Let $X$ be a normed vector space. $f:X\to X$ is defined to be completely continuous when $\overline{f(B)}$ is compact whenever $B$ is a bounded set of X.
I want to show that $I-f$ is proper, that means, if $K\subset X$ is compact, then $(I-f)^{-1}(K)$ is also compact.
My attempt:
Since normed spaces are metric spaces, compactness and sequential compactness are equivalent.
Let $K\subset X$ be compact.
Let $(x_{n})_{n}$ be a sequence in $(I-f)^{-1}(K)$. I want to show that it has a convergent subsequence. We have that $(I-f)(x_{n})$ is a sequence in $K$, which is compact, then there is a convergent subsequence $(I-f)(x_{n_{k}})\rightarrow x$. (*)
Also, since $f$ is completely continuous, we have that $\overline{f\{(x_{n_{k}}): k\in \mathbb{N}\}}$ is compact. Since $f(x_{n_{k}})\in \overline{f\{(x_{n_{k}}): k\in \mathbb{N}\}}$, for all k, there is another convergent subsequence $f(x_{n_{k_{l}}})\rightarrow y$.
But from (*) we can say that $(I-f)(x_{n_{k_{l}}})\rightarrow x$
Since $(I-f)(x_{n_{k}})=x_{n_{k}}-f(x_{n_{k}})$, I can show that $x_{n_{k_{l}}}\rightarrow x+y$, and then we are done, since $(x_{n_{k_{l}}})_{l}$ is a convergent subsequence of our initial sequence.
However, in order for the bold text to be correct, I have to show that $\{(x_{n_{k}}): k\in \mathbb{N}\}$ is bounded. I know that $(I-f)(x_{n_{k}})$ is bounded, but this doesn't seem to help. Is there a way to show this boundedness?
Is there a different approach to prove the statement?