Prove that if $f:A\to B$ is uniformly continuous on $A$ and $g$ is uniformly continuous on $B$, then $g(f(x))$ is uniformly continuous on $A$

Question

Suppose that $f\colon A \to B$ is uniformly continuous on $A$ and $g$ is uniformly continuous on $B$. Show that $g \circ f$ is uniformly continuous on $A$.

I tried to use the definition of uniformly continuous but it doesn't work.

Answer 1

Here is a slightly more formal answer. Throughout, I implicitly assume $\;\varepsilon,\delta,\ldots > 0\;$, $\;x,y \in A\;$, and $\;w,z \in B\;$: this simplifies the formulae.

By the definition of uniform continuity, you are asked to prove $$ (0) \;\;\; \langle \forall \varepsilon :: \langle \exists \delta :: \langle \forall x,y :: |x-y| < \delta \;\Rightarrow\; |(g \circ f)(x)-(g \circ f)(y)| < \varepsilon \rangle \rangle \rangle $$ given that \begin{align} (1) \;\;\; & \langle \forall \varepsilon_1 :: \langle \exists \delta_1 :: \langle \forall x,y :: |x-y| < \delta_1 \;\Rightarrow\; |f(x)-f(y)| < \varepsilon_1 \rangle \rangle \rangle \\ (2) \;\;\; & \langle \forall \varepsilon_2 :: \langle \exists \delta_2 :: \langle \forall w,z :: |w-z| < \delta_2 \;\Rightarrow\; |g(w)-g(z)| < \varepsilon_2 \rangle \rangle \rangle \\ \end{align} (I have chosen different names for the quantified variables to reduce confusion in what follows.)

So starting with most complex part of $(0)$, which is $\;|(g \circ f)(x)-(g \circ f)(y)| < \varepsilon\;$, and working backwards, we have for any $\;\varepsilon\;$ \begin{align} & |(g \circ f)(x)-(g \circ f)(y)| < \varepsilon \\ \equiv & \;\;\;\;\;\text{"definition of $\;\circ\;$"} \\ & |g(f(x))-g(f(y))| < \varepsilon \\ \Leftarrow & \;\;\;\;\;\text{"by $(2)$ for $\;\varepsilon_2 := \varepsilon\;$, for some $\;\delta_2\;$"} \\ & |f(x)-f(y)| < \delta_2 \\ \Leftarrow & \;\;\;\;\;\text{"by $(1)$ for $\;\varepsilon_1 := \delta_2\;$, for some $\;\delta_1\;$"} \\ & |x-y| < \delta_1 \\ \end{align} In other words, there is a $\;\delta\;$ (namely, $\;\delta_1\;$) such that $$\langle \forall x,y :: |x-y| < \delta \;\Rightarrow\; |(g \circ f)(x)-(g \circ f)(y)| < \varepsilon \rangle$$ In yet other words, this proves $(0)$.

Answer 2

Let $X,Y,Z$ be metric spaces, $f:X\to Y$ and $g:Y\to Z$ two arbitrary functions. In general, one has the relation $$\omega_{g\, \circ\, f}\leq \omega_{g} \circ \omega_{ f} \, , $$ where $\omega_f$ stands for the modulus of continuity of $f$. To conclude, just recall that the modulus of continuity is always zero at zero, but is continuous at zero if, and only if, the function is uniformly continuous.

Answer 3

So is this the usual "composition of uniformly continuous functions is uniformly continuous"? The uniform continuity of $f$ means that $\forall \epsilon_1 >0$ there exists $\delta_1 >0$ such that $|x-y| < \delta_1 $ implies $|f(x)-f(y)|< \epsilon_1$. Analogously, the uniform continuity of $g$ means that $\forall \epsilon_2 >0$ there exists $\delta_2 >0$ such that $|w-z| < \delta_2 $ implies $|g(w)-g(z)|< \epsilon_2$. So, once you have chosen $\epsilon_2$, you can find $\delta_2$ as prescribed, and then find a suitable $\delta_1$ accordingly, by imposing $\epsilon_1 \leq \delta_2$, thus obtaining $|g \circ f(x)-g \circ f(y)|< \epsilon_2$ whenever $|x-y| < \delta_1$. It's sort of like nesting two $\epsilon - \delta$ arguments. This is morally the thesis, just write it down a bit more formally if you need to.