If a continuous function $f: \mathbb{C} \rightarrow \mathbb{C}$ satisfies $f(z) \rightarrow 0$ as $|z| \rightarrow \infty$, then $f$ is uniformly continuous on $\mathbb{C}$.
Should I be thinking about the Riemann sphere here? I have no clue what my intuition should I have be. Any helpful comments are appreciated. Just trying to brush up on my complex analysis, but I have forgotten everything.
$f(z)\to 0$ as $|z|\to\infty$ implies for each $\varepsilon>0$ there exists $R>0$ such that $|f(z)|<\varepsilon/2$ for $|z|>R$. Since the disk $D_R=\{z\in\Bbb{C}:|z|\le 2R\}$ is compact, $f$ is uniformly continuous on $D_R$. Therefore, for each $\varepsilon>0$ there is $\delta>0$ such that $|f(z)-f(w)|<\varepsilon$ for all $z,w\in D_R$ satisfy that $|z-w|<\delta$.
For each $\varepsilon>0$, let $R$ and $\delta$ are positive numbers given above. Let $z$ and $w$ are complex numbers satisfy that $|z-w|<\min(\delta,R/2)$. If $|z|,|w|\le 2R$ then $|f(z)-f(w)|<\varepsilon$ because of the definition of $\delta$. If not, then $|z|>2R$ or $|w|>2R$. Without loss of generality we assume that $|z|>2R$. Since $|z-w|<R/2$, we get $|w|>3R/2$ and $|f(z)-f(w)|\le |f(z)|+|f(w)|<\varepsilon$.
