The Liouville theorem for a class of Fourier multipliers and its connection to coupling

The classical Liouville property says that all bounded harmonic functions in Rn$\mathbb {R}^n$ , that is, all bounded functions satisfying Δf=0$\Delta f = 0$ , are constant. In this paper, we obtain necessary and sufficient conditions on the symbol of a Fourier multiplier operator m(D)$m(D)$ , such that the solutions f$f$ to m(D)f=0$m(D)f=0$ are Lebesgue a.e. constant (if f$f$ is bounded) or coincide Lebesgue a.e. with a polynomial (if f$f$ is polynomially bounded). The class of Fourier multipliers includes the (in general non‐local) generators of Lévy processes. For generators of Lévy processes, we obtain necessary and sufficient conditions for a strong Liouville theorem where f$f$ is positive and grows at most exponentially fast. As an application of our results above, we prove a coupling result for space‐time Lévy processes.

and we establish a further probabilistic interpretation of the Liouville property for Lévy generators in terms of coupling and space-time harmonic functions.
1.The proof of the Liouville theorem from [3] revisited We will need a few notions from probability theory and stochastic processes, which can be found in Sato [18] or Jacob [10], [11] and [12], but the essential ingredient is the structure of the infinitesimal generator, see below.A Lévy process is a stochastic process (X t ) t≥0 with values in R n and sample paths t → X t (ω) which are for almost all ω rightcontinuous with finite left-hand limits; moreover the random variables X t k − X t k−1 , 0 = t 0 < t 1 < • • • < t m , m ∈ N, are stochastically independent (independent increments) and each increment X t k −X t k−1 has the same distribution as X t k −t k−1 (stationary increments).The fact that we are looking at a process with independent and stationary increments means that the distribution of X t (for any fixed t > 0) characterizes the whole process; moreover, X t is necessarily infinitely divisible, so that its characteristic function (inverse Fourier transform) is of the form where the characteristic exponent ψ(ξ) is uniquely given by the Lévy-Khintchine formula 1 − e iy•ξ + iy • ξ ν(dy) + |y|≥1 1 − e iy•ξ ν(dy).
Since Lévy processes are Markov processes, their transition behaviour can be described by a transition semigroup P t f (x) = Ef (X t + x) which, in turn, is uniquely characterized by the infinitesimal generator Lf := d dt P t f t=0 (in the Banach space C ∞ (R n ) of all continuous functions vanishing at infinity, say).Now the key point is the following observation: Fact.The infinitesimal generator L = L ψ of a Lévy process with characteristic exponent ψ(ξ) is on C ∞ c (R n ) a Fourier multiplier operator with symbol −ψ(ξ), i.e.
and, combining this with (3), Note that we get Our first aim is to give a purely analytic proof of the following result.
Theorem 1 (Liouville; [1], [3]).Let ψ be the characteristic exponent of a Lévy process and denote by ψ(D) the corresponding Fourier multiplier operator.Suppose [3,Theorem 3.2]).However, ψ may be continuous but nowhere differentiable (see [3,Remark 3.3]), in which case defining the product of ψ and the distribution f is by no means trivial.
Our analytic proof of Theorem 1 is based on the following standard result, which is known from the proof of Wiener's Tauberian theorem, cf.Rudin [17,Theorem 9.3].
then the set contains the support of the tempered distribution f .[19,Lemma 3.4] or the discussion in Example 6 below.Thus, the dual pairing in ( 6) is well-defined. Suppose It is easy to see that Applying Wiener's theorem (Theorem 2) with we conclude that the support of the tempered distribution f is contained in {0}.Therefore, there exist see, e.g.[17,Thorems 6.24 and 6.25].Inverting the Fourier transform, and using the assumption that f is bounded, shows that f ≡ c 0 Lebesgue a.e.
For the converse direction, we assume the contrary, i.e. that the zero-set {η ∈ R n | ψ(η) = 0} contains some γ = 0. Then f (x) := e iγ•x satisfies ψ(D x )e iγ•x = e iγ•x ψ(γ) = 0 for all x ∈ R n .Thus, f is a non-constant solution, and we are done.We note in passing that since ψ(−γ) = ψ(γ), −γ ∈ {ψ = 0}, and we can even get a real-valued solution: Remark 3.With a bit more effort, see [3] or [1], one can show that all bounded solutions in the converse direction of Theorem 1 are necessarily periodic.Since {ψ = 0} is a closed subgroup of the additive group (R n , +), the periodicity group of all bounded solutions is given by the dual lattice {ψ = 0} [⊥] := {x ∈ R n | e iγ•x = 1, ∀γ ∈ {ψ = 0}}.The proof in [1] actually shows that a Lévy generator ψ(D) has the Liouville property if, and only if, The above proof of Theorem 1 extends without change to Fourier multiplier operators that map a) κ is a linear combination of terms of the form ab, where a is the Fourier transform of a finite Borel measure on R n , and all partial derivatives ∂ α ξ b with |α| ≤ n + 1 are polynomially bounded. Indeed: * ϕ, and it follows from Young's inequality A particular example is the characteristic exponent of a Lévy process The last term in the above formula is the Fourier transform of a finite Borel measure.The smoothness of ψ 0 follows immediately from the differentiation lemma for parameter-dependent integrals and the observation that the integrand of the formally differentiated function ∂ α ξ ψ 0 can be bounded by const |y| 2 , which is ν-integrable over {y ∈ R n : 0 < |y| < 1}.This bound also shows that The proof of this assertion goes along the lines of Part a).Functions of this type appear naturally in positivity questions related to generalized functions, see e.g.Gelfand & Vilenkin [7,Chapter II.4] or Wendland [22].Some authors call the function −κ (under suitable additional conditions on c α 's) a conditionally positive definite function.Note that s = 1 is just the Lévy-Khintchine formula (3).

The Liouville theorem for polynomially bounded functions
We are now going to show that our argument used in the proof of Theorems 1 and 4 extends to polynomially bounded functions f in (6).
To simplify the presentation, we use the function Λ(x) = (1 + |x| 2 ) 1/2 as well as the following function spaces.Let β ≥ 0, and We need the following analogue of Theorem 2 for the pair (L then the set contains the support of the tempered distribution f . Proof.Pick any ξ 0 ∈ R n \ Z(Y ).There exists some g ∈ Y such that g(ξ 0 ) = 1.Since g is continuous, there is a neighbourhood To prove the theorem, it is sufficient to show that f = 0 in V , or, equivalently, that f , v = 0 for every v ∈ S(R n ) whose Fourier transform v has its support in V .Since It is easy to see that Re u(ξ) > 1/2 for all ξ ∈ R n .Then there exists some w ∈ A β such that w = 1/ u, see, e.g.[5, Theorems 1.41 and 2.11] or [6, Theorem 1 [17,Theorem 9.3].Unfortunately, this version does not seem sufficient for the proof of the equality f * This yields We can now state and prove the Liouville theorem for polynomially bounded functions.
Theorem 8 (Liouville property for polynomially bounded functions).Let m ∈ C(R n ) be such that the Fourier multiplier operator , then f coincides Lebesgue a.e. with a polynomial of degree at most ⌊β⌋.
we can follow the argument line-by line up to the point where we get Again, we invert the Fourier transform and use the polynomial boundedness of f to see that f coincides Lebesgue a.e. with a polynomial of degree less or equal than ⌊β⌋.
The converse statement follows from that in Theorem 4. Indeed, if every f ∈ L ∞ −β (R n ) satisfying (12) coincides Lebesgue a.e. with a polynomial, then the same is true for any Since the only polynomials contained in L ∞ (R n ) are constants, one can apply the converse statement in Theorem 4.
Example 9.In this example we discuss the multipliers from Example 6 in the setting of Theorem 8.The following conditions ensure that κ(D) maps κ is a linear combination of terms of the form ab, where a = F µ, µ is a finite Borel measure on R n such that Λ β (y)µ(dy) < ∞, and all partial derivatives ∂ α ξ b with |α| ≤ ⌊n + β⌋ + 1 are polynomially bounded. Indeed: * ϕ, and it follows from Young's and Peetre's inequalities With a bit more effort, one can actually show that Λ β (y) µ(dy) < ∞ is also a necessary condition for a(D) to map A particular example is the characteristic exponent κ = ψ of a Lévy process such that the Lévy measure has finite moments of order β (cf.Example 6a)).
Proof.Pick k ∈ N such that 2k is greater than the degree of p.
The following result shows that the polynomial f appearing in Theorem 8 has degree at most 1 in the case where n = 1 and m is the characteristic exponent of a Lévy process.
Corollary 11.Let ψ : R → C be the characteristic exponent of a Lévy process with Lévy triplet (Q, b, ν).Suppose there exists a β ≥ 0 such that is a weak solution of the equation ψ(D)f = 0, then f (x) = c 1 x + c 0 with some constants c 1 and c 0 .
Proof.It follows from Theorem 8 that f is a polynomial of degree at most [β].According to Lemma 2.4 in [15], the degree of f is less than or equal to 2. The lemma deals with solutions of the equation ψ(D)f = const.In the case of ψ(D)f = 0, its proof (see, in particular, the last paragraph of the proof) shows that the degree of f is actually less than or equal to 1.
The above result does not hold in the multi-dimensional case n ≥ 2. Note that in the case n = 1, f ′′ = 0 =⇒ f (x) = ax+b, while in the case n = 2, ∆f = 0 has polynomial solutions of any degree, e.g.Re(x 1 + ix 2 ) k , k ∈ N.

The Liouville theorem for slowly growing functions
Theorem 7 covers bounded functions f , while Theorem 8 is about functions whose growth is compared with the growth of a polynomial.This leaves a gap where f grows slower than a polynomial, e.g. at a logarithmic scale.To deal with this case, we need the notion of a measurable, locally bounded, submultiplicative function, i.e. a locally bounded measurable function h : R n → (0, ∞) satisfying h(x + y) ≤ ch(x)h(y) for some c ≥ 1 and all x, y ∈ R n .
Fix some locally bounded submultiplicative h that satisfies, in addition, The condition ( 14) implies the so-called GRS (Gelfand-Raikov-Shilov)-condition, see [2] and [6] for details.We replace the pair (L The family A h is a convolution algebra and, due to the GRS-condition, an element of u ∈ A h is invertible if, and only if, it is invertible in A 1 , where A 1 is the algebra with h = 1.One can replace the Peetre inequality by h(x − y) ≤ ch(x)h(−y), which is a direct consequence of the submultiplicativity of h, and then get by iteration h(x − y) ≤ c 2 h(x)h(−(y−z))h(−z).This allows one to repeat the arguments in the proof of Theorem

one can apply this version of Theorem 7 with
).This results in the following analogue of Theorem 8 for slowly growing functions.
Theorem 12 (Liouville property for slowly growing functions).Let m ∈ C(R n ) be such that the Fourier multiplier operator Conversely, if f coincides Lebesgue a.e. with a polynomial for every Theorem 12 can be used for functions of the form h(x) = Λ β (x) log(e+|x|) α for α, β ≥ 0. If we use h(x) = log(e + |x|), we see that in the setting of Theorem 12 every solution m(D)f = 0 is a.e.constant.
We want to point out that without the boundedness condition ( 14) the function f is not necessarily a tempered distribution.For example, if we choose h(x) = e a|x| γ , then f need not be a tempered distribution, and our method above would not work.In the next section we discuss functions, which might not define a tempered distributions, but are positive.The condition that m(D) maps ) is essential for Theorem 12.Recently we learned from M. Kwasnicki (private communication) and the paper by T. Grzywny & M. Kwasnicki [8, Theorem 1.9.(c) & Theorem 4.1] that there is a multiplier given by a Lévy process, i.e. m = ψ, admitting a very slowly growing non-constant function u such that ψ(D)u = 0. Note that this multiplier does not satisfy the mapping property required for Theorem 12.

The Liouville theorem for rapidly growing functions
We will now turn to the case where the solution of ψ(D)f = 0 is locally bounded and positive.It is well known that This is usually called the strong Liouville property.One cannot expect this property to hold for general Fourier multiplier operators discussed in Sections 2 and 3 or even for higher order partial differential operators.Indeed, is a non-constant nonnegative polynomially bounded solution of the equation ∆ 2 f = 0.
Here, ∆ 2 = ψ(D), ψ(ξ) = |ξ| 4 , and (cf.Theorem 14 below).So, we consider in this section Fourier multiplier operators ψ(D) that generate positivity preserving operator semigroups e −tψ(D) t≥0 , i.e. such that ψ are the characteristic exponents of Lévy processes, see Section 1 and Example 6.Even within this class of operators, the Laplacian is a special case, and the strong Liouville property does not hold for more general second order partial differential operators L without restrictions on the growth rate of a solution f of the equation Lf = 0. Let where Q ∈ R n×n is a positive semi-definite matrix, and b ∈ R n \ {0}.Then Lf = 0 has non-constant nonnegative solutions.Indeed, if Q = 0, there exists c 0 ∈ R n such that c 0 • Qc 0 > 0. Since b = 0, there exists c in a neighbourhood of c 0 such that c • Qc > 0 and b Then f > 0 is non-constant and is positive, non-constant, and It follows from the above that one needs to put appropriate boundedness restrictions on f .Note also that while local boundedness of f is sufficient to ensure that f, ψ(D)ϕ is well-defined when ψ(D) is a local operator, i.e. a partial differential operator, one needs boundedness restrictions on f to ensure that ψ(D)f has a meaning when ψ(D) is non-local.
Our current proof is a refinement of our result in [3,Theorem 17], and we focus here on the role of the upper bound.The key ingredient in the proof of [3,Theorem 17] is the equivalence of −ψ(D)f = 0 and e −tψ(D) f = f , which is used to get a Choquet representation of all positive solutions f ≥ 0. In order to define ψ(D)f or e −tψ(D) f as a distribution, we need a bound f ≤ g and an integrability condition on the measure ν appearing in the Lévy-Khintchine representation (3) of ψ, see the discussion in [3].Here we will concentrate on the role of the upper bound g in the proof of the strong Liouville property which was glossed-over in our presentation in [3,Theorem 17]; this explains, in particular, in which directions ψ can be extended from Let g : R n → [1, ∞) be a locally bounded, measurable submultiplicative function.We need to describe the directional growth behaviour of g.For ω Since ln g ω is subadditive, the infimum is, in fact, a limit; moreover, the function β : Applying [13, Chapter II, Theorem 1.3] to the function v(t) := g ω (ln t), t > 0, we conclude that g ω (r) ≥ e β(ω)r for all r > 0, (16) and that for every ǫ > 0, there is some r ǫ > 0 such that g ω (r) ≤ e (β(ω)+ǫ)r for all r > r ǫ .

Coupling
In this section we want to establish a connection between our Livouville theorem (Theorem 1), the coupling property of Lévy processes, and the notion of space-time harmonic functions.Throughout, (X t ) t≥0 is a Lévy process starting at 0 with characteristic exponent ψ, see Section 1.
A coupling of the Lévy process (X t ) t≥0 with values in R n is any Markov process ((Z x t , Z y t )) t≥0 taking values in R 2n such that each of the marginal processes (Z z t ) t≥0 has the same finite dimensional distributions as (X z t ) t≥0 , X z t := X t +z, z = x, y.The coupling is usually realized on a new probability space (Ω, P = P (x,y) , F ).The Lévy process (X t ) t≥0 has the (exact) coupling property, if for some coupling with x = y the trajectories of the processes (Z x t ) t≥0 and (Z y t ) t≥0 meet with probability one in finite time; this is the coupling time τ = τ x,y .Intuitively, both processes (Z x t ) t≥0 and (Z y t ) t≥0 run on the same probability space, move together and cannot (statistically) be distinguished from each other or from (X t + z) t≥0 .
Coupling techniques provide powerful tools to study the regularity of the semigroup x → P t f (x) = Ef (X x t ), the existence of invariant (stationary) measures for (X x t ) t≥0,x∈R n and many further properties, see the discussion in [21].It was shown in [20,Theorem 4.1] that a Lévy process has the coupling property if, and only if, the transition probability p t (dy) of the Lévy process (X t ) t≥0 has an absolutely continuous component for some t ≥ 0.
If (X x t ) t≥0,x∈R n is a Lévy process (or a general Markov process with generator A x ), the space-time process ((s + t, X x t )) t≥0,(s,x)∈[0,∞)×R n is again a Lévy process (resp., Markov process), and its semigroup is given by Q t u(s, x) = E(s + t, X x t ).Thus, the infinitesimal generator is of the form d ds − ψ(D x ) (resp.d ds + A x ), and we are naturally led to the notion of space-time harmonic functions and the space-time Liouville property.Definition 15.Let (X x t ) t≥0,x∈R n be a Lévy process.a) A function for every t, s ≥ 0 and every x ∈ R n .b) The process has the space-time Liouville property, if every measurable and bounded space-time harmonic function is constant.
a) The space-time Liouville property is known to be equivalent to the exact coupling property for Markov processes, see [21,Theorem 4.1,p. 205].b) We call a function f : R n+1 → R satisfying f (s, x) = E x f (s + t, X t ) for every t ≥ 0 and (s, x) ∈ R n+1 also space-time harmonic.c) Notice that the notions of 'space-time harmonicity' and 'exact coupling' are pointwise defined notions, which do not allow for exceptional sets.This is the main difficulty when we want to compare our notion of the Liouville property and the space-time Liouville property.
For a Lévy triplet (b, Q, ν) with Lévy process X = (X t ) t≥0 in R n and a vector η ∈ R n we introduce the notation This is the vector in the Lévy triplet of η • X t , see [18,Proposition 11.10].We need a few auxiliary lemmas preparing the proof of the main result of this section, Theorem 22.This implies that Y has the strong Feller property.That the symbol of Y is given by ( 20) is a direct consequence of the subordination of the process ((t, X t )) t≥0 : E e iτ St+iξ•X S t = E Ee iτ r+iξXr r=St = E e −(−iτ +ψ X (ξ))St = e −tf S (−iτ +ψ X (ξ)) .
Theorem 22.Let (X t ) t≥0 be a strong Feller Lévy process with characteristic exponent ψ X .Then the following assertions are equivalent: a) (X t ) t≥0 has the (exact) coupling property, b) (X t ) t≥0 has the space-time Liouville where S = (S t ) t≥0 is a 1/2-stable subordinator.( 3 By Lemma 21, (S t , X St ) has again the strong Feller property and hence, we choose u to be continuous, and the equality P t u(s, x) = u(s, x) holds pointwise for every (s, x) ∈ R × R n ; in particular, u is a bounded and continuous function.Setting f (t, x) := u(t, x) it is clear that f is spacetime harmonic, hence constant.c)⇒b): We have to show that any space-time harmonic function f : [0, ∞) × R n → R is constant.Fix such an f and construct, as in Lemma 20, its unique extension to the negative real line; this extension is still space-time harmonic.By a similar argument as before we see that f (s, x) = Ef (s + S t , x + X St ) pointwise for every (s, x) ∈ R × R n , where S = (S t ) t≥0 is again a 1/2-stable subordinator.We conclude that f is continuous.The characteristic exponent of the process (S t , X St ) is given by −iτ + ψ X (ξ) by Lemma 21.As (t, X t ) t≥0 has the Liouville property, we know that iτ − ψ X (ξ) = 0 if, and only if, (τ, ξ) = (0, 0), from which we conclude that −iτ + ψ X (ξ) = 0 if, and only if, (τ, ξ) = (0, 0).In view of Theorem 1, (S t , X St ) t≥0 has the Liouville property; therefore, f is a.e.constant, and, as f is continuous, f is constant.

Example 6 .
Let us give a few examples of multipliers satisfying the key assumption of Theorem 4. The following multipliers κ are such that κ thus we get polynomial boundedness of ψ 0 and its derivatives.A fully worked-out proof can be found in [3, Lemma 4] as well as [10, Lemma 3.6.22,Theorem 3.7.13].b) Any κ of the form

Lemma 17 .
Let ψ : R d → C be the characteristic exponent of a Lévy process with Lévy triplet (b, Q

0
property, c) (t, X t ) t≥0 has the Liouville property (as in Theorem 1),d) {ξ ∈ R n | ψ X (ξ) ∈ iR} = {0}.Proof.a)⇔b) is due to Thorisson [21, Theorem 4.5, p. 205].c)⇔d) is due to Theorem 1, (the proof of) Lemma 21 for f s (λ) = λ and Remark 19.b)⇒c): let u be a bounded measurable function such that d ds − ψ X (D x ) u = 0 in the sense of distributions.If P t is the semigroup generated by d ds − ψ(D x ), we know from the relation between semigroup and generator thatP t u(s, x) = u(s, x) + t 0 P r d ds − ψ X (D x ) u(s, x) dr = u(s, x)t > 0 and all (s, x) ∈ R×R n in the sense of distributions.Since u(s, x) = Eu(s+t, x+X t ) does not depend on t > 0, we haveu(s, x) = ∞ Eu(s + r, x + X τ )=u(s,x) P(S t ∈ dr) = Eu(s + S t , x + X St ),