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How to show that $ \max(f,g) = \frac{f+g+|f-g|}{2} $

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    Just check that it works for any two numbers x and y. Assume the three possibilities xy and calculate the value of the definition (with x and y insteas of f and g). If the thing is supposed to refer to functions f and g I would assume they take on values in the reals, or some simple thing in which < is defined (so also |x|).2012-10-18

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Although you’re proving a fact about functions, you can do it by looking at individual function values: you need to show that for each $x$ in the domain of $f$ and $g$,

$\max\{f(x),g(x)\}=\frac12\Big(f(x)+g(x)+|f(x)-g(x)|\Big)\;.$

It suffices to show that if $a$ and $b$ are any real numbers, then

$\max\{a,b\}=\frac12\Big(a+b-|a-b|\Big)\;.\tag{1}$

To see what’s going on, start by drawing pictures, one for $a and one for $a>b$. In each case $\frac12(a+b)$, the arithmetic mean of $a$ and $b$, is the midpoint of the interval between $a$ and $b$, $|a-b|$ is the length of that interval, and $\frac12|a-b|$ is the distance from the midpoint to each end. Once you’ve seen that, it should at least be intuitively clear why $(1)$ is true, even if you still have to work a bit to prove it.

The most straightforward way to prove it is to break the result into two cases, $a\le b$ and $a>b$. In each case you can say exactly what $\max\{a,b\}$ is, and in each case you can simplify the expression $\frac12\big(a+b-|a-b|\big)$ greatly by getting rid of the absolute value; when you do all this, you’ll find that in each of the two cases the lefthand and righthand sides of $(1)$ are indeed equal.

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    @André: That’s exactly as it should be: if you subtract half the distance between $a$ and $b$ from their midpoint, you will of course get the smaller of the two.2013-01-02