An extension of the Liouville theorem for Fourier multipliers to sub-exponentially growing solutions

We study the equation $m(D)f = 0$ in a large class of sub-exponentially growing functions. Under appropriate restrictions on $m \in C(\mathbb{R}^n)$, we show that every such solution can be analytically continued to a sub-exponentially growing entire function on $\mathbb{C}^n$ if and only if $m(\xi) \not= 0$ for $\xi \not= 0$.


Introduction
The classical Liouville theorem for the Laplace operator ∆ := n k=1 ∂ 2 ∂x 2 k on R n says that every bounded (polynomially bounded) solution of the equation ∆f = 0 is in fact constant (is a polynomial).Recently, similar results have been obtained for solutions of more general equations of the form m(D)f = 0, where m(D) := F −1 m(ξ)F , and are the Fourier and the inverse Fourier transforms, see [1,2,3,12], and the references therein.Namely, it was shown that, under appropriate restrictions on m ∈ C(R n ), the implication f is bounded (polynomially bounded) and m(D)f = 0 =⇒ f is constant (is a polynomial) holds if, and only if, m(ξ) = 0 for ξ = 0.Much of this research has been motivated by applications to infinitesimal generators of Lévy processes.
In this paper, we study solutions of m(D)f = 0 that can grow faster than any polynomial.Of course, one cannot expect such solutions to have a simple structure, not even in the case of ∆f = 0 in R 2 , see, e.g., [22, Ch.I, § 2].We consider sub-exponentially growing solutions whose growth is controlled by a submultiplicative function, cf.(1), satisfying the Beurling-Domar condition (3), and we show that, under appropriate restrictions on m ∈ C(R n ), every such solution admits analytic continuation to a sub-exponentially growing entire function on C n if, and only if, m(ξ) = 0 for ξ = 0, see Corollary 4.4.Results of this type have been obtained for solutions of partial differential equations with constant coefficients by A. Kaneko and G.E. Šilov, see [17,18,27], [7, Ch. 10, Sect.2, Theorem 2], and Section 5 below.
Keeping in mind applications to infinitesimal generators of Lévy processes, we do not assume that m is the Fourier transform of a distribution with compact support, so our setting is different from that in, e.g., [6], [16,Ch. XVI].
The paper is organized as follows.In Chapter 2, we consider submultiplicative functions satisfying the Beurling-Domar condition.For every such function g, we introduce an auxiliary function S g , see (14), (15), which appears in our main estimates.Chapter 3 contains weighted L p estimates for entire functions on C n , which are a key ingredient in the proof of our main results in Chapter 4. Another key ingredient is the Tauberian theorem 4.1, which is similar to [3,Thm. 7] and [24,Thm. 9.3].The main difference is that the function f in Theorem 4.1 is not assumed to be polynomially bounded, and hence it might not be a tempered distribution.So, we avoid using the Fourier transform f = F f and its support (and non-quasianalytic type ultradistributions).Although we are mainly interested in the case m(ξ) = 0 for ξ = 0, we also prove a Liouville type result for m with compact zero set {ξ ∈ R n | m(ξ) = 0}, see Theorem 4.3.Finally, we discuss in Section 5 A. Kaneko's Liouville type results for partial differential equations with constant coefficients, cf.[17,18], which show that the Beurling-Domar condition is in a sense optimal in our setting.

Submultiplicative functions and the Beurling-Domar condition
Let g : R n → (0, ∞) be a locally bounded, measurable submultiplicative function, i.e. a locally bounded measurable function satisfying g(x + y) Cg(x)g(y) for all x, y ∈ R n , where the constant C ∈ [1, ∞) does not depend on x and y.Without loss of generality, we will always assume that g 1, as otherwise we can replace g with g + 1.Also, replacing g with Cg, we can assume that g(x + y) g(x)g(y) for all x, y ∈ R n . ( A locally bounded submultiplicative function is exponentially bounded, i.e.
We will say that g satisfies the Beurling-Domar condition if If g satisfies the Beurling-Domar condition, then it also satisfies the Gelfand-Raikov-Shilov condition while g(x) = e |x|/ log(e+|x|) satisfies the latter but not the former condition, see [10].It is also easy to see that g(x) = e |x|/ log γ (e+|x|) satisfies the Beurling-Domar condition if, and only if, γ > 1.The function satisfies the Beurling-Domar condition for any a, s, t 0 and b ∈ [0, 1), see [10].

Estimates for entire functions
Let 1 p ∞ and let ω : R n → [0, ∞) be a measurable function such that ω > 0 Lebesgue almost everywhere.We set ∞) be a locally bounded, measurable submultiplicative function satisfying the Beurling-Domar condition (3).Let ϕ be a measurable function such that for almost every see ( 14), (15), where the constant C g < ∞ depends only on g.
Applying (1) again, one gets The latter inequality can be rewritten as follows Hence, It follows from ( 14) that for y 1 > 1, So, where c g := M + Ig(1) π .Using Jensen's inequality, one gets where Then ϕ(z) := e w(z) is analytic in z for Im z > 0 and continuous up to R, and see (15), and ) , which shows that the factor e Sg(y 1 )y 1 in the right-hand side of ( 24) is optimal in this case. Clearly, where g(x) := g(Ax) and A ∈ O(n) is an arbitrary orthogonal matrix, see ( 14), ( 26) and (5). where and the constant C α ∈ (0, ∞) depends only on α and g.

Main results
We will use the notation g(x) := g(−x), x ∈ R n .It follows from submultiplicativity of g that L 1 g (R n ) is a convolution algebra.Taking y − x in place of y in (1) and rearranging, one gets Using this inequality, one can easily show that Suppose the set is bounded, and Proof.In order to prove the equality f = f * u, it is sufficient to show that Since the set of functions h with compactly supported Fourier transforms h is dense in If Z(Y ) = ∅, let V be an open neighbourhood of Z(Y ) such that u = 1 in V .Similarly to the above, there exists Since Y is a linear subspace, for every continuous, and there is a neighbourhood V η of η such that | v η (ξ) − 1| < 1/2 for all ξ ∈ V η .Similarly to the above, there exists , where σ Then there exists υ ∈ A g such that υ = 1/σ, see [5,Thm. 1.53].Since u 0 (1 − u) = 0 and (1 − u R ) w = 0, one has It now follows from (34) and (35) that If Z(Y ) = ∅, one can take u = 0, and the equality f = f * u means that f = 0.
For a bounded set E ⊂ R n , let conv(E) denote its closed convex hull, and H E denote its support function: Clearly, H E is positively homogeneous and convex: for all x, y ∈ R n and τ 0 we have For every positively homogeneous convex function H, Then f admits analytic continuation to an entire function f : see ( 14), (15), where the constant C α ∈ (0, ∞) depends only on α and g.
Proof.Take any ε > 0. There exists with some constant c ε ∈ (0, ∞).So, u satisfies the conditions of Corollary 3.4 with g in place of g, and Since  (2), and So, (41) follows from Theorem 3.3.
Proof.Denote by For the converse direction, we assume the contrary, i.e. that the zero-set {η ∈ R n | m(η) = 0} contains some γ ∈ K, see (39).Then there exists a So, f does not satisfy (45).
where the constant C α ∈ (0, ∞) depends only on α and g.
holds for every ε > 0 with a constant M ε ∈ (0, ∞) that depends only on ε, m, and g, then Proof.The only part that does not follow immediately from Theorem 4.3 is that f = 0 in the case {η ∈ R n | m(η) = 0} = ∅.In this case, one can take the same Y as in the proof of Theorem 4.
) is satisfied if m is a linear combination of terms of the form ab, where a = F µ, µ is a finite complex Borel measure on R n such that R n g(y) |µ|(dy) < ∞, and b is the Fourier transform of a compactly supported distribution.Indeed, it is easy to see that b A particular example is the characteristic exponent of a Lévy process (this is a stochastic process with stationary and independent increments, such that the trajectories are right-continuous with finite left limits, see e.g.Sato [25]) where b ∈ R n , Q ∈ R n×n is a symmetric positive semidefinite matrix, and ν is a measure on R n \ {0} such that 0<|y|<1 |y| 2 ν(dy) + |y| 1 g(y) ν(dy) < ∞.More generally, one can take with s ∈ N, c α ∈ R, and a measure ν on R n \ {0} such that 0<|y|<1 |y| 2s ν(dy) + |y| 1 g(y) ν(dy) < ∞. (As usual, for any α ∈ N n 0 and ξ ∈ R n , we define α! := n 1 α k ! and ξ α := n 1 ξ α k k .)Functions of this type appear naturally in positivity questions related to generalised functions (see, e.g.[9, Ch.II, §4] or [28,Ch. 8]).Some authors call the function −m for such an m (under suitable additional conditions on the c α 's) a conditionally positive definite function.
Remark 4.6.We are mostly interested in super-polynomially growing weights as polynomially growing ones have been dealt with in our previous paper [3].Nevertheless, it is instructive to look at the behaviour of the factor e Sg(|y|)|y| for typical super-polynomially, polynomially, and sub-polynomially growing weights.
Remark 4.7.If g is polynomially bounded in Corollary 4.4, then it follows from (46) and (49) that f is a polynomially bounded entire function on C n , hence a polynomial, see, e.g.[21,Cor. 1.7].The fact that f is a polynomial in this case was established in [3] and [12].
Since the Taylor series of cos w contains only even powers of w, cos(i √ z) is an analytic ) is a solution of the elliptic partial differential equation It follows from (50) that , where g(x) = e a|x| 1/2 with a = 1+ε √ 2 κ 1/2 ε .Clearly, the analytic continuation of f to C 2 is given by the formula Finally, see (7), letting (−∞, 0) ∋ y 2 → −∞, we arrive at

Concluding remarks
Corollary 4.4 shows that sub-exponentially growing solutions of m(D)f = 0 admit analytic continuation to entire functions on C n .It is well known that no growth restrictions are necessary in the case when m(D) is an elliptic partial differential operator with constant coefficients, and every solution of m(D)f = 0 in R n admits analytic continuation to an entire function on C n , see [23,6].see, e.g.[26, §7].Then the same argument as in the proof of [19,Cor. 8.2] shows that f admits continuation to an analytic function in the ball x ∈ C n : |x| < (1 + C −2 m ) −1/2 R .Note that in the case of the Laplacian, one can take C m = 1 and c m = (1+C −2 m ) −1/2 = 1 √ 2 , which is the optimal constant for harmonic functions, see [13].
Let us return to equations in R n .Below, m(ξ) will always denote a polynomial with {ξ ∈ R n | m(ξ) = 0} ⊆ {0}.For non-elliptic partial differential operators m(D), one needs to place growth restrictions on solutions of m(D)f = 0 to make sure that they admit analytic continuation to entire functions on C n .
We say that a function f defined on R n (or C n ) is of infra-exponential growth, if for every ε > 0, there exists Let µ : [0, ∞) → [0, ∞) be an increasing function, which increases to infinity and satisfies µ(t) At + B, t 0 for some A, B > 0, and Then, under additional restrictions on µ, every solution f of m(D)f = 0 that has growth O(e εµ(|x|) ) for every ε > 0 admits analytic continuation to an entire function of infra-exponential growth on C n , see [18].It is easy to see that (51) is equivalent to the Beurling-Domar condition (3) for g(x) := e εµ(|x|) .One cannot replace O(e εµ(|x|) ) with O(e ε|x| ) in the above result without placing a restriction on the complex zeros of m.If there exists δ > 0 such that m(ζ) has no complex zeros in | Im ζ| < δ, | Re ζ| > δ −1 , (52) then every solution of m(D)f = 0 that, together with its partial derivatives up to the order of m(D), is of infra-exponential growth on R n , admits analytic continuation to an entire function of infra-exponential growth on C n , see [17,18].
On the other hand, if for every δ > 0, (52) contains complex zeros of m(ζ), then m(D)f = 0 has a solution in C ∞ all of whose derivatives are of infra-exponential growth on R n , but which is not entire infra-exponential in C n .The proof of the latter result in [17,18] is not constructive, and the author writes: "Unfortunately we cannot present concrete examples of such solutions"; however, it is not difficult to construct, for any ε > 0, a solution in C ∞ all of whose derivatives have growth O(e ε|x| ), but which is not real-analytic.Indeed, according to the assumption, there exist complex zeros ) Choosing a subsequence, we can assume that ω Then, for every multi-index α and every x ∈ R n , . Further, On the other hand, f is not real-analytic.Before we prove this, note that formally putting x − itω 0 , t > 0 in place of x in the right-hand side of (55), one gets a divergent series.Indeed, its terms can be estimated as follows For any j > ε −1 , there exists ℓ j ∈ N such that It is clear that ℓ j → ∞ as j → ∞, see (53).Note that Clearly, |ξ j | 6ℓ j πj for sufficiently large j, see (56).Hence, one has the following estimate for the directional derivative ∂ ω 0 (−i∂ ω 0 ) ℓ j f (0) = −C(10ℓ j ) ℓ j + (2e) −(ℓ j +1) ℓ 2ℓ j j .Hence, (−i∂ ω 0 ) ℓ j f (0) ℓ 3 2 ℓ j j for all sufficiently large j, which means that f is not real-analytic in a neighbourhood of 0.
The operator m(D) in the previous example is not hypoelliptic.If m(D) is hypoelliptic, then every solution of m(D)f = 0, such that |f (x)| Ae a|x| , x ∈ R n , for some constants A, a > 0, admits analytic continuation to an entire function of order one on C n , see [11,§4,Cor. 2].For elliptic operators, this result can be strengthened: every solution of m(D)f = 0, such that |f (x)| Ae a|x| β , x ∈ R n , for β 1 and some constants A, a > 0, admits analytic continuation to an entire function of order β on C n , see [11,§4,Cor. 3].Let us show that for every β > 1 there exists a semi-elliptic operator m(D), see [16,Thm. 11.1.11],and a C ∞ solution of m(D)f = 0, all of whose derivatives have growth O(e a|x| β ), but which does not admit analytic continuation to an entire function on C n .A simple example of such a semi-elliptic operator is ∂ 2 It is easy to see that the last series is uniformly convergent on compact subsets of Π 1 .So, f admits analytic continuation to Π 1 .On the other hand, f (iy 1 , 0) → ∞ as y 1 → 1 − 0. Indeed, So, f (iy 1 , 0) → ∞ as y 1 → 1 − 0.

4 .
Let g : R n → [1, ∞) be a locally bounded, measurable submultiplicative function satisfying the Beurling-Domar condition (3).Let ϕ : C n → C be an entire function such that log |ϕ(z)| = O(|z|) for |z| large, z ∈ C n , and that the restriction of ϕ to R n belongs to L p g ±1 (R n ), 1 p ∞. Then for every multi-index α ∈ Z n + and every see[5, Thm.1.52 and 2.11], it is enough to prove (37) for such h.Further, f, h = f * h (0).So, we have to show only thatf * w = f * u * w (38) for every w ∈ L 1 g (R n) with compactly supported Fourier transform w.Take any such w and choose R > 0 such that the support of w lies in B R := {ξ ∈ R n : |ξ| R}.It is clear that g satisfies the Beurling-Domar condition.Then there exists u R ∈ L 1 g (R n ) such that 0 u R 1, u R (ξ) = 1 for |ξ| R, and u R (ξ) = 0 for |ξ| R + 1, see[5, Lemma 1.24].

Corollary 4 . 4 .
Let g : R n → [1, ∞) be a locally bounded, measurable submultiplicative function satisfying the Beurling-Domar condition (3) and let m ∈ C(R n ) be such that the Fourier multiplier operator 3, note that Z(Y ) = ∅ and apply Theorem 4.1 to conclude that f = 0. (It is instructive to compare this result to [18, Proposition 2.2].)Remark 4.5.The condition that m

Remark 5 . 1 .
The latter result has a local version similar to Hayman's theorem on harmonic functions, see [13, Thm.1]: for every elliptic partial differential operator m(D) with constant coefficients there exists a constant c m ∈ (0, 1) such that every solution of m(D)f = 0 in the ball {x ∈ R n : |x| < R} of any radius R > 0 admits continuation to an analytic function in the ball {x ∈ C n : |x| < c m R}.Indeed, let m 0 (D) = |α|=N a α D α be the principal part of m(D) = |α| N a α D α .There exists C m > 0 such that |α|=N a α (a + ib) α = 0, a, b ∈ R n =⇒ |a| C m |b|,