Kf\cdot v \ dx =\frac{1}{\gamma}\int_{\Omega}K^*v\cdot f \ dx.$$ Thus, since $v=K^*v$, $$Kf\equiv f,$$ but for some reason, the preceding equality states $$\frac{1}{\gamma}(f,K^*v)= \frac{1}{\gamma}(f,v),$$ which suggests that in the original inequality, he did not follow this procedure since he could have done this in one step. Where have I gone wrong here. What is the explanation for this equality?