# Riesz projection on polynomials

Consider a trigonometric polynomial of degree ${n}$ with complex coefficients, represented as a Laurent polynomial ${L(z) = \sum\limits_{k=-n}^n c_k z^k}$ where ${|z|=1}$. The Riesz projection of ${L}$ is just the regular part of ${L}$, the one without negative powers: ${R(z) = \sum\limits_{k=0}^n c_k z^k}$. Let’s compare the supremum norm of ${L}$, ${ \|L\|=\max\limits_{|z|=1} |L(z)|}$, with the norm of ${R}$.

The ratio ${\|R\|/\|L\|}$ may exceed ${1}$. By how much?

The extreme example for ${n=1}$ appears to be ${L(z) = 2z + 4 - z^{-1}}$, pictured below together with ${R(z)=2z+4}$. The polynomial ${L}$ is in blue, ${R}$ is in red, and the point ${0}$ is marked for reference.

Since ${R(z)=2z+4}$ has positive coefficients, its norm is just ${R(1)=6}$. To compute the norm of ${L}$, let’s rewrite ${|L(z)|^2 = L(z)L(z^{-1})}$ as a polynomial of ${x=\mathrm{Re}\,(z)}$. Namely, ${|L(z)|^2 = -2z^2 + 4z + 21 + 4z^{-1} - 2z^{-2}}$ which simplifies to ${27 - 2(2x-1)^2}$ in terms of ${x}$. Hence ${\|L\| = \sqrt{27}}$ and ${\|R\|/\|L\| = 6/\sqrt{27} = 2/\sqrt{3}\approx 1.1547}$.

The best example for ${n=2}$ appears to be vaguely binomial: ${L(z) = 2z^2 + 4z + 6 - 4z^{-1} + z^{-2}}$. Note that the range of ${R}$ is a cardioid.

Once again, ${R(z) = 2z^2 + 4z + 6}$ has positive coefficients, hence ${\|R\| = R(1) = 12}$. And once again, ${|L(z)|^2}$ is a polynomial of ${x=\mathrm{Re}\,(z)}$, specifically ${|L(z)|^2 = 81 - 8(1-x^2)(2x-1)^2}$. Hence ${\|L\| = 9}$ and ${\|R\|/\|L\| = 12/9 = 4/3\approx 1.3333}$.

I do not have a symbolic candidate for the extremal polynomial of degree ${n= 3}$. Numerically, it should look like this:

Is the maximum of ${\|R\|/\|L\|}$ 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: ${\|R\|_4 \le \|L\|}$ where ${\|R\|_p^p = \int_0^1 |R(e^{2\pi i t})|^p\,dt}$.

The theorem is not just about polynomials: it says the Riesz projection is a contraction (has norm ${1}$) as an operator ${L^\infty\to L^4}$.

Proof. Let ${S=L-R}$, the singular part of ${L}$. The polynomial ${R-S}$ differs from ${L}$ only by the sign of the singular part, hence ${\|R-S\|_2 = \|L\|_2}$ by Parseval’s theorem.

Since ${S^2}$ consists of negative powers of ${z}$, while ${R^2}$ does not contain any negative powers, these polynomials are orthogonal on the unit circle. By the Pythagorean theorem, ${\|R^2-S^2\|_2 \ge \|R^2\|_2 = \|R\|_4^2}$. On the other hand, ${R^2-S^2 = (R+S)(R-S)=L(R-S)}$. Therefore, ${\|R^2-S^2\|_2 \le \|L\| \|R-S\|_2 = \|L\| \|L\|_2 \le \|L\|^2}$, completing the proof.

This is so neat. And the exponent ${4}$ is best possible: the Riesz projection is not a contraction from ${L^\infty}$ to ${L^p}$ when ${p>4}$ (the Marzo-Seip paper has a counterexample).

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