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 $S$. $\blacksquare$