I have a question about a proof in Evans "Partial Differential Equation, Theorem 5 (Third Existence Theorem for weak solution) in Chapter 6.2. They have shown that the equation
$$Lu=\lambda u+f \tag{1}$$
in some open bounded set $U$ and zero on the boundary has a weak solution for every $f\in L^2(U)$ if and only if $\lambda\notin \Sigma$. The proof uses Fredholm alternative to show the equivalence of a weak solution of $(1)$ and that $\frac{\gamma}{\gamma+\lambda}$ is not an eigenvalue of $K$. I do not see the equivalence between $\frac{\gamma}{\gamma+\lambda}$ is not an eigenvalue of $K$ and $\lambda\notin \Sigma$. It would be appreciated if someone could explain the connection. Thanks in advance.
hulik
EDIT: The bilinear form associated with the operator $L$ is defined as $B[u,v]:= \int_U \sum_{i,j=1}^n a^{ij}u_{x_i}v_{x_j}+\sum_{i=1}^nb^iu_{x_i}v+cvudx$, where $u,v\in H^1_0(U)$, then we have proved that there are constants $\alpha,\beta>0$ and $\gamma\ge 0$ such that $|B[u,v]|\le \alpha\|u\|\|v\|$, where the norm is in the space $H^1_0(U)$ and $\beta\|u\|^2\le B[u,u]+\gamma\|u\|^2_{L^2}$. Furthermore we define $L_\gamma u := Lu+\gamma u$ associated bilinear form is $B_\gamma[u,v]=B[u,v]+\gamma(u,v)$, where $(\cdot,\cdot)$ is the $L^2(U)$ inner product. They proved that for every $g\in L^2$ there is a unique weak solution $u\in H^1_0(U)$, i.e. $B_\gamma[u,v]=(g,v)$ for all $v\in H^1_0(U)$. They write $u=L_\gamma^{-1}g$ whenever the above formula holds and then define $K$ to be $Ku:=\gamma L^{-1}_\gamma u$, which is a compact operator from $L^2(U)\to L^2(U)$. They call $\Sigma$ the (real) spectrum of $L$.
Best Answer
You have to use Theorem 6 in the Appendix D: For this compact operator we have that $\sigma (K) -\{0\}=\sigma_p (K) -\{0\}$. This is either finite or it is a sequence tending to $0$. Now we have $Lu=\lambda u$ if and only if $(L+\gamma Id)u=(\lambda +\gamma) u$ and this by definition $u=L_\gamma^{-1}(\lambda+\gamma)u$ hence by definition of $K$: $\gamma u = K(\lambda+\gamma)u$ or $\frac{\gamma}{\lambda+\gamma}u=Ku$. (Eigenvalue of $K$). The equivalence between the equation $Lu=\lambda u$ and $\frac{\gamma}{\lambda+\gamma}$ not an eigenvalue of $K$ is given in Evans.
cheers
math