# Hankel operator with symbol a Blaschke product

If $B={\prod}_j \varphi_j$ is a Blaschke product (finite or infinite) of Blaschke factors $\varphi_j(w)=\frac{w-\alpha_j}{1-\overline{\alpha_j}w}$ with $|\alpha_j|>1$, is it true that the norm of the Hankel operator (in Hardy spaces on the unit disk) $||H_B||$ is equal to one? I think I have proved it for a Blaschke factor but I do not see how to generalize it (if it is possible).

## 1 Answer

It is straightforward to see that $$\|H_B\|=\|P_-B|H^2\|\le \|B\|_\infty=1.$$ Here $P_-$ is the projection onto $H_2^\bot$ On the other hand, $B^*h$ is analytic for every $h\in H^\infty$, and from the maximum modulus principle it follows that $$\|1-B^*h\|_\infty\ge |1-\underbrace{B^*(z_i)}_{=0}h(z_i)|=1$$ where $z_i$ is one of the zeros of $B^*$, for example, $1/\bar{\alpha_i}$. Then Nehari theorem implies that $$\|H_B\|=\inf_{h\in H^\infty}\|B-h\|_\infty=\inf_{h\in H^\infty}\|1-B^*h\|_\infty\ge 1.$$ P.S. It is more common to call $B^*$ a Blaschke product - an analytic unimodular function with zeros inside the disc.

• By $B^*$ you mean $\bar{B}$? – Babyblog Oct 19 '17 at 10:07
• I have two more questions. If I consider $f(w)=w^n/B(w)$, where $B$ is a "common" Blaschke factor with $d$ poles $\beta_j \in \mathbb{D}$, can we have the explicit form of the optimal $h_{opti} \in H^{\infty}$ in Nehari's theorem: $\|H_f\| = \inf_{h\in H^{\infty}} \| f-h \|$? Is it true that if $f$ is rational then $h_{opti}$ is also rational? – Babyblog Oct 19 '17 at 12:34
• @Babyblog If $f$ is a rational function without zeros on the unit circle then the optimal $h$ exists and is also rational. There is an algorithm how to construct $h_{opt}$. Briefly, take the antianalytic part of $f$, do a state space realization, solve two Lyapunov equations and build $h_{opt}$ by Schmidt pair. The algorithm for continuous time is writen here, Ch, 5,6. – A.Γ. Oct 26 '17 at 8:30
• @Babyblog For discrete-time (the unit disc) I do not have any link in mind right now, but it is quite straightforward to adapt from the continuous time (e.g. using the bilinear transformation). MATLAB can do it numerically within "robust" toolbox. – A.Γ. Oct 26 '17 at 8:33