This is the definition of a modifying function I've got to work with:

In this problem, a function $\phi :[0,\infty)\rightarrow [0,\infty)$ is called a modifying function if

  • (a) $\phi (0)=0$
  • (b) $\phi $ is strictly increasing
  • (c) $\phi$ is subadditive; i.e. $\phi (s+t) \leq \phi (s)+\phi (t)$ for all $s,t \in [0,\infty)$.

The idea is to show that a modifying function which is continuous at 0 is uniformly continuous.

My question: Is the following proof correct or is at least the reasoning correct?

If $\phi$ is continuous at $0$, then $0<|t-0|<\delta \Rightarrow |\phi (t)-\phi(0)|<\epsilon $, or using (a) and that $\phi$ is positive, $t<\delta \Rightarrow \phi (t)<\epsilon $.

Then we have that for any $s,t\in [0,\infty)$: $s<\frac{\delta}{2} \Rightarrow \phi (s)<\frac{\epsilon}{ 2} $ and $t<\frac{\delta}{2} \Rightarrow \phi (t)<\frac{\epsilon}{ 2}$.

But then $\phi (s+t)\leq \phi (s) + \phi(t) < \epsilon$, and since $s+t$ is an arbitrary number in $[0,\infty) $ that is larger than $s$ or $t$, $\phi$ must be uniformly continuous. (Because $\frac{\epsilon}{2}$ is the smallest epsilon we'll ever need to find a $\delta$ for; once we "match" $\frac{\epsilon}{2}$ with a $\delta$ this $\delta$ will automatically work for all other numbers)


Prove that $$ |\phi(t)-\phi(s)| \le \phi(|t-s|). $$

If $\phi$ is continuous in $0$, then for all $\epsilon>0$ exists $\delta>0$ such that $$ |t-s|<\delta \Rightarrow \phi(|t-s|)<\epsilon $$ and from the first inequality you get the uniform continuity.

  • $\begingroup$ @Oliver Can I do like this?: After proving the above it follows that if $t,s\neq 0$,$|t-s|>0\Rightarrow 0<|\phi (t) - \phi (s)|<\phi (|t-s|).$ Can I just set my delta to be $\min \lbrace \phi (|t-s|),\epsilon_0 \rbrace$ (where $\epsilon_0$ is the epsilon at zero where we know $\phi$ is continuous) $\endgroup$ – john.abraham Mar 19 '13 at 13:47

I am not sure I follow your reasoning. Especially not the part where you say that $s + t$ is an arbitrary number in $[0,\infty)$ as you have chosen them both to be less than $\delta/2$. Further I am not sure that $s < \delta/2$ implies that $\phi(s) < \epsilon /2$ follows from $\phi(s) < \epsilon$ whenever $s < \delta$. If this is so you should give an argument for this!You should also try to proove that your conclusion implies that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\phi(x) - \phi(y)| < \epsilon$ whenever $|x - y| < \delta$ is satisfied.

If I was to solve this I would try to prove the hint given above by Emanuele and use this + the continuity at $0$ to prove the uniform continuity.

Here is a hint that might be of use to you: if $s-t \ge 0 $ then $\phi(s) = \phi(s-t + t) \le \phi(s-t) + \phi(t) $ by $(c)$ and recall that $|s-t| = max(s-t,t-s)$.


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.