The Uniform Continuity of Composite Functions on Metric Spaces

The Uniform Continuity of Composite Functions on Metric Spaces

Recall from the Uniform Continuity of Functions on Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are two metric spaces, $A \subseteq S$, and $f : A \to T$ then $f$ is said to be uniformly continuous on $A$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in A$ with $d_S(x, y) < \delta$ we have that:

(1)
\begin{align} \quad d_T(f(x), f(y)) < \epsilon \end{align}

We will now show that the composition of uniformly continuous functions is also uniformly continuous.

 Theorem 1: Let $(S, d_S)$, $(T, d_T)$, and $(U, d_U)$ be metric spaces. Let $f : S \to T$ and $g : T \to U$. Then if $f$ is uniformly continuous on $S$ and $g$ is uniformly continuous on $f(T)$ then $g \circ f : S \to U$ is uniformly continuous on $S$.
• Proof: Let $\epsilon > 0$ be given. We want to find a $\delta > 0$ such that if $x, y \in S$ and $d_S(x, y) < \delta$ then:
(2)
\begin{align} \quad d_U(g(f(x)), g(f(y))) < \epsilon \end{align}
• Since $g$ is uniformly continuous on $f(S)$ we have that there exists a $\delta_1 > 0$ such that if $f(x), f(y) \in S$ and $d_T(f(x), f(y)) < \delta_1$ then:
(3)
\begin{align} \quad d_U(g(f(x)), g(f(y))) < \epsilon \quad (*) \end{align}
• Since $f$ is uniformly continuous on $S$ we have that there exists a $\delta_2 > 0$ such that if $x, y \in S$ and $d_S(x, y) < \delta_2$ then:
(4)
\begin{align} \quad d_T(f(x), f(y)) < \delta_1 \quad (**) \end{align}
• Let $\delta = \delta_2$. Then for all $x, y \in S$ we have that $(**)$ holds, and since $(**)$ holds we have that $(*)$ holds. Thus $g \circ f$ is uniformly continuous on $S$. $\blacksquare$