# Is function $f$ also uniformly continuous?

I've been thinking on the following problem lately:

Let $(X,d)$ be a metric space and $f_1,f_2,...,f_n: X \rightarrow \mathbb{R}$ and

$f(x) = \max\{f_1(x),f_2(x),...,f_n(x) \}$,$x\in X$

If the functions $f_1,f_2,...,f_n$ are uniformly continuous, is $f$ also uniformly continuous?

I feel like it is(since it's obvious), just having difficulties proving it.

Can I get any help starting my proof?

Thank you!

In choosing an $x,y$ with $|x-y|<\delta$, if the same $f_n$ is realizes the maximum for both $x$ and $y$, then we're done by the uniform continuity of $f_n$, otherwise, if $f_m(x)$ is the max for $x$ and $f_n(y)$ is max for $y$, then $f_m(t)-f_n(t)$ is zero for some $t$ between $x$ and $y$ by intermediate value. So then use $|x-y|<|x-t|+|t-y|$ by triangle inequality.
• Let $\delta:=\operatorname{min}\{\delta_1,\delta_2\}$, where $\delta_1$ is small enough that $|f_m(x)-f_m(y)|<\epsilon/2$ whenever $|x-y|<\delta_1$, and $\delta_2$ does the same for $f_n$. These exist by uniform continuity of $f_m,f_n$. Let $|x-y|<\delta$, then $$|f_{max}(x)-f_{max}(y)|=|f_m(x)-f_n(y)|=|f_m(x)-f_m(t)+f_n(t)-f_n(y)|\leq|f_m(x)-f_m(t)|+|f_n(t)-f_n(y)|<\epsilon/2+\epsilon/2$$ – T.J. Gaffney Dec 15 '14 at 5:31
• The last step follows since $|x-t|<|x-y|<\delta$. Same for $|t-y|$. – T.J. Gaffney Dec 15 '14 at 5:37