Let $U, V$ be Banach spaces and $L(U,V)$ be the space of bounded linear operators from $U$ to $V$. An operator $T \in L(U,V)$ is said to be a Fredholm operator if dim $N(T) < +\infty$ and $\mathrm{codim} R(T) < +\infty$. In this case, we define the index of a Fredholm operator $T$ as the integer $$ \mathrm{ind}(T) = \mathrm{dim} N(T) - \mathrm{codim} R(T). $$ Denote by $\mathrm{Fred}_0 (U,V)$ the set of the Fredholm operators with index zero from $U$ to $V$.
Given $\lambda \in \mathbb{R}$, consider the operator $$T := -\Delta - \lambda I : C^{2}(\Omega) \rightarrow C(\overline{\Omega})$$ I'm trying to prove that $T$ is a Fredholm operator with $\mathrm{ind}(T) = 0$, for all $\lambda$. I already know that the operator $-\Delta : C^{2}(\Omega) \rightarrow C(\overline{\Omega})$ is in $\mathrm{Fred}_0(C^{2}, C)$, since its kernel is trivial by maximum principle and is surjective by Lax Milgram theorem and regularity theory. In addition, I know that $\mathrm{Fred}_0(U,V)$ is an open subset of $L(U,V)$. So, as $-\Delta \in \mathrm{Fred}_0(C^{2}, C)$, for small values of $\lambda$, $-\Delta - \lambda I$ is in a ball centered in $-\Delta$ which is contained in $\mathrm{Fred}_0(C^{2}, C)$. I don't know if there's another justification to conclude this with arbitrary values of $\lambda$.
I would appreciate any help.