1
$\begingroup$

Let $X, Y, Z$ be subsets of $\mathbb R$. 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 $X$.

$\endgroup$
  • $\begingroup$ What have you tried? As a tip: the proof is very similar to proving the same thing of continuous functions. $\endgroup$ – J. David Taylor May 5 '16 at 6:39
0
$\begingroup$

We want to find $\delta$ such that for any $x,y\in X$ and $\epsilon>0$, $|x-y|<\delta$ implies $|g\circ f(x)-g\circ f(y)|<\epsilon$.

But since $g$ is uniformly continuous on $Y$, there is $\delta_1$ such that if two points in $Y$ (which are equal to some $f(x),f(y)$ where $x,y\in X$) satisfy $|f(x)-f(y)|<\delta$, then $|g(f(x))-g(f(y))|<\epsilon$.

Now we need to make sure that $f(x)$ and $f(y)$ are sufficiently close; I'll let you fill in the details.

Overview: we can pick $x,y\in X$ making $f(x),f(y)\in Y$ sufficiently close to make $g(f(x)), g(f(y))$ arbitrarily close.

$\endgroup$
  • $\begingroup$ not sure i am stuck $\endgroup$ – Ben May 5 '16 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.