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My friend asked me this question and I gave him a sketch of proof. My idea is that to construct a function $$h = \begin{cases} f-g & \textrm{if $f \ge g$}\\ 0 & \textrm{if $f < g$} \end{cases}$$ and show that $h$ is uniformly continuous. Then since $\max(f, g) = g + h$, so it is uniformly continuous.

He believed that I oversimplified, and show me this site.

The proof of this statement is on the second page. I completely don't see why the third (and fourth) case should be consider, since there is a theorem that for a continuous function $f$, if $f(x_0) > 0$, then there exist a nbhd $U(x_0)$ of $x_0$ s.t. $f(x) > 0$ for all $x \in U(x_0)$. In other words, I can just consider the nbhd s.t. $f(x) > g(x)$ if $f(x_0) > g(x_0)$. For a larger $\epsilon$, I can simply maintain the $\delta$ and everything is fine.

So my questions are:

  1. Is my sketch correct?
  2. Why do we need to consider 4 cases as shown in the "answer"?
Martin Sleziak
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MonkeyKing
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  • Your sketch looks reasonable, but I think if you try to carry it out, you will find it is a fair amount of work to carefully check that $h$ is uniformly continuous, and you will probably have to consider some cases when estimating $h(x)-h(y)$ according to whether or not $f(x) \ge g(x)$, $f(y) \ge g(y)$, etc. You will also need to do some work in verifying that $g+h$ is uniformly continuous. – Nate Eldredge Mar 25 '15 at 04:06
  • As for the "four cases" in the answer, if given $x_0$ you choose $\delta$ less than the radius of $U(x_0)$, then $\delta$ depends on $x_0$. For a different $x_0$, you might have to choose a smaller $\delta$. The whole point of uniform continuity is to be able to choose a single $\delta$ that works for every $x_0$. Your argument would show continuity but not uniform continuity. – Nate Eldredge Mar 25 '15 at 04:09
  • My motivation of constructing $h$ is that I only need to think about non-zero part, where $f \ge g$. I think doing that $\epsilon / 2$ kind of thing works well. And there is a theorem saying $f + g$ is uniformly continuous providing $f$ and $g$ are uniformly continuous. If my friend is not allowed to use this, using that $\epsilon / 2$ kind of thing to prove is also easy. – MonkeyKing Mar 25 '15 at 04:16
  • Oh I get it. Seems if I use definition to prove, 4 cases is necessary. – MonkeyKing Mar 25 '15 at 04:19

2 Answers2

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It is much simpler if you use the formula (proved here) $$\max(f, g) = \frac{1}{2}\left(f + g + |f - g| \right)$$ and then we can reason in the following manner. Sum and difference of two uniformly continuous functions is uniformly continuous. Hence both $(f + g)$ and $(f - g)$ are also uniformly continuous. If we note the inequality $||a| - |b|| \leq |a - b|$, then we get that modulus of an uniformly continuous function is also uniformly continuous. Hence $|f - g|$ is uniformly continuous. By sum property $h = (f + g + |f - g|)/2$ is also uniformly continuous.

Community
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Paramanand Singh
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(The proof proposed by @Paramanand is quick and elegant, but most textbooks proceed in the reverse order: first we show that $\max(f,g)$ is uniformly continuous, and then we show that $|f|$ is uniformly continuous because $|f| = \max(f,-f)$. Thus, I propose an alternative sketch/idea of proof below.)

Let be $f,g : X \to \mathbb{R}$ two uniformly continuous functions, let be $f \oplus g : X \to \mathbb{R}^2$ the direct sum of $f$ and $g$ (i.e., $f \oplus g(x) = (f(x), g(x))$ for all $x \in \mathbb{R}$), and let be $h : \mathbb{R}^2 \to \mathbb{R}$ the maximum function (i.e., $h(x,y) = \max(x,y)$). You can simply see the function $\max(f,g)$ as the composition $h \circ (f \oplus g)$. If you already know that:

  • the composition of two uniformly functions is uniformly continuous
  • the direct sum of two uniformly functions is uniformly continuous

then you're done. (And this is usually the case: those results are usually introduced before the result about $\max(f,g)$. If this is not the case, the two properties I mentioned above are not too difficult to prove and will be a pretty good exercise!)

Philopolis
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