Mate–Nevai–Totik Theorem for Krein Systems

Abstract

We prove the Cesàro boundedness of eigenfunctions of the Dirac operator on the half-line with a square-summable potential. The proof is based on the theory of Krein systems and, in particular, on the continuous version of a theorem by A. Máté, P. Nevai and V. Totik from 1991.

Appendix

1.1 Proof of Theorem F

Recall that $\Phi$ is an even function such that $|\Phi (x)(1 + x^2)|$ is bounded and $\sigma$ is a measure satisfying (1.5). We need to prove that

$$\begin{aligned} \left( \Phi _r * \sigma \right) (x)\rightarrow \sigma '(x)\int _{{\mathbb {R}}}\Phi (t)dt, \quad r\rightarrow \infty , \end{aligned}$$

(6.1)

for almost every $x\in {\mathbb {R}}$, where $\Phi _r(t) = r\Phi (rt)$. By a restriction of $\sigma$ to some interval one can assume that $\sigma$ is finite. Now to prove (6.1) one can deal with the density and the singular part of $\sigma$ individually: (6.1) for the density follows from the basic property of the approximate identity, see Theorem 8.15 in [13]; for the singular part one can adapt the proof of Lemma 5.4 in [14].

1.2 Proof of Lemma 3

We need to prove that

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{f_r(x)D(x)}{x - z}\,dx&= f_r(z)D(z), \nonumber \\ \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{f_r(x)D(x)}{x + z}\,dx&= 0. \end{aligned}$$

(6.2)

We will need the following

Lemma 7

For every point $z\in {\mathbb {C}}_+$ the following inequality holds:

$$\begin{aligned} |f_r(z)D(z)|\leqslant K_r(0,0)^{-\frac{1}{2}}(4\pi \mathop {\mathrm {Im}}\nolimits (z))^{-\frac{1}{2}}, \end{aligned}$$

in particular, for any $c > 0$ the function $f_rD$ is bounded in the half-plane $\{z:\mathop {\mathrm {Im}}\nolimits (z) > c\}$.

Proof

$$\begin{aligned}&| f_r(z)| = \left| \frac{K_r(0,z)}{K_r(0,0)}\right| = \left| \frac{\int _0^r \overline{P(s,0)}P(s,z) \,ds}{\int _0^r \left| P(s,0)\right| ^2 ds}\right| \\&\quad \leqslant \frac{\left( \int _0^r \left| P(s,0)\right| ^2 ds\right) ^{\frac{1}{2}}\left( \int _0^r \left| P(s,z)\right| ^2 ds\right) ^{\frac{1}{2}}}{\int _0^r \left| P(s,0)\right| ^2 ds} \leqslant K_r(0,0)^{-\frac{1}{2}}\left( \frac{|\Pi (z)|^2}{2\mathop {\mathrm {Im}}\nolimits (z)} \right) ^{\frac{1}{2}} \\&\quad {\mathop {=}\limits ^{(2.9)}} K_r(0,0)^{-\frac{1}{2}}\left( \frac{1}{2\mathop {\mathrm {Im}}\nolimits (z)\cdot 2\pi |D(z)|^2} \right) ^{\frac{1}{2}}\\&\quad = K_r(0,0)^{-\frac{1}{2}}\left( 4\pi \mathop {\mathrm {Im}}\nolimits (z) \right) ^{-\frac{1}{2}} |D(z)|^{-1}. \end{aligned}$$

The second inequality on the second line follows from Corollary 3. Multiplying the last inequality by |D(z)|, we obtain the required. $\square$

Remark

If the case when $f_r D\in H^p({\mathbb {C}}_+)$ for some $p\geqslant 1$, the conclusion of Lemma 3 is known (see p. 116 in [15]). It is not our case but function $f_rD$ satisfies a “boundedness” condition from Lemma 7.

Proof of Lemma 3

We have already mentioned that the function

$$\begin{aligned} \frac{\sqrt{2\pi }^{-1}\Pi ^{-1}}{\lambda + i} = \frac{D}{\lambda + i} \end{aligned}$$

is in the Hardy space $H^2({\mathbb {C}}_+)$. Hence, for a fixed $z\in {\mathbb {C}}_+$ the function $\frac{D}{\lambda + z}$ also belongs to $H^2({\mathbb {C}}_+)$. The function $f_r$ belongs to the Paley-Wiener space and consequently to $H^2({\mathbb {C}}_+)$. Therefore both integrals in the lemma are absolutely convergent and define analytic functions in ${\mathbb {C}}_+$. Moreover, $f_r\frac{D}{\lambda + z}$ is a multiplication of two functions from $H^2({\mathbb {C}}_+)$ and consequently it belongs to $H^1({\mathbb {C}}_+)$. Integral of an $H^1({\mathbb {C}}_+)$ function over the real line equals 0 (see p. 122 in [15] or Lemma 3.7 in [14]). Thus, the second equality in lemma is proved.

By Lemma 7, the function $G_{\varepsilon }(\lambda ) = f_r(\lambda + \varepsilon i)D(\lambda + \varepsilon i)$ is bounded in the upper half-plane ${\mathbb {C}}_+$. Proceeding as above, we deduce that for all $z\in {\mathbb {C}}_+$

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{G_{\varepsilon }(x)}{x + z}\,dx&= 0. \end{aligned}$$

Next, the following chain of equalities holds:

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{f_r(x + \varepsilon i)D(x + \varepsilon i)}{x - z}\,dx&= \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{G_{\varepsilon }(x)}{x - z}\,dx\nonumber \\ =\frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{G_{\varepsilon }(x)}{x - z}\,dx&- \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{G_{\varepsilon }(x)}{x + z}\,dx = \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\,dx.\nonumber \\ \end{aligned}$$

(6.3)

The latter integral can be calculated by the calculus of residues. Take a large positive number R and consider a contour $C_R$ consisting of the segment $[-R, R]$ and the half-circle $\Gamma _R$ connecting points R and $-R$ in the upper half-plane.

$$\begin{aligned} G_{\varepsilon }(z)&= \text {Res}_z\left( \frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\right) = \frac{1}{2\pi i}\int \limits _{C_R}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}dx\nonumber \\&=\frac{1}{2\pi i}\int \limits _{[-R, R]}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}dx + \frac{1}{2\pi i}\int \limits _{\Gamma _R}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}dx. \end{aligned}$$

(6.4)

The first term in the latter sum has a limit as $R\rightarrow \infty$.

$$\begin{aligned} \lim _{R\rightarrow \infty } \frac{1}{2\pi i}\int \limits _{[-R, R]}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\,dx = \frac{1}{2\pi i}\int \limits _{{\mathbb {R}}}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\,dx. \end{aligned}$$

Since the function $G_{\varepsilon }$ is bounded, the second term tends to zero as $R\rightarrow \infty$:

$$\begin{aligned} \left| \frac{1}{2\pi i}\int _{\Gamma _R}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\,dx\right|&\leqslant \frac{1}{2\pi }\int _0^{\pi }\left| \frac{2zG_{\varepsilon }(Re^{i\theta })}{R^2e^{2i\theta } - z^2}\right| R\, d\theta \\&\leqslant \frac{R|z|}{R^2 - |z|^2}\sup \nolimits _{\theta }|G_{\varepsilon }(Re^{i\theta })| =O\left( \frac{1}{R}\right) . \end{aligned}$$

We can therefore pass to the limit $R\rightarrow \infty$ in (6.4) and obtain

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{2zG_{\varepsilon }(x)}{x^2 -z^2}\,dx = G_{\varepsilon }(z). \end{aligned}$$

Substituting this into (6.3) we get

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{f_r(x + \varepsilon i)D(x + \varepsilon i )}{x - z}\,dx&= G_{\varepsilon }(z)= f_r(z + i\varepsilon )D(z + i\varepsilon ). \end{aligned}$$

To prove equality (6.2) it only remains to pass to the limit as $\varepsilon$ tends to 0. The right part of the equality tends to $f_r(z)D(z)$ so we need to evaluate how the left side differs from the integral in the lemma.

$$\begin{aligned}&\left| \int _{{\mathbb {R}}}\frac{f_r(x + \varepsilon i)D(x + \varepsilon i )}{x - z}\,dx - \int _{{\mathbb {R}}}\frac{f_r(x )D(x )}{x - z}\,dx\right| \\&\quad = \left| \int _{{\mathbb {R}}}f_r(x +i\varepsilon )\frac{D(x + \varepsilon i ) - D(x)}{x - z}\,dx + \int _{{\mathbb {R}}}D(x)\frac{f_r(x +i\varepsilon )- f_r(x)}{x - z}\,dx\right| \\&\quad \leqslant \int _{{\mathbb {R}}}\left| f_r(x +i\varepsilon )\frac{D(x + \varepsilon i ) - D(x)}{x - z}\right| \,dx + \int _{{\mathbb {R}}}\left| D(x)\frac{f_r(x +i\varepsilon )- f_r(x)}{x - z}\right| \,dx \\&\quad \leqslant \left\| f_r(x + \varepsilon i) \right\| _{2}\cdot \left\| \frac{x + i}{x - z} \right\| _{\infty }\left\| \frac{D(x + \varepsilon i) - D(x)}{x + i} \right\| _{2} \\&\qquad +\left\| f_r(x + i\varepsilon ) - f_r(x) \right\| _{2}\cdot \left\| \frac{x + i}{x - z} \right\| _{\infty }\left\| \frac{D(x)}{x + i} \right\| _{2} \\&\quad \leqslant C\left( \left\| \frac{D(x + \varepsilon i) - D(x)}{x + i} \right\| _{2} + \left\| f_r(x + i\varepsilon ) - f_r(x) \right\| _{2}\right) \\&\quad \leqslant C\left( \left\| \frac{D(x+\varepsilon i )}{x + i}- \frac{D(x+\varepsilon i)}{x+\varepsilon i+i} \right\| _{2} + \left\| \frac{D(x+\varepsilon i)}{x+\varepsilon i +i} - \frac{D(x)}{x+i} \right\| _{2} \right. \\&\qquad \left. + \left\| f_r(x+i\varepsilon ) - f_r(x) \right\| _{2}\right) . \end{aligned}$$

The first term can be bounded in the following way:

$$\begin{aligned} \left\| \frac{D(x + \varepsilon i )}{x + i}- \frac{D(x + \varepsilon i)}{x + \varepsilon i + i} \right\| _{2} = \varepsilon \left\| \frac{D(x + \varepsilon i)}{(x + \varepsilon i + i)(x + i)} \right\| _{2} \leqslant \varepsilon \left\| \frac{D(x)}{x + i} \right\| _{2}\rightarrow 0. \end{aligned}$$

From a general property of functions in $H^2({\mathbb {C}}_+)$ (see Theorem 3.1 in [14] or Section 19.2 in [17]), it follows that the second and the third terms in the bracket tend to zero as $\varepsilon \rightarrow 0$. $\square$

1.3 Proof of (2.18)

Proof

We need to prove that for every $\delta > 0$ the following holds:

$$\begin{aligned} \int \limits _{-\infty }^{\infty }\left| H_r(x - \delta i)\right| ^2\,dx < 2\pi r, \end{aligned}$$

where $H_r(x) = \frac{e^{-irx} -1 }{x}$. Recall the formula of the Fourier transform of $\frac{1}{x^2 + 1}$:

$$\begin{aligned} {\mathcal {F}}\left( \frac{1}{x^2 + 1}\right) (w) = \int \limits _{-\infty }^{\infty }\frac{e^{iwx}}{x^2 + 1}\,dx = \pi e^{-|w|}. \end{aligned}$$

With that in mind, the integral can be easily estimated, we have

$$\begin{aligned} \int \limits _{-\infty }^{\infty }\left| H_r(x - \delta i)\right| ^2\,dx= & {} \int \limits _{-\infty }^{\infty }\frac{\left| e^{-r\delta }e^{irx} - 1\right| ^2}{x^2 + \delta ^2}\,dx \\= & {} \int \limits _{-\infty }^{\infty }\frac{e^{-2r\delta } + 1 - e^{-r\delta }(e^{irx} + e^{-irx})}{x^2 + \delta ^2} \,dx \\= & {} \frac{1}{\delta }\left[ (e^{-2r\delta } + 1)\int \limits _{-\infty }^{\infty }\frac{\,dy}{y^2 + 1} - e^{-r\delta }\int \limits _{-\infty }^{\infty }\frac{e^{ir\delta y} + e^{-ir\delta y}}{y^2 + 1}\,dy\right] \\= & {} \frac{1}{\delta }\left[ (e^{-2r\delta } + 1)\pi - 2e^{-r\delta }\pi e^{-r\delta }\right] = \pi \frac{1 - e^{-2r\delta }}{\delta } < 2\pi r. \end{aligned}$$

$\square$