This simple identity hold for any two real numbers :

Indeed, if realizes the maximum, then both and have the same sign as . After opening the absolute value signs, we get to cancel out.

So, (1) represents , also known as the norm, as the sum of absolute values of linear functions. Let’s try the same with . Since the right hand side of (1) is just the average of over all possible choices of signs, the natural thing to do is to average over all eight choices. The sign in front of can be taken to be , which simplifies the average to

Does this formula give ? Instead of trying random numbers, let’s just plot the unit ball for the norm given by (2). If the identity works, it will be a cube. I used Maple:

```
with(plots): f:=(abs(x+y+z)+abs(x+y-z)+abs(x-y-z)+abs(x-y+z))/4:
implicitplot3d(f=1,x=-1/4..1/4,y=-1/4..1/4,z=-1/4..1/4,grid=[25,25,25]);
```

Close, but no cube. Luckily, this is my favorite Archimedean solid, the cuboctahedron.

Although all terms of (2) look exactly the same, the resulting shape has both triangular and square faces. Where does the difference of shapes come from?

More importantly, is (2) really the best we can do? Is there some other sum of moduli of linear functions that will produce ?

*— No. *

Even if negative coefficients are allowed?

*— Even then. * (But you can come arbitrarily close.)

What if we allow integrals with respect to an arbitrary (signed) measure, as in

*— Still no. * But if is allowed to be a distribution of higher order (an object more singular than a measure), then a representation exists (W. Weil, 1976). Yes, one needs the theory of distributions to write the maximum of three numbers as a combination of linear functions.

I’ll only prove that there is no identity of the form

Indeed, such an identity amounts to having an isometric embedding . The adjoint operator is a submetry meaning that it maps the unit ball of onto the unit ball . The unit ball of is just a cube; all of its faces are centrally symmetric, and this symmetry is preserved by linear maps. But is an octahedron, with triangular faces. A contradiction.

An aside: what if instead of averaging over all choices (i.e., unimodular real coefficients) we take the average over all unimodular **complex** coefficients? This amounts to . I expected something nice from this norm, but

it’s a strange shape whose equation involves the complete elliptic integral of second kind. Yuck.