Let $K:E \rightarrow E$ be a compact operator on a normed space $E$. Let $R^q$ be the range of $(\lambda I-K)^q$ where $q$ is such that $R^n = R^q$ for all $n \geq q$. Also, the restriction $(\lambda I - K)|_{R^q}$ has an inverse. Prove that the inverse of this restriction $(\lambda I-K)|_{R^q}^{-1}$ is continuous.
Since these are linear operators, continuous is equivalent to bounded. So, $(\lambda I -K)$ is continuous (because $K$ is compact), then bounded. Hence, there's a $C > 0$ such that $||(\lambda I - K)x|| \leq C ||x||$. The only hint I have is that I've to prove that there's a $C_2>0$ such that $$||(\lambda I - K)x|| \geq C_2 ||x||.$$ But I don't see why this proves the statement. Since I'm working with the restriction to $R^q$, $x \in R^q$ so $(\lambda I -K)x = (\lambda I -K)(\lambda I -K)^q y$ for some $y \in E$. But since $R^q = R^{q+1}$, $(\lambda I -K)x = (\lambda I -K)^q z$ for some $z \in E$. And now I'm stuck. Then, I tried to write $x = (\lambda I-K)^{-1}[(\lambda I-K)x]$ but I didn't get anything. Any hint? Thanks
I suppose $\lambda$ is assumed to be nonzero, correct? If so, $\lambda I-K$ is a Fredholm operator on $R^q$ whose index is zero. Since the index of a Fredholm operator $T$ is defined by $$\text{ind}(T)= \text{dim}(\text{Ker}(T)) - \text{codim}(\text{Ran}(T)), $$ and since $\lambda I-K$ is onto $R^q$ by the assumption that $R^q=R^{q+1}$, we deduce that $\lambda I-K$ is one-to-one, and hence invertible on $R^q$.
It is also well known that the range of a Fredholm operator is always closed, so $R^q$ is closed, and hence a Banach space. Finally, since $\lambda I-K$ is invertible on $R^q$, the open mapping Theorem implies that $(\lambda I-K)^{-1}$ is continuous on $R^q$.
PS: I realize that there might be some rather sophisticated results being used here, so please do not hesitate to ask if you'd like me to be more explicit with any of the above statements.