## Visualizing the Hardy norms on polynomials

The Hardy norm of a polynomial ${f}$ is $\|f\|_p = \left( \int_0^{1} |f(e^{2\pi it})|^p \,dt \right)^{1/p}$

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

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

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.

## Orthonormal bases formed by translation of sequences

The standard orthonormal basis (ONB) in the Hilbert space ${\ell^2(\mathbb N_0)}$ consists of the vectors
(1, 0, 0, 0, …)
(0, 1, 0, 0, …)
(0, 0, 1, 0, …)

Let S be the forward shift operator: ${S(x_0, x_1, \dots) = (0, x_0, x_1, \dots)}$. The aforementioned ONB is precisely the orbit of the first basis vector ${e_0}$ under the iteration of S. Are there any other vectors x whose orbit under S is an ONB?

If one tries to construct such x by hand, taking some finite linear combination of ${e_k}$, this is not going to work. Indeed, if the coefficient sequence has finitely many nonzero terms, then one of them, say, ${c_m}$, is the last one. Then ${S^m x}$ is not orthogonal to ${x}$ because the inner product is ${\overline{c_0} c_m}$ and that is not zero.

However, such vectors x exist, and arise naturally in complex analysis. Indeed, to a sequence ${(c_n)\in\ell^2(\mathbb N_0)}$ we can associate a function ${f(z) = \sum_{n=1}^\infty c_n z^n}$. The series converges in the open unit disk to a holomorphic function which, being in the Hardy space ${H^2}$, has boundary values represented by an square-integrable function on the unit circle ${\mathbb T}$. Forward shift corresponds to multiplication by ${z}$. Thus, the orthogonality requires that for every ${k\in \mathbb N}$ the function ${z^kf}$ be orthogonal to ${f}$ in ${L^2(\mathbb T)}$. This means that ${|f|^2}$ is orthogonal to all such ${z^k}$; and since it’s real, it is orthogonal to ${z^k}$ for all ${k\in \mathbb Z\setminus \{0\}}$ by virtue of conjugation. Conclusion: |f| has to be constant on the boundary; specifically we need |f|=1 a.e. to have a normalized basis. All the steps can be reversed: |f|=1 is also sufficient for orthogonality.

So, all we need is a holomorphic function f on the unit disk such that almost all boundary values are unimodular and f(0) is nonzero; the latter requirement comes from having to span the entire space. In addition to the constant 1, which yields the canonical ONB, one can use

• A Möbius transformation ${f(z)=(z-a)/(1-\bar a z)}$ where ${a\ne 0}$.
• A product of those (a Blaschke product), which can be infinite if the numbers ${a_k}$ converge to the boundary at a sufficient rate to ensure the convergence of the series.
• The function ${f(z) = \exp((z+1)/(z-1))}$ which is not a Blaschke product (indeed, it has no zeros) yet satisfies ${|f(z)|=1}$ for all ${z\in \mathbb T\setminus \{-1\}}$.
• Most generally, an inner function which is a Blaschke product multiplied by an integral of rotated versions of the aforementioned exponential function.

Arguably the simplest of these is the Möbius transformation with ${a=1/2}$; expanding it into the Taylor series we get $-\frac12 +\sum_{n=1}^\infty \frac{3}{2^{n+1}} z^n$
Thus, the second simplest ONB-by-translations after the canonical one consists of
(-1/2, 3/4, 3/8, 3/16, 3/32, 3/64, …)
(0, -1/2, 3/4, 3/8, 3/16, 3/32, …)
(0, 0, -1/2, 3/4, 3/8, 3/16, …)
and so on. Direct verification of the ONB properties is an exercise in summing geometric series.

What about the exponential one? The Taylor series of ${\exp((z+1)/(z-1))}$ begins with $\frac{1}{e}\left(1 - 2z +\frac23 z^3 + \frac23 z^4 +\frac25 z^5 + \frac{4}{45} z^6 - \frac{10}{63} z^7 - \frac{32}{105}z^8 - \cdots \right)$
I don’t know if these coefficients in parentheses have any significance. Well perhaps they do because the sum of their squares is ${e^2}$. But I don’t know anything else about them. For example, are there infinitely many terms of either sign?

Geometrically, a Möbius transform corresponds to traversing the boundary circle once, a Blaschke product of degree n means doing it n times, while the exponential function, as well as infinite Blaschke products, manage to map a circle onto itself so that it winds around infinitely many times.

Finally, is there anything like that for the backward shift ${S^*(x_0, x_1, x_2, \dots) = (x_1, x_2, \dots)}$? The vector ${(S^*)^k x}$ is orthogonal to ${x}$ if and only if ${x}$ is orthogonal to ${S^kx}$, so the condition for orthogonality is the same as above. But the orbit of any ${\ell^2(\mathbb N_0)}$ vector under ${S^*}$ tends to zero, thus cannot be an orthonormal basis.