The Hardy norm of a polynomial is

with the usual convention

(This applies more generally to holomorphic functions, but polynomials suffice for now.) The quantity makes sense for but is not actually a norm when .

When restricted to polynomials of degree , the Hardy norm provides a norm on , which we can try to visualize. In the special case the Hardy norm agrees with the Euclidean norm by Parseval’s theorem. But what is it for other values of ?

Since the space has real dimensions, it is hard to visualize unless is very small. When , the norm we get on is just the Euclidean norm regardless of . The first nontrivial case is . That is, we consider the norm

Since the integral on the right does not depend on the arguments of the complex numbers , we lose nothing by restricting attention to . (Note that this would not be the case for degrees .)

One easy case is since we can find for which the triangle inequality becomes an equality. For the values numerics will have to do: we can use them to plot the unit ball for each of these norms. Here it is for :

And :

And which is pretty close to the rotated square that we would get for .

In the opposite direction, brings a surprise: the Hardy -norm is strictly convex, unlike the usual -norm.

This curve already appeared on this blog in Maximum of three numbers: it’s harder than it sounds where I wrote

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

Well, that was 6 years ago; with time I grew to appreciate this shape and the elliptic integrals.

Meanwhile, surprises continue: when , the Hardy norm is an actual norm, with a convex unit ball.

Same holds for :

And even for :

And here are all these norms together: from the outside in, the values of are 0.01, 0.2, 0.5, 1, 2, 4, 10, 100. Larger makes a larger Hardy norm, hence a smaller unit ball: the opposite behavior to the usual -norms .

I think this is a pretty neat picture: although the shapes look vaguely like balls, the fact that the range of the exponent is instead of means there is something different in the behavior of these norms. For one thing, the Hardy norm with turns out to be almost isometrically dual to the Hardy norm with : this became the subject of a paper and then another one.

Are there parameter ranges for which the curves are not strictly convex or not smooth?

Strictly convex in the range 0 < p < infinity. Not strictly convex when p = 0 or p = infinity (those are squares).

Apart from the extreme cases 0 and infinity, the smoothness is "one derivative more than the function |x|^p on the real line". That is, the curves are infinitely smooth when p is a positive even integer, are in C^{p, 1} class when p is an odd integer, and are in C^{k+1, a} class when p = k + a with integer k and 0 < a < 1. The issue of smoothness exists only on the diagonals |a|=|b| which is the only place where the expression in the absolute value |a + b*exp(2pi i t)| can turn into zero. Away from diagonals, the curves are infinitely smooth for all 0 < p < infinity.

The reason for "one derivative more" is that something equivalent to |x|^p is being integrated.