Consider $\Omega \subset \mathbb{R}^{N}$ a smooth domain and Let $\lambda$ an eigenvalue of $-\Delta$. Define the operator $$ Lu = -\Delta u - \lambda u. $$ Now I will use the Theorem 3, page 319 from Evans-2010: Consider $\gamma$ a constant such that for each $g \in L^{2}(\Omega)$ there's a unique weak solution $u_g \in H^{1}_{0}(\Omega)$ of the problem $$ (P) \begin{cases} L_{\gamma} u := Lu + \gamma u = g, \,\,\Omega \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,u = 0, \partial \Omega \end{cases}. $$ In this way, we can define the operator $T: L^{2}(\Omega) \rightarrow L^{2}(\Omega)$, with $T(H^{1}_{0}(\Omega)) \subset L^{2}(\Omega)$ such that $T(g)$ is the weak solution of the problem $(P)$. In other words, if we consider $B_{\gamma}$ the bilinear form associated to the problem $(P)$ we must have the equation $$ B_{\gamma}(T(g), v) = (g ,v)_{L^{2}}, \forall v \in H^{1}_{0}(\Omega). $$ In adition consider the problem $$ (\ast) \begin{cases} L^{\ast} v = 0, \Omega \\ v = 0, \partial \Omega . \end{cases} $$ and define $$ N^\ast = \{v \in H^{1}_0(\Omega) : v \text{ is weak solution of }(\ast) \}. $$ I would like to know if $N^{\ast} = \ker(L)$ ?
I know how to prove that $v \in N^{\ast}$ if and only if $v - K^{\ast} v = 0$, where $K := \gamma\,T$ is a linear, continous and comapact map. So my question becomes: Is it true that $$ \ker(L) = \{v \in H^{1}_0(\Omega) : v - K^{\ast} v = 0\} ? \tag1 $$
What I tried: I tried to prove that $K = K^{\ast}$ as follows. $$ (u , K(v))_{L^{2}} = \gamma (u , T(v))_{L^{2}} = \gamma B_{\gamma}(T(u), T(v))_{L^{2}} = \gamma B_{\gamma}(T(v), T(u))_{L^{2}} = \gamma (v , T(u))_{L^{2}}, $$ which implies, $$ (u , K(v))_{L^{2}} = \gamma (v , T(u))_{L^{2}} = (v , K(u))_{L^{2}} = (K(u),v)_{L^{2}}. $$ However, the conclusion of $K = K^{\ast}$ seems wrong. But if this really works, once we have in a certain way $T = L^{-1}_{\gamma}$, we would conclude that \begin{eqnarray} \{v \in H^{1}_0(\Omega) : v - K^{\ast} v = 0\} &=& \{v \in H^{1}_0(\Omega) : v - K v = 0\} \\ & = & \{v \in H^{1}_0(\Omega) : v = \gamma L_{\gamma}^{-1}(v) \} \\ &=& \{v \in H^{1}_0(\Omega) : v = L_{\gamma}^{-1}(\gamma v) \} \\ &=& \{v \in H^{1}_0(\Omega) : L_{\gamma}(v) = \gamma v \} \\ & = & \{v \in H^{1}_0(\Omega) : L(v) + \gamma v = \gamma v \}, \end{eqnarray} so we would have $$ \{v \in H^{1}_0(\Omega) : v - K^{\ast} v = 0\} = \ker(L). $$
What I affirmed in (1) makes any sense? Also, how to prove that precisely that $T = L^{-1}_{\gamma}$ ?