A New Proof of Nakazi Theorem

TL;DR

This paper presents a novel proof of Nakazi's theorem by leveraging recent advancements in the algebraic properties of truncated Toeplitz operators, providing new insights into the theorem's foundation.

Contribution

The paper introduces a new proof of Nakazi's theorem utilizing recent results on truncated Toeplitz operators, offering a fresh perspective and potentially simplifying the original proof.

Findings

New proof of Nakazi's theorem established.

Connections between Toeplitz operators and Nakazi's theorem clarified.

Potential for further research in operator theory highlighted.

Abstract

We give a new proof of an important theorem by Nakazi using recent results by Sarason in his seminal paper on agebraic properties of truncated Toeplitz operators.

Equations

$$C(g)(z)=u(z)\overline{zg(z)},g\in {\mathcal{K}}_u^2.$$

$$T_{\phi }(uh)=T_{\bar{f}+zg}(uh)=T_{\bar{c}}(uh)+T_{zg}(uh).$$

$$P_{H^2}(\bar{f}(uh))=P_{H^2}((\overline{f_1+f_2+f_3})(uh))$$

$$=P_{H^2}(\overline{c}\widetilde{k_0^u}zh+\widetilde{f_2}zh+\overline{l}h).$$

$$P_{H^2}(\bar{f}(uh))=\overline{c}\widetilde{k_0^u}zh+\widetilde{f_2}zh+P_{H^2}(\overline{l}h)$$

$$P_{H^2}(\bar{f}u)=\overline{c}\widetilde{k_0^u}z+\widetilde{f_2}z+P_{H^2}(\overline{l})$$

$$⟨P_{H^2}(\bar{f}u),\widetilde{f_2}z⟩$$

$$⟨P_{H^2}(\bar{f}u),\widetilde{f_2}z⟩={||\widetilde{f_2}z||}^2$$

$$⟨P_{H^2}(\bar{f}(u{h}_N)),k_0^u⟩=⟨\overline{a_N},k_0^u⟩=\overline{a_N}⟨1,k_0^u⟩=\overline{a_N}{||k_0^u||}^2\ne 0$$

$$P_{H^2}(\bar{f}u)=P_{H^2}(\overline{k_0^u}u+\overline{ul}u)=\widetilde{k_0^u}z+P_{H^2}(\bar{l})$$

$$P_{H^2}(\bar{f}u)=(u-u(0))+P_{H^2}\bar{l}=(u-u(0))+P_c\bar{l}$$

$$⟨P_{H^2}(\bar{f}u),k_0^u⟩=⟨(u-u(0))+c_2,k_0^u⟩=⟨u,k_0^u⟩+⟨c_2-u(0),k_0^u⟩=(c_2-u(0)){||k_0^u||}^2\ne 0$$

Full text

1. Introduction

We follow the notations of Sarason [2]. Let $\mathbb{D}$ denote the open unit disc, $H^{2}$ the Hardy space on $\mathbb{D}$, the Hilbert space of holomorphic functions with square summable Taylor series around [math]. For an inner function $u$, denote by $K^{2}_{u}$ the orthocomplement $H^{2}\ominus uH^{2}$ of the shift invariant subspace $uH^{2}$. All functions in Hardy space are identified with their radial limits defined on the unit circle and hence the space $H^{2}$ is identified as a subspace of $L^{2}$ of the unit circle with the normalized arc length measure. So, any input to a function, say $z$ in this paper will denote a point on the unit circle and thus will satisfy $|z|=1$. $P_{H^{2}}$ will denote the projection from $L^{2}$ to $H^{2}$.

For an $L^{\infty}$ function $\varphi$ on the unit circle, we write the multiplication operator with symbol $\varphi$ on $L^{2}$ as $M_{\varphi}$ where $M_{\varphi}(f)=\varphi f$, $f\in L^{2}$ . We will represent the Toeplitz operator with symbol $\varphi$ by $T_{\varphi}$, i.e., $T_{\varphi}=P_{H^{2}}M_{\varphi}|_{H^{2}}$. Let $P_{c}$ denote the projection operator from $H^{2}$ to the one-dimensional space of constant functions.

We will be using a tool called conjugation denoted by $C$, which acts on $\mathcal{K}_{u}^{2}$ as

[TABLE]

Conjugation is an antiunitary involution. Conjugation exhibits many other remarkable properties (see [2, p. 495]). In this paper, we will content ourselves by using the fact that conjugation acts on an element of $\mathcal{K}_{u}^{2}$ and gives back an element of $\mathcal{K}_{u}^{2}$. $Cf$ will routinely be denoted by $\tilde{f}$.

If $k_{w}$ denotes the reproducing kernel $k_{w}(z)=(1-z{\overline{w}})^{-1}$ on the Hardy space, then its projection $P_{u}k_{w}$ on $\mathcal{K}_{u}^{2}$ is denoted by $k_{w}^{u}$. Hence $k_{0}^{u}=P_{u}k_{0}=P_{u}1$ as $k_{0}=1$, and $\tilde{k_{0}^{u}}=Ck_{0}^{u}$.

Before moving forward, it will be worthwhile to mention that whenever $\mathcal{K}_{u}^{2}$ is non trivial (i.e., is a proper non zero subspace of $H_{2}$), then $k_{0}^{u}\neq 0$ and consequently $\tilde{k_{0}^{u}}\neq 0$. Indeed, as $k_{0}^{u}(z)=1-\overline{u(0)}u(z)$, (see [2, p. 494]), if $k_{0}^{u}$ is constant, then $u$ has to be constant, which can never give rise to a non trivial $\mathcal{K}_{u}^{2}$. It is clear that if the inner function $u$ is not constant, then $\mathcal{K}_{u}^{2}$ is non trivial and hence $\tilde{k_{0}^{u}}\neq 0$.

A new proof of the following theorem will be the main content of this paper. Our technique relies heavily on results from a recent paper by Sarason (see [2]). We state here the theorem by Takahiko Nakazi (see [1]) which he proved in much more generality.

Theorem 1.1** (Nakazi Theorem).**

Let $\varphi$ be an $L^{\infty}$ function on the unit circle. Decompose $\varphi$ as $\bar{f}+zg$, where $f,g\in H^{2}$. Then, for any non constant inner function $u$, $uH^{2}$ is invariant under $T_{\varphi}$ if and only if $f$ is a constant.

Recall that by Beurling’s theorem, any closed subspace of $H^{2}$ that is invariant under the forward shift is of the form $uH^{2}$ for some inner function $u$. Thus, a non trivial invariant subspace of the shift is invariant under a Toeplitz operator if and only if the symbol of the Toeplitz operator is analytic.

2. Proof of Nakazi Theorem

Proof.

Pick an element from $uH^{2}$ say $uh$, where $h\in H^{2}$. We will first show that if $f(z)=c$, where $c$ is a constant, then, $uH^{2}$ is invariant under $T_{\varphi}$. So, we compute:

[TABLE]

Clearly, the second term in the last expression i.e., $T_{zg}(uh)$ lies in $uH^{2}$ as $T_{zg}(uh)=P_{H^{2}}(u(zgh))=u(zgh)\in uH^{2}$. Similarly, the first term in (1) i.e., $T_{\bar{c}}(uh)=P_{H^{2}}(\bar{c}uh)=\bar{c}uh\in uH^{2}$. Thus, $T_{\bar{c}}(uh)+T_{zc}(uh)\in uH^{2}$ and hence $T_{\varphi}(uh)\in uH^{2}$. This shows that $uH^{2}$ is invariant under $T_{\varphi}$.

Now, we will show that $uH^{2}$ is not invariant under $T_{\varphi}$ whenever $f$ is non constant and vanishes at [math], i.e., if $f$ is orthogonal to the constant functions. Decompose such an $f$ as $f_{1}+f_{2}+f_{3}$ according to the decomposition $H^{2}=\mathcal{K}_{u}^{2}\oplus uH^{2}=span\{k_{0}^{u}\}\oplus(\mathcal{K}_{u}^{2}\ominus span\{k_{0}^{u}\})\oplus uH^{2}$ of the space, i.e., $f_{1}\in span\{k_{0}^{u}\}$, $f_{2}\in\mathcal{K}_{u}^{2}\ominus span\{k_{0}^{u}\}$ and $f_{3}\in uH^{2}$. Thus $f_{1}=ck_{0}^{u},c\in\mathbb{C}$ and $f_{3}=ul,l\in H^{2}$. Our strategy to prove the claim will be as follows. For any $h\in H^{\infty}$, $T_{\varphi}(uh)=P_{H^{2}}(\bar{f}(uh))+P_{H^{2}}(zg(uh))$. Since $P_{H^{2}}((zg)uh)=u(zgh)\in uH^{2}$. Thus, if we show that $P_{H^{2}}(\bar{f}(uh))\not\in uH^{2}$ for some $h\in H^{\infty}$, then $T_{\varphi}(uh)\not\in uH^{2}$, hence proving our claim.

Case 1: $\mathbf{f_{2}\neq 0}$

Note that, for any $h\in H^{\infty}$, we have

[TABLE]

The last equality follows from the definition of conjugation and the fact that an inner function is 1 a.e. Observe that $\overline{c}\tilde{k_{0}^{u}}zh$ and $\tilde{f_{2}}zh$ lie in $H^{2}$. So,

[TABLE]

If we take $h=1$, we get

[TABLE]

Since $\overline{l}$ is coanalytic, $P_{H^{2}}(\overline{l})=c_{1}$, a constant. Also, note that $\tilde{f_{2}}z\in\mathcal{K}_{u}^{2}$ as $\tilde{f_{2}}\in\mathcal{K}_{u}^{2}\ominus\text{span}\{\tilde{k_{0}^{u}}\}$ (see remark in [2, p. 512]). Hence we have

[TABLE]

Note that, $\langle c_{1},\tilde{f_{2}}z\rangle=0$ and, $\langle\overline{c}\tilde{k_{0}^{u}}z,\tilde{f_{2}}z\rangle=\langle\overline{c}\tilde{k_{0}^{u}},\bar{z}\tilde{f_{2}}z\rangle=\langle\overline{c}\tilde{k_{0}^{u}},\tilde{f_{2}}\rangle=0$ where the first equality follows from the fact that $M_{z}^{\star}=M_{\bar{z}}$. The second equality follows from the fact that $z\bar{z}=1$ and the third equality follows as $\tilde{f_{2}}\in\mathcal{K}_{u}^{2}\ominus\text{span}\{\tilde{k_{0}^{u}}\}$. So, we get,

[TABLE]

which is a non-zero because of the assumption that $f_{2}\neq 0$. Thus $P_{H^{2}}(\bar{f}u)\not\in uH^{2}$ as $\tilde{f_{2}}z\in{uH^{2}}^{\perp}$ and hence the claim.

Case 2: $\mathbf{f_{2}=0}$

Now we shall assume that $f_{2}=0$ i.e., $f=f_{1}+f_{3}=ck_{0}^{u}+ul$ with all the symbols same as in Case 1.

If $c=0$, then equation (2) gives that $P_{H^{2}}(\bar{f}(uh))=P_{H^{2}}(\overline{l}h)$. Since $l\in H^{2}$, $l$ admits a series expansion of the form $l=\sum_{n=N}^{\infty}a_{n}z^{n}$, where $a_{N}$ is non zero and $N\geq 0$. So, $\overline{l}=\sum_{n=N}^{\infty}\overline{a_{n}}z^{-n}$ and therefore $\overline{l}z^{N}=\sum_{n=N}^{\infty}\overline{a_{n}}z^{-n+N}=\sum_{m=0}^{\infty}\overline{a_{m+N}}z^{-m}$. If we take $h_{N}(z)=z^{N}$, then $P_{H^{2}}(\overline{l}h_{N})=\overline{a_{N}}\neq 0$. Hence,

[TABLE]

Since $k_{0}^{u}\in\mathcal{K}_{u}^{2}={uH^{2}}^{\perp}$, we get that $P_{H^{2}}(\bar{f}(uh_{N}))\not\in uH^{2}$ which proves the claim that $uH^{2}$ is not invariant under $T_{\varphi}$.

If $c\neq 0$, we can assume without loss of generality that $f=k_{0}^{u}+ul$. Note that for $f$ to belong to $H^{2}\ominus\text{span}\{1\}$, we need that $uP_{c}(l)\neq 1-k_{0}^{u}$, which gives $P_{c}(l)\neq\overline{u(0)}$, which follows from the fact that $k_{0}^{u}(z)=1-\overline{u(0)}u(z)$ [2, p. 494]. From equation (2), we have

[TABLE]

Now, we shall make an important observation that $\tilde{k_{0}^{z}}=T_{z}^{\star}(u)$ [2, p. 495]. And $T_{z}^{\star}(u)=\frac{u(z)-u(0)}{z}$, which is direct from definition of backward shift $T_{z}^{\star}$ [2, p. 492]. Thus, we get that $\tilde{k_{0}^{u}}z=u(z)-u(0)$. So, (3) becomes,

[TABLE]

The last equality follows from the fact that $\bar{l}$ is co analytic. Call $c_{2}=P_{c}\bar{l}$. Clearly, $c_{2}\neq u(0)$ as is shown above. Now, taking inner product with $k_{0}^{u}\in\mathcal{K}_{u}^{2}$,

[TABLE]

as $(c_{2}-u(0))\neq 0$ from last paragraph. Hence $P_{H^{2}}(\bar{f}u)\not\in uH^{2}$ which proves the claim that $uH^{2}$ is not invariant under $T_{\varphi}$ in this case as well.

From the above two cases, it follows that for all $f$ which are in the orthocomplement of the constant function $1$, $uH^{2}$ is not invariant under $T_{\varphi}$. Since $f\in H^{2}$ can be written as $f=f_{4}+f_{5}$ where $f_{4}\in\text{span}\{1\}$ and $f_{5}\in H^{2}\ominus\text{span}\{1\}$, we have $\varphi=\bar{f}+zg=(\overline{f_{4}+f_{5}})+zg=\bar{f_{4}}+(\bar{f_{5}}+zg)$. Hence, we get $T_{\varphi}=T_{\bar{f_{4}}}+T_{\bar{f_{5}}+zg}$. Note that $T_{\bar{f_{4}}}$ leaves $uH^{2}$ invariant. And, if ${f_{5}}\neq 0$, then $T_{\bar{f_{5}}+zg}$ does not leave $uH^{2}$ invariant as we have already shown using Cases 1 and 2. Therefore, $uH^{2}$ is invariant under $T_{\varphi}$ if and only if $f$ is a constant, i.e., $f_{5}=0$. ∎