Consider a Banach space $(X, ||\cdot||)$ and a compact mapping $f: \overline{B}_1(0) \rightarrow X$ such that $||f(x)||\leq 1$ for all $||x|| = 1$, where $\overline{B}_1(0)$ denotes the closed unit ball in $X$. Show that $f$ has a fixed point.
My attempt:
Since $f$ is compact, then $f(B_1(0))$ is relatively compact in $X$. In particular, it is bounded, hence there exists $r>0$ s.t. $f(B_1(0)) \subset \overline{B}_r(0)$. Define $R: = \max\{r, 1\} > 0$. Thus, $f(\overline{B}_1(0)) \subset \overline{B}_R(0)$.
My idea now, was to apply the Schauder's fixed point theorem to an auxiliary compact self-mapping $g$ on $\overline{B}_R(0)$ to establish the existence of a fixed point of $g$. The problem is that I'm not able to find such an auxiliary function which permits to conclude the existence of a fixed point of $f$.
Any suggestions? Thanks in advance!
Best Answer
Apply Schauder's theorem to $$g : x \in \overline{B}_1(0)\mapsto \frac{f(x)}{\max \lbrace 1, ||f(x)||\rbrace} \in \overline{B}_1(0)$$
There exists $x_0 \in \overline{B}_1(0)$ such that $$\frac{f(x_0)}{\max \lbrace 1, ||f(x_0)||\rbrace} = x_0$$
If $||x_0|| = 1$, then by the assumption on $f$, one must have $||f(x_0)|| \leq 1$, so $\max \lbrace 1, ||f(x_0)|| \rbrace = 1$ and $f(x_0)=x_0$.
If $||x_0|| < 1$, then $\max \lbrace 1, ||f(x_0)|| \rbrace > ||f(x_0)||$, then again $\max \lbrace 1, ||f(x_0)|| \rbrace = 1$ and $f(x_0)=x_0$.