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}:

hardy4
p = 4

And {p=10}:

hardy10
p = 10

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

hardy100
p = 100

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

hardy1
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.

hardy05
p = 0.5

Same holds for {p=1/5}:

hardy02
p = 0.2

And even for {p=1/100}:

hardy001
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}}.

hardy_all
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.

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
{\displaystyle -\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
{\displaystyle \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.