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<p\le \infty} 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}:

p = 4

And {p=10}:

p = 10

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

p = 100

In the opposite direction, {p=1} brings a surprise: the Hardy {1}-norm is strictly convex, unlike the usual {1}-norm.

p = 1

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.

p = 0.5

Same holds for {p=1/5}:

p = 0.2

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}}.

All together now

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. 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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