# Continuity of $\max$ function

Given continuous functions $f,g: \mathbb{R} \to \mathbb{R}$, in order to prove that $\max(f(x),g(x))$ is continuous, a standard trick is to rewrite it as a linear combination of continuous functions:
$$\max(f(x),g(x)) = \frac{1}{2} (f(x) + g(x) + |f(x) – g(x)| )$$
Is there any sort of motivation for why one might come up with this particular combination of continuous functions? I can see that it works, but without knowing this fact beforehand , what might lead you to consider writing $\max(f(x),g(x) )$ in the above manner?

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What I would do is this:
$$\max(f(x),g(x))=g(x)+\max(f(x)-g(x),0),$$
so this reduces us to write an expression for $\max(f(x),0)$. If we write $f=f^+-f^-$ (positive and negative parts), we have $\max(f(x),0)=f^+(x)$. We also have $|f|=f^++f^-$, so $f^+=(f+|f|)/2$. In the end,
$$\max(f(x),0)=\frac{f(x)+|f(x)|}2.$$
If we now go back to the initial case,
\begin{align}
\max(f(x),g(x))&=g(x)+\max(f(x)-g(x),0)\\ \ \\ &=g(x)+\frac{f(x)-g(x)+|f(x)-g(x)|}2\\ \ \\ &=\frac{f(x)+g(x)+|f(x)-g(x)|}2
\end{align}

Given two real numbers $a$ and $b$ there is a quadratic equation having these two numbers as solutions, namely
$$(x-a)(x-b)=x^2-p x + q =0\ ,$$
where $p=a+b$ and $q=ab$. The two coefficients $p$ and $q$ encode the multiset $\{a,b\}$ in a reversible way, insofar as we can write
$$\{a,b\}=\left\{{p-\sqrt{p^2-4q}\over2},{p+\sqrt{p^2-4q}\over2}\right\}\ .$$
Now on the right side the two numbers appear in increasing order. Therefore
$$\max\{a,b\}={1\over2}\left(p+\sqrt{p^2-4q}\right)={1\over2}\bigl(a+b+|a-b|\bigr)\ .$$

I think I’d start by considering the easier case where one of $f$, $g$ was constant and zero. If I’ve proved that $\max(h(x),0)$ is continuous for all continuous $h$ and want to generalise, I’ll see that I can write
$$\max(f(x),g(x)) = \max (f(x)-g(x), 0) + g(x).$$
This formula breaks the symmetry between $f$ and $g$ that we started with, so it’s natural also to write down $$max(f(x), g(x) = \max( g(x)-f(x), 0) + f(x)$$
and we recover the symmetry by adding the two expressions up:
$$2 max (f(x),g(x)) = f(x) + g(x) +\left(\max (f(x)-g(x), 0) + \max( g(x)-f(x), 0)\right).$$
And now the expression inside the bracket looks familiar, being exactly $|f(x)-g(x)|$.

Indeed you really have to break up $|f(x)|$ into these two parts to prove that the absolute value of a continuous function is continuous. So the magic formula doesn’t really make for a shortcut in the proof, even if you’re told about it without motivation.

We’re given two numbers, $x$ and $y$, and want to figure out the bigger one. Now we probably have to define a function piecewise. However, there already is one simple piecewise defined function, which is also closely related to comparing numbers, namely the absolute value function. I think this is the biggest leap you have to take.

When you notice that $x\ge y$ iff $x-y\ge 0$ iff $|x-y|=x-y$, and similarly $x\le y$ iff $|x-y|=y-x$, you’re basically done. In the first case adding $y$, the smaller number, will give $|x-y|+y=x$ and in the second case adding $x$, again the smaller number, gives $|x-y|+x=y$. If you add both numbers you just get the bigger number times two, and you’ve managed to hide the selection process inside the absolute value function!