# Prove that the composition of uniformly continuous functions is uniformly continuous.

Let $$X,Y,Z$$ be subsets of $$\textbf{R}$$. Let $$f:X\rightarrow Y$$ be a function which is uniformly continuous on $$X$$, and let $$g:Y\rightarrow Z$$ be a function which is uniformly continuous on $$Y$$. Show that the function $$g\circ f:X\rightarrow Z$$ is uniformly continuous on $$X$$.

MY ATTEMPT

Since both $$g$$ is uniformly continuous, for every $$\varepsilon > 0$$, there is a $$\delta_{1} > 0$$ such that for every $$w,z\in Y$$ we have that \begin{align*} |w - z| \leq \delta_{1} \Longrightarrow |g(w) - g(z)| \leq \varepsilon \end{align*} On the other hand, $$f$$ is uniformly continuous too. Then for every $$\delta_{1} > 0$$, there is a $$\delta > 0$$ such that for every $$x,y\in X$$ we have that \begin{align*} |x - y| \leq \delta \Longrightarrow |f(x) - f(y)| \leq \delta_{1} \end{align*}

Hence, if we make the substitution $$w = f(x)$$ and $$z = f(y)$$, we conclude that for every $$\varepsilon > 0$$, there is a $$\delta > 0$$ such that \begin{align*} |x - y| \leq \delta \Longrightarrow |g(f(x)) - g(f(y))| \leq \varepsilon \end{align*} thence we conclude that $$g\circ f$$ is uniformly continuous as well.

Could someone provide any comments on the proposed solutions?

• If function maps cauchy sequence to cauchy sequence doesn't imply function is uniformly continuous Commented May 1, 2020 at 18:48
• Indeed, you are right. Uniformly continuous functions map Cauchy sequence onto Cauchy sequences, but the converse in not true in general. Thanks for the comment. I have edited my answer. Commented May 1, 2020 at 18:52

Say $$\delta_g$$ and $$\varepsilon_g$$ in the first line, and $$\delta_f$$ and $$\varepsilon_f$$ in the second.
Then you only need to consider $$\varepsilon_f\leq \delta_g$$ and take, as you say, $$w=f(x),z=f(y)$$ to conclude.
I'm just pointing out the fact that the $$\epsilon_f$$ has to meet some requirements in order for $$g$$ to do as you desire. You can't say "for every $$\delta$$" everywhere because it's not really true. They are related.