2
$\begingroup$

Prove that a uniformly continuous function of a uniformly continuous function is uniformly continuous.

In other words, is it saying that for a given uniformly continuous function $f$ and $g$ such that $f: X \to Y$ and $g: Y\to Z$ then $g \circ f$ is also uniformly continuous?

If it is then since $f$ is uniformly continuous then for $\delta >0$ and for all $x,y$ we have $|x-y| < \delta$ implies $|f(x) - f(y)| <\epsilon$. And for $\delta'>0$ we have for all $y,z$ $|y-z|<\delta'$ then $|g(y)-g(z)|<\epsilon$. So now we have to prove that $g\circ f$ is uniformly continuous.

So I have to prove $|g(f(x)) - g(f(y))| <\epsilon$. How can I show that? I know that I can make $f(x) = y$ since it is defined there but can I make $f(y) = z$ even though it is not defined for $z$? I don't think I can.

$\endgroup$
4
$\begingroup$

Fix $\varepsilon>0$. Since $g$ is uniformly continuous there is a $\delta'>0$ such that $|g(x')-g(y')|<\varepsilon$ whenever $|x'-y'|<\delta'$. But since $f$ is uniformly continuous as well, there is a $\delta>0$ such that $|f(x)-f(y)|<\delta'$ whenever $|x-y|<\delta$. By putting $x'=f(x)$ and $y'=f(y)$ we conclude: $|g(f(x))-g(f(y))|<\varepsilon$ whenever $|x-y|<\delta$, and this shows, that even $g\circ f$ is uniformly continuous.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So $y'$ is an element of $Z$? $\endgroup$ – Tom Oct 28 '13 at 4:45
  • $\begingroup$ No, $x,y\in X$ and $x',y'\in Y$. $\endgroup$ – sranthrop Oct 28 '13 at 11:23

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.