Consider a trigonometric polynomial of degree with complex coefficients, represented as a Laurent polynomial where . The Riesz projection of is just the regular part of , the one without negative powers: . Let’s compare the supremum norm of , , with the norm of .
The ratio may exceed . By how much?
The extreme example for appears to be , pictured below together with . The polynomial is in blue, is in red, and the point is marked for reference.
Since has positive coefficients, its norm is just . To compute the norm of , let’s rewrite as a polynomial of . Namely, which simplifies to in terms of . Hence and .
The best example for appears to be vaguely binomial: . Note that the range of is a cardioid.
Once again, has positive coefficients, hence . And once again, is a polynomial of , specifically . Hence and .
I do not have a symbolic candidate for the extremal polynomial of degree . Numerically, it should look like this:
Is the maximum of attained by polynomials with real, rational coefficients (which can be made integer)? Do they have some hypergeometric structure? Compare with the Extremal Taylor polynomials which is another family of polynomials which maximize the supremum norm after elimination of some coefficients.
Riesz projection as a contraction
To have some proof content here, I add a 2010 theorem by Marzo and Seip: where .
The theorem is not just about polynomials: it says the Riesz projection is a contraction (has norm ) as an operator .
Proof. Let , the singular part of . The polynomial differs from only by the sign of the singular part, hence by Parseval’s theorem.
Since consists of negative powers of , while does not contain any negative powers, these polynomials are orthogonal on the unit circle. By the Pythagorean theorem, . On the other hand, . Therefore, , completing the proof.
This is so neat. And the exponent is best possible: the Riesz projection is not a contraction from to when (the Marzo-Seip paper has a counterexample).