A rational deformation of the monomial basis of the Hardy space

The Hardy space $H^{2}$ has the familiar orthonormal basis \[ 1,z,z^{2},z^{3},\ldots. \] Here is a one-parameter family of rational orthonormal bases that reduces to the monomial basis when the parameter is zero.

Fix $0\leq r<1$, and consider the degree-two Blaschke product \[ f_{r}\left(z\right)=z\frac{z-r}{1-rz}. \] We write \[ \mathbb{N}_{0}=\left\{ 0,1,2,\ldots\right\} . \] For every finite set $E\subseteq\mathbb{N}_{0}$, define \[ \psi_{r,E}\left(z\right)=\left(1-r^{2}\right)^{\left|E\right|/2}\prod_{j\in E}\frac{f^{\circ j}_{r}\left(z\right)}{1-rf^{\circ j}_{r}\left(z\right)}, \] where $f^{\circ0}_{r}\left(z\right)=z$, and the empty product is understood to be $1$.

The first few functions are \[ 1,\qquad\sqrt{1-r^{2}}\frac{z}{1-rz},\qquad\sqrt{1-r^{2}}\frac{f_{r}\left(z\right)}{1-rf_{r}\left(z\right)},\qquad\left(1-r^{2}\right)\frac{z}{1-rz}\frac{f_{r}\left(z\right)}{1-rf_{r}\left(z\right)}. \] There is no visible reason from these formulas that the functions should be orthogonal. The explanation comes from an elementary decomposition of the Hardy space associated with an inner function.

The two basic functions

Put \[ b_{r}\left(z\right)=\frac{z-r}{1-rz}, \] so that \[ f_{r}=zb_{r}. \] For an inner function $\theta$, write \[ K_{\theta}=H^{2}\ominus\theta H^{2}. \] The usual product decomposition \[ K_{\theta\varphi}=K_{\theta}\oplus\theta K_{\varphi} \] gives \[ K_{f_{r}}=K_{z}\oplus zK_{b_{r}}. \] The space $K_{z}$ is spanned by the constant function $1$, while $K_{b_{r}}$ is spanned by \[ \frac{\sqrt{1-r^{2}}}{1-rz}. \] It follows that \[ u_{0}\left(z\right)=1,\qquad u_{1}\left(z\right)=\sqrt{1-r^{2}}\frac{z}{1-rz} \] form an orthonormal basis of $K_{f_{r}}$.

These two functions generate the whole basis.

Words in two functions

The standard decomposition \[ H^{2}=\bigoplus^{\infty}_{m=0}f^{m}_{r}K_{f_{r}} \] shows that \[ \left\{ u_{k}f^{m}_{r}:k\in\left\{ 0,1\right\} ,\ m\geq0\right\} \] is an orthonormal basis of $H^{2}$.

Define \[ S_{k}h\left(z\right)=u_{k}\left(z\right)h\left(f_{r}\left(z\right)\right),\qquad k\in\left\{ 0,1\right\} . \] Since \[ S_{k}z^{m}=u_{k}f^{m}_{r}, \] the operators $S_{0}$ and $S_{1}$ are isometries with orthogonal ranges, and their ranges together span $H^{2}$.

We may therefore repeat the construction. For every $n\geq1$, the functions \[ S_{\alpha_{0}}S_{\alpha_{1}}\cdots S_{\alpha_{n-1}}z^{m},\qquad\alpha_{j}\in\left\{ 0,1\right\} ,\quad m\geq0, \] form an orthonormal basis of $H^{2}$.

A direct calculation gives \[ S_{\alpha_{0}}S_{\alpha_{1}}\cdots S_{\alpha_{n-1}}z^{m}=\left(\prod^{n-1}_{j=0}u_{\alpha_{j}}\left(f^{\circ j}_{r}\left(z\right)\right)\right)\left(f^{\circ n}_{r}\left(z\right)\right)^{m}. \] Write \[ w_{\alpha}\left(z\right)=\prod^{n-1}_{j=0}u_{\alpha_{j}}\left(f^{\circ j}_{r}\left(z\right)\right). \] The resulting orthonormal basis is \[ \left\{ w_{\alpha}\left(f^{\circ n}_{r}\right)^{m}:\alpha\in\left\{ 0,1\right\} ^{n},\ m\geq0\right\} . \]

The functions with $m\geq1$ span \[ f^{\circ n}_{r}H^{2}. \] Indeed, \[ w_{\alpha}\left(f^{\circ n}_{r}\right)^{m}=f^{\circ n}_{r}w_{\alpha}\left(f^{\circ n}_{r}\right)^{m-1}, \] and the functions on the right after the factor $f^{\circ n}_{r}$ is removed form an orthonormal basis of $H^{2}$. It follows that the functions with $m=0$, \[ \left\{ w_{\alpha}:\alpha\in\left\{ 0,1\right\} ^{n}\right\} , \] form an orthonormal basis of \[ H^{2}\ominus f^{\circ n}_{r}H^{2}. \]

Because $u_{0}=1$, a word $\alpha=\left(\alpha_{0},\ldots,\alpha_{n-1}\right)$ is determined by the set \[ E_{\alpha}=\left\{ j:\alpha_{j}=1\right\} . \] Substituting the formula for $u_{1}$ gives \[ w_{\alpha}\left(z\right)=\left(1-r^{2}\right)^{\left|E_{\alpha}\right|/2}\prod_{j\in E_{\alpha}}\frac{f^{\circ j}_{r}\left(z\right)}{1-rf^{\circ j}_{r}\left(z\right)}=\psi_{r,E_{\alpha}}\left(z\right). \] This proves the finite-level statement.

Completeness

It remains to see that the finite-level spaces fill all of $H^{2}$.

Since \[ f_{r}\left(z\right)=zb_{r}\left(z\right), \] we have \[ f^{\circ\left(n+1\right)}_{r}\left(z\right)=f^{\circ n}_{r}\left(z\right)b_{r}\left(f^{\circ n}_{r}\left(z\right)\right). \] Thus \[ f^{\circ\left(n+1\right)}_{r}H^{2}\subseteq f^{\circ n}_{r}H^{2}, \] and hence the spaces \[ H^{2}\ominus f^{\circ n}_{r}H^{2} \] form an increasing sequence.

The iterates $f^{\circ n}_{r}$ converge locally uniformly to zero. To see this, fix $0<\rho<1$. Since \[ \left|b_{r}\left(z\right)\right|<1 \] for $z\in\mathbb{D}$, we have \[ q_{\rho}=\max_{\left|z\right|\leq\rho}\left|b_{r}\left(z\right)\right|<1. \] Moreover, \[ \left|f_{r}\left(z\right)\right|=\left|z\right|\left|b_{r}\left(z\right)\right|\leq\left|z\right|. \] It follows by induction that \[ \left|f^{\circ n}_{r}\left(z\right)\right|\leq q^{n}_{\rho}\left|z\right|,\qquad\left|z\right|\leq\rho. \]

Now suppose that \[ h\in\bigcap_{n\geq1}f^{\circ n}_{r}H^{2}. \] For every $n$, write \[ h=f^{\circ n}_{r}h_{n}. \] Multiplication by an inner function is an isometry, so \[ \left\Vert h_{n}\right\Vert _{H^{2}}=\left\Vert h\right\Vert _{H^{2}}. \] The point-evaluation estimate for $H^{2}$ gives \[ \left|h\left(z\right)\right|\leq\left|f^{\circ n}_{r}\left(z\right)\right|\frac{\left\Vert h\right\Vert _{H^{2}}}{\sqrt{1-\left|z\right|^{2}}}. \] Letting $n$ tend to infinity gives $h\left(z\right)=0$ for every $z\in\mathbb{D}$. Therefore \[ \bigcap_{n\geq1}f^{\circ n}_{r}H^{2}=\left\{ 0\right\} . \] Consequently, \[ \overline{\bigcup_{n\geq1}\left(H^{2}\ominus f^{\circ n}_{r}H^{2}\right)}=H^{2}. \] The functions $\psi_{r,E}$, with $E$ finite, therefore form an orthonormal basis of $H^{2}$.

Binary digits and the monomial basis

The indexing by finite subsets has a simple interpretation. For a nonnegative integer $N$, write its binary expansion as \[ N=\sum_{j\in E_{N}}2^{j}. \] We may then index the basis by the nonnegative integers by setting \[ \psi_{r,N}=\psi_{r,E_{N}}. \]

When $r=0$, \[ f_{0}\left(z\right)=z^{2} \] and hence \[ f^{\circ j}_{0}\left(z\right)=z^{2^{j}}. \] Therefore \[ \psi_{0,E}\left(z\right)=\prod_{j\in E}z^{2^{j}}=z^{\sum_{j\in E}2^{j}}. \] In the integer indexing, \[ \psi_{0,N}\left(z\right)=z^{N}. \]

Thus the ordinary monomial basis is the $r=0$ member of this family. For $r>0$, the binary digits remain, but each power $z^{2^{j}}$ is replaced by the same rational factor evaluated along the orbit of $f_{r}$.