# Showing that a modifying function which is continuous at 0 is uniformly continuous

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.
• @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) – 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 $|s-t| = max(s-t,t-s)$.