Normal Truncated Toeplitz Operators

Abstract

The characterization of normal truncated Toepltiz operators is first given by Chalendar and Timotin. We give an elementary proof of their result without using the algebraic properties of truncated Toeplitz operators.

1 Introduction

Let $\mathbb D$ be the open unit disk in the complex plane. Let $L^2$ denote the Lebesgue space of square integrable functions on the unit circle $\partial \mathbb D$. The Hardy space $H^2$ is the subspace of analytic functions on $\mathbb D$ whose Taylor coefficients are square summable. Then it can also be identified with the subspace of $L^2$ of functions whose negative Fourier coefficients vanish. Let P and $P^{\perp }$ be the orthogonal projections from $L^2$ to $H^2$ and $[H^2]^\perp$, respectively. Here $[H^2]^{\perp }$ is the orthogonal complement of $H^2$ in $L^2$. For $f\in L^\infty$, the space of essentially bounded Lebesgue measurable functions on $\partial \mathbb D$, the Toeplitz operator $T_f$ with symbol $f\in L^\infty$ is defined by

$$\begin{aligned} T_fh=P(fh), \end{aligned}$$

for $h\in H^2$.

An analytic function $\theta$ is called an inner function if $|\theta |=1$ a.e. on $\mathbb T$. For each non-constant inner function $\theta$, the so-called model space is

$$\begin{aligned} K_\theta =H^2\ominus \theta H^2. \end{aligned}$$

It is a reproducing kernel Hilbert space with reproducing kernels

$$\begin{aligned} k_w^{\theta }(z)=\frac{1-\overline{\theta (w)}\theta (z)}{1-\bar{w}z}. \end{aligned}$$

Let $P_\theta$ denote the orthogonal projection from $L^2$ onto $K_\theta$,

$$\begin{aligned} P_{\theta }f=P f-\theta P(\bar{\theta }f). \end{aligned}$$

(1.1)

For $\varphi \in L^2$, the truncated Toeplitz operator $A_\phi$ is defined by

$$\begin{aligned} A^{\theta }_\varphi f=P_\theta (\varphi f), \end{aligned}$$

on the dense subset $K_\theta \cap H^{\infty }$ of $K_\theta$. In particular, $K_\theta \cap H^{\infty }$ contains all reproducing kernels $k_w^{\theta }$. The operator $A^{\theta }_\varphi$ may be extended to a bounded operator on $K_\theta$ even for unbounded symbols $\varphi$. The symbol $\varphi$ is never unique and it is proved in [2] that

$$\begin{aligned} A^{\theta }_\varphi =0 \end{aligned}$$

if and only if

$$\begin{aligned} \varphi \in \theta H^2+\overline{\theta H^2}. \end{aligned}$$

If $\theta (0)=0$, then $A^{\theta }_\varphi$ has a unique symbol

$$\begin{aligned} \varphi \in K_\theta +\overline{K_\theta }. \end{aligned}$$

The set of all bounded truncated Toeplitz operators is denoted by ${\mathcal T}_\theta$.

Recall that a bounded operator T on a Hilbert space ${\mathcal H}$ is normal if $T^*T=TT^*.$ The characterization of normal truncated Toepltiz operators is first given by Chalendar and Timotin using the algebraic properties of truncated Toeplitz operators obtained by Sarason [2] and Sedlock [3].

Theorem 1.1

[1, Theorem 6.2] Let $\theta$ be a non-constant inner function vanishing at 0. Then $A^\theta _\varphi$ is normal if and only if one of the following holds

• (1) $A^\theta _\varphi$ belongs to ${\mathscr {B}}_\theta ^\alpha$, for some unimodular constant $\alpha$.

• (2) $A^\theta _\varphi$ is a linear combination of a self-adjoint truncated Toeplitz operator and the identity.

Here ${\mathscr {B}}_\theta ^\alpha$ is a class of truncated Toeplitz operators introduced in [3]. In this note, we give an elementary proof of their result.

2 Proof of the Main Result

In this section we offer a proof of our characterization of normal truncated Toepltiz operators $A^{\theta }_{\varphi }$. We begin with some reduction. Notice that for any constant C, $A^{\theta }_{\varphi +C}=A^{\theta }_{\varphi }+CI,$ which implies $A^{\theta }_{\varphi }$ is normal if and only if $A^{\theta }_{\varphi +C}$ is normal. Thus we may assume, without losing of generality, that $\varphi (0)=0$.

For $a\in \mathbb D$, let $u_a$ be the Möbius transform

$$\begin{aligned} u_a(z)=\frac{z-a}{1-\bar{a}z}. \end{aligned}$$

The Crofoot transform is the unitary operator $J: K_\theta \rightarrow K_{ u_a\circ \theta }$ defined by

$$\begin{aligned} J(f)=\frac{\sqrt{1-|a|^2}}{1-\bar{a}\theta }f. \end{aligned}$$

It is proved in [2] that

$$\begin{aligned} J{\mathcal T}_\theta J^*={\mathcal T}_{u_a\circ \theta }. \end{aligned}$$

Taking $a=\theta (0)$, we see that it is sufficient to consider the normal truncated Toeplitz operators for $\theta (0)=0$. In this case, constant functions are in $K_{\theta }$. Write $\varphi =\varphi _1+\overline{\varphi _2}$, where $\varphi _1, \varphi _2$ are in $K_{\theta }$. We may also assume $\varphi _1(0)=\varphi _2(0)=0$.

It is easy to see that

$$\begin{aligned} (A^{\theta }_{\varphi })^*=A^{\theta }_{\bar{\varphi }}. \end{aligned}$$

Our approach to characterizing normal truncated Toeplitz operators starts with a computation of

$$\begin{aligned} ||A^{\theta }_{\varphi }u||^2-||(A^{\theta }_{\varphi })^*u||^2. \end{aligned}$$

Lemma 2.1

Let $\theta$ be a non-constant inner function. Suppose

$$\begin{aligned} \varphi =\varphi _1+\overline{\varphi _2}, \end{aligned}$$

where $\varphi _1, \varphi _2$ are in $K_{\theta }$. Then for every $u\in K_\theta \cap H^{\infty }$,

$$\begin{aligned}&||A^{\theta }_{\varphi }u||^2-||(A^{\theta }_{\varphi })^*u||^2\\&\quad =||P^{\perp }(\bar{\theta }\varphi _1 u)||^2-||P(\bar{\varphi _1}u) ||^2-( ||P^{\perp }(\bar{\theta }\varphi _2 u)||^2-||P(\bar{\varphi _2}u) ||^2). \end{aligned}$$

Proof

By (1.1), we have for every $u\in K_\theta \cap H^\infty$

$$\begin{aligned} A^{\theta }_{\varphi }u&=P_{\theta }(\varphi u)\\&=P(\varphi u)-\theta P(\bar{\theta }\varphi u)\\&=\varphi _1 u+P(\bar{\varphi _2}u)-\theta P(\bar{\theta }\varphi _1 u+\bar{\theta }\bar{\varphi _2} u )\\&=\varphi _1 u-\theta P(\bar{\theta }\varphi _1 u) +P(\bar{\varphi _2}u). \end{aligned}$$

Then

$$\begin{aligned} ||A^{\theta }_{\varphi }u||^2&=||(\varphi _1 u-\theta P(\bar{\theta }\varphi _1 u))+P(\bar{\varphi _2}u) ||^2\\&=||\varphi _1 u-\theta P(\bar{\theta }\varphi _1 u)||^2+||P(\bar{\varphi _2}u) ||^2+2\, \text{ Re } \langle \varphi _1 u-\theta P(\bar{\theta }\varphi _1 u), P(\bar{\varphi _2}u) \rangle \\&=||\bar{\theta }\varphi _1 u||^2-||P(\bar{\theta }\varphi _1 u)||^2+||P(\bar{\varphi _2}u) ||^2\\&\quad +\,2\, \text{ Re } \langle \varphi _1 u-\theta P(\bar{\theta }\varphi _1 u), P(\bar{\varphi _2}u) \rangle \\&=||P^{\perp }(\bar{\theta }\varphi _1 u)||^2+||P(\bar{\varphi _2}u) ||^2+2\, \text{ Re } \langle \varphi _1 u-\theta P(\bar{\theta }\varphi _1 u), P(\bar{\varphi _2}u)\rangle . \end{aligned}$$

And

$$\begin{aligned} \langle \varphi _1 u-\theta P(\bar{\theta }\varphi _1 u), P(\bar{\varphi _2}u) \rangle&=\langle \varphi _1 u, P(\bar{\varphi _2}u) \rangle -\langle \theta P(\bar{\theta }\varphi _1 u), P(\bar{\varphi _2}u) \rangle \\&=\langle \varphi _1 u, \bar{\varphi _2}u \rangle -\langle P(\bar{\theta }\varphi _1 u), \bar{\theta }P(\bar{\varphi _2}u) \rangle \\&=\langle \varphi _1 u, \bar{\varphi _2}u \rangle -\langle P(\bar{\theta }\varphi _1 u), \bar{\theta }u\bar{\varphi _2}-\bar{\theta }P^{\perp }(\bar{\varphi _2}u)\rangle \\&=\langle \varphi _1 u, \bar{\varphi _2} u \rangle . \end{aligned}$$

Thus

$$\begin{aligned} ||A^{\theta }_{\varphi }u||^2=||P^{\perp }(\bar{\theta }\varphi _1 u)||^2+||P(\bar{\varphi _2}u) ||^2+2\, \text{ Re } \langle \varphi _1 u, \bar{\varphi _2} u \rangle . \end{aligned}$$

(2.1)

Similarly

$$\begin{aligned} ||(A^{\theta }_{\varphi })^*u||^2=||A^{\theta }_{\varphi _2+\bar{\varphi _1}} u||^2=||P^{\perp }(\bar{\theta }\varphi _2 u)||^2+||P(\bar{\varphi _1}u) ||^2+2\, \text{ Re } \langle \varphi _2 u, \bar{\varphi _1}u\rangle . \end{aligned}$$

(2.2)

Subtracting (2.2) from (2.1), we get the desired identity. $\square$

For $w\in \mathbb D$, let

$$\begin{aligned} k_w(z)=\frac{1}{1-\bar{w}z} \end{aligned}$$

be the reproducing kernel of $H^2$.

First we show that if $A^{\theta }_{\varphi }$ is normal then $\varphi _1/\varphi _2$ is a unimodular function.

Lemma 2.2

Let $\theta$ be a non-constant inner function vanishing at 0. Suppose $\varphi =\varphi _1+\overline{\varphi _2}$, where $\varphi _1, \varphi _2$ are in $K_{\theta }$, and $\varphi _1(0)=\varphi _2 (0)=0$. If $A^{\theta }_{\varphi }$ is normal then

$$\begin{aligned} |\varphi _1|=|\varphi _2|, \end{aligned}$$

a.e. on $\mathbb T$.

Proof

By Lemma 2.1, $A^{\theta }_{\varphi }$ is normal implies

$$\begin{aligned} ||P^{\perp }(\bar{\theta }\varphi _1 u)||^2-||P(\bar{\varphi _1}u) ||^2=||P^{\perp }(\bar{\theta }\varphi _2 u)||^2-||P(\bar{\varphi _2}u) ||^2, \end{aligned}$$

(2.3)

for every $u\in K_\theta \cap H^{\infty }$. Take $u=1$, we get

$$\begin{aligned} ||P^{\perp }(\bar{\theta }\varphi _1)||^2-||P(\bar{\varphi _1}) ||^2= ||P^{\perp }(\bar{\theta }\varphi _1)||^2-||P(\bar{\varphi _2}) ||^2. \end{aligned}$$

Since

$$\begin{aligned} P^{\perp }(\bar{\theta }\varphi _j)=\bar{\theta }\varphi _j, \end{aligned}$$

(2.4)

and

$$\begin{aligned} P(\bar{\varphi _j})=0, \end{aligned}$$

(2.5)

we have

$$\begin{aligned} ||\varphi _1||=||\varphi _2||. \end{aligned}$$

(2.6)

Next we consider the reproducing kernels of $K_\theta$:

$$\begin{aligned} k_w^{\theta }(z)=\frac{1-\overline{\theta (w)}\theta (z)}{1-\bar{w}z}, \end{aligned}$$

and take $u=u_w =k_w^{\theta }+1$ in (2.3). Using (2.4) and (2.5), we have

$$\begin{aligned} ||P^{\perp }(\bar{\theta }\varphi _j u_w)||^2= & {} ||P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta })||^2+||P^{\perp }(\bar{\theta }\varphi _j)||^2+2\text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta }), P^{\perp }(\bar{\theta }\varphi _j )\rangle \\= & {} ||P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta })||^2+||\bar{\theta }\varphi _j||^2+2\text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta }), \bar{\theta }\varphi _j \rangle , \end{aligned}$$

and

$$\begin{aligned} ||P(\bar{\varphi _j}u_w)||^2= & {} ||P(\bar{\varphi _j}k_w^{\theta })||^2+||P(\bar{\varphi _j})||^2+2\text{ Re }\,\langle P(\bar{\varphi _j}k_w^{\theta }), P(\bar{\varphi _j})\rangle \\= & {} ||P(\bar{\varphi _j}k_w^{\theta })||^2. \end{aligned}$$

This together with Lemma 2.1 and (2.6) implies

$$\begin{aligned} \text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _1 k_w^{\theta }), \bar{\theta }\varphi _1\rangle =\text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _2 k_w^{\theta }), \bar{\theta }\varphi _2 \rangle . \end{aligned}$$

(2.7)

Since

$$\begin{aligned} k_w^{\theta }=(1-\overline{\theta (w)}\theta )k_w, \end{aligned}$$

we get

$$\begin{aligned} P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta })= & {} P^{\perp }(\bar{\theta }\varphi _j(1-\overline{\theta (w)}\theta ) k_w)=P^{\perp }(\bar{\theta }\varphi _jk_w)-\overline{\theta (w)}P^{\perp }(\varphi _j k_w)\\= & {} P^{\perp }(\bar{\theta }\varphi _jk_w). \end{aligned}$$

Hence

$$\begin{aligned} \quad&\text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _j k_w^{\theta }), \bar{\theta }\varphi _j\rangle \\&=\text{ Re }\,\langle P^{\perp }(\bar{\theta }\varphi _jk_w), \bar{\theta }\varphi _j\rangle \\&=\text{ Re }\,\langle \bar{\theta }\varphi _jk_w, \bar{\theta }\varphi _j\rangle \\&=\text{ Re }\,\int _{0}^{2\pi }\frac{ |\varphi _j(e^{it})|^2}{1-\bar{w}e^{it}}\frac{dt}{2\pi }\\&=\int _{0}^{2\pi } |\varphi _j(e^{it})|^2 \Big (\text{ Re }\,\frac{1}{1-\bar{w}e^{it}}\Big )\frac{dt}{2\pi }\\&=\frac{1}{2}\int _{0}^{2\pi } |\varphi _j(e^{it})|^2 \Big (1+\text{ Re }\,\frac{1+\bar{w}e^{it}}{1-\bar{w}e^{it}}\Big )\frac{dt}{2\pi }\\&=\frac{1}{2}||\varphi _j||^2+\frac{1}{2}\int _{0}^{2\pi } |\varphi _j(e^{it})|^2 \Big (\text{ Re }\,\frac{1+\bar{w}e^{it}}{1-\bar{w}e^{it}}\Big )\frac{dt}{2\pi }\\&=\frac{1}{2}(||\varphi _j||^2+\widehat{|\varphi _j|^2}(w)). \end{aligned}$$

The last equality holds because

$$\begin{aligned} \text{ Re }\,\frac{1+\bar{w}e^{it}}{1-\bar{w}e^{it}} \end{aligned}$$

is the Poisson kernel at w. Here $\widehat{|\varphi _j|^2}$ is the harmonic extension of the function $|\varphi _j|^2$. It follows from (2.7) and (2.6) that

$$\begin{aligned} \widehat{|\varphi _1|^2}(w)=\widehat{|\varphi _2|^2}(w). \end{aligned}$$

Let $w\rightarrow \zeta \in \mathbb T$ nontangentially, we see that

$$\begin{aligned} |\varphi _1|=|\varphi _2|, \end{aligned}$$

a.e. on $\mathbb T$. $\square$

Let U is the unitary operator on $L^2$ defined by

$$\begin{aligned} Uh(z)=\bar{z}\tilde{h}(z), \end{aligned}$$

where $\tilde{h}(z)=h(\bar{z})$. Let $V_\theta$ be the operator

$$\begin{aligned} V_\theta h=P(\theta h), \end{aligned}$$

for $h\in L^2$. Consider the decomposition

$$\begin{aligned}{}[H^2]^{\perp }=\bar{\theta }K_\theta \oplus \bar{\theta }[H^2]^{\perp }. \end{aligned}$$

It is easy to check that $V_\theta$ maps $\bar{\theta }K_\theta$ onto $K_\theta$, and maps $\bar{\theta }[H^2]^{\perp }$ to 0. Thus $V_\theta$ maps $[H^2]^\perp$ onto $K_\theta$. Since U maps $H^2$ onto $[H^2]^\perp$, we see that

$$\begin{aligned} V_\theta U: H^2\rightarrow K_\theta \end{aligned}$$

is also onto.

We shall use the following identity.

Lemma 2.3

Let $\theta$ be an inner function and let g be in $H^2$. Then for every function $f\in H^\infty$

$$\begin{aligned} ||P(\bar{g}V_\theta U f)||=||P^\perp (\bar{\theta } g f^*)||, \end{aligned}$$

where $f^*(z)=\overline{f(\bar{z})}$.

Proof

Notice that for all $h\in L^2$, we have

$$\begin{aligned} (Uh)^*=U(h^*) \end{aligned}$$

and

$$\begin{aligned} (Ph)^*=P(h^*). \end{aligned}$$

Thus

$$\begin{aligned} P(\bar{g}V_\theta U f)&=P(\bar{g}P(\theta U f))=P(\bar{g}\theta U f)=P(\bar{z}\theta \bar{g}\tilde{f})\\&=PU((\bar{\theta }g)^*f)=P(U(\bar{\theta }g f^*))^*\\&=(PU(\bar{\theta }g f^*))^*=(UP^\perp (\bar{\theta }g f^*))^*. \end{aligned}$$

Here we used $PU=UP^\perp$ in the last equality. Since $||h||=||h^*||,$ for all $h\in L^2$ and U is an isometry, we get the desired identity. $\square$

The following result is well-known (see e.g. [4, Lemma 8]).

Theorem 2.1

If $f\in H^2$, then for every $w\in \mathbb D$,

$$\begin{aligned} P(\bar{f}k_w) = \overline{f(w)} k_w. \end{aligned}$$

Now we can prove the main result.

Theorem 2.2

Let $\theta$ be a non-constant inner function vanishing at 0. Suppose $\varphi =\varphi _1+\overline{\varphi _2}$, where $\varphi _1, \varphi _2$ are in $K_{\theta }$. Then $A^{\theta }_{\varphi }$ is normal if and only if either

$$\begin{aligned} \varphi _2-\varphi _2(0)=\alpha (\varphi _1-\varphi _1(0)) \end{aligned}$$

or

$$\begin{aligned} \varphi _2-\varphi _2(0)=\alpha \theta (\overline{\varphi _1}-\overline{\varphi _1(0)}), \end{aligned}$$

for some unimodular constant $\alpha$.

Proof

We may assume $\varphi _1(0)=\varphi _2(0)=0$. Sufficiency follows easily from Lemma 2.1.

Suppose $A^{\theta }_{\varphi }$ is normal. By (2.3) and Lemma 2.2, we have

$$\begin{aligned} ||P(\bar{\varphi _1}u)||^2+||P(\bar{\theta }\varphi _1 u)||^2=||P(\bar{\varphi _2}u)||^2+||P(\bar{\theta }\varphi _2 u)||^2, \end{aligned}$$

(2.8)

for every $u\in K_\theta \cap H^\infty$. According to the discussion before Lemma 2.3, if we write $u=V_\theta Uf$, where $f\in H^\infty$, (2.8) is equivalent to

$$\begin{aligned} ||P(\bar{\varphi _1}V_\theta Uf)||^2+||P(\bar{\theta }\varphi _1 V_\theta Uf)||^2=||P(\bar{\varphi _2}V_\theta Uf)||^2+||P(\bar{\theta }\varphi _2 V_\theta Uf)||^2, \end{aligned}$$

for every $f\in H^\infty$. Using Lemma 2.3 and that $f\mapsto f^*$ is a bijection on $H^\infty$, we have

$$\begin{aligned} ||P^\perp (\bar{\theta }\varphi _1 f)||^2+||P^\perp (\bar{\varphi _1}f)||^2=||P^\perp (\bar{\theta }\varphi _2 f)||^2+||P^\perp (\bar{\varphi _2}f)||^2, \end{aligned}$$

(2.9)

for every $f\in H^\infty$. By Lemma 2.2,

$$\begin{aligned} ||\bar{\theta }\varphi _1 f||=||\bar{\theta }\varphi _2 f||, \end{aligned}$$

and

$$\begin{aligned} ||\bar{\varphi _1}f||=||\bar{\varphi _2}f||. \end{aligned}$$

We see that (2.9) implies

$$\begin{aligned} ||P(\bar{\theta }\varphi _1 f)||^2+||P(\bar{\varphi _1}f)||^2=||P(\bar{\theta }\varphi _2 f)||^2+||P(\bar{\varphi _2}f)||^2, \end{aligned}$$

(2.10)

for every $f\in H^\infty$.

Take $f=k_w$ in (2.10). By Theorem 2.1, we get

$$\begin{aligned} |\varphi _1(w)|^2+|(\theta \bar{\varphi _1})(w)|^2=|\varphi _2(w)|^2+|(\theta \bar{\varphi _2})(w)|^2, \end{aligned}$$

(2.11)

for every $w\in \mathbb D$. Here $(\theta \bar{\varphi _1})(w)$ means $\langle \theta \bar{\varphi _1}, k_w\rangle$.

On the other hand, using Lemma 2.2, we have

$$\begin{aligned} {\varphi _1(w)}(\theta \bar{\varphi _1})(w)= & {} \langle \varphi _1(\theta \bar{\varphi _1}), k_w\rangle =\langle \theta |\varphi _1|^2, k_w\rangle =\langle \theta |\varphi _2|^2, k_w\rangle \nonumber \\= & {} \langle \varphi _2(\theta \bar{\varphi _2}), k_w\rangle ={\varphi _2(w)}(\theta \bar{\varphi _2})(w). \end{aligned}$$

(2.12)

for every $w\in \mathbb D$.

Multiplying both sides of (2.11) by $|\varphi _2(w)|^2$ and using (2.12), we have

$$\begin{aligned} |\varphi _1(w)\varphi _2(w)|^2+|\varphi _2(w)(\theta \bar{\varphi _1})(w)|^2&=|\varphi _2(w)|^4+|\varphi _2(w)(\theta \bar{\varphi _2})(w)|^2\\&=|\varphi _2(w)|^4+|\varphi _1(w)(\theta \bar{\varphi _1})(w)|^2, \end{aligned}$$

which is equivalent to

$$\begin{aligned} (|\varphi _1(w)|^2-|\varphi _2(w)|^2)(|(\theta \bar{\varphi _1})(w)|^2-|\varphi _2(w)|^2)=0. \end{aligned}$$

Thus for every $w\in \mathbb D$, either

$$\begin{aligned} |\varphi _1(w)|=|\varphi _2(w)|, \end{aligned}$$

or

$$\begin{aligned} |\varphi _2(w)|=|(\theta \bar{\varphi _1})(w)|. \end{aligned}$$

Then it follows from the properties of analytic functions that either

$$\begin{aligned} \varphi _1=\alpha \varphi _2, \end{aligned}$$

or

$$\begin{aligned} \varphi _2=\alpha \theta \bar{\varphi _1}, \end{aligned}$$

for some unimodular constant $\alpha$. $\square$

Remark 2.1

The characterization given in Theorem 2.2 is equivalent to that in Theorem 1.1. In fact, if we write $\varphi =\varphi _1+\bar{\varphi _2}+\varphi (0)$, where $\varphi _1, \varphi _2$ are in $K_\theta \cap zH^2$, it is shown in [1, Section 5] that $A^\theta _\varphi \in {\mathscr {B}}_\theta ^\alpha$ if and only if $\theta \bar{\varphi _2}=\alpha \varphi _1$.