Questions About Extreme Points

Abstract

We discuss the geometry of the unit ball—specifically, the structure of its extreme points (if any)—in subspaces of $L^1$ and $L^\infty$ on the circle that are formed by functions with prescribed spectral gaps. A similar issue is considered for kernels of Toeplitz operators in $H^\infty$.

1 Introduction

Given a Banach space $X=(X,\Vert \cdot \Vert )$, we write

$$\begin{aligned} \text {ball}(X):=\{x\in X:\,\Vert x\Vert \le 1\}. \end{aligned}$$

An element x of $\text {ball}(X)$ is said to be an extreme point thereof if it is not expressible as $x=\frac{1}{2}(u+v)$ with two distinct points $u,v\in \text {ball}(X)$. Clearly, every extreme point x of $\text {ball}(X)$ satisfies $\Vert x\Vert =1$.

In what follows, the role of X is played by certain function spaces on the circle ${\mathbb T}:=\{z\in {\mathbb C}:|z|=1\}$ which are defined in spectral terms. First of all, letting m stand for the normalized arc length measure on ${\mathbb T}$, we introduce the (Lebesgue) spaces $L^p=L^p({\mathbb T},m)$ in the usual way, and we denote the standard $L^p$ norm by $\Vert \cdot \Vert _p$. Further, we recall that the Fourier coefficients of a function $f\in L^1$ are given by

$$\begin{aligned} \widehat{f}(k):=\int _{\mathbb T}\overline{\zeta }^kf(\zeta )\,dm(\zeta ),\qquad k\in {\mathbb Z}, \end{aligned}$$

and the set

$$\begin{aligned} \text {spec}\,f:=\{k\in {\mathbb Z}:\,\widehat{f}(k)\ne 0\} \end{aligned}$$

is called the spectrum of f.

For $1\le p\le \infty$, the Hardy space $H^p$ is then defined by

$$\begin{aligned} H^p:=\{f\in L^p:\,\text {spec}\,f\subset {\mathbb Z}_+\}, \end{aligned}$$

where ${\mathbb Z}_+:=\{0,1,2,\dots \}$. (We also introduce the notation ${\mathbb Z}_-:={\mathbb Z}\setminus {\mathbb Z}_+$ for future reference.) As usual, we may view elements of $H^p$ as holomorphic functions on the open unit disk when convenient; see [15, Chapter II] for the underlying theory and basic properties of $H^p$ spaces.

More generally, given a nonempty set $\Lambda \subset {\mathbb Z}_+$, we consider the lacunary (or punctured) Hardy spaces

$$\begin{aligned} H^p(\Lambda ):=\{f\in L^p:\,\text {spec}\,f\subset \Lambda \},\qquad 1\le p\le \infty , \end{aligned}$$

normed by $\Vert \cdot \Vert _p$ as before. We are concerned with the extreme points of $\text {ball}(H^p(\Lambda ))$, so only the endpoint exponents $p=1$ and $p=\infty$ are of interest. Indeed, for $1<p<\infty$, the uniform convexity of $L^p$ implies that every unit-norm function is extreme.

In the classical setting, it is well known that the extreme points of $\text {ball}(H^1)$ are precisely the outer functions $\mathcal F\in H^1$ with $\Vert \mathcal F\Vert _1=1$, whereas the extreme points of $\text {ball}(H^\infty )$ are the functions $f\in H^\infty$ satisfying $\Vert f\Vert _\infty =1$ and

$$\begin{aligned} \int _{\mathbb T}\log (1-|f|)\,dm=-\infty . \end{aligned}$$

(1)

Both results can be found in [4]; alternatively, see [15, Chapter IV] or [17, Chapter 9].

Recently, the author was able to establish the corresponding extreme point criteria in $H^p(\Lambda )$, with $p=1,\infty$, under the hypothesis that the underlying set $\Lambda$ is either small or large. Precisely speaking, it was assumed that either

$$\begin{aligned} \#\Lambda <\infty \end{aligned}$$

(2)

or

$$\begin{aligned} \#({\mathbb Z}_+\setminus \Lambda )<\infty . \end{aligned}$$

(3)

In the case of $H^1(\Lambda )$, the extreme points of the unit ball were described in [9] under condition (2), and in [10, 11] under condition (3); the case of $H^\infty (\Lambda )$ was treated in [12] for both types of $\Lambda$’s.

Little seems to be known about the extreme points in $H^1(\Lambda )$ and $H^\infty (\Lambda )$ when neither (2) nor (3) holds. The questions we ask below are largely motivated by our curiosity in this regard. Sometimes, however, we find it natural to adopt a more general viewpoint. Namely, letting $\Lambda$ be a subset of ${\mathbb Z}$ (not necessarily of ${\mathbb Z}_+$), we extend our attention to the lacunary $L^p$ spaces

$$\begin{aligned} L^p_\Lambda :=\{f\in L^p:\,\text {spec}\,f\subset \Lambda \},\qquad p=1,\infty , \end{aligned}$$

with norm $\Vert \cdot \Vert _p$.

2 Questions, Problems, and a Bit of Discussion

Here are some of the questions that puzzle us.

Question 1

Given a set $\Lambda \subset {\mathbb Z}$, which unit-norm functions from $L^1_\Lambda$ (if any) are extreme points for $\text {ball}(L^1_\Lambda )$? Also, what are the extreme points of $\text {ball}(L^\infty _\Lambda )$?

Clearly, of concern are the cases that do not reduce to the existing results on $H^p(\Lambda )$ as described above. Furthermore, it may well happen for a suitable $\Lambda$ that $\text {ball}(L^1_\Lambda )$ has no extreme points at all. (A classical example is provided by taking $\Lambda ={\mathbb Z}$, in which case $L^1_\Lambda$ becomes the “full” $L^1$.) In fact, the mere existence of extreme points seems to present a nontrivial problem, which we now state and discuss in some detail.

Question 2

For which sets $\Lambda \subset {\mathbb Z}$ does $\text {ball}(L^1_\Lambda )$ possess an extreme point? In particular, for which sets $\Lambda$ of the form

$$\begin{aligned} \Lambda =E\cup {\mathbb Z}_+,\quad \text {with }E\subset {\mathbb Z}_-, \end{aligned}$$

(4)

does this happen?

Our interest in this last class of sets reflects an attempt to interpolate, so to speak, between $H^1$ and $L^1$ (i.e., between the cases $E=\emptyset$ and $E={\mathbb Z}_-$), where two different things occur. Namely, the unit ball has plenty of extreme points in the former case, and none at all in the latter.

Now, let us say that a set $\Lambda \subset {\mathbb Z}$ is periodic if there is a positive integer n such that

$$\begin{aligned} \Lambda +n=\Lambda \end{aligned}$$

(5)

(as usual, $\Lambda +n$ stands for $\{k+n:k\in \Lambda \}$). For instance, any arithmetic progression in ${\mathbb Z}$ is obviously periodic.

To introduce another type of sets that we need here, we first recall the notation $M({\mathbb T})$ for the space of all finite Borel complex measures on ${\mathbb T}$. Also, for $\mu \in M({\mathbb T})$, we let $\text {spec}\,\mu$ denote the set of those indices $k\in {\mathbb Z}$ for which $\widehat{\mu }(k):=\int _{\mathbb T}\overline{z}^kd\mu$ is nonzero. Finally, a subset $\Lambda$ of ${\mathbb Z}$ is said to be a Riesz set, written as $\Lambda \in \mathcal R$, if every measure $\mu \in M({\mathbb T})$ with $\text {spec}\,\mu \subset \Lambda$ is absolutely continuous with respect to m.

The classical F. and M. Riesz theorem (see, e.g., [15, Chapter II]) tells us that ${\mathbb Z}_+\in \mathcal R$. A deeper study and further examples of Riesz sets can be found in [16, Part One, Chapter 1]. Among these examples are the $\Lambda$’s given by (4), where E is one of the following sets:

$$\begin{aligned} \{-2^k:k\in {\mathbb N}\},\qquad \{-k^2:k\in {\mathbb N}\},\qquad \{-p:\,p\,\,\text {prime}\}. \end{aligned}$$

The next result provides a bit of information on Question 2 (but is a far cry from answering it completely).

Theorem 2.1

Let $\Lambda \subset {\mathbb Z}$. If either $\Lambda$ is periodic or $\#({\mathbb Z}\setminus \Lambda )<\infty$, then $\text {ball}(L^1_\Lambda )$ has no extreme points. On the other hand, if $\Lambda \in \mathcal R$ then $\text {ball}(L^1_\Lambda )$ does possess extreme points.

Proof

Suppose that $\Lambda$ is periodic, so that (5) holds for some $n\in {\mathbb N}$. Now let $f\in L^1_\Lambda$ be an arbitrary function with $\Vert f\Vert _1=1$. To show that f is not an extreme point of $\text {ball}(L^1_\Lambda )$, it suffices to find a real-valued function $h\in L^\infty$ such that $fh\in L^1_\Lambda$ and h is nonconstant on the set $\{\zeta \in {\mathbb T}:f(\zeta )\ne 0\}$. (The existence of such an h is actually equivalent to the statement that f is nonextreme for $\text {ball}(L^1_\Lambda )$. We refer to [14, Chapter V, Section 9] or [11, Lemma 2.1], where the equivalence is proved in the context of $H^1$ and its subspaces; the case of a general subspace in $L^1$ is similar.) One possible choice is

$$\begin{aligned} h(z)=\text {Re}(z^n)=\frac{1}{2}\left( z^n+\overline{z}^n\right) ,\qquad z\in {\mathbb T}. \end{aligned}$$

Indeed, the assumption that $\text {spec}\,f\subset \Lambda$ implies, in conjunction with (5), that

$$\begin{aligned} \text {spec}\,(z^nf)\subset \Lambda \quad \text {and}\quad \text {spec}\,(\overline{z}^nf)\subset \Lambda . \end{aligned}$$

Hence $\text {spec}\,(fh)\subset \Lambda$, so that $fh\in L^1_\Lambda$.

Now suppose that ${\mathbb Z}\setminus \Lambda$ is a finite set, say, of cardinality N. Thus,

$$\begin{aligned} {\mathbb Z}\setminus \Lambda =\{k_1,\dots ,k_N\}, \end{aligned}$$

(6)

where the $k_j$’s are pairwise distinct integers. Once again, given an arbitrary unit-norm function f in $L^1_\Lambda$, we prove that f is a nonextreme point of $\text {ball}(L^1_\Lambda )$ by constructing a real-valued function $h\in L^\infty$ that satisfies $fh\in L^1_\Lambda$ and is nonconstant on the support of f. In fact, we claim that for a suitable nonzero vector

$$\begin{aligned} \alpha =(\alpha _1,\dots ,\alpha _{2N+1})\in {\mathbb R}^{2N+1}, \end{aligned}$$

(7)

the function

$$\begin{aligned} h_\alpha (z):=\text {Re}\left( \sum _{j=1}^{2N+1}\alpha _jz^j\right) ,\qquad z\in {\mathbb T}, \end{aligned}$$

does the job. To check this, we associate with each vector (7) the numbers

$$\begin{aligned} \gamma _\nu (\alpha ):=\widehat{\left( fh_\alpha \right) }(k_\nu ),\qquad \nu =1,\dots ,N, \end{aligned}$$

(8)

and consider the linear map $S:{\mathbb R}^{2N+1}\rightarrow {\mathbb R}^{2N}$ defined by

$$\begin{aligned}S\alpha = \left( \text {Re}\,\gamma _1(\alpha ),\,\text {Im}\,\gamma _1(\alpha ),\dots , \text {Re}\,\gamma _N(\alpha ),\,\text {Im}\,\gamma _N(\alpha )\right) . \end{aligned}$$

The rank of S is of course bounded by 2N, and we deduce from the rank-nullity theorem (see, e.g., [2, p. 63]) that the kernel of S has dimension at least 1; in particular, the kernel is nontrivial. Now, if $\alpha \in {\mathbb R}^{2N+1}$ is a nonzero vector with $S\alpha =0$, then the numbers (8) are all null, whence $fh_\alpha \in L^1_\Lambda$. Also, the function $h_\alpha$ (which is obviously real-valued and bounded) is then nonconstant on any set $\mathcal E\subset {\mathbb T}$ with $m(\mathcal E)>0$. Our claim is thereby verified.

Finally, suppose that $\Lambda \in \mathcal R$. Consider the space $C:=C({\mathbb T})$ of all continuous functions on ${\mathbb T}$, and put

$$\begin{aligned} C^\Lambda :=\{f\in C:\,\text {spec}\,f\subset \widetilde{\Lambda }\}, \end{aligned}$$

where

$$\begin{aligned} \widetilde{\Lambda }:=\{-k:\,k\in {\mathbb Z}\setminus \Lambda \}. \end{aligned}$$

As usual, we identify the dual of C with $M:=M({\mathbb T})$, the functional induced by a measure $\mu \in M$ being $g\mapsto \int _{\mathbb T}g\,d\mu$. The dual of the quotient space $C/C^\Lambda$ is then $(C^\Lambda )^\perp$, the annihilator of $C^\Lambda$ in M. On the other hand,

$$\begin{aligned} (C^\Lambda )^\perp =\{\mu \in M:\,\text {spec}\,\mu \subset \Lambda \}. \end{aligned}$$

This last set of measures embeds in $L^1$ (the $\mu$’s involved are absolutely continuous with respect to m because $\Lambda \in \mathcal R$), so it coincides with $L^1_\Lambda$. Consequently, we have

$$\begin{aligned} (C/C^\Lambda )^*=(C^\Lambda )^\perp =L^1_\Lambda . \end{aligned}$$

The existence of extreme points in $\text {ball}(L^1_\Lambda )$ is now guaranteed by the Krein–Milman theorem; see, e.g., [17, Chapter 9]. $\square$

Our next question is motivated by the conjecture—or perhaps a vague feeling—that if $\Lambda \subset {\mathbb Z}_+$ and if $\Lambda$ contains “most” of ${\mathbb Z}_+$, then the extreme points of $\text {ball}(H^1(\Lambda ))$ are “not too far” from being outer functions. Indeed, when $\Lambda$ is all of ${\mathbb Z}_+$, our space is just $H^1$ and its extreme points are precisely the outer functions of norm 1; see [4]. Furthermore, it was shown in [11] (see also [10]) that if ${\mathbb Z}_+\setminus \Lambda$ is a finite set, say with $\#({\mathbb Z}_+\setminus \Lambda )=N$, and if f is an extreme point of $\text {ball}(H^1(\Lambda ))$, then the inner factor of f is necessarily a finite Blaschke product with at most N zeros. In light of these facts, it seems tempting to conjecture that when ${\mathbb Z}_+\setminus \Lambda$ is appropriately “thin” (or “sparse”) in ${\mathbb Z}_+$, the inner factors corresponding to the extreme points of $\text {ball}(H^1(\Lambda ))$ are still fairly “tame,” in some sense or another. It would be nice to have a rigorous result to that effect.

Question 3

Suppose that F is a suitably sparse (infinite) subset of ${\mathbb Z}_+$, and let $\Lambda ={\mathbb Z}_+\setminus F$. What can we say about the inner factors of functions that arise as extreme points of $\text {ball}(H^1(\Lambda ))$? To be more specific, what happens when F is $\{2^k:\,k\in {\mathbb Z}_+\}$ or $\{2^{2^k}:\,k\in {\mathbb Z}_+\}$?

On the other hand, the case of $H^1(\Lambda )$ where $\Lambda$ (rather than ${\mathbb Z}_+\setminus \Lambda$) is a sparse—say, Hadamard lacunary—subset of ${\mathbb Z}_+$ is also worth studying; that would provide a natural extension to what was done in [9].

Turning to the $L^\infty$ part of Question 1, we now make a few observations pertaining to that setting. First we show that if $\Lambda$ is obtained from ${\mathbb Z}$ by removing a finite number of elements, then the extreme points in $L^\infty _\Lambda$ are precisely the unimodular functions, just as it happens for $L^\infty (=L^\infty _{\mathbb Z})$.

Proposition 2.2

Suppose that $\Lambda \subset {\mathbb Z}$ and $\#({\mathbb Z}\setminus \Lambda )<\infty$. In order that a function $f\in L^\infty _\Lambda$ with $\Vert f\Vert _\infty =1$ be an extreme point of $\text {ball}(L^\infty _\Lambda )$, it is necessary and sufficient that $|f|=1$ a.e. on ${\mathbb T}$.

Proof

The sufficiency is obvious, since $L^\infty _\Lambda \subset L^\infty$ and every unimodular function is an extreme point of $\text {ball}(L^\infty )$.

To prove the necessity, let (6) be an enumeration of ${\mathbb Z}\setminus \Lambda$. Now suppose f is a unit-norm function in $L^\infty _\Lambda$ that satisfies $|f|<1$ on a set of positive measure on ${\mathbb T}$. We then define $g:=1-|f|$, so that g is a non-null function in $L^\infty$; clearly, we also have $g\ge 0$ a.e. on ${\mathbb T}$. Further, with each vector

$$\begin{aligned} \beta =(\beta _0,\beta _1,\dots ,\beta _N)\in {\mathbb C}^{N+1} \end{aligned}$$

we associate the polynomial

$$\begin{aligned} p_\beta (z):=\sum _{j=0}^N\beta _jz^j,\qquad z\in {\mathbb T}, \end{aligned}$$

and consider the linear map $T:{\mathbb C}^{N+1}\rightarrow {\mathbb C}^N$ that acts by the rule

$$\begin{aligned} T\beta =\left( \widehat{(gp_\beta )}(k_1),\dots ,\widehat{(gp_\beta )}(k_N)\right) . \end{aligned}$$

The rank of T being obviously bounded by N, we invoke the rank-nullity theorem to conclude that the kernel of T is nontrivial.

Now, if $\beta \in {\mathbb C}^{N+1}$ is a nonzero vector with $T\beta =0$, then the corresponding polynomial $p=p_\beta$ is non-null and satisfies $gp\in L^\infty _\Lambda {\setminus }\{0\}$. We may assume in addition that $\Vert p\Vert _\infty =1$, which yields

$$\begin{aligned} |f\pm gp|\le |f|+g|p|\le |f|+g=1 \end{aligned}$$

almost everywhere on ${\mathbb T}$. Consequently, $f+gp$ and $f-gp$ are two distinct points of $\text {ball}(L^\infty _\Lambda )$, and the identity

$$\begin{aligned} f=\frac{1}{2}(f+gp)+\frac{1}{2}(f-gp) \end{aligned}$$

shows that f fails to be extreme for the ball. $\square$

At the same time, it is not hard to produce a set $\Lambda \subset {\mathbb Z}$ with $\sup \Lambda =\infty$ and $\inf \Lambda =-\infty$ for which $\text { ball}(L^\infty _\Lambda )$ has a much richer supply of extreme points. To this end, we first introduce a bit of terminology. Following [16], we say that a set $\Lambda (\subset {\mathbb Z})$ is a $\mathcal D$-set if it has the following property: whenever $\mu \in M({\mathbb T})$ is a measure with $\text {spec}\,\mu \subset \Lambda$ whose total variation $|\mu |$ assigns zero mass to a set of positive m-measure (length) on ${\mathbb T}$, we have $\mu =0$.

As a classical example of a $\mathcal D$-set, we mention ${\mathbb Z}_+$; indeed, an $H^1$ function that vanishes on a set $\mathcal E\subset {\mathbb T}$ with $m(\mathcal E)>0$ must be null. For more sophisticated examples, we refer the reader to [16, Part One, Chapter 1]. In particular, it is shown there that if $E=\{-n^k:\,k\in {\mathbb N}\}$ with an integer $n\ge 2$, then $E\cup {\mathbb Z}_+$ is a $\mathcal D$-set.

Proposition 2.3

Let $\Lambda$ be a $\mathcal D$-set. Suppose further that $f\in L^\infty _\Lambda$ is a function with $\Vert f\Vert _\infty =1$ for which

$$\begin{aligned} m\left( \{\zeta \in {\mathbb T}:\,|f(\zeta )|=1\}\right) >0. \end{aligned}$$

(9)

Then f is an extreme point of $\text {ball}(L^\infty _\Lambda )$.

Proof

We want to check that the only function $g\in L^\infty _\Lambda$ satisfying

$$\begin{aligned} \Vert f+g\Vert _\infty \le 1\quad \text {and}\quad \Vert f-g\Vert _\infty \le 1 \end{aligned}$$

(10)

is $g\equiv 0$. Since

$$\begin{aligned} |f|^2+|g|^2=\frac{1}{2}\left( |f+g|^2+|f-g|^2\right) , \end{aligned}$$

it follows from (10) that $|g|^2\le 1-|f|^2$ a.e. on ${\mathbb T}$. Consequently, $g=0$ a.e. on

$$\begin{aligned} \mathcal E_f:=\{\zeta \in {\mathbb T}:\,|f(\zeta )|=1\}, \end{aligned}$$

while (9) tells us that $m(\mathcal E_f)>0$. The desired conclusion that $g\equiv 0$ is now ensured by the hypothesis that $\Lambda$ is a $\mathcal D$-set. (To see why, identify g with the measure $\mu _g\in M({\mathbb T})$ given by $d\mu _g=g\,dm$. Note also that

$$\begin{aligned} \text {spec}\,\mu _g=\text {spec}\,g\subset \Lambda \end{aligned}$$

and use the identity $|\mu _g|(\mathcal E_f)=\int _{\mathcal E_f}|g|\,dm=0$ to deduce that $\mu _g$, and hence g, is null.) We are done. $\square$

We mention in passing that, by a theorem of Amar and Lederer (see [1]), the unit-norm $H^\infty$ functions that obey (9) are precisely the exposed points of $\text {ball}(H^\infty )$. (Recall that, for a Banach space X, a point x in $\text {ball}(X)$ is said to be exposed for the ball if there exists a functional $\phi \in X^*$ of norm 1 such that the set $\{y\in \text {ball}(X):\phi (y)=1\}$ equals $\{x\}$. It is well known, and easily shown, that every exposed point is extreme.) The following question might be of interest in this connection.

Question 4

Does there exist a set $\Lambda \subset {\mathbb Z}$ such that the extreme points of $\text {ball}(L^\infty _\Lambda )$ are characterized, among the unit-norm functions $f\in L^\infty _\Lambda$, by condition (9)?

From (9), we now turn to the weaker condition (1) which characterizes the extreme points of $\text {ball}(H^\infty )$. This time, we ask whether the criterion remains unchanged for suitably perturbed $H^\infty$-spaces of the form $L^\infty _\Lambda$, provided that $\Lambda$ is “not too different” from ${\mathbb Z}_+$.

Question 5

For which sets $F\subset {\mathbb Z}_+$ is it true that (1) characterizes the extreme points f of $\text {ball}(H^\infty ({\mathbb Z}_+\setminus F))$? Also, for which sets $E\subset {\mathbb Z}_-$ does (1) characterize the extreme points f of $\text {ball}(L^\infty _\Lambda )$, where $\Lambda =E\cup {\mathbb Z}_+$?

The function f to be tested is, of course, always assumed to be a unit-norm element of the space in question. Now, if $\#F<\infty$, then the corresponding extreme point criterion is indeed given by (1) (see [12, Theorem 2.1]), and a similar fact is true if $\#E<\infty$. The same criterion should apply when F (resp., E) is appropriately sparse in ${\mathbb Z}_+$ (resp., ${\mathbb Z}_-$), and we would like to see a reasonably sharp sparseness condition that ensures this.

We note, however, that taking F to be the set of odd positive integers, we get ${\mathbb Z}_+\setminus F=2{\mathbb Z}_+$ and the extreme points f of $\text {ball}(H^\infty (2{\mathbb Z}_+))$ are again described by (1) (see [12] for a more detailed discussion of this example). Thus, F need not be any thinner than ${\mathbb Z}_+\setminus F$ in this situation.

Going back to our description of the extreme points of $\text {ball}(H^1(\Lambda ))$ and $\text {ball}(H^\infty (\Lambda ))$, as obtained previously in the cases (2) and (3), we now want to extend these results in yet another direction.

Question 6

What happens to the results just mentioned, as well as to their $L^p_\Lambda$ versions, in higher dimensions (say, on ${\mathbb T}^d$ in place of ${\mathbb T}$)? Also, what happens when passing from ${\mathbb T}$ to ${\mathbb R}$ (or ${\mathbb R}^d$)?

Of course, the lacunary Hardy spaces $H^p(\Lambda )$ (resp., the $L^p_\Lambda$ spaces) on the torus ${\mathbb T}^d$ should be defined appropriately in terms of a given set of multi-indices $\Lambda \subset {\mathbb Z}^d_+$ (resp., $\Lambda \subset {\mathbb Z}^d$). In particular, the analogue of (3) should now read $\#({\mathbb Z}^d_+\setminus \Lambda )<\infty$.

Moving to the real line, we fix a closed set $\Lambda \subset {\mathbb R}$ and define $L^p_\Lambda =L^p_\Lambda ({\mathbb R})$ with $p=1,\infty$ as the space of all functions $f\in L^p({\mathbb R})$ whose Fourier transform $\widehat{f}$ vanishes on ${\mathbb R}\setminus \Lambda$ (when $p=\infty$, we interpret $\widehat{f}$ in the sense of distributions). Now, as a natural counterpart of (2), we may impose the condition that $\Lambda$ be a compact set of positive length; the corresponding Paley–Wiener type spaces $L^p_\Lambda$ are actually of special interest. In the simplest case where $\Lambda$ is an interval, the extreme (and exposed) points of $\text {ball}(L^1_\Lambda ({\mathbb R}))$ were characterized in [6]. A similar study of the “second simplest” case, where $\Lambda$ is made up of two disjoint intervals, was recently carried out in [18] (also in the $L^1$ setting), and little—if anything—is known for more general Paley–Wiener spaces of the $L^1_\Lambda$ type.

When $\Lambda$ is contained in ${\mathbb R}_+:=[0,\infty )$, we call $L^p_\Lambda ({\mathbb R})$ a lacunary Hardy space on ${\mathbb R}$ and we denote it by $H^p_{\mathbb R}(\Lambda )$. The usual Hardy spaces on ${\mathbb R}$ are thus $H^p_{\mathbb R}:=H^p_{\mathbb R}({\mathbb R}_+)$; the elements of $H^p_{\mathbb R}$ are precisely the functions in $L^p({\mathbb R})$ whose Poisson integral extension to the upper half-plane is holomorphic there. We now mention a simple situation where the extreme points of $\text {ball}(H^\infty _{\mathbb R}(\Lambda ))$ are easy to describe. Namely, this happens when ${\mathbb R}_+\setminus \Lambda$ is a bounded set, a condition that can be viewed as an analogue of (3). The next result provides a counterpart to [12, Theorem 2.1], where the disk version was treated.

Proposition 2.4

Suppose $\Lambda$ is a closed subset of ${\mathbb R}_+$ such that ${\mathbb R}_+\setminus \Lambda$ is bounded. Assume also that $f\in H^\infty _{\mathbb R}(\Lambda )$ and $\Vert f\Vert _\infty =1$. Then f is an extreme point of $\text {ball}(H^\infty _{\mathbb R}(\Lambda ))$ if and only if

$$\begin{aligned} \int _{\mathbb R}\frac{\log (1-|f(t)|)}{1+t^2}\,dt=-\infty . \end{aligned}$$

(11)

Proof

The “if” part follows from the inclusion $H^\infty _{\mathbb R}(\Lambda )\subset H^\infty _{\mathbb R}$, coupled with the fact that the extreme points of $\text {ball}(H^\infty _{\mathbb R})$ are characterized by (11).

To prove the “only if” part, assume that (11) fails and let $G(\in H^\infty )$ be the outer function with modulus $1-|f|$ on ${\mathbb R}$. Observe further that, for a suitably large number $A>0$, the function

$$\begin{aligned} g(x)=g_A(x):=e^{iAx}G(x),\qquad x\in {\mathbb R}, \end{aligned}$$

will be in $H^\infty _{\mathbb R}(\Lambda )$; indeed, the spectrum of g is contained in $[A,\infty )$. Since

$$\begin{aligned} |f\pm g|\le |f|+|g|=|f|+|G|=1 \end{aligned}$$

a.e. on ${\mathbb R}$, the identity

$$\begin{aligned} f=\frac{1}{2}(f+g)+\frac{1}{2}(f-g) \end{aligned}$$

shows that f is not an extreme point of $\text {ball}(H^\infty _{\mathbb R}(\Lambda ))$. $\square$

Our last question deals with a different type of subspaces in $H^\infty$ (we are back to ${\mathbb T}$ now), where the structure of extreme points seems to be unclear. Given a function $\varphi$ in $L^\infty =L^\infty ({\mathbb T})$, we put

$$\begin{aligned} K_p(\varphi ):=\{f\in H^p:\,\overline{z\varphi f}\in H^p\},\qquad 1\le p\le \infty , \end{aligned}$$

so that $K_p(\varphi )$ is the kernel in $H^p$ of the Toeplitz operator with symbol $\varphi$.

Question 7

Let $\varphi \in L^\infty$ and assume that $K_\infty (\varphi )\ne \{0\}$. What are the extreme points of $\text {ball}(K_\infty (\varphi ))$?

When $\varphi =\overline{\theta }$ for an inner function $\theta$, $K_\infty (\varphi )$ becomes the model subspace $H^\infty \cap \theta \overline{z}\overline{H^\infty }$, and the problem of determining its extreme points was posed earlier in [8]. Furthermore, if $\varphi (z)=\overline{z}^{N+1}$ for some $N\in {\mathbb Z}_+$, then $K_\infty (\varphi )$ coincides with $H^\infty (\Lambda _N)$, where $\Lambda _N:=\{0,1,\dots ,N\}$, and is formed by the polynomials of degree at most N. In this last case, the extreme points are known (see [7] or [12]). On the other hand, the extreme points of $\text {ball}(K_1(\varphi ))$ admit a neat description for a general $\varphi \in L^\infty$; this can be found in [5].

We remark, in conclusion, that there are related geometric concepts—such as exposed or strongly extreme points of the unit ball—which are also worth studying in the context of lacunary $H^p$ or $L^p$ spaces, as well as in $K_p(\varphi )$, with $p=1,\infty$. In fact, even for the usual (nonlacunary) $H^1$, the structure of its exposed points is far from being understood; the case of $H^1(\Lambda )$ is touched upon in [9, 11] for the sets $\Lambda$ that obey (2) or (3). As regards strongly extreme points, we refer to [3] for the definition and a characterization of these in the classical $H^p$ setting; see also [13] for further results involving subspaces of $H^\infty$.