Laplace transforms and valuations

Laplace transforms and valuations

Jin Li

and Dan Ma

1

Department of Mathematics, Shanghai University, Shanghai 200444, China, [email protected]

2

Institut f¨ ur Diskrete Mathematik und Geometrie, Technische Universit¨ at Wien, Wiedner Hauptstraße 8–10/104, 1040 Wien, Austria,

[email protected]

3

Department of Mathematics, Shanghai Normal University, Shanghai 200234, China, [email protected]

Abstract

It is proved that the classical Laplace transform is a continuous valuation which is positively GL(n) covariant and logarithmic translation covariant. Conversely, these properties turn out to be sufficient to characterize this transform.

1 Introduction

Let f : [0,∞)→R be a measurable function. The Laplace transform of f is given by Lf(s) =

Z ∞

0

e^−stf(t)dt, s ∈R

whenever the integral converges. In the 18th century, Euler first considered this transform to solve second-order linear ordinary differential equations with constant coefficients. One hundred years later, Petzval and Spitzer named this transform after Laplace. Doetsch initiated systematic investigations in 1920s. The Laplace transform now is widely used for

2010Mathematics subject classification. 44A10, 52A20, 52B45

Key words and phrases. Laplace transform, valuation, GL(n) covariance, logarithmic translation covariance.

solving ordinary and partial differential equations. Therefore, it is a useful tool not only for mathematicians but also for physicists and engineers (see, for example, [7]).

The Laplace transform has been generalized to the multidimensional setting in order to solve ordinary and partial differential equations in boundary value problems of several variables (see, for example, [6]). Let f be a compactly supported function that belongs to L¹(Rⁿ). The multidimensional Laplace transform of f is defined as

Lf(x) = Z

Rⁿ

f(y)e^−x·ydy, x∈Rⁿ.

The Laplace transform is also considered on Kⁿ_n, the set ofn dimensional convex bodies (i.e., compact convex sets) in Rⁿ. The Laplace transform of K ∈ K_nⁿ is defined by

LK(x) =L(1K)(x) = Z

K

e^−x·ydy, x∈Rⁿ,

where 1K is the indicator function of K. Making use of the logarithmic version of this transform, Klartag [19] improved Bourgain’s esimate on the slicing problem (or hyperplane conjecture), which is one of the main open problems in the asymptotic theory of convex bodies. It asks whether every convex body of volume 1 has a hyperplane section through the origin whose volume is greater than a universal constant (see also [20] for more information).

Noticing that both Laplace transforms are valuations, we aim at a deeper understanding on these classical integral transforms. A function z defined on a lattice (Γ,∨,∧) and taking values in an abelian semigroup is called a valuation if

z(f∨g) +z(f∧g) = z(f) +z(g) (1.1) for allf, g∈Γ. A functionzdefined on some subset Γ0 of Γ is called a valuation on Γ0 if (1.1) holds whenever f, g, f∨g, f∧g ∈Γ₀. Valuations were a key ingredient in Dehn’s solution of Hilbert’s Third Problem in 1901. They are closely related to dissections and lie at the very heart of geometry. Here, valuations were considered on the space of convex bodies in Rⁿ, denoted byKⁿ. Perhaps the most famous result is Hadwiger’s characterization theorem which classifies all continuous and rigid motion invariant real valued valuations on Kⁿ. Klain [15]

provided a shorter proof of this beautiful result based on the following characterization of the volume.

Theorem 1.1 ( [14,15]). Suppose µ is a continuous rigid motion invariant and simple valuation on Kⁿ. Then there exists c∈R such that µ(K) = cV_n(K), for all K ∈ Kⁿ. Here, V_n is the n dimensional volume.

Other important later contributions can be found in [14,18,34,35]. For more recent results, we refer to [1,2,8–13,16,17,22,24–26,30,32,37,38,40–42,46].

With the first result of this paper, we characterize the Laplace transform on convex bodies.

Theorem 1.2. A map Z : Kⁿ_n → C(Rⁿ) is a continuous, positively GL(n) covariant and logarithmic translation covariant valuation if and only if there exists a constant c ∈R such that

ZK =cLK for every K ∈ Kⁿ_n.

Throughout this paper, without further remark, we briefly write positively GL(n) covariant as GL(n) covariant; see Section2for definitions of GL(n) covariance and logarithmic translation covariance. We call Z : Kⁿ_n → C(Rⁿ) continuous if for every x ∈ Rⁿ, we have ZK_i(x) → ZK(x) whenever K_i → K with respected to the Hausdorff metric, where K_i, K ∈ Kⁿ_n.

Notice that LK(0) = V_n(K) holds for all K ∈ Kⁿ_n. Thus, this characterization is a generalization of Theorem 1.1.

Valuations are also considered on spaces of real valued functions. Here, we take the pointwise maximum and minimum as the join and meet, respectively. Since the indicator functions of convex bodies provide a one-to-one correspondence with convex bodies, valuations on function spaces are generalizations of valuations on convex bodies. Valuations on function spaces have been studied since 2010. Tsang [43] characterized real valued valuations on L^p-spaces. Kone [21] generalized this characterization to Orlicz spaces. As for valuations on Sobolev spaces, Ludwig [27,28] characterized the Fisher information matrix and the Lutwak-Yang-Zhang body. Other recent and interesting characterizations can be found in [3–5,29,33,36,44,45].

With the second result of this paper, we characterize the Laplace transform on functions based on Theorem 1.2 and the natural connection between indicator functions and convex bodies. LetL¹_c(Rⁿ) denote the space of compactly supported functions that belong toL¹(Rⁿ).

We call z : L¹_c(Rⁿ) → C(Rⁿ) continuous if for every x ∈ Rⁿ, we have z(f_i)(x) → z(f)(x) whenever f_i →f in L¹(Rⁿ).

Theorem 1.3. A map z :L¹_c(Rⁿ)→C(Rⁿ) is a continuous, positivelyGL(n) covariant and logarithmic translation covariant valuation if and only if there exists a continuous function h on R with the properties that

h(0) = 0 (1.2)

and that there exists a constant γ ≥0 that

|h(α)| ≤γ|α| (1.3)

for all α ∈R, such that

z(f) = L(h◦f) for every f ∈L¹_c(Rⁿ).

If we further assume that z is 1-homogeneous, that is, z(sf) = sz(f) for all s ∈ R and f ∈L¹_c(Rⁿ), then we obtain the Laplace transform.

Corollary 1.4. A map z : L¹_c(Rⁿ) → C(Rⁿ) is a continuous, 1-homogeneous, positively GL(n) covariant and logarithmic translation covariant valuation if and only if there exists a constant c∈R such that

z(f) =cLf, for every f ∈L¹_c(Rⁿ).

2 Preliminaries and Notation

Our setting is the n-dimensional Euclidean space Rⁿ with the standard basis {e₁, . . . , e_n}, wheren ≥1. The convex hull of a setA is denoted by [A] and the convex hull of a setAand a point x∈Rⁿ will be briefly written as [A, x] instead of [A,{x}]. A hyperplane is an n−1 dimensional affine space in Rⁿ. The unit cube Cⁿ =P

1≤i≤n[o, e_i] and the standard simplex Tⁿ= [o, e₁, . . . , e_n] are two important convex bodies in this paper.

The Hausdorff distance ofK, L∈ Kⁿ is

d(K, L) = inf{ε >0 :K ⊂L+εB, L⊂K +εB}.

The norm on the spaceL¹_c(Rⁿ) is the ordinary L¹ norm which is denoted byk·k.

A map z :L¹_c(Rⁿ)→C(Rⁿ) is called GL(n) covariant if

z(f◦φ⁻¹)(x) = |detφ|z(f)(φ^tx) (2.1) for all f ∈ L¹_c(Rⁿ), φ ∈ GL(n) and x ∈ Rⁿ. In this paper, we actually deal with positive GL(n) covariance, that is (2.1) is supposed to hold for all φ ∈ GL(n) that have positive determinant. Also, a map z :L¹_c(Rⁿ)→C(Rⁿ) is called logarithmic translation covariant if

z(f(· −t))(x) =e^−x·tz(f)(x)

for all f ∈L¹_c(Rⁿ) and t, x∈Rⁿ. This definition is motivated by the relation logL(f(· −t))(x) = −x·t+ logLf(x)

(see Theorem 3.1).

A map Z :K_nⁿ→C(Rⁿ) is called GL(n) covariant if Z(φK)(x) = |detφ|ZK(φ^tx)

for all K ∈ Kⁿ_n, φ ∈ GL(n) and x ∈ Rⁿ. Also, a map z : Kⁿ_n → C(Rⁿ) is called logarithmic translation covariant if

Z(K +t)(x) = e^−t·xZK(x)

for all K ∈ K_nⁿ and t, x∈Rⁿ. Again, it is motivated by the relation logL(K+t)(x) =−t·x+ logLK(x)

(see Theorem 3.3). If a valuation vanishes on lower dimensional convex bodies, we call it simple.

As we will see in Lemma3.2, if z :L¹_c(Rⁿ)→C(Rⁿ) is continuous, GL(n) covariant and logarithmic translation covariant, then Z : K_nⁿ → C(Rⁿ) defined by ZK = z(1K) is also continuous, GL(n) covariant and logarithmic translation covariant, respectively.

For the constant zero function, ifz :L¹_c(Rⁿ)→C(Rⁿ) is GL(n) covariant and logarithmic translation covariant, then

z(0) ≡0. (2.2)

Indeed, z(0)(φ^tx) =z(0)(x) for any φ∈GL(n). Let x=e₁. We have that z(0) is a constant function on Rⁿ\ {0}. The continuity of the functionz(0) now gives that z(0) ≡conRⁿ for a constant c∈R. Since z is also logarithmic translation covariant, z(0)(x) =e^−t·xz(0)(x) for any x, t ∈Rⁿ. Hence z(0)≡0.

3 Laplace transforms

In this section, we study some properties of Laplace transforms.

Theorem 3.1. Let h be a continuous function on R satisfying (1.2) and (1.3). If a map z :L¹_c(Rⁿ)→C(Rⁿ) satisfies

z(f)(x) = Z

Rⁿ

(h◦f)(y)e^−x·ydy

for every x ∈Rⁿ and f ∈L¹_c(Rⁿ), then z is a continuous, GL(n) covariant and logarithmic translation covariant valuation.

In particular, if h(α) = α for all α ∈ R, the Laplace transform L on L¹_c(Rⁿ) is a continuous, GL(n) covariant and logarithmic translation covariant valuation.

Proof. Let f, g∈L¹_c(Rⁿ) and E ={x∈Rⁿ:f(x)≤g(x)}. Then z(f ∨g)(x) =

Z

Rⁿ

h◦(f∨g)(y)e^−x·ydy

= Z

E

(h◦g)(y)e^−x·ydy+ Z

Rⁿ\E

(h◦f)(y)e^−x·ydy for every x∈Rⁿ. Similarly, we have

z(f ∧g)(x) = Z

Rⁿ

h◦(f∧g)(y)e^−x·ydy

= Z

E

(h◦f)(y)e^−x·ydy+ Z

Rⁿ\E

(h◦g)(y)e^−x·ydy for every x∈Rⁿ. Thus,

z(f ∨g)(x) +z(f∧g)(x)

= Z

Rⁿ

(h◦f)(y)e^−x·ydy+ Z

Rⁿ

(h◦g)(y)e^−x·ydy

=z(f)(x) +z(g)(x) for every x∈Rⁿ.

Next, we are going to show thatzis continuous. Letf ∈L¹_c(Rⁿ) and let{f_i}be a sequence in L¹_c(Rⁿ) that converges to f in L¹(Rⁿ). We will show the continuity of z by showing that for every subsequence {z(f_ij)(x)} of {z(f_i)(x)}, there exists a subsequence {z(f_ijk)(x)} that converges to z(f)(x) for every x∈Rⁿ.

Let{f_ij} be a subsequence of {f_i} and y ∈Rⁿ. Then, for everyx ∈Rⁿ, the sequence of functions y 7→ f_ij(y)e^−x·y converges to the function y 7→ f(y)e^−x·y as j → ∞ with respect to the L¹ norm. It follows that there exists a subsequence {f_ijk} of {f_ij} and a nonnegative function F_x ∈L¹(Rⁿ) such that

(i) f_ijk(y)e^−x·y →f(y)e^−x·y almost every y with respect to Lebesgue measure;

(ii) |f_ijk(y)|e^−x·y ≤F_x(y) almost every y with respect to Lebesgue measure (see [23, Section 2.7]). Since h is continuous, we obtain

h◦f_ijk →h◦f a.e.

Also since h satisfies (1.3), we have

|h◦f_ijk(y)| ≤γ|f_ijk(y)| ≤γF_x(y)e^x·y.

Note that R

RⁿF_x(y)e^x·y·e^−x·ydy <∞. We conclude from the dominated convergence theorem that

z(f)(x) = Z

Rⁿ

(h◦f)(y)e^−x·ydy= lim

k→∞

Z

Rⁿ

(h◦f_ijk)(y)e^−x·ydy= lim

k→∞z(f_ijk)(x).

Moreover, forφ ∈GL(n),

z(f◦φ⁻¹)(x) = Z

Rⁿ

(h◦f◦φ⁻¹)(y)e^−x·ydy

=|detφ|

Z

Rⁿ

(h◦f)(w)e^−x·(φw)dw

=|detφ|

Z

Rⁿ

(h◦f)(w)e^−(φtx)·wdw

=|detφ|z(f)(φ^tx) for every x∈Rⁿ and f ∈L¹_c(Rⁿ). Finally, let t∈Rⁿ. Then

z(f(· −t))(x) = Z

Rⁿ

(h◦f)(y−t))e^−x·ydy

= Z

Rⁿ

(h◦f)(w)e^−x·(w+t)dw

=e^−x·t Z

Rⁿ

(h◦f)(w)e^−x·wdw

=e^−x·tz(f)(x) for every x∈Rⁿ and f ∈L¹_c(Rⁿ).

Next, we turn to the Laplace transform on convex bodies.

Lemma 3.2. If z : L¹_c(Rⁿ) → C(Rⁿ) is a continuous, GL(n) covariant and logarithmic translation covariant valuation, then for any α∈R, Z :Kⁿ_n →C(Rⁿ) defined by

ZK =z(α1K)

is a continuous, GL(n) covariant and logarithmic translation covariant valuation on K_nⁿ. Proof. For K, L, K∪L, K∩L∈ K_nⁿ, we have

Z(K∪L) +Z(K∩L) = z(α1K∪L) +z(α1K∩L)

=z((α1K)∨(α1L)) +z((α1K)∧(α1L))

=z(α1K) +z(α1L)

=ZK+ZL.

Also, for a sequence {K_i} in Kⁿ_n that converges to K ∈ Kⁿ_n as i → ∞, we have kα1Ki−α1Kk →0 as i→ ∞. Indeed, for any 0< ε≤1, we have

K_i ⊂K+εB, K ⊂K_i+εB

for sufficiently large i. Hence (K_i\K)∪(K\K_i)⊂ {x: ∃y∈bdK, s.t.d(x, y)≤ε}, where bdK is the boundary of K. Hence,

Z

Rⁿ

|α1Ki(y)−α1K(y)|dy ≤ |α|V_n({x: ∃ y∈bdK, s.t.d(x, y)≤ε})

≤ |α| ·2εS(K+B)

for sufficiently large i. Here, S denotes the surface area. By the continuity of z on L¹_c(Rⁿ), we obtain

Z(K_i) =z(α1Ki)→z(α1K) = ZK as i→ ∞. Moreover, for each φ∈GL(n) and K ∈ Kⁿ_n, we have

Z(φK) = z(α1φK) = z(α1K ◦φ⁻¹)

=|detφ|z(α1K)◦φ^t=|detφ|ZK ◦φ^t. Finally, for each t, x∈Rⁿ, we have

Z(K+t)(x) =z(α1K+t)(x) = z(α1K(· −t))(x)

=e^−t·xz(α1K)(x) =e^−t·xZK(x).

The following theorem directly follows from the definition of the Laplace transform, Theorem 3.1 and Lemma 3.2.

Theorem 3.3. The Laplace transform on Kⁿ_n is a continuous, GL(n) covariant and logarithmic translation covariant valuation.

4 Characterizations of Laplace transforms

In this section, we first characterize the Laplace transform on Kⁿ_n as a continuous, GL(n) covariant and logarithmic translation covariant valuation. Afterwards, via an approach developed from Tsang’s in [43], we further characterize the Laplace transform onL¹_c(Rⁿ).

4.1 The Laplace transform on convex bodies

We first need to extend the valuation to Kⁿ.

Lemma 4.1. If Z : K_nⁿ → C(Rⁿ) is a continuous, GL(n) covariant and logarithmic translation covariant valuation, then Z :Kⁿ→C(Rⁿ) defined by

ZK(x) =

(ZK(x), dimK =n, 0, dimK < n

is a simple, GL(n) covariant and logarithmic translation covariant valuation on Kⁿ.

Proof. The GL(n) covariance and the logarithmic translation covariance are trivial. It remains to show that

ZK(x) = Z(K∩H⁺)(x) +Z(K∩H⁻)(x), x∈Rⁿ (4.1) for every hyperplaneH (whenn = 1,His a single point) such thatK, K∩H⁺, K∩H⁻ ∈ Kⁿ_n. Since Z is logarithmic translation covariant, we can assume w.l.o.g. that o ∈ (intK∩H).

We can further assume that e_n ⊥ H and e_n ∈ H⁺ due to the GL(n) covariace of Z. For a fixed K, note that±se_n∈K for sufficiently smalls >0. Hence the valuation property ofZ shows that

ZK(x) +Z[K∩H,±se_n](x) =Z[K∩H⁺,−se_n](x) +Z[K ∩H⁻, se_n](x) (4.2) for every x∈Rⁿ and sufficiently small s >0. The GL(n) covariance of Z gives that

Z[K∩H,±se_n](x) = sZ[K ∩H,±e_n](x₁e₁+· · ·+xn−1en−1 +sx_ne_n), where x=x₁e₁+· · ·+x_ne_n∈Rⁿ. Since

lim

s→0⁺Z[K∩H,±e_n](x₁e₁+· · ·+xn−1en−1+sx_ne_n)

=Z[K∩H,±e_n](x₁e₁+· · ·+xn−1en−1), we have

lim

s→0⁺Z([K∩H,±se_n])(x)→0. (4.3)

Now combing (4.2) and (4.3) with the continuity of Z, we get that (4.1) holds true.

Next we consider ZCⁿ.

Lemma 4.2. If Z :Kⁿ →C(Rⁿ) is a simple, GL(n) covariant and logarithmic translation covariant valuation, then there exists a constant c∈R such that

ZCⁿ(re₁) = cLCⁿ(re₁) = c Z

Cⁿ

e^−re1·ydy, (4.4)

for every r∈R. Proof. First note that

Z

Cⁿ

e^−re1·ydy= 1

r(1−e^−r). (4.5)

For s > 0, let ψ_s ∈ GL(n) such that ψ_se₁ =se₁ and ψ_se_k =e_k for 2 ≤k ≤ n. For integers p, q >0, since Z is simple, we have

Z(ψ_q/pCⁿ)(e₁) =

q−1

X

j=0

Z

ψ_1/pCⁿ+je₁ p

(e₁).

Also since Z is GL(n) and logarithmic translation covariant, we have q

pZCⁿ q

pe₁

= 1 p

q−1

X

j=0

e^−j/pZCⁿ e₁

p

= 1 pZCⁿ

e1

p

1−e^−q/p 1−e^−1/p. In particular, if q =p, we have

ZCⁿ e1

p

= p(1−e^−1/p)

1−e⁻¹ ZCⁿ(e₁).

Combining the two formulas above with (4.5), and lettingc= ^ZC_1−e^n(e−1¹⁾, (4.4) holds forr=q/p.

Now since ZCⁿ is a continuous function on Rⁿ, (4.4) holds forr≥0.

Forr <0. Repeating the same process for −e₁, we obtain q

pZCⁿ

−q pe₁

= 1 p

q−1

X

j=0

e^j/pZCⁿ

−e₁ p

= 1 pZCⁿ

−e1

p

1−e^q/p 1−e^1/p

and

ZCⁿ

−e₁ p

= p(1−e^1/p)

1−e ZCⁿ(−e₁).

Combining the two formulas above and letting c⁰ =−^ZC_1−e^n(−e1), we have ZCⁿ(re₁) = c⁰

Z

Cⁿ

e^−re1·ydy

for r =−q/p. The continuity of the functionZCⁿ gives that c=c⁰ and thus (4.4) holds for r ≤0.

Now we consider valuations on polytopes. LetPⁿ denote the set of polytopes in Rⁿ and let Z :Pⁿ →C(Rⁿ) be a valuation. The inclusion-exclusion principle states that Z extends uniquely to U(P), the set of finite unions of polytopes, with

Z(P₁∪ · · · ∪P_m) = X

1≤j≤m

(−1)^j−1 X

1≤i₁<···<i_j≤m

Z(P_i1 ∩ · · · ∩P_ij) for every P₁, . . . , P_m∈ Pⁿ (see [31] or [39, Theorem 6.2.1 and Theorem 6.2.3]).

Lemma 4.3. If Z :Pⁿ →C(Rⁿ) is a simple, GL(n) covariant and logarithmic translation covariant valuation such that

ZCⁿ(re₁) = 0 (4.6)

for every r∈R, then

ZTⁿ(re₁) = 0 for every r∈R.

Proof. The case n= 1 is trivial. We only consider n ≥2.

First we prove that ZTⁿ(o) = 0. Since Cⁿ = S

1≤i1<···<in≤n{x ∈ Rⁿ : 0 ≤ x_i1 ≤ · · · ≤ x_in ≤ 1}, and all the sets {x ∈ Rⁿ : 0 ≤ x_i1 ≤ · · · ≤ x_in ≤ 1} are GL(n) transform (with positive determinant) images of Tⁿ, the valuation property, simplicity, and GL(n) covariance of Z combined with (4.6) give that

ZTⁿ(o) = 0.

Next we deal with the case r 6= 0. Let g(m, s) = Z(mTⁿ)(se₁) for s ∈ R and integer m ≥0. For integer k≥1, denote M_kⁿ =kTⁿ∩Cⁿ. Note that when k ≥n, we have

M_kⁿ=Cⁿ. (4.7)

For 1 ≤k ≤n−1,

kTⁿ∪Cⁿ=

n

[

j=1

(kTⁿ∩ {x∈Rⁿ :x_j ≥1})

!

∪Cⁿ (4.8)

Denote T_j =kTⁿ∩ {x∈Rⁿ:x_j ≥1}. We have

T_j = (k−1)Tⁿ+e_j, and

T_j1 ∩ · · · ∩T_ji = (k−i)Tⁿ+e_j1 +· · ·+e_ji

for i ≤ k and 1 ≤ j1 < · · · < ji ≤ n. Hence, the valuation property (after extension), inclusion-exclusion principle, simplicity and logarithmic translation covariance of Z, combined with (4.6) and (4.8), give that

Z(M_kⁿ)(se₁) = Z(kTⁿ)(se₁)−Z(kTⁿ∪Cⁿ)(se₁)

=Z(kTⁿ)(se₁)−

k−1

X

i=1

(−1)ⁱ⁻¹ X

1≤j1<···<ji≤n

Z(T_j1 ∩ · · · ∩T_ji)(se₁)

!

=Z(kTⁿ)(se₁)−

k−1

X

i=1

(−1)ⁱ⁻¹

n−1 i−1

e^−s+

n−1 i

Z((k−i)Tⁿ)(se₁)

!

=

k−1

X

i=0

(−1)ⁱa_i(s)g(k−i, s), (4.9)

where 1≤k≤n−1,ai(s) = ⁿ⁻¹_i−1

e^−s+ ⁿ⁻¹_i

for 1≤i≤n−1, anda0(s) = 1.

For non negative integersk₁, . . . , k_n satisfying k=k₁+· · ·+k_n≤m−1, we have mTⁿ∩(Cⁿ+k₁e₁+· · ·+k_ne_n) = M_m−kⁿ +k₁e₁ +· · ·+k_ne_n.

Form ≥n, applying the valuation property, simplicity, logarithmic translation covariance of

Z, (4.6), (4.7) and (4.9), we have g(m, s) =Z(mTⁿ)(se1)

=

m−1

X

k=0

X

k1+···+kn=k, k1,...,kn≥0

Z(mTⁿ∩(Cⁿ+k₁e₁+· · ·+k_ne_n))(se₁)

=

m−1

X

k=0

X

k1+···+kn=k, k1,...,kn≥0

Z(M_m−kⁿ +k₁e₁+· · ·+k_ne_n)(se₁)

=

m−1

X

k=0 k

X

k1=0

e^−k1s X

k2+···+kn=k−k1, k2,...,kn≥0

Z(M_m−kⁿ )(se₁)

=

m−1

X

k=0

Z(M_m−kⁿ )(se₁)

k

X

k1=0

e^−k1s

k−k₁+n−2 n−2

=

m−1

X

k=m−n+1

m−k−1

X

i=0

(−1)ⁱa_i(s)g(m−k−i, s)

! _k X

k1=0

e^−k1s

k−k₁+n−2 n−2

!

=

n−1

X

j=1

b_j(m, s)g(j, s), (4.10)

where b_j(m, s) =

m−j

P

k=m−n+1

(−1)^m−k−jam−k−j(s)

k

P

k1=0

e^{−k1s k−k}_n−2¹⁺ⁿ⁻²

for 1≤j ≤n−1.

SinceZ is GL(n) covariant, we have

g(m, s) =mⁿg(1, ms).

Hence the equation (4.10) gives that g(1, ms) =

n−1

X

j=1

(j/m)ⁿb_j(m, s)g(1, js). (4.11) For any fixed r6= 0, taking s =r/m in (4.11), we get

g(1, r) =

n−1

X

j=1

(j/m)ⁿbj(m, r/m)g(1, jr/m). (4.12)

Since g(1,·) is a continuous function and g(1,0) = ZTⁿ(o) = 0, if we can show that (j/m)ⁿb_j(m, r/m) is finite when m → ∞, then g(1, r) = 0 which gives the desired result for r 6= 0.

Indeed, for sufficiently large m, (m+n)/m≤ 2 and a_i(r/m), 1 ≤ i≤ n−1 are smaller than a constant N >0. Hence, for 1≤j ≤n−1,

|(j/m)ⁿb_j(m, r/m)| ≤(n/m)ⁿ

m−j

X

k=m−n+1

a_m−k−j(r/m)

k

X

k1=0

e^−k1r/m

k−k₁+n−2 n−2

≤(n/m)ⁿ

m−j

X

k=m−n+1

am−k−j(r/m)

k

X

k1=0

e^−k1r/m(m+n)ⁿ⁻²

= (m+n)ⁿ⁻²(n/m)ⁿ

m−j

X

k=m−n+1

a_m−k−j(r/m)|1−e^−(k+1)r/m|

|1−e^−r/m|

≤2ⁿ⁻²nⁿN(1/m)²

m−j

X

k=m−n+1

|1−e^−(k+1)r/m|

|1−e^−r/m|

≤2ⁿ⁻²nⁿN(1/m)²(n−j) max{1, e^−r}

|1−e^−r/m| .

Note that (1/m)²_|1−e−r/m¹ | → 0 when m → ∞. Hence (j/m)ⁿb_j(m, r/m) → 0 when m →

∞.

For 0 < λ < 1, let Hλ be the hyperplane through the origin with normal vector (1− λ)e₁−λe₂. Since Z :Pⁿ→C(Rⁿ) is a simple valuation,

ZTⁿ(x) = Z(Tⁿ∩H_λ⁻)(x) +Z(Tⁿ∩H_λ⁺)(x), x∈Rⁿ. (4.13) We define φ1, φ2 ∈GL(n) by

φ₁e₁ =λe₁+ (1−λ)e₂, φ₁e₂ =e₂, φ₁e_i =e_i, for 3≤i≤n, and

φ₂e₁ =e₁, φ₂e₂ =λe₁+ (1−λ)e₂, φ₂e_i =e_i, for 3≤i≤n.

Note that Tⁿ∩H_λ⁻ = φ₁Tⁿ, Tⁿ∩H_λ⁺ = φ₂Tⁿ. The GL(n) covariance of Z and valuation relation (4.13) show that

ZTⁿ(x) =λZTⁿ(φ^t₁x) + (1−λ)ZTⁿ(φ^t₂x)

Let f(·) = ZTⁿ(·). We have

f(x) =λf(φ^t₁x) + (1−λ)f(φ^t₂x) (4.14) for every 0< λ <1 andx= (x₁, . . . , x_n)^t∈Rⁿ, whereφ^t₁x= (λx₁+ (1−λ)x₂, x₂, x₃, . . . , x_n)^t and φ^t₂x= (x₁, λx₁+ (1−λ)x₂, x₃, . . . , x_n)^t.

Lemma 4.4. Let n≥2 and let the function f ∈C(Rⁿ) satisfy the following properties.

(i) f satisfies the functional equation (4.14);

(ii) For every even permutation π and x∈Rⁿ, f(x) =f(πx).

If f(re₁) = 0 for every r∈R, then

f(x) = 0 for every x∈Rⁿ.

Proof. We prove the statement by induction on the number m of coordinates ofx not equal to zero. By property (ii), we can assume that the first m coordinates of x are not equal to zero.

It is trivial that the statement is true for m = 1. Assume that the statement holds true for m−1. We want to show that

f(x₁e₁+· · ·+x_me_m) = 0 (4.15) for all the x1, . . . , xm not zero. For x1 > x2 > 0 or 0 > x2 > x1, taking x = x1e1+x3e3 +

· · ·+x_me_m, λ= ^x_x²

1 in (4.14), we get f(x₁e₁+x₃e₃+· · ·+x_me_m)

= x₂ x1

f(x₂e₁+x₃e₃+· · ·+x_me_m) +

1−x₂ x1

f(x₁e₁+x₂e₂+x₃e₃+· · ·+x_me_m). (4.16) For x₂ > x₁ >0 or 0> x₁ > x₂, taking x=x₂e₂+x₃e₃+· · ·+x_me_m, 1−λ= ^x_x¹

2 in (4.14), we get

f(x₂e₂+x₃e₃+· · ·+x_me_m),

=

1−x₁ x₂

f(x₁e₁+x₂e₂+x₃e₃+· · ·+x_me_m) + x₁

x₂f(x₁e₂+x₃e₃+· · ·+x_me_m). (4.17)

Forx₁ >0> x₂ orx₂ >0> x₁, taking 0< λ= _x^x2

2−x1 <1 andx=x₁e₁+x₂e₂+x₃e₃+· · ·+ x_me_m in (4.14), we get

f(x₁e₁+x₂e₂+x₃e₃+· · ·+x_me_m)

= x₂

x₂−x₁f(x₂e₂+x₃e₃+· · ·+x_me_m) + −x₁

x₂−x₁f(x₁e₁+x₃e₃+· · ·+x_me_m). (4.18) Now, combined with the induction assumption and the continuity of f, (4.16), (4.17) and (4.18) show that (4.15) holds true.

Proof of Theorem 1.2. By Theorem3.3,cLis a continuous, GL(n) covariant and logarithmic translation covariant valuation on K_nⁿ.

Now we turn to the reverse statement. Since Z : Kⁿ_n → C(Rⁿ) is a continuous, GL(n) covariant and logarithmic translation covariant valuation, Lemma 4.1 allows us to extend this valuation to a simple, GL(n) covariant and logarithmic translation covariant valuation on Kⁿ. Hence Lemma 4.2 gives that there exists a constant c∈R such that

ZCⁿ(re₁) = cLCⁿ(re₁) for every r ∈R. Now defineZ⁰ :Pⁿ →C(Rⁿ) by

Z⁰P(x) =ZP(x)−cLP(x), x∈Rⁿ.

It is easy to see thatZ⁰is also a simple, GL(n) covariant and logarithmic translation covariant valuation on Kⁿ. Also Z⁰Cⁿ(re₁) = 0 for every r ∈ R. Applying Lemma 4.3 (for Z⁰) and Lemma 4.4 (for f =Z⁰Tⁿ) we get

Z⁰Tⁿ = 0.

Now using the inclusion-exclusion principle and the GL(n) covariance and the simplicity of Z⁰ again, we have

Z⁰P = 0

for every P ∈ Pⁿ since every P ∈ Pⁿ can be dissected into finite pieces of GL(n) (with positive determinant) transforms and translations of Tⁿ. Hence

ZP =cLP

for every P ∈ Pⁿ. Since both Z and L are continuous on K_nⁿ, ZK =cLK

for every K ∈ K_nⁿ.

4.2 Laplace transforms on functions

We first consider indicator functions of Borel sets.

Lemma 4.5. If a map z : L¹_c(Rⁿ) → C(Rⁿ) is a continuous, GL(n) covariant and logarithmic translation covariant valuation, then there exists a continuous function h on R satisfying (1.2) and (1.3) such that

z(α1E)(x) = h(α) Z

Rⁿ

1E(y)e^−x·ydy for every α∈R, x∈Rⁿ and bounded Borel set E ⊂Rⁿ.

Proof. For any α∈R, define Z_α:Kⁿ_n →C(Rⁿ) by Z_αK =z(α1K)

for every K ∈ K_nⁿ. Lemma 3.2 shows that Z_α is a continuous, GL(n) and logarithmic translation covariant valuation on K_nⁿ.

Therefore, by Theorem 1.2, there exists a functionh ∈R such that z(α1K)(x) =Z_αK(x) =h(α)LK(x)

=h(α) Z

K

e^−x·ydy=h(α) Z

Rⁿ

1K(y)e^−x·ydy

for every K ∈ Kⁿ_n and x ∈ Rⁿ. The function h is continuous since z is continuous and kα_i1K−α1Kk →0 wheneverα_i →α. z(0) = 0 (see (2.2)) gives thath(0) = 0. In particular, this representation holds on Par(n), the set of finite union of cubes, by the inclusion-exclusion principle.

Now we consider Borel sets in Rⁿ. For each bounded Borel set E ⊂ Rⁿ, there exists a sequence {K_i} in Par(n) such that α1Ki → α1E in L¹_c(Rⁿ) for every α ∈ R as i → ∞.

Moreover, for every x∈Rⁿ, 0≤ lim

i→∞

Z

Rⁿ

(1Ki(y)−1E(y))e^−x·ydy

≤ lim

i→∞

Z

Rⁿ

|1Ki(y)−1E(y)|e^−x·ydy

≤max

y∈E e^−x·y lim

i→∞

Z

Rⁿ

|1Ki(y)−1E(y)|dy

= 0.

Thus, the continuity of z gives that z(α1E)(x) = lim

i→∞z(α1Ki)(x) =h(α) lim

i→∞

Z

Rⁿ

1Ki(y)e^−x·ydy

=h(α) Z

Rⁿ

1E(y)e^−x·ydy.

The last step is to show thath satisfies (1.3). Ifhdoes not satisfy (1.3), then there exists a sequence {α_j} in R\ {0} (since h satisfies (1.2)) such that

|h(α_j)|>2ⁱ|α_j|. (4.19)

Set E_j = [0,2^−j/|α_j|]×[0,1]ⁿ⁻¹. We have Z

Ej

dy = 2^−j

|α_j|. Let g_j =α_j1Ej and f ≡0. Clearlyg_j, f ∈L¹_c(Rⁿ). Since

Z

Rⁿ

|gj(y)|dy=|αj| Z

Ej

dy = 2^−j →0

when j → ∞. Hence kg_j −fk →0. The continuity of z now implies that z(f)(o) = lim

j→∞z(g_j)(o)dy.

On the other hand, z(f)(o) = 0 (see (2.2)) and the above statement gives that z(g_j)(o) = h(α_j)

Z

Rⁿ

1Ej(y)dy However, since h satisfies (4.19), we obtain

0 =|z(f)(o)|= lim

j→∞|z(g_j)(o)|= lim

j→∞|h(α_j)|

Z

Ej

dy

= lim

j→∞

|h(α_j)|

2^j|α_j|

= lim sup

j→∞

|h(α_j)|

2^j|α_j|

≥1.

It is a contradiction. Hence h satisfies (1.3).

Next, we deal with simple functions

Lemma 4.6. Let z :L¹_c(Rⁿ)→C(Rⁿ)be a valuation. Suppose that there exists a continuous function h on R satisfying (1.2) and (1.3) such that

z(α1E)(x) = h(α) Z

Rⁿ

1E(y)e^−x·ydy

for every α∈R, x∈Rⁿ and bounded Borel set E ⊂Rⁿ. Then z(g) =

Z

Rⁿ

(h◦g)(y)e^−x·ydy for every simple function g ∈L¹_c(Rⁿ).

Proof. Letg ∈L¹_c(Rⁿ) be a simple function. We can writeg =Pm

i=1α_i1Ei, whereE₁, . . . , E_m are disjoint bounded Borel sets and α₁, . . . , α_m ∈R. Hence

z(g) =zX^m

i=1

α_i1Ei

=z((α₁1E1)∨ · · · ∨(α_m1Em)) =

m

X

i=1

z(α_i1Ei), (4.20) where the last equation is the inclusion-exclusion principle for valuations on the lattice (L¹_c(Rⁿ),∨,∧).

Sinceh◦g =Pm

i=1h(αi)1Ei, by (4.20), we obtain z(g) =

m

X

i=1

z(αi1Ei) =

m

X

i=1

h(αi) Z

Rⁿ

1Ei(y)e^−x·ydy

= Z

Rⁿ m

X

i=1

h(α_i)1Ei(y)e^−x·ydy= Z

Rⁿ

(h◦g)(y)e^−x·ydy.

Finally, we are ready to prove Theorem1.3.

Proof of Theorem 1.3. Theorem3.1shows thatf 7→ L(h◦f) is a continuous, GL(n) covariant and logarithmic translation covariant valuation. It remains to show the reverse statement.

For a nonnegative functionf ∈L¹_c(Rⁿ), there exists an increasing sequence of nonnegative simple functions {g_k} ⊂ L¹_c(Rⁿ) such that g_k ↑ f pointwise. The monotone convergence theorem gives that kg_k−fk → 0. Note that every function f ∈ L¹_c(Rⁿ) can be written as f =f₊−f−, where

f+=

(f(x), x∈ {f ≥0}

0, x∈ {f <0}, f− =

(0, x∈ {f ≥0}

−f(x), x∈ {f <0}.

Hence the above statement gives that there exists a sequence of simple functions {g_k} ⊂ L¹_c(Rⁿ) such that gk →f pointwise and kgk−fk → 0 by the triangle inequality. Moreover,

the increasing sequence |g_k(x)| ↑ |f(x)|for everyx∈Rⁿ. Due to the continuity of z, Lemma 4.5 and Lemma4.6, we have

z(f)(x) = lim

k→∞z(g_k)(x) = lim

i→∞

Z

Rⁿ

(h◦g_k)(y)e^−x·ydy, (4.21) where h is a continuous function satisfying (1.2) and (1.3). Therefore,

|h◦g_k| ≤γ|g_k| ≤γ|f|.

The dominated convergence theorem, the continuity of h, and (4.21) now yield z(f) = lim

i→∞

Z

Rⁿ

(h◦g_k)(y)e^−x·ydy

= Z

Rⁿ

(h◦f)(y)e^−x·ydy

=L(h◦f).

Acknowledgement

The authors wish to thank the referee for valuable suggestions and careful reading of the original manuscript. The work of the first author was supported in part by the National Natural Science Foundation of China (11671249) and the Shanghai Leading Academic Discipline Project (S30104). The work of the second author was supported in part by the European Research Council (ERC) Project 306445 and the Austrian Science Fund (FWF) Project P25515-N25.

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