MATHEMATICS

V. P. GURARII

ON THE SPECTRUM OF GROWING FUNCTIONS

(Presented by Academician S. N. Bernstein on 7 IV 1958)

We consider the normed ring (L_\varphi) of functions* (f(x)) with norm

[
|f|=\int_{-\infty}^{\infty}|f(x)|\varphi(x)\,dx,
]

where (\varphi(x)) satisfies the conditions**: 1) (\varphi(x)\geqslant 1);

2) (\varphi(x+y)\leqslant \varphi(x)\varphi(y)); 3)

[
\int_1^\infty \frac{\psi(x)}{x}\,dx<\infty
\quad
\left(\psi(x)=\sup_{|t|\geq x\geq 1}\frac{\ln\varphi(t)}{t}\right).
]

Denote by (M_\varphi) the space conjugate to (L_\varphi); it consists of the functions (g(x)) for which

[
\operatorname{vrai\,max}\frac{|g(x)|}{\varphi(x)}<\infty.
]

A point (\lambda) of the real axis will be called a point of the spectrum (\Lambda_g^{\mathrm{I}}) of the function (g(x)\in M_\varphi), if at this point the Fourier transforms of all functions (f(x)\in L_\varphi) satisfying the equality

[
\int_{-\infty}^{\infty} f(x-t)g(x)\,dx=0
\quad (-\infty<t<\infty)
\tag{1}
]

vanish.

Following Beurling ((^2)), a point (\lambda) of the real axis will be called a point of the spectrum (\Lambda_g^{\mathrm{II}}) of the function (g(x)\in M_\varphi), if for all (\varepsilon>0) the inequality

[
\lim_{y\to+0}\int_{\lambda-\varepsilon}^{\lambda+\varepsilon}|U_g(x,y)|\,dx>0
]

holds, where

[
U_g(x,y)=\int_{-\infty}^{\infty} g(t)e^{-y|t|}e^{itx}\,dt
]

is the harmonic Fourier transform of the function (g(x)) in the half-plane (y>0). The function (U_g(x,y)) exists for (y>0), since from 2) and 3) it follows that (\psi(x)\to 0) as (x\to\infty). If (\lambda) does not belong to the spectrum (\Lambda_g^{\mathrm{II}}), then the functions

[
G^{+}(z)=\int_0^\infty g(t)e^{itz}\,dt,\qquad \operatorname{Im} z>0;
\tag{2}
]

[
G^{-}(z)=-\int_{-\infty}^{0} g(t)e^{itz}\,dt,\qquad \operatorname{Im} z<0,
\tag{2′}
]

* For a similar ring, see ((^1)).

** Condition 3) may be replaced by the following: 3)

[
\int_{-\infty}^{\infty}\frac{\ln\varphi^(x)}{1+x^2}\,dx<\infty,
\quad
\text{where }
\varphi^(x)=\sup_{|t|\leq x}\varphi(t).
]

are analytic continuations of one another through the interval ((\lambda-\varepsilon,\lambda+\varepsilon)). In this paper we investigate the question of the relation between the spectra (\Lambda_g^{\mathrm I}) and (\Lambda_g^{\mathrm{II}}).

Theorem 1. Let (\varphi(x)) satisfy conditions 1), 2), 3) and let (g(x)\in M_\varphi); then (\Lambda_g^{\mathrm I}=\Lambda_g^{\mathrm{II}}).

Proof. First we prove that (\Lambda_g^{\mathrm{II}}\supset \Lambda_g^{\mathrm I}). Indeed, let (\lambda) not be a point of the spectrum (\Lambda_g^{\mathrm{II}}), i.e., let the functions (G^+(z)) and (G^-(z)) admit analytic continuation through some segment ([\lambda-\varepsilon,\lambda+\varepsilon]). From 3) there follows the existence of a function (f_1(t)\in L_\varphi) whose Fourier transform (F_1(x)) is equal to zero outside the interval ((\lambda-\varepsilon,\lambda+\varepsilon)) and (F_1(\lambda)\ne 0). Parseval’s equality gives

[
\int_0^\infty f(x-t)g(x)e^{-yx}\,dx
=
\int_{\lambda-\varepsilon}^{\lambda+\varepsilon}
F_1(x)G^+(x+iy)e^{itx}\,dx,
]

[
-\int_{-\infty}^0 f(x-t)g(x)e^{yx}\,dx
=
\int_{\lambda-\varepsilon}^{\lambda+\varepsilon}
F_1(x)G^-(x-iy)e^{itx}\,dx
\qquad (f(x)=f_1(-x)).
]

As (y\to +0) the limits of the right-hand sides are equal; consequently,

[
\int_0^\infty f(x-t)g(x)\,dx
=
-\int_{-\infty}^0 f(x-t)g(x)\,dx,
]

whence the validity of equality (1) follows and, therefore, (\lambda\notin\Lambda_g^{\mathrm I}).

The proof that (\Lambda_g^{\mathrm{II}}\subset \Lambda_g^{\mathrm I}) is based on two lemmas.

Lemma 1. If the function (\varphi(x)) satisfies conditions 1)—3), then for the function

[
M(y)=\int_1^\infty \varphi(x)e^{-2xy}\,dx
]

the condition

[
\int_0^k \ln\ln M(y)\,dy<\infty,
\qquad
\text{where }
k=\sup_{1\le x<\infty}\frac{\ln\varphi(x)}{x}.
]

is satisfied.

Proof. Without loss of generality, we shall assume the function (\psi(x)) to be continuous. Denote by (\alpha(y)) the function inverse to (\psi(x)). It is clear that (\alpha(y)) decreases monotonically, and (\alpha(y)\to\infty) as (y\to 0). We shall show that

[
\int_0^k \ln\alpha(y)\,dy<\infty .
\tag{3}
]

Indeed, after integration by parts we shall have:

[
\int_\varepsilon^k \ln\alpha(y)\,dy
\le
\int_k^\varepsilon \frac{y}{\alpha(y)}\,d\alpha(y)
+
k\ln\alpha(k)
\le
\int_1^\infty \frac{\psi(x)}{x}\,dx
+
k\ln\alpha(k)
<\infty
]

for any (\varepsilon>0). The function (M(y)) can be estimated in the following way:

[
M(y)
\le
\int_1^{\alpha(y)} e^{x\psi(x)}e^{-2xy}\,dx
+
\int_{\alpha(y)}^\infty e^{x\psi(x)}e^{-2xy}\,dx
\le
]

[
\le
\int_1^{\alpha(y)} e^{kx}\,dx
+
\int_0^\infty e^{-xy}\,dx
\le
\frac{1}{k}e^{k\alpha(y)}+\frac{1}{y}.
]

The last inequality, together with (3), proves the lemma.

Lemma 2 (Levinson*). Let (M(y)) be a positive monotonically decreasing function on the interval ((0,1)), with (M(y)\to\infty) as (y\to 0). Let (f(z)) be a function analytic in the rectangle (|x|\leq a,\ |y|\leq b), and let, in this rectangle, (|f(z)|<M(|y|)).

If

[
\int_0^1 \ln \ln M(y)\,dy<\infty,
]

then there exists a constant (M), depending only on (M(y)) and (\delta>0), such that in the rectangle (|x|\leq a-\delta,\ |y|\leq b)

[
|f(x+iy)|<M .
]

We shall now prove that (\Lambda_g^{\mathrm I}\supset \Lambda_g^{\mathrm{II}}). Suppose that relation (1) holds and that the Fourier transform of the function (f(x)) is different from zero on the interval ((a,b)). It is necessary to show that the functions (G^{+}(z)) and (G^{-}(z)), defined by the equalities (2) and ((2')), are analytic continuations of one another through the interval ((a,b)). To this end we construct a family of functions (\mu_\delta(x)\in L_\varphi) satisfying the conditions: (1^\circ). (|\mu_\delta(x)|\leq 1). (2^\circ). (\lim_{\delta\to 0}\mu_\delta(x)=1). (3^\circ). The Fourier transform of the function (\mu_\delta(x)) vanishes outside the interval ((-\delta,\delta)).

If (\lambda\in(a,b)) and (\delta) is so small that the interval ((\lambda-\delta,\lambda+\delta)\subset(a,b)), then from Wiener’s generalized theorem ((^1)) and from 3) it follows that the equation

[
f*h=\mu_\delta(t)e^{it\lambda}
]

has a solution (h(t)\in L_\varphi). Multiplying (1) by (h(t)) and integrating with respect to (t) from (-\infty) to (\infty), after changing the order of integration we obtain:

[
\int_{-\infty}^{\infty}\mu_\delta(t)g(t)e^{it\lambda}\,dt=0
]

or

[
\int_{0}^{\infty}\mu_\delta(t)g(t)e^{it\lambda}\,dt
=
-\int_{-\infty}^{0}\mu_\delta(t)g(t)e^{it\lambda}\,dt
\qquad (a+\delta<\lambda<b-\delta).
\tag{4}
]

The functions

[
G_\delta^{+}(z)=\int_{0}^{\infty}\mu_\delta(t)g(t)e^{itz}\,dt,
\qquad \operatorname{Im} z\geq 0;
]

[
G_\delta^{-}(z)=-\int_{-\infty}^{0}\mu_\delta(t)g(t)e^{itz}\,dt,
\qquad \operatorname{Im} z\leq 0,
\tag{5}
]

are analytic and bounded in their domains. Equality (4) means that (G_\delta^{+}(\lambda)=G_\delta^{-}(\lambda)), (a+\delta<\lambda<b-\delta), i.e. that the functions (G_\delta^{+}(z)) and (G_\delta^{-}(z)) form a single analytic function (G_\delta(z)) in the plane cut along the rays (-\infty<x<a+\delta,\ b-\delta<x<\infty).

We shall prove that the family (G_\delta(z)), as (\delta\to 0), converges in the interval (a<a'<x<b'<b). To this end note that in the rectangle (a+\delta<x<b-\delta,\ |y|\leq d) the function (G_\delta(z)) is analytic. We show that in the inner rectangle (D:\ a'<x<b',\ |y|\leq d), where ((a',b')\subset(a,b)), the estimate

[
|G_\delta(z)|<M
\quad \text{for all } \delta<\min(b-b',\,a'-a).
\tag{6}
]

* See ((^3)), p. 127. Beurling ((^2)) calls this lemma Sjöberg’s lemma.

From the integral representation (5) of the function (G_\delta(z)) we obtain, for (y>0),

[
|G_\delta(x+iy)|<K_g\int_0^\infty \varphi(t)e^{-yt}\,dt\leq K_g M\left(\frac y2\right)+C.
]

An analogous inequality is obtained for (y<0).
By Lemma 1, (M_1(y)=K_g M(y/2)+C) satisfies the condition

[
\int_0^{2k}\ln\ln M_1(y)\,dy<\infty .
]

This condition ensures the possibility of applying Lemma 2 and, thus, inequality (6) is satisfied. Since in the upper and lower half-planes (G_\delta^+(z)) and (G_\delta^-(z)) converge respectively to (G^+(z)) and to (G^-(z)), it follows by the well-known Stieltjes theorem that (G_\delta(z)) converges in the rectangle (D) to some function (G(z)), and

[
G(z)=G^+(z),\quad \operatorname{Im} z<0;\qquad
G(z)=G^-(z),\quad \operatorname{Im} z<0 .
]

Theorem 1 is proved. From it the following theorem is obtained.

Theorem 2*. If (\varphi(x)) satisfies the conditions of Theorem 1, (f(x)\in L_\varphi), (g(x)\in M_\varphi), and if the Fourier transform of the function (f(x)) does not vanish outside the interval ((-h,h)), then from equality (1) it follows that (g(x)) coincides with a certain entire function of finite degree everywhere on the real axis except, possibly, on a set of measure zero. Moreover,

[
|g(z)|<A\varphi(x)\exp h|y|.
\tag{7}
]

Proof. The Laplace transform (G(z)) of the function (g(x)) is, by Theorem 1, an analytic function outside the interval ([-h',h]), where (h'<h). Take an interval ((a,b)) not intersecting the segment ([-h',h']), and choose the rectangle (D) as was done in the proof of Theorem 1. Since

[
|G(cz)|=|G(cx+icy)|\leq M_1(cy)\leq M_1(y)
]

for any (c>1), it follows that (|G(cx)|