# Existence of fixed point for a compact mapping from the closed unit ball

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 $$f$$.

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 $$f(x_0)=x_0$$.