Documenta Math. 235 Laplace Transform Representations and Paley–Wiener Theorems for Functions on Vertical Strips

We consider the problem of representing an analytic function on a vertical strip by a bilateral Laplace transform. We give a Paley-Wiener theorem for weighted Bergman spaces on the existence of such representa- tions, with applications. We generalise a result of Batty an d Blake, on ab- scissae of convergence and boundedness of analytic functions on halfplanes, and also consider harmonic functions. We consider analytic continuations of Laplace transforms, and uniqueness questions: if an analytic function is the Laplace transform of functions f1, f2 on two disjoint vertical strips, and extends analytically between the strips, when is f1 = f2? We show that this is related to the uniqueness of the Cauchy problem for the heat equation with complex space variable, and give some applications, including a new proof of a Maximum Principle for harmonic functions.


Introduction and notation
We are concerned with Laplace transforms: for an analytic function F on {a < Re(z) < b}, we would like to know when for some h, in some sense: either as an absolutely convergent Lebesgue integral, or as the L 2 or tempered distribution Fourier transform of e −xt h(t).Our normalisation of the Fourier transform is In Section 2 we give a fairly general Paley-Wiener theorem which guarantees the existence of such an h for analytic functions F in certain weighted Bergman spaces, with applications.In Section 3 we generalise a result of C. Batty and M. D. Blake concerning bounded functions on halfplanes; we obtain the same result, but under weaker assumptions, as well as a similar result for harmonic functions.
In Section 4 we consider the uniqueness problem, which is important because analytic functions can sometimes be represented by Laplace transforms of different h on disjoint vertical strips.We obtain an explicit formula for analytic continuation under quite mild conditions, and relate this to the heat equation.Thus uniqueness theorems on the heat equation immediately give uniqueness theorems for boundary values of harmonic functions; see Corollaries 4.5, 4.6.Finally, Sections 5, 6 contain some longer proofs.
Let C + = {z ∈ C : Re(z) > 0} and R + = {t ∈ R : t 0}.For any Banach space E, 1 p < ∞ and F ∈ Hol(C + , E), define The set of all F with F < ∞ is the Hardy space H p (C + , E).When E = C we write simply H p (C + ).We mainly use the case p = 2 with E a Hilbert space.The classical Paley-Wiener Theorem says that L : L 2 (R + , E) → H 2 (C + , E) is a unitary operator from L 2 onto H 2 , provided that E is a Hilbert space:

Documenta Mathematica 15 (2010) 235-254
For any h : R → E strongly measurable, define Assume also that: Then: ∃ h such that F = Lh on Ω, and ) for every a < c < b, so h is given by the standard Bromwich Inversion Formula in the sense of L 2 (R, E) Fourier transforms.
The paper [11] proves this result in the special case a = 0, b = ∞ and v(x) = x r with r 0, and gives some applications.However, their method is different and probably cannot be generalised (the conformal transformation 1−z  1+z induces an isometric isomorphism with a weighted Bergman space on the disc, for which (z n ) n 0 is an orthogonal basis).Other related results and examples are given in Section 2 of [19].For the converse: first, let a < α < β < b.We must show that F is bounded on {x + iy : α x β}.Let r > 0 be sufficiently small, so that a < α − r < α < β < β + r < b.Fix ϕ ∈ E * and consider F ϕ (z) = ϕ(F (z)).We have the following result, which is a substitute for the lack of subharmonicity of |F ϕ | p when p < 1. See Lemma 2, p. 172 of [14], there attributed to Hardy and Littlewood; the proof is given also on p. 185 of [23]: with some C p < ∞. (This is true more generally for harmonic functions in several variables.The case p 1 is trivial by the Mean Value Property).By assumption, , independently of λ with α Re(λ) β.By (2) and (1), we now have |F ϕ (λ)| K ϕ E * , so indeed F is bounded on {α x β} as required.Second, suppose that ∞ −∞ F (x+iy) 2 dy < ∞ for x = α, β, where a < α < β < b.Thus for each Y > 0, Cauchy's Integral Formula gives But F is bounded on R Y , uniformly in Y , by above; so we can let Y → ∞ for each fixed λ to obtain z−(α+ω) dz, as a function of ω ∈ C + , is the Szegö projection of the L 2 (iR, E) function F (iy + α) onto the Hardy space H 2 (C + , E), and so by the Paley-Wiener Theorem it can be represented as Lf 1 for some and so This shows that e −ct h(t) ∈ L 2 (R, E) for each α < c < β.Now v > 0 a.e., so ∞ −∞ F (x + iy) 2 dy < ∞ for a.e.x ∈ (a, b).So choose sequences (α j ) ց a and (β j ) ր b such that this holds with x = α j , β j .Then F = Lh j on {α j < Re(λ) < β j } for each j.By uniqueness of the Fourier transform we must have h j ≡ h 1 = h a.e.So finally F = Lh on {a < Re(λ) < b}, and Plancherel's Theorem gives Similarly, with the Hausdorff-Young theorem and Paley-Wiener theorem for H p , we can easily obtain the following result: Then there exists some h such that F = Lh and We can consider Dirichlet-type norms also; for example: This is obvious, since In the case R = +∞, we have g(t) = 0 for all t < 0.
BM O(R) is the very important Bounded Mean Oscillation space, discussed in [1], [16], [23] and many other books, which often serves as a useful substitute for L ∞ (R).
For locally integrable f : R → C we have I ranges over all bounded intervals of R, and |I| is the length.Proof: The existence of g is immediate from Theorem 2.1, if we consider G(z) = F (z)/z and take, e.g.v(x) = x/(1 + x 3 ).The estimates follow from Plancherel's Theorem: For the estimate with t > 0, let T > 1/R and consider For the BM O result, let 0 < c < R. Then F (c+iy) c+iy is an L 2 function of y ∈ R, with We have Now apply Lemma 2.5 below to get e −ct g BMO(R) K for some K independent of c.For c = 0 and c = R, choose sequences c j ց 0 and c j ր R and use Dominated Convergence: for each interval I, (e −cjt g) I → g I or (e −Rt g) I as appropriate.Then |e −cjt g − (e −cjt g) I | I also converges appropriately; since the BM O norm is given by a supremum over all I, we have the result.
Taking Laplace transforms and using (the easy half of) Theorem 2.1 gives But this says exactly that |L(βf [13].But also Lf ∈ H 2 (C + ), and so Lf is the Poisson integral of its boundary function f .Hence f = U ∈ BM O as required.
In Theorem 3.1 below we obtain further results on g, assuming extra conditions on F (decay behaviour on a vertical line).

Results assuming decay on a vertical line
The following theorem generalises the main result of [3].
Let E be a Banach space, and let F ∈ Hol(Ω, E) be bounded.Assume that ∃ 0 < c < R, 0 < δ 1 and ν > 1 such that Then there exists some continuous g : R → E with F (z) = zLg(z) for all z ∈ Ω, such that In the case R = +∞, we have g(t) = 0 for all t 0. Also g satisfies local Hölder estimates: there is some M < ∞ such that The proof is given in Section 5. Of course we can get additional information about |ϕ(g(t))| 2 by applying Corollary 2.4 above to ϕ • F .In [3] the main result was the estimate (6) for the case R = +∞ only, assuming the much stronger condition [3] also explains that ( 8) is not sufficient in the case p = 1.Under assumption (8), we would have g(t) = t 0 f (s) ds.By increasing r if necessary and using Hölder's inequality, we could take 1 < p 2 without loss of generality.Then the Hausdorff-Young Theorem would give (5) The estimate ( 6) is best possible in general, even under the extra assumption (8), as shown in [2].Additionally (7), which is a conclusion of our theorem, would follow automatically from the assumption (8).In the case R = +∞, we have a similar result for harmonic functions: Theorem 3.2 Let F : C + → E be a bounded harmonic function, where E is a Banach space.Assume that (5) holds with c > 0. Then: there exist g j : R + → E continuous, j = 1, 2, such that and g 1 , g 2 satisfy the same Hölder estimate (7) from Theorem 3.1.
See Section 6 for the proof.Unfortunately, the case R < ∞ is unsatisfactory.For example, there is no function g such that z + a = zLg(z), with a ∈ C constant.Thus 2Re(z) = z + z is harmonic and bounded on {0 < Re(z) < 1} but cannot be written as zLg 1 (z) + zLg 2 (z) for any functions g 1 , g 2 .

Uniqueness conditions
It is natural to consider uniqueness: if Lf 1 = Lf 2 on {a < Re(z) < b}, in any reasonable sense, then f 1 = f 2 by uniqueness of Fourier transforms.However, this does not answer the following: so that F = Lf j on {a j < Re(z) < b j } for some (uniquely determined) f 1 , f 2 , by Theorem 2.1.When do we have In contrast to Laurent series on concentric annuli {r j < |z| < R j }, it is possible to have f 1 = f 2 .The paper [24] considers Then G is entire, and bounded on {|Im(z)| > π/2 + δ} for each δ > 0. Define F (z) = −iG(iz).By Cauchy's Theorem as in [24] we obtain So F is entire and represented by different bilateral Laplace transforms on {x > π/2}, {x < −π/2}, even though (using Plancherel's Theorem) Thus by rescaling, for any ǫ > 0 the "gap" {|x| < ǫ} is "unsafe": crossing the gap can change the Laplace transform function.However, we shall prove below that the gap {x = 0} can be safely crossed under quite mild restrictions.First we derive an explicit formula for analytic continuation of Laplace transforms.
Theorem 4.2 Let Ω = {z : a < Re(z) < b} and F ∈ Hol(Ω, E), with E a Banach space.Assume that a < c < b, κ > 0, and Define F σ ∈ Hol(C, E), for sufficiently small σ > 0, by , so that For each fixed z ∈ C, F σ (z) is the integral of G(λ, z) over the contour c + iR, which converges for 1/2σ 2 > κ by condition (9).Since G(λ, z) is an analytic function of λ, we can use Cauchy's Theorem with the same contour as in Theorem 3.1.Pick Let Y be large, much larger than δ, and consider the contours uniformly for z ∈ Σ, where G dλ → 0 rapidly as σ → 0, uniformly in z, as long as Y is large enough.By condition (9) again, also Γ ′ G dλ → 0 rapidly as σ → 0, uniformly for z ∈ Σ.Finally, the integral over Γ(x) is a standard Gaussian convolution approximation to F (z): After Y is chosen, F (t+iu) is then bounded on the tall, narrow rectangle |t−ω locally uniformly for z ∈ Ω, for some measurable h satisfying Proof: We use the Hausdorff-Young Theorem.Set for p ′ = (1 − 1/p) −1 the conjugate exponent to p.This is well-defined for a.e.t ∈ R. Now h(t)e −ct (y)g(y) dy = h(t)e −ct g(t) dt for every Schwartz function g.Now put F (c + iy) ∼ he −ct (y) in the definition of F σ,c (z) and calculate.
Notice that ( 10) is just a weak kind of Laplace transform representation for F .It says that a particular Abelian summability method assigns the value F (z) to the formal integral " ∞ −∞ e −zt h(t) dt", even though this integral may diverge.See [17] for much more on these topics; unfortunately the classical results given there appear to be inadequate for our problem.

Corollary 4.4
Let Ω = {z : a < Re(z) < b} and F ∈ Hol(Ω).Suppose that there exist a < c 1 < c 2 < b and f 1 , f 2 such that as Fourier transforms of f j (t)e −cj t ∈ L 2 (R).Define Proof: By Corollary 4.3, equation (10) Because this convergence is locally uniform, we have the required continuity of H. Finally, the complex heat equation ∂ 2 H ∂z 2 = ∂H ∂v follows immediately by differentiating under the integral sign.
The letter t is normally used for the time variable, but we use v = σ 2 /2 (for variance, with an extra factor of 2).Now we can apply known results on the heat equation.The papers [6], [30] prove many results about functions on discs.The following corollaries are closely related (after applying a conformal transformation), but our proofs are easier and quite different.

Corollary 4.6
Let Ω = {x + iy : 0 < x < 1}.Let E be a Banach space, F : Ω → E continuous, harmonic on Ω, and F ∈ L ∞ (∂Ω).Suppose that A, B, r > 0 satisfy This is a Maximum Principle, similar in some ways (but quite different in other ways) to the Phragmén-Lindelöf theorems.
Proof: By considering ϕ • F for each ϕ ∈ E * , it is enough to consider the case E = C; by considering Re(F ), Im(F ), we can take E = R. Now let F be the unique bounded harmonic function with F = F on ∂Ω.For example, we could obtain F by conformal mapping and the well-known Poisson Formula for the disc.By considering F − F , we only need to prove the special case where F is real-valued, and zero on ∂Ω.
By the Schwarz Reflection Principle, we can extend F to be harmonic on C + and continuous on iR, by defining , where dist means distance.We have F = g + ḡ for some g ∈ Hol(C + ).Now By applying Corollary 4.5 to h repeatedly on the domains {|x − n| < 1 − ǫ} (with trivial rescaling), we obtain for all n = 1, 2, 3, . ... So L(Re u) is bounded and analytic on {Re(z) > 1/2}, with a zero at each n, and thus identically zero everywhere by the Blaschke condition for zero sequences of Hardy space functions.Thus Re u = 0 almost everywhere.
We remark, omitting the details, that Corollary 4.6 can be used to prove Corollary 4.5, so they are equivalent: given an analytic

proof of Theorem 3.1
We first prove (6).First consider the scalar case E = C.By Theorem 2.1 applied to F (z)/z, we see immediately that F (z)/z = Lg(z) for some g, given by If R = +∞ then Theorem 2.1 also gives g(t) ≡ 0 for t 0. But c+iR |dz| < ∞ by Hölder's inequality, so in fact g : R → C is continuous (after changing g on a set of measure zero).

The estimate |g(t)|
M (1 + t) for t > 0 was already proved in [3] for the special case R = +∞ and F ∈ L q (c + iR) for some q > 1.But that proof needed only the estimate |z|>κ for some 0 < ǫ < 1, which follows from ( 5) by Hölder's inequality.The proof also applies without change when R < ∞.For t < 0, we can simply apply the result to F (R − z).The Hölder estimate (7) follows by direct calculation: we have By Hölder's inequality, this is

Documenta Mathematica 15 (2010) 235-254
Here t is small, A, B > 0 depend on c, and |e λ − 1| ≪ |λ| for λ bounded; note also that ν ′ > α.Since (α − 1)/ν ′ = δ/ν, we obtain (7) as required, in the special case where E = C. Now let E be a general Banach space.A standard Closed Graph Theorem argument shows that for all ϕ ∈ E * , with some constant K < ∞.For each ϕ ∈ E * we consider ϕ(F (z)) and apply the scalar-valued case, to obtain a continuous g ϕ : R → C such that Examining the above proof carefully, we find that the constant M in ( 6) is bounded by an absolute constant multiple of with some M ′ < ∞ for t > 0, and similarly for t < 0, so we can define g : R → E * * by g(t)(ϕ) = g ϕ (t).As usual, regard E ⊆ E * * via the canonical embedding.We also have a similar estimate to (7) for g ϕ (t + s) − g ϕ (s) = [g(t + s) − g(s)](ϕ), which gives (7) with g(t + s) − g(s) E * * instead of • E .Crucially, this also shows that g : R → E * * is continuous.But now ϕ(F (z)) = z(Lg ϕ )(z) = [zLg(z)](ϕ), so that F (z) = zLg(z), considered as an E * * -valued function; note that Lg converges because we have an estimate for g(t) E * * .Thus all is finished, except that g(t) ∈ E * * instead of E. Put which is well-defined and continuous because c+iR Hence H(t) = H(0) + t 0 g(τ ) dτ as an E * * -valued integral, so H ′ (t) = g(t) for all t ∈ R, again by continuity of g : R → E * * .Thus finally g(t) ∈ E as required, because H is E-valued.Now F : C + → E is harmonic; so there exist analytic F j ∈ Hol(C + , E), with j = 1, 2, such that F (z) = F 1 (z) + F 2 (z).The functions F 1 , F 2 are unique up to additive constants.We will show that F 1 , F 2 can be chosen to satisfy (5) in Theorem 3.1, and that F 1 , F 2 are almost bounded (with only logarithmic unboundedness); the result will then follow by a similar proof to Theorem 3.1.F is bounded, so we can represent F on {Re(z) > c} by its Poisson integral: We use the standard theory of the Weighted Hilbert Transform, found in [22], [16] and many other sources.The famous Muckenhoupt weight condition w ∈ A ν (R) for w : R → C w f L ν (w) for some constant C w < ∞ depending only on w.
For our problem, we easily check that w(y) = |y| −ς satisfies w ∈ A ν (R) for any 0 < ς < 1. Assume for the moment that E = C. Define where . This is true for almost every α, and F (c + iα), H(α) are both in L ν (|α| −(1−δ) ), and thus so is G 1 (c + iα).Now we have c+iR |G1| ν |z| 1−δ dy < ∞, in the special case E = C.In general, we get the same for ϕ • G 1 , for each ϕ ∈ E * .This is the estimate (5) we need in Theorem 3.1, for G 1 instead of F .Theorem 3.1 does not apply to G 1 because G 1 may be unbounded.However, the unboundedness is at most logarithmic, by two simple calculations: To estimate g j (t) for large t > 0, we use the same method as Theorem 3.1 (which in fact is the method used in [3]), but with additional logarithmic estimates.As usual, consider ϕ • F for each ϕ ∈ E * .In the contour integral formula for t large and some 0 < ǫ < 1, which is ≪ t 2 upon taking κ = exp(ct/ǫ).
with the same norm is not hard: by Fubini's Theorem, b x=a e −xt h(t) 2 L 2 (R,E) v(x) dx < ∞.Thus e −xt h(t) ∈ L 2 for a.e.x ∈ (a, b), because v > 0 a.e.Now the Plancherel Theorem can be applied to the function e −xt h, for a.e.x, and integrating with v(x)dx gives the result.