I have to prove continuity of inverse operator.
Let $T$ be an operator on X (a Banach Space), and $\lambda \in C$. Show that $T-\lambda I$ has a continuous inverse (not necessarily defined on the whole space) if and only if there exists a constant $c > 0$ such that
$c||u|| \le ||(T − \lambda I)u||, u \in dom(T)$.
I think that if inverse exists, then $\lambda \in \rho (T)$, so we have to prove that:
$\lambda _n \rightarrow \lambda \Rightarrow R(\lambda _n)-R(\lambda) \rightarrow 0$
$ ||R(\lambda _n)-R(\lambda)||= ||(\lambda _n - \lambda)R(\lambda)R(\lambda _n)||$
Now I'm using inequality.
$||(\lambda _n - \lambda)R(\lambda)R(\lambda _n)|| \le \frac{1}{c}||(T-\lambda I)(\lambda _n -\lambda)R(\lambda)R(\lambda _n)||=\frac{1}{c}||(\lambda _n - \lambda)R(\lambda _n)||$
I don't know if it's all correct and how to find boundedness of $R(\lambda _n)$.
I think the following theorem solves the problem:
Let $X$ be a Banach-space, $A \in \mathscr{B}(X)$ (i.e. A is a bounded linear operator). Then the following statements are equivalent:
The $1 \implies 2$ can be proven with the the fact that $\lVert Ax \rVert \leqslant \lVert A \rVert \lVert x \rVert$.
For $2 \implies 1$, we can show that $\text{ran}(A)$ is closed with taking $x \in \overline{\text{ran}(A)}$, and picking a sequence in $X$ so that $Ax_n \to x$. Then we have that $$\lVert x_n -x_m\rVert\leqslant \frac{1}{C}\lVert Ax_n-Ax_m \rVert$$ which implies that $x_n$ is Cauchy, so $x_n \to x'$ for some $x' \in X$. But the continuity of $A$ implies that $Ax_n \to Ax'$, so $Ax'=x$, so $x \in \text{ran}(A)$.
But $\text{ran}(A)$ is everywhere dense and closed, which implies that $\text{ran}(A)=X$, so $A$ is surjective. And $A$ is also injective, because if we assume that $x\neq x'$, but $Ax=Ax'$, we get a contradiction: $$0=\lVert A(x'-x)\rVert \geqslant C \lVert x-x' \rVert \neq 0$$ This implies that $A$ is bijective, so Banach's linear homeomorphism theorem guarantees that $A$ is invertible.