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.

Laurent polynomial in blue, its regular part in red

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.

Laurent polynomial in blue, its regular part in red

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:

Laurent polynomial in blue, its regular part in red

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).

Measurability of Banach space valued functions

There is only an indirect proof of the existence of a function {f\colon [0, 1]\to \mathbb R} that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of {f} is a Banach space: just let {b(t)} be the sequence of the binary digits of {t}, considered as an element of the sequence space {\ell_\infty}.

Why is {b} not measurable? Recall that a Banach space-valued function {f} is (Bochner) measurable iff there is a sequence of simple functions {\sum v_k \chi_{A_k}} (finite sum, measurable {A_k}, arbitrary vectors {v_k}) that converges to {f} almost everywhere. This property implies that, with an exception of a null set, the range of {f} lies in the separable subspace spanned by all the vectors {v_k} used in the sequence of simple functions. But {b} has the property {\|b(t)-b(s)\|=1} whenever {t\ne s}, so the image of any uncountable set under {b} is nonseparable.

Another way to look at this: on the interval [0, 1) the function {b} is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of {\ell_\infty} under {b}.

The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces {L_p(0, 1; X)} where {X} is a Banach space and {1\le p<\infty}. (So, {f} belongs to this space iff it is Bochner measurable and the {L^p} norm of {\|f\|\colon [0, 1]\to [0, \infty)} is finite.) In general we do not have the expected relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} with {1/p+1/q=1}. The natural isometric embedding of {L_q(0, 1; X^*)} into {L_p(0, 1; X)^*} is still there: any {g\in L_q(0, 1; X^*)} acts on {L_p(0, 1; X)} by {f\mapsto \int \langle f(t), g(t) \rangle\, dt}. But unless {X^*} has the Radon–Nikodym property, these are more bounded linear functionals on {L_p(0, 1; X)}.

To construct such a functional, let {b_n(t)} be the {n}-th binary digit of {t}. Given {f\in L_1(0, 1; \ell_1)}, write it in coordinates as {(f_1, f_2, \dots)} and define {\varphi(f) = \sum_n \int_0^1 f_n b_n}. This is a bounded linear functional, since {|\varphi(f)|\le \sum_n \int_0^1 |f_n| = \|f\|}. But there is no function {g\in L_\infty(0, 1; \ell_\infty)} that represents it, i.e., {\varphi(f) = \int_0^1 \langle f(t), g(t)\rangle \,dt = \sum_n \int_0^1 f_n g_n }. Indeed, if such {g} existed then by considering {f} with only one nonzero coordinate, we find that {g_n} must be {b_n}, using the duality {L_1^* =  L_\infty} in the scalar case. But the function {[0, 1]\to \ell_\infty} with the components {(b_1, b_2, \dots)} is not measurable, as shown above.

This example, which applies to all {1\le p<\infty}, also serves as a reminder that the duality relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} depends on the dual space {X^*} having the Radon-Nikodym property (RNP), not {X} itself. Indeed, {X=\ell_1} has the RNP; its dual does not.

The importance of {X^*} having the RNP becomes clear once one tries to follow the usual proof of {L_p^*=L_q}. Given {\varphi\in L_p(0,1;X)^*}, we can define an {X^*}-valued measure {\tau} on {[0, 1]} by {\tau(A)(v) = \varphi( v\chi_A)} where {A\subset [0, 1]} is Lebesgue measurable and {v\in X}. This measure has reasonable finiteness and continuity properties coming from {\varphi} being a bounded functional. Still, the existence of a density {g\colon [0, 1]\to X^*} of the measure {\tau} depends on the structure of the Banach space {X^*}.