In euclidean space, $\forall x\in E:\|f(x)\|\le \|x\|$ implies $\ker(f-id)\oplus \mathrm{Im}(f-id)=E$ 
Le $E$ be an euclidean space, $f\in\mathscr L(E)$, such as $\forall
 x\in E:\|f(x)\|\le \|x\|$. Show that $\ker(f-id)\oplus
 \mathrm{Im}(f-id)=E$.

I've tried to show that $\ker(f-id)\perp \mathrm{Im}(f-id)$: let $x\in\ker(f-id)$ and $y=f(s)-s \in \mathrm{Im}(f-id)$.
We have: $(x|y) = (x|f(s)-s) = (x|f(s)) - (x|s)$.
Since $f(x)=x$, we have $(x|y) = (f(x)|f(s)) - (x|s)$.
But $f$ is not an isometry, so I can't say $(f(x)|f(s)) = (x|s)$. I don't see neither how to use finite dimension of $E$ => ok, I've already noticed that I only have to prove that $\ker(f-id)\cap \mathrm{Im}(f-id)=\{0\}$.
 A: Since $\text{dim Im}(f-\text{Id})+\text{dim ker}(f-\text{Id})=\text{dim} E$ by dimension theorem, it suffices to show that $\text{Im}(f-\text{Id})\cap \text{ker}(f-\text{Id})=\{0\}$. Let $v\in\text{ker}(f-\text{Id})\cap\text{Im}(f-\text{Id})$. Then $f(v)=v$ and $f(w)-w=v$ for some $w\in E$. We have
\begin{align*}
f(w)-w&=v\\
f^2(w)-f(w)&=v\\
\vdots&\\
f^n(w)-f^{n-1}(w)&=v
\end{align*} 
By induction we have $f^n(w)=nv+w$. So $\|f^n(w)\|\leq \|w\|$ for all $n$ implies
\begin{align*}
\|nv+w\|^2&\leq\|w\|^2\\
n\|v\|^2+2\langle v, w\rangle&\leq 0
\end{align*}
If $v\neq 0$ then for large enough $n$ the LHS of the last line must become positive. So $v$ is forced to be 0.
A: Let $V=\{x:f(x)=x\}$. Suppose there is $x\notin V$ such that $f(x)-x\in V$. Let $x^*$ be the projection of $x$ onto $V$. We have
$$\operatorname{dist}(f(x),V)\le \|f(x)-x^*\|=\|f(x)-f(x^*)\|\le \|x-x^*\| = \operatorname{dist}(x,V) $$
But since $f(x)-x\in V$, both $f(x)$ and $x$ have equal distance to $V$. Thus, equality holds throughout. In particular,  $x^*$ is the projection of  $f(x)$ onto $V$. This implies $f(x)-x\perp V$, hence $f(x)-x=0$, a contradiction.
