Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 22, 1997, 339–348

ENTIRE FUNCTIONS OF ORDER ≤ 1, WITH BOUNDS ON BOTH AXES

Yngve Domar

Uppsala University, Department of Mathematics Box 480, S-75106 Uppsala, Sweden; yngve@math.uu.se

Abstract. Results are proved on the existence of a not identically vanishing entire function f, of order ≤ 1, and satisfying log |f(x)| ≤ k(x), log |f(ix)| ≤ a|x|, x ∈ R, where k is a given odd function and a a positive constant. Existence is proved if k ∈ C^{2}(R) with |xk′′(x)| ≤ 2aπ^{−1} , x ∈ R, or, for every a, if k is convex on R^{+} , with k′ bounded. The second result is used to produce non-trivial translation invariant subspaces of the corresponding weighted l^{p}(Z).

0. Introduction

Throughout this paper a denotes a positive number and k a continuous realvalued function on R.

Question. Is there a not identically vanishing entire function f, of order ≤ 1, such that

(0.1)  log |f(x)| ≤ k(x),  x ∈ R,
(0.2)  log |f(iy)| ≤ a|y|,  y ∈ R.

A famous result of A. Beurling and P. Malliavin [1, Theorem I] gives the following:

Theorem (Beurling–Malliavin). The answer is yes for every a if k, outside some compact interval, is absolutely continuous with bounded derivative, and

∫_{−∞}^{∞} \frac{min(0, k(x))}{1 + x^{2}} dx > −∞.

This paper contains some simple complementary results in the case when k is an odd function. We will prove the following theorems.

Theorem 1. Let k be odd on R, differentiable outside some compact interval I, with absolutely continuous derivative. Then the answer is yes if

(0.3)  |xk′′(x)| ≤ \frac{2a}{π},  x ∉ I.

Theorem 2. Let k be odd on R, and convex (concave) with bounded derivative for sufficiently large positive values of the variable. Then the answer is yes, for every a.

1991 Mathematics Subject Classification: Primary 30D15; Secondary 47A15.

These three theorems can be combined to more general results, since positive answers for the pairs (a_{1}, k_{1}) and (a_{2}, k_{2}) give positive answers for the pairs (a_{1} + a_{2}, k_{1} + k_{2}) and (a_{1} + a_{2}, k_{1} − k_{2}).

We will give some comments to Theorem 1. Take a = \frac{1}{2}π and

k(x) = x log |x|, x ≠ 0, k(0) = 0.

Then

(0.4)  xk′′(x) = 1,  x ≠ 0.

By Theorem 1 the answer in this case is yes. This can also be seen directly by choosing

f(z)^{−1} = z exp(z)Γ(z).

But if b > 1 and

k(x) ≥ bx log |x|, x > 0,

then the answer is no, for a = \frac{1}{2}π. For if f satisfies the conditions in the question, then

g(z) = f(z) exp(−bz log(−z)),

with the principal branch of log, is analytic and of order at most 1 in the left halfplane. It is bounded on the negative real axis and vanishes of exponential order at ∞ and −∞ on the imaginary axis. A standard Pragmén–Lindelöf argument shows that g is bounded in the left half-plane. Hence a classical corollary of Jensen’s formula gives

(0.5)  ∫_{R} log |g(iy)|(1 + y^{2})^{−1} dy > −∞,

a contradiction.

Theorems 1 and 2 are proved in Sections 1 and 2, respectively. Section 3 contains essentially a discussion on the sharpness of Theorem 2 and of Theorem 2′, a slightly more general variant of it. In Section 4, Theorem 2 is applied to produce non-trivial invariant subspaces with respect to translation in certain weighted l^{p}(Z).

Entire functions of order ≤ 1, with bounds on both axes 341

1. Proof of Theorem 1

Let u be a convex function on [0, ∞) with u(0) = 0 and

∫_{1}^{∞} u(x)x^{−3} dx < ∞.

Observe that the integrand has a summable negative part. The Poisson formula for quadrants can be applied to extend u continuously to a function U in the closed right half-plane with boundary value 0 on the imaginary axis and harmonic inside the first and fourth quadrants.

Lemma. U is subharmonic in the open right half-plane.

Proof of the lemma. It suffices to show that U satisfies the mean value inequality locally at each point in the open half-plane. Hence it is enough to find, for every x_{o} > 0, a harmonic function U_{o}, such that U_{o} ≤ U in the open half-plane, and U_{o}(x_{o}) = U(x_{o}).

By the convexity of u there is a function l of the form

l(x) = cx + d, x ∈ R,

such that

l(x_{o}) = u(x_{o}), l(x) ≤ u(x), x ≥ 0.

Take

U_{o}(x + iy) = l(x), x ≥ 0.

In each of the two quadrants, the harmonic function U_{o} is given by the corresponding Poisson integral, and since U_{o} ≤ U on the boundaries, U_{o} ≤ U in the half-plane. Since U_{o}(x_{o}) = U(x_{o}), the lemma is proved.

To prove the theorem, let us first compare two functions k_{1} and k_{2} , each satisfying the conditions of the theorem, and such that k′′_{1} = k′′_{2} outside some compact interval. Then

k_{1}(x) − k_{2}(x) − bx, x ∈ R,

is bounded, for some real constant b. It follows from this that if f is an entire function, giving a positive answer of the question for (a, k_{1}), then, for some constant C, Cf(z) exp(bz), z ∈ C, gives a positive answer for (a, k_{2}). Hence we can without loss of generality assume that (0.3) holds if x ≠ 0. We can obviously choose a = 1, hence

(1.1)  |xk′′(x)| ≤ \frac{2}{π} ,  x ≠ 0.

As a consequence of the growth restriction on k, we can use Poisson’s formula in each quadrant to produce a continuous function U in C, harmonic in each quadrant, 0 on the imaginary axis and with values k(x), x ∈ R. By the symmetry, U is harmonic in the upper and the lower half-plane. Let V be the function obtained, if we perform the corresponding extension of 2π^{−1}x log |x|, x ∈ R. It is easy to see that for z = x + iy ∈ C

V (z) = Re(\frac{2}{π} z log z) + |y|.

(0.4) and (1.1) show that

\frac{2}{π} x log |x| − k(x)  and  k(x) + \frac{2}{π} x log |x|

are convex, for x > 0. By the lemma, both V − U and U + V are subharmonic in the open right half-plane. This means that in distribution sense

(1.2)  −ΔV ≤ ΔU ≤ ΔV.

in the open right half-plane. Let m denote Lebesgue measure on R, considered as a Borel measure in the plane. Since Δ|y| = 2m, the fact that V is odd shows that

ΔV = 2(sign)m + v,

where the right hand member stands for the measure m, multiplied with the function with values 2 sign x, x ∈ R, plus a distribution v with support in 0. If v ≠ 0, we can find a sequence {ϕ_{n}} of test functions, such that ⟨ΔV, ϕ_{n}⟩ → 1, while ⟨V, Δϕ_{n}⟩ → 0, as n → ∞, a contradiction. Hence v = 0, and we obtain

ΔV = 2(sign)m.

Since ΔU is odd, (1.2) shows that

ΔU = gm + w,

where g is an odd Lebesgue measurable function, satisfying

−2 ≤ g(x) ≤ 2, x ∈ R,

and w is a distribution with support in 0. Exactly as above we see that w = 0, hence

ΔU = gm.

Let us now form the function W, defined by

W (x + iy) = U(x + iy) + |y|, x + iy ∈ C.

This is a subharmonic function in C, since, with h = g + 2,

ΔW = ΔU + 2m = hm ≥ 0.

As a subharmonic function, W is of order ≤ 1, and it satisfies the conditions

W (x) = k(x), x ∈ R, W (iy) = |y|, y ∈ R. Our theorem is proved if we can modify W to a function W_{o} = log |f|, where

f is entire of order ≤ 1, and satisfies (0.1) and (0.2).

To construct W_{o} we use a procedure of atomizing the measure ΔW. The method was introduced by B. Kjellberg [5, pp. 39–41]. A generalization of it, the KKK method, is described in W. Hayman [3, Ch. 10.5].

Let us first observe that

(1.3)  0 ≤ h ≤ 4.

Let H be the absolutely continuous function on R with H′ = h, and H(0) = 0. Let

A(x) = 2π[(2π)^{−1}H(x)] − H(x), x ∈ R,

where [ ] denotes the integral part. The distributional derivative A′ of A is then a Borel measure on R. We think of it as a measure on C, supported by the real axis. Since Δ log equals 2π times the Dirac distribution at 0, we know that if T is upper semicontinuous on C, and ΔT = A′, then W + T is a function of the form log |f_{o}|, where f_{o} is entire.

It suffices to choose T so that it can be estimated from above by an expression of the type C log(|z| + 2), z ∈ C. For since H is unbounded, f_{o} has infinitely many zeros, and hence we can divide away finitely many of them in order to obtain an entire function of order ≤ 1 and satisfying (0.1) and (0.2).

A convenient choice is

(1.4) T (z) = (2π)^{−1} ∫_{|t|<1} log |z − t| dA(t) + (2π)^{−1} ∫_{|t|≥1} log|1 − \frac{z}{t} | dA(t).

It follows from (1.3) that there is a constant C_{1} , such that the contribution from the set [−1, 1] ∪ [x − 1, x + 1] to (1.4) is bounded above by C_{1} log(|z| + 2). In the remaining intervals, the integrand has, for fixed z, just one interior extremal point, and its value varies in some interval

[−C_{2} log(|z| + 2), C_{2} log(|z| + 2)].

A partial integration shows that the integral over these intervals converges and is bounded by an expression of the form C_{3} log(|z| + 2) since A is a bounded function. It is easy to see that T is upper semicontinous and that ΔT = A′. Hence the theorem is proved.

2. Proof of Theorem 2

We can restrict ourselves to the case when k is convex on [0, ∞). We represent k as a difference k = k_{1} − k_{2} , where

(2.1)  k_{1}(x) = ∫_{1/2}^{3/2} k(xy) dy,  x ∈ R.

Then k_{1} and k_{2} are odd, and k_{2} ≥ 0 on R^{+} , by the convexity. k_{1} fulfils the differentiability assumption of Theorem 1, and k_{2} the differentiability assumption of the Beurling–Malliavin theorem. Furthermore k′_{2} is bounded. We will prove

(2.2)  lim_{x→∞} x k′′_{1}(x) = 0

and

(2.3)  ∫_{2}^{∞} k_{2}(x)x^{−2} dx < ∞.

Then Theorem 1, applied to (\frac{1}{2}a, k_{1}), and the Beurling–Malliavin theorem, applied to (\frac{1}{2}a, k_{2}), prove our theorem.

For x ≥ 0, we obtain from (2.1)

k′′_{1}(x) = x^{−3} ∫_{x/2}^{3x/2} t^{2} dk′(t),

and since k′ increases on R^{+} ,

0 ≤ k′′_{1}(x) ≤ 3x^{−1}(k′(∞) − k′(\frac{1}{2}x)), x ∈ R^{+}.

Since k′(∞) is finite, (2.2) is proved.

Partial integrations give, for x > 0,

8xk_{2}(x) = ∫_{x/2}^{x}(x − 2t)^{2} dk′(t) + ∫_{x}^{3x/2}(3x − 2t)^{2} dk′(t) ≤ x^{2} ∫_{x/2}^{3x/2} dk′(t).

Hence

∫_{2}^{∞} x^{−2}k_{2}(x) dx ≤ ∫_{1}^{∞} ∫_{2t/3}^{2t}(8x)^{−1} dx dk′(t) ≤ ∫_{1}^{∞} dk′(t) = k′(∞) − k′(1) < ∞,

and (2.3) is proved.

3. Comments on Theorem 2

Theorem 2 can obviously be stated in the following slightly more general form.

Theorem 2′. Let k be odd on R, absolutely continuous in some interval [b, ∞), and with its derivative equivalent to a function of bounded variation. Then the answer is yes, for every a.

Take any a and any positive function ϱ on R^{+} , such that ϱ(x) → ∞, as x → ∞. We will prove that Theorem 2′ is no longer true if the condition of bounded variation is changed to the condition that k′ is equivalent to a bounded function, satisfying

(3.1)  ∫_{b}^{x} |dk′(t)| = O(ϱ(x)),  x → ∞,

and

(3.2)  ∫_{−x}^{x} \frac{max(0, k(t))}{1 + t^{2}} dt = O(ϱ(x)),  x → ∞.

A consequence of this is that, in Theorem 2, the condition of boundedness of k′ can not be weakened to the condition

k′(x) = O(ϱ(x)), x → ∞.

Let us first form the harmonic function

u(z) = Re(−iz log(−iz)) = −y log \frac{1}{|z|} − ix Arg(−iz).

in the closed upper half-plane. As always we choose the principal branch of log. Let l be the odd continuous function on R, vanishing on (0, \frac{1}{2}) and (\frac{3}{2}, ∞), and such that l(1) = 1, and l is linear in the complementary intervals. Put

k_{o}(x) = ∑_{n=−∞}^{∞} 3^{n}l(3^{−n}x), x ∈ R.

Extend k_{o} by Poisson’s formula for the first quadrant to a function U_{o}, harmonic inside, and with U_{o}(iy) = ay, y ∈ R^{+} . A standard comparison between the Poisson integrals U_{o}(z) and u(z − 1) in the first quadrant shows that

U_{o}(z) = 1 + Au(z − 1) + O(|z − 1|) = 1 − By log \frac{1}{|z − 1|} + O(|z − 1|), z → 1,

where A and B are certain positive constants. Extend U_{o} to a function in C, even in y, odd in x. Then there is a r, 0 < r < \frac{1}{2} , such that, for every x ∈ [1−r, 1+r], the mean value of U_{o}, taken over a circular disc with center x and radius r, is < U_{o}(x) − r. Since

U_{o}(3z) = 3U_{o}(z), z ∈ C,

we have, for every integer n and every x ∈ [3^{n}(1−r), 3^{n}(1+r)], that the mean value of U_{o}, taken over a circular disc with center x and radius 3^{n}r, is < U_{o}(x) − 3^{n}r.

Now we are ready for the final construction, which does not depend on a. It is possible to choose a strictly increasing infinite sequence {n_{ν}}, of non-negative integers such that k, defined by

k(x) = ∑_{ν} 3^{n_{ν}} l(3^{−n_{ν}} x), x ∈ R,

satisfies (3.1) and (3.2). Obviously k is absolutely continuous and k′ is bounded. Extend k by Poisson’s formula to a continuous function U, harmonic in each quadrant, and satisfying U(iy) = a|y|, y ∈ R. Since U ≤ U_{o} in the right halfplane we we can conclude, for every ν and every x ∈ [3^{n_{ν}} (1 − r), 3^{n_{ν}} (1 + r)], that the mean value of U taken over a circular disc with center x and radius 3^{n_{ν}} r, is

< U_{o}(x) − 3^{n_{ν}} r = k(x) − 3^{n_{ν}} r.

If f is an entire function, giving a positive answer for (a, k), then log |f| ≤ U in C, and hence, by the mean value inequality for log |f|, we have for every ν that

log |f(x)| ≤ k(x) − 3^{n_{ν}} r, x ∈ [3^{n_{ν}} (1 − r), 3^{n_{ν}} (1 + r)].

Define g by

g(z) = f(z)f(−z), z ∈ C.

g is an entire function of exponential type, and log |g| ≤ 0 on R. On every interval of the form [3^{n_{ν}} (1 − r), 3^{n_{ν}} (1 + r)] we have

log |g| ≤ −3^{n_{ν}} r,

and hence

∫_{1}^{∞} x^{−2} log |g(x)| dx = −∞.

As in the introduction this gives a contradiction.

Observe that the results here and in the introduction on non-existence of functions log |f|, satisfying (0.1) and (0.2), where f is entire of order ≤ 1, are valid as well with log |f| exchanged to an arbitrary subharmonic function of order ≤ 1, not ≡ −∞.

4. Translation-invariant subspaces of l^{p}(w, Z)

Let {k(n)}, n ∈ Z, be a real odd sequence, such that {k(n + 1) − k(n)}, z ∈ Z, is bounded and

∑_{−∞}^{∞} |k(n + 1) − 2k(n) + k(n − 1)| < ∞.

Put exp k(n) = w_{n}, w = {w_{n}}, n ∈ Z. For a given p, 1 ≤ p < ∞, l^{p}(w, Z) denotes the Banach space of complex sequences c = {c_{n}}, n ∈ Z, with

∥c∥^{p} = ∑_{−∞}^{∞} |c_{n}|^{p}w_{n}^{p} < ∞.

In l^{p}(w, Z) translation, defined by {c_{n}} ↦ {c_{n−1}}, is an bounded operator with a bounded inverse. We will use Theorem 2 to construct a non-trivial subspace of l^{p}(w, Z), invariant under translation.

Extend k to an odd continuous function on R, linear in each interval in R \ Z. Since k can be written k_{1} − k_{2} , where k_{1} and k_{2} satisfy the conditions of Theorem 2, there is for every given a an entire function f giving a positive answer to the question for the pair (a, k). Take any a in the interval (0, \frac{1}{2}π). It follows from the proofs of Theorems 1 and 2, that we can assume that f has infinitely many zeros. After dividing away two of these zeros, if necessary, we can assume that

|f(x)| ≤ (1 + x^{2})^{−1} exp(k(x)), x ∈ R.

Let us, for any given integer m, form the entire function g_{m}, defined by

g_{m}(z) = f(z)f(m − z), z ∈ C.

A comparison between log |f| and convenient linear functions in each quadrant shows that log |g_{m}(iy)|−2a|y|, y ∈ R, is bounded above. Since g_{m} is bounded on the real axis and of order ≤ 1, it is thus of exponential typ ≤ 2a. g_{m} is summable on the real axis, and a theorem of Paley and Wiener [6, Theorem V] shows that the Fourier transform of the restriction of g_{m} to R vanishes outside [−2a, 2a]. Since g_{m}(x)(1 + x^{2}), x ∈ R is summable and g_{m} is continuous, we can apply the Poisson summation formula (see for instance Y. Katznelson [4]) to the function

x ↦ g_{m}(x) exp(πix),

and since the Fourier transform of this function vanishes at all integer multiples of 2πi we obtain

∑_{n∈Z} f(n)(−1)^{n}f(m − n) = 0.

For every p, {f(m − n)}, n ∈ Z, belongs to l^{p}(w, Z), and {f(n)(−1)^{n}}, n ∈ Z, belongs to the dual Banach space. Obviously we can choose f such that it does not vanish identically at the integer points. Then the closed linear span of the translates of {f(−n)} is a non-trivial closed translation-invariant subspace.

348 Yngve Domar

Theorems 1 and 2′ can be used in a similar way. For a particular case, see Example 2 in §2 of [2].

References

[1] Beurling, A., and P. Malliavin: On Fourier transforms of measures with compact support. - Acta Math. 107, 1962, 291–309.

[2] Borichev, A.A.: The generalized Fourier transformation, Titchmarsh’s theorem and asymptotically holomorphic functions. - Leningrad Math. J. 1, 1990, 825–857.

[3] Hayman, W.: Subharmonic Functions, II. - Academic Press, London, 1989.

[4] Katznelson, Y.: An Introduction to Harmonic Analysis. - Dover, New York, 1968.

[5] Kjellberg, B.: On certain integral and harmonic functions. - Appelbergs, Uppsala, 1948.

[6] Paley, R.E.A.C., and N. Wiener: Fourier transforms in the complex domain. - Amer. Math. Soc. Colloq. Publ. 19, New York, 1934.

Received 4 December 1995