1 Introduction

Let \({\mathbb {D}}\) be the unit disc in \({\mathbb {R}}^2\). The Paley–Wiener space \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) is the subspace of \(L^2({\mathbb {R}}^2)\) comprised of functions f whose Fourier transforms \({\widehat{f}}\) are supported in \(\overline{{\mathbb {D}}}\). For a tempered distribution \(\varphi \), we consider the Hankel operator \({\mathbf {H}}_\varphi \) defined by the equation

$$\begin{aligned} \widehat{{\mathbf {H}}_\varphi f}(\eta ) = \int _{{\mathbb {D}}} {\widehat{f}}(\xi ) {\widehat{\varphi }}(\xi +\eta )\,\mathrm{d}\xi , \qquad \eta \in {\mathbb {D}}, \end{aligned}$$
(1)

on the dense subset of \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) comprised of functions f such that \({\widehat{f}}\) is smooth and compactly supported in \({\mathbb {D}}\).

We are interested in the characterization of the symbols \(\varphi \) such that \({\mathbf {H}}_\varphi \) extends by continuity to a bounded operator on \({{\,\mathrm{PW}\,}}({\mathbb {D}})\). If \(\varphi \) is in \(L^\infty ({\mathbb {R}}^2)\), then clearly

$$\begin{aligned} \Vert {\mathbf {H}}_\varphi f\Vert _2 \le \Vert f\Vert _2 \Vert \varphi \Vert _\infty . \end{aligned}$$
(2)

Since \(\xi +\eta \) is in \(2{\mathbb {D}}\) whenever \(\xi \) and \(\eta \) are in \({\mathbb {D}}\), \({\mathbf {H}}_\varphi = {\mathbf {H}}_\psi \) for any \(\psi \) such that the restrictions of \({\widehat{\psi }}\) and \({\widehat{\varphi }}\) to \(2{\mathbb {D}}\) coincide (as distributions in \(2{\mathbb {D}}\)). We thus find that

$$\begin{aligned} \Vert {\mathbf {H}}_\varphi \Vert \le \inf \big \{\Vert \psi \Vert _\infty \,:\, {\widehat{\psi }}\,\big |_{2{\mathbb {D}}} = {\widehat{\varphi }}\,\big |_{2{\mathbb {D}}}\big \}. \end{aligned}$$
(3)

We say that the Hankel operator \({\mathbf {H}}_\varphi \) has a bounded symbol if the quantity on the right hand side of (3) is finite. We have just demonstrated that if \({\mathbf {H}}_\varphi \) has a bounded symbol, then \({\mathbf {H}}_\varphi \) is bounded. We wish to explore the converse.

Question

Does every bounded Hankel operator on \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) have a bounded symbol?

In the classical one-dimensional setting, where the role of \({\mathbb {D}}\) is played by the half-line \({\mathbb {R}}_+ = [0,\infty )\), Nehari [6] gave a positive answer to this question. We therefore refer to affirmative answers to analogous questions as Nehari theorems. Our question for \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) was first raised implicitly by Rochberg [9, Sec. 7], after he had proved that Nehari’s theorem holds for the Paley–Wiener space \({{\,\mathrm{PW}\,}}(I)\) of a finite interval \(I \subseteq {\mathbb {R}}\).

It was conditionallyFootnote 1 shown in [1] that the Nehari theorem holds for the Paley–Wiener space \({{\,\mathrm{PW}\,}}({\mathbb {P}})\) of any convex polygon \({\mathbb {P}}\). However, in view of C. Fefferman’s negative resolution [3] of the disc conjecture for the Fourier multiplier of a disc, it would not be surprising to see differing results for \({{\,\mathrm{PW}\,}}({\mathbb {P}})\) and \({{\,\mathrm{PW}\,}}({\mathbb {D}})\).

The main purpose of the present note is to establish the following.

Theorem 1

There is a bounded Hankel operator on \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) which does not have a bounded symbol.

Minor modifications of our proof show that if \({\mathbb {P}}_n\) is an n-sided regular polygon, then the optimal constant in the inequality

$$\begin{aligned} \inf \big \{\Vert \psi \Vert _\infty \,:\, {\widehat{\psi }}\,\big |_{2{\mathbb {P}}_n} = {\widehat{\varphi }}\,\big |_{2{\mathbb {P}}_n}\big \} \le C_n\Vert {\mathbf {H}}_\varphi \Vert _{{{\,\mathrm{PW}\,}}({\mathbb {P}}_n)} \end{aligned}$$

satisfies \(C_n \ge c_\varepsilon n^{1/2 - \varepsilon }\) for any fixed \(\varepsilon > 0\). Here, \(c_\varepsilon >0\) denotes a constant which depends only on \(\varepsilon \). Conversely, the conditional argument of [1] yields that \(C_n \le cn\) for some absolute constant \(c > 0\). Analogous estimates for Fourier multipliers associated with polygons were considered in [2].

Finally, let us remark that Ortega-Cerdà and Seip [7] have shown that Nehari’s theorem also fails for (small) Hankel operators on the infinite-dimensional torus. However, Helson [4] proved that if the Hankel operator is in the Hilbert–Schmidt class \(S_2\), then it is induced by a bounded symbol. We are led to the following.

Question

Does every Hankel operator on \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) in \(S_2\) have a bounded symbol?

In this context, we mention that Peng [8] has characterized when \({\mathbf {H}}_\varphi \) is in the Schatten class \(S_p\), for \(1 \le p \le 2\), in terms of the membership of \(\varphi \) in certain Besov spaces adapted to \(2{\mathbb {D}}\). In particular, \({\mathbf {H}}_\varphi \) is in \(S_2\) if and only if

$$\begin{aligned} \int _{2{\mathbb {D}}} |{\widehat{\varphi }}(\xi )|^2 (2-|\xi |)^{3/2}\,\mathrm{d}\xi < \infty . \end{aligned}$$

2 Proof of Theorem 1

If the Nehari theorem were to hold for \({{\,\mathrm{PW}\,}}({\mathbb {D}})\), there would by the closed graph theorem exist an absolute constant \(C < \infty \) such that

$$\begin{aligned} \inf \big \{\Vert \psi \Vert _\infty \,:\, {\widehat{\psi }}\,\big |_{2{\mathbb {D}}} = {\widehat{\varphi }}\,\big |_{2{\mathbb {D}}}\big \} \le C\Vert {\mathbf {H}}_\varphi \Vert \end{aligned}$$
(4)

for every bounded Hankel operator on \({{\,\mathrm{PW}\,}}({\mathbb {D}})\). To prove Theorem 1, we will construct a sequence of symbols which demonstrates that no such \(C<\infty \) can exist.

We begin with an upper bound for \(\Vert {\mathbf {H}}_\varphi \Vert \). Guided by the following lemma, our plan is to construct \(\varphi \) such that \({\mathbf {H}}_\varphi \) admits an orthogonal decomposition. For a symbol \(\varphi \), define

$$\begin{aligned} D_\varphi = \big \{\eta \in {\mathbb {D}}\,:\, \xi +\eta \in {{\,\mathrm{supp}\,}}{{\widehat{\varphi }}} \,\text { for some }\, \xi \in {\mathbb {D}}\big \}. \end{aligned}$$

Lemma 2

Suppose that \(\varphi = \varphi _1 + \varphi _2\) and that \(D_{\varphi _1} \cap D_{\varphi _2} = \emptyset \). Then,

$$\begin{aligned} {\mathbf {H}}_\varphi = {\mathbf {H}}_{\varphi _1} \oplus {\mathbf {H}}_{\varphi _2}. \end{aligned}$$

Proof

Let f be any function in \({{\,\mathrm{PW}\,}}({\mathbb {D}})\) such that \({\widehat{f}}\) is smooth and compactly supported in \({\mathbb {D}}\) . Since \({\mathbf {H}}_\varphi f = {\mathbf {H}}_{\varphi _1} f + {\mathbf {H}}_{\varphi _2} f\) by linearity of the integral (1), it is sufficient to demonstrate that \({\mathbf {H}}_{\varphi _1} f \perp {\mathbf {H}}_{\varphi _2} f\). It follows directly from the definition of the Hankel operator (1) that

$$\begin{aligned} {{\,\mathrm{supp}\,}}{\widehat{{\mathbf {H}}_{\varphi _1} f}} \subseteq D_{\varphi _1} \qquad \text {and} \qquad {{\,\mathrm{supp}\,}}{\widehat{{\mathbf {H}}_{\varphi _2} f}} \subseteq D_{\varphi _2}. \end{aligned}$$

By the assumption that \(D_{\varphi _1} \cap D_{\varphi _2} = \emptyset \), we therefore conclude that

$$\begin{aligned} \langle {\mathbf {H}}_{\varphi _1} f, {\mathbf {H}}_{\varphi _2} f \rangle = \langle \widehat{{\mathbf {H}}_{\varphi _1} f}, \widehat{{\mathbf {H}}_{\varphi _2} f} \rangle = 0. \end{aligned}$$

\(\square \)

In particular, if \(D_{\varphi _1} \cap D_{\varphi _2}= \emptyset \), then

$$\begin{aligned} \Vert {\mathbf {H}}_\varphi \Vert = \max (\Vert {\mathbf {H}}_{\varphi _1}\Vert ,\Vert {\mathbf {H}}_{\varphi _2}\Vert ). \end{aligned}$$

Let us next explain the construction of \(\varphi \). Consider a radial smooth bump function \({\widehat{b}}\) which is bounded by 1, equal to 1 on \(\frac{1}{2}{\mathbb {D}}\) and compactly supported in \({\mathbb {D}}\). For a real number \(0<r < 1/2\), set \({\widehat{b}}_r(\xi ) = {\widehat{b}}(\xi /r)\). Note that

$$\begin{aligned} \Vert {\widehat{b}}_r \Vert _1 \le \pi r^2. \end{aligned}$$
(5)

For \(j=1,2,\ldots ,n\), we let \({\widehat{\varphi }}_j\) be the function obtained by translating \({\widehat{b}}_r\) by \(2-r\) units in the direction \(\theta _j = 2 \pi (j-1)/n\), as measured with respect to the positive \(\xi _1\)-axis in the \(\xi _1\xi _2\)-plane. We set

$$\begin{aligned} \varphi = \varphi _1 + \varphi _2 + \cdots + \varphi _n. \end{aligned}$$
(6)

Since \(0< r < 1/2\), it is clear that \({{\,\mathrm{supp}\,}}{{\widehat{\varphi }}} \subseteq 2{\mathbb {D}} \setminus {\mathbb {D}}\). Let \(r_0 = 1-\frac{1}{\sqrt{2}}=0.29\ldots \).

Fig. 1
figure 1

Plots of D(w) and the corresponding disc sector from the proof of Lemma 3, for \(w=1.1\), \(w=1.5\), and \(w = 1.8\)

Full size image

Lemma 3

If \(n \ge 2\) and \(r=\min (r_0,(2/n)^2)\), then

$$\begin{aligned}D_{\varphi _j} \cap D_{\varphi _k} = \emptyset \end{aligned}$$

for every \(1 \le j \ne k \le n\).

Proof

Throughout this proof, we identify \({\mathbb {R}}^2\) with \({\mathbb {C}}\). We consider first a simpler situation. For a point w in \(2 {\mathbb {D}} \setminus {\mathbb {D}}\), let

$$\begin{aligned} D(w) = \big \{\eta \in {\mathbb {D}}\,:\, \xi + \eta = w\, \text { for some }\, \xi \in {\mathbb {D}}\big \}. \end{aligned}$$

In other words, D(w) is the intersection of the discs defined by \(|\xi |<1\) and \(|w-\xi |<1\). To find the intersection of the corresponding circles, we set \(\xi = e^{i\theta }\) and let \(\theta ^{\pm }\) denote the solutions of the equation

$$\begin{aligned} 1 = |w-e^{i\theta }| \qquad \Longleftrightarrow \qquad \theta ^{\pm } = \arg {w} \pm \arccos \left( \frac{|w|}{2}\right) . \end{aligned}$$

Let \(P_0\) denote the origin, \(P_\pm \) the points \(e^{i\theta ^\pm }\), and \(P_w\) the point w. The law of cosines implies that the angle \(\angle P_0 P_\pm P_w\) is greater than or equal to \(\pi /2\) if and only if \(|w|\ge \sqrt{2}\). If this holds, then the intersection of the two discs is contained in the disc sector defined by the origin and the two points \(P_\pm \). See Fig. 1.

Suppose therefore that \(|w|\ge \sqrt{2}\) and set \(I(w) = (\theta ^-,\theta ^+)\). If \(\xi \) is in D(w), we have just seen that \(\arg {\xi }\) is in I(w). It follows that if \(w_1\) and \(w_2\) are points in \(2{\mathbb {D}} \setminus \sqrt{2}{\mathbb {D}}\), then

$$\begin{aligned} I(w_1) \cap I(w_2) = \emptyset \qquad \implies \qquad D(w_1) \cap D(w_2) = \emptyset . \end{aligned}$$
(7)

Our goal is now to estimate

$$\begin{aligned}I_{\varphi _j} = \bigcup _{w \in {{\,\mathrm{supp}\,}}{{\widehat{\varphi }}_j}} I(w). \end{aligned}$$

Since \({{\,\mathrm{supp}\,}}{{\widehat{\varphi }}_j}\) is contained in a disc with center \((2-r) e^{i\theta _j}\) and radius r, straightforward geometric arguments show that if w is in \({{\,\mathrm{supp}\,}}{{\widehat{\varphi }}_j}\), then

$$\begin{aligned} |w| \ge 2(1-r) \qquad \text {and} \qquad |\arg {w}-\theta _j| \le \arctan \left( \frac{r}{2-r}\right) . \end{aligned}$$

To ensure that \(|w|\ge \sqrt{2}\) we require that \(r \le r_0 = 1-\frac{1}{\sqrt{2}}\). Moreover, if \(\theta ^\pm \) correspond to the point w as above, then

$$\begin{aligned} |\theta ^\pm -\theta _j| \le \arccos (1-r) + \arctan \left( \frac{r}{2-r}\right) \le 2 \sqrt{r} + r \le 3 \sqrt{r}.\ \end{aligned}$$

Here, we used that \(2-r \ge 1\) and that \(\arctan {r}\le r\) for \(0\le r \le 1\). This shows that

$$\begin{aligned} I_{\varphi _j} \subseteq \left( \theta _j-3\sqrt{r},\theta _j+3\sqrt{r}\right) . \end{aligned}$$

Since \(|\theta _j-\theta _k| \ge 2 \pi /n\) for every \(1 \le j \ne k \le n\) and since \(\pi >3\), it follows that if we choose \(r = \min (r_0,(\frac{2}{n})^2)\), then we guarantee that \(I_{\varphi _j} \cap I_{\varphi _k}= \emptyset \) for every \(1 \le j \ne k \le n\). The proof is completed by appealing to (7). \(\square \)

Let \(\varphi \) be as in (6), with \(n\ge 2\) and \(r=\min (r_0,(2/n)^2)\). It then follows from Lemmas 2, 3, (2), and (5) that

$$\begin{aligned} \Vert {\mathbf {H}}_\varphi \Vert = \Vert {\mathbf {H}}_{\varphi _j}\Vert \le \Vert \varphi _j\Vert _\infty \le \Vert {\widehat{\varphi }}_j\Vert _1 = \Vert {\widehat{b}}_r \Vert _1 \le \pi r^2. \end{aligned}$$
(8)

A lower bound for the left hand side in (4) will be established through duality.

Lemma 4

Suppose that \({\widehat{f}}\) is smooth and compactly supported in \(2 {\mathbb {D}}\). Then,

$$\begin{aligned} \frac{|\langle {\widehat{f}}, {\widehat{\varphi }} \rangle |}{\Vert f\Vert _1} \le \inf \big \{\Vert \psi \Vert _\infty \,:\, {\widehat{\psi }}\,\big |_{2{\mathbb {D}}} = {\widehat{\varphi }}\,\big |_{2{\mathbb {D}}}\big \}. \end{aligned}$$

Proof

Obviously,

$$\begin{aligned} \frac{|\langle f, \psi \rangle |}{\Vert f\Vert _1} \le \Vert \psi \Vert _\infty , \end{aligned}$$

and when \({\widehat{f}}\) is supported in \(2{\mathbb {D}}\) and \({\widehat{\psi }}|_{2{\mathbb {D}}} = {\widehat{\varphi }}|_{2{\mathbb {D}}}\), we have that

$$\begin{aligned} \langle f, \psi \rangle = \langle {\widehat{f}}, {\widehat{\psi }} \rangle = \langle {\widehat{f}},{\widehat{\varphi }} \rangle . \end{aligned}$$

\(\square \)

We now need to choose a test function f adapted to the symbol \(\varphi \) of (6). It turns out that \(f=f_1+f_2+\cdots +f_n\), where \(f_j = \varphi _j\) for \(j=1,2,\ldots ,n\), will do. By our choice of \(n\ge 2\) and \(r=\min (r_0,(2/n)^2)\), it is clear that \({{\,\mathrm{supp}\,}}{{\widehat{f}}_j} \cap {{\,\mathrm{supp}\,}}{{\widehat{f}}_k}=\emptyset \) for every \(1 \le j \ne k \le n\), since the converse statement would contradict Lemma 3.

Exploiting this, we find that

$$\begin{aligned} |\langle f, \varphi \rangle | = \Vert f\Vert _2^2 = \Vert {\widehat{f}}\Vert _2^2 = n \Vert {\widehat{b}}_r\Vert _2^2 \ge \frac{\pi }{4} n r^2. \end{aligned}$$
(9)

To get an upper bound for \(\Vert f\Vert _1\), we split the integral at some \(R>0\),

$$\begin{aligned} \Vert f\Vert _1 = \int _{|x| \le R} |f(x)|\,\mathrm{{d}}x + \int _{|x| > R} |f(x)|\,\mathrm{{d}}x = I_1 + I_2. \end{aligned}$$

For the first integral, we use the Cauchy–Schwarz inequality,

$$\begin{aligned} I_1 \le \sqrt{\pi } R \bigg (\int _{|x| \le R} |f(x)|^2\,\mathrm{{d}}x\bigg )^\frac{1}{2} \le \sqrt{\pi } R \Vert f\Vert _2 = \sqrt{\pi } R \Vert {\widehat{f}}\Vert _2 \le \pi R \sqrt{n} r,\end{aligned}$$

where we again exploited that \({{\,\mathrm{supp}\,}}{{\widehat{f}}_j} \cap {{\,\mathrm{supp}\,}}{{\widehat{f}}_k} = \emptyset \) for \(1 \le j \ne k \le n\). For the second integral, we note that b is rapidly decaying, since \({\widehat{b}}\) is smooth and compactly supported. In particular, for every \(\kappa \ge 1\), there is a constant \(A_\kappa \) such that

$$\begin{aligned} \int _{|x| > \varrho } |b(x)| \,\mathrm{{d}}x \le \frac{A_\kappa }{\varrho ^{\kappa -1}}, \end{aligned}$$
(10)

holds for every \(\varrho >0\). We constructed \(\widehat{f_j}\) by translating \({\widehat{b}}_r\) by \(2-r\) units in direction \(\theta _j\), so there is a unimodular function \(g_j\) such that

$$\begin{aligned} f_j(x) = g_j(x) b_r(x) = g_j(x) r^2 b(rx). \end{aligned}$$

Thus \(|f(x)| \le n r^2 b(rx)\) and (10), with \(\varrho = Rr\), yields

$$\begin{aligned} I_2 \le n \int _{|x|> R} r^2 |b(rx)|\,\mathrm{{d}}x = n \int _{|x| > rR} |b(x)|\,\mathrm{{d}}x \le A_\kappa \frac{n}{(Rr)^{\kappa -1}}. \end{aligned}$$

Combining our estimates for \(I_1\) and \(I_2\) and choosing \(R = n^{1/(2\kappa )}/r\), we find that

$$\begin{aligned} \Vert f\Vert _1 = I_1 + I_2 \le (\pi + A_\kappa ) n^{1/2+1/(2\kappa )}. \end{aligned}$$
(11)

Inserting the estimates (9) and (11) into Lemma 4, we obtain

$$\begin{aligned} \frac{\pi r^2 n^{1/2-1/(2\kappa )}}{4(\pi + A_\kappa )} \le \inf \big \{\Vert \psi \Vert _\infty \,:\, {\widehat{\psi }}\,\big |_{2{\mathbb {D}}} = {\widehat{\varphi }}\,\big |_{2{\mathbb {D}}}\big \}. \end{aligned}$$
(12)

Final part of the proof of Theorem 1

To finish the proof of Theorem 1, we combine (8) and (12) to conclude that the constant C in (4) must satisfy

$$\begin{aligned} \frac{n^{1/2-1/(2\kappa )}}{4(\pi +A_\kappa )} \le C \end{aligned}$$

for any fixed \(\kappa \ge 1\) and every integer \(n\ge 2\). Choosing some \(\kappa >1\) and letting \(n\rightarrow \infty \), we obtain a contradiction. \(\square \)