Let $f: \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$ a continuous application and let $0\leq \alpha <1$ and $\beta >0$ numbers such that $||f(x)||\leq \alpha ||x|| + \beta$ for every $x \in \mathbb{R}^{m}$.
i) If $c=\frac{\beta}{1-\alpha}$, prove that $f(\overline{B}(0,c)) \subset \overline{B}(0,c)$
ii) Prove that $f$ has a fixed point $x^{*} \in \mathbb{R}^{m}$ with $||x^{*}|| \leq\frac{\beta}{1-\alpha}$
My attempt
i) Let $b \in f(\overline{B}(0,c))$. Defining $b=f(a)$, this means $a \in\overline{B}(0,c)$, and $||a||\leq c$ and by definition of $c$: $||a||\leq\frac{\beta}{1-\alpha}$. Then using the statement $||f(x)||\leq \alpha ||x|| + \beta$ for every $x \in \mathbb{R}^{m}$:
$||f(a)||\leq \alpha ||a|| + \beta \leq \alpha \frac{\beta}{1-\alpha} + \beta \rightarrow ||f(a)||\leq\frac{\beta}{1-\alpha}$ and this means $||b||\leq c$ or $b \in \overline{B}(0,c)$. Finally, as $b$ is arbitrary, we can say that $f(\overline{B}(0,c)) \subset \overline{B}(0,c)$.
ii) I'm not sure what's supposed to be the space that I have to prove is closed, bounded and convex in order to use Fixed Point Theorem, maybe $\overline{B}(0,c)$?
Your proof of i) is fine and your idea for ii) also. Use the fixed point theorem on $K := \bar B(0,c)$. Note that i) makes sure, that $f$ maps $K$ to $K$, that is, is a selfmap of $K$.