# Visualizing the Hardy norms on polynomials

The Hardy norm of a polynomial ${f}$ is

${\displaystyle \|f\|_p = \left( \int_0^{1} |f(e^{2\pi it})|^p \,dt \right)^{1/p} }$

with the usual convention

${\displaystyle \|f\|_\infty = \sup_{[0, 1]} |f(e^{2\pi it})| }$

(This applies more generally to holomorphic functions, but polynomials suffice for now.) The quantity ${\|f\|_p}$ makes sense for ${0 but is not actually a norm when ${p<1}$.

When restricted to polynomials of degree ${d}$, the Hardy norm provides a norm on ${\mathbb C^{d+1}}$, which we can try to visualize. In the special case ${p=2}$ the Hardy norm agrees with the Euclidean norm by Parseval’s theorem. But what is it for other values of ${p}$?

Since the space ${\mathbb C^{d+1}}$ has ${2d+2}$ real dimensions, it is hard to visualize unless ${d}$ is very small. When ${d=0}$, the norm we get on ${\mathbb C^1}$ is just the Euclidean norm regardless of ${p}$. The first nontrivial case is ${d=1}$. That is, we consider the norm

${\displaystyle \|(a, b)\|_p = \left( \int_0^{1} |a + b e^{2\pi it}|^p \,dt \right)^{1/p} }$

Since the integral on the right does not depend on the arguments of the complex numbers ${a, b}$, we lose nothing by restricting attention to ${(a, b)\in \mathbb R^2}$. (Note that this would not be the case for degrees ${d\ge 2}$.)

One easy case is ${\|(a, b)\|_\infty = \sup_{[0, 1]} |a+be^{2\pi i t}| = |a|+|b|}$ since we can find ${t}$ for which the triangle inequality becomes an equality. For the values ${p\ne 2,\infty}$ numerics will have to do: we can use them to plot the unit ball for each of these norms. Here it is for ${p=4}$:

And ${p=10}$:

And ${p=100}$ which is pretty close to the rotated square that we would get for ${p=\infty}$.

In the opposite direction, ${p=1}$ brings a surprise: the Hardy ${1}$-norm is strictly convex, unlike the usual ${1}$-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 ${p=1/2 < 1}$, the Hardy norm is an actual norm, with a convex unit ball.

Same holds for ${p=1/5}$:

And even for ${p=1/100}$:

And here are all these norms together: from the outside in, the values of ${p}$ are 0.01, 0.2, 0.5, 1, 2, 4, 10, 100. Larger ${p}$ makes a larger Hardy norm, hence a smaller unit ball: the opposite behavior to the usual ${p}$-norms ${(|a|^p + |b|^p)^{1/p}}$.

I think this is a pretty neat picture: although the shapes look vaguely like ${\ell^p}$ balls, the fact that the range of the exponent is ${[0, \infty] }$ instead of ${[1, \infty]}$ means there is something different in the behavior of these norms. For one thing, the Hardy norm with ${p=1}$ turns out to be almost isometrically dual to the Hardy norm with ${p=4}$: this became the subject of a paper and then another one.

## 2 thoughts on “Visualizing the Hardy norms on polynomials”

1. Thomas Jahn says:

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

1. 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.

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