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)