Positive definite radial kernels on homogeneous trees and products
Ignacio Vergara
math.FA math.OA Primary 43A35, Secondary 46L07, 47B10, 05C63


Positive definite radial kernels]Positive definite radial kernels on
 homogeneous trees and products



I. Vergara]Ignacio Vergara

Institute of Mathematics of the Polish Academy of Sciences,
 ul. Śniadeckich 8, 00-656 Warszawa, Poland.



ivergara@impan.pl












Primary 43A35; Secondary 46L07, 47B10, 05C63








We give a new proof of a classical result which provides a one-to-one correspondence between positive definite radial kernels on a homogeneous tree and finite Borel measures on the interval [-1,1]. Our methods allow us to find a new characterisation in terms of positive trace-class operators on ℓ_2. Furthermore, we extend both characterisations to finite products of homogeneous trees. The proof relies on a formula for the norm of radial Schur multipliers, in the spirit of Haagerup–Steenstrup–Szwarc, and a variation of the Hamburger moment problem.


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    December 30, 2023
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§ INTRODUCTION


Let X be a nonempty set. We say that a function ψ:X× X→ is positive definite (or that it is a positive definite kernel on X) if, for every finite family x_1,...,x_n∈ X and every z_1,...,z_n∈, we have

    ∑_i,j=1^nψ(x_i,x_j)z_iz_j ≥ 0.


Positive definite kernels are particular cases of Schur multipliers. Let ℓ_2(X) be the Hilbert space of square-summable, complex-valued functions on X. A function ψ:X× X→ is said to be a Schur multiplier on X if, for every bounded operator T∈ℬ(ℓ_2(X)), there exists another bounded operator M_ψ(T)∈ℬ(ℓ_2(X)) whose matrix coefficients are given by

    ⟨ M_ψ(T)δ_y,δ_x⟩ = ψ(x,y) ⟨ Tδ_y,δ_x⟩,  ∀ x,y∈ X.

In this case, M_ψ defines a completely bounded map on ℬ(ℓ_2(X)), and we denote by ψ_cb its completely bounded norm, which coincides with its norm as a bounded operator on ℬ(ℓ_2(X)). See <cit.> for more details.

Positive definite kernels can be characterised by the complete positivity of the associated map (<ref>).


Let X be a non-empty set and let ψ:X× X→ be a function which is constant on the diagonal. Then ψ is positive definite if and only if the map M_ψ given by (<ref>) is completely positive.


For a proof of this fact, see e.g. <cit.>. For more on completely positive maps, see <cit.>. 

If now X is (the set of vertices of) a connected graph, we say that ψ:X× X→ is radial if it depends only on the distance between each pair of vertices. In other words, if there exists φ:→ such that

    ψ(x,y)=φ(d(x,y)),  ∀ x,y∈ X.

Here the distance d(x,y) is given by the length of the shortest path between x and y. We can thus define the following subsets of ℓ_∞():

    ℛ(X)    ={φ:→ | φ defines a radial Schur multiplier on X},
    ℛ_+(X)    ={φ:→ | φ defines a positive definite radial kernel on X}.

It follows that ℛ(X) is a linear subspace of ℓ_∞(), and ℛ_+(X) is a convex cone in ℛ(X).

Let 2≤ q≤∞ and let 𝒯_q denote the (q+1)-homogeneous tree (the choice of this somewhat strange notation will become clearer afterwards). It was proven by Haagerup, Steenstrup and Szwarc <cit.> that a function φ:→ belongs to ℛ(𝒯_q) if and only if the Hankel matrix

    H=(φ(i+j)-φ(i+j+2))_i,j∈

is of trace class. See <cit.> for details on trace-class operators. Observe that the aforementioned characterisation is independent of q. In other words, ℛ(𝒯_q)=ℛ(𝒯_∞) for all 2≤ q<∞. Since ℛ_+(𝒯_q)⊂ℛ(𝒯_q), the elements of ℛ_+(𝒯_q) must at least satisfy this trace-class condition. 

Fix 2≤ q<∞ and let (P_n^(q))_n∈ be the family of polynomials defined recursively as follows:

    P_0^(q)(x)=1,    P_1^(q)(x)=x,

and

    P_n+1^(q)(x)=(1+1q)xP_n^(q)(x)-1qP_n-1^(q)(x),  ∀ n≥ 1.

For z∈ fixed, the function n↦ P_n^(q)(z) corresponds to an eigenfunction of the Laplace operator on 𝒯_q, with eigenvalue z. Arnaud <cit.> proved that a function φ:→ belongs to ℛ_+(𝒯_q) if and only if there exists a positive Borel measure μ on [-1,1] such that

    φ(n)=∫_-1^1 P_n^(q)(s) dμ(s),

for all n∈. See also <cit.>. Haagerup and Knudby <cit.> gave a new proof of this fact for odd values of q, and extended it to q=∞. In the latter case, the sequence (P_n^(∞)) is given by

    P_n^(∞)(x)=x^n,  ∀ n∈.

In this paper, we provide a new proof of this characterisation by means of the Haagerup–Steenstrup–Szwarc formula for radial Schur multipliers. We prove the following.


Let 2≤ q≤∞ and let φ:→. The following are equivalent:


  a) The function φ belongs to ℛ_+(𝒯_q).

  b) The operator B=(B_i,j)_i,j∈ given by

    B_i,j   =φ(i+j)-φ(i+j+2),       (q=∞)
    
    B_i,j   =∑_k=0^min{i,j}1/q^k(φ(i+j-2k)-φ(i+j-2k+2)),       (q<∞)

is of trace class and positive. Moreover, the following limits

    l_0   =lim_n→∞φ(2n),     l_1   =lim_n→∞φ(2n+1)

 exist, and

    |l_1|≤ l_0.


  c) There exists a positive Borel measure on [-1,1] such that

    φ(n)=∫_-1^1 P_n^(q)(s) dμ(s),

for all n∈.



The equivalence (a)⇔(b) will follow from the characterisation of ℛ(𝒯_q) proven in <cit.>. The proof of (b)⇒(c) is inspired by Hamburger's solution to the moment problem <cit.>. Observe that when q=∞, B coincides with the Hankel matrix H defined in (<ref>).

Since ℛ(𝒯_q)=ℛ(𝒯_∞) for all 2≤ q<∞, a natural question is what is the relation between ℛ_+(𝒯_q) and ℛ_+(𝒯_r) for q≠ r. Observe that there is a sequence of isometric embeddings

    𝒯_2↪𝒯_3↪⋯↪𝒯_∞.

Such embeddings can be defined as follows. Choose any two vertices x∈𝒯_q, y∈𝒯_q+1, and map x↦ y. Then inductively map neighbours to neighbours. Since the degree of 𝒯_q+1 is bigger, there is always enough room for this procedure, and since there are no cycles, the resulting map is isometric. This yields a sequence of inclusions

    ℛ_+(𝒯_2)⊃ℛ_+(𝒯_3)⊃⋯⊃ℛ_+(𝒯_∞).

Moreover, since the condition (<ref>) involves only finitely many points, these isometric embeddings allow one to see that

    ℛ_+(𝒯_∞)=⋂_q≥ 2ℛ_+(𝒯_q).

As a consequence of Theorem <ref>, we obtain the following.


For all 2≤ q<∞, we have

    ℛ_+(𝒯_q+1)⊊ℛ_+(𝒯_q).



In the proof of Corollary <ref>, we construct explicit examples of elements φ∈ℛ_+(𝒯_q)∖ℛ_+(𝒯_q+1), namely φ(n)=P_n^(q)(0), where (P_n^(q)) is the family of polynomials defined in (<ref>).

We also study positive definite kernels on products of trees. Given a family of connected graphs X_1,...,X_N, the product

    X=X_1×⋯× X_N

is the graph whose vertices are N-tuples (x_1,...,x_N), with x_i∈ X_i, and there is an edge between (x_1,...,x_N) and (y_1,...,y_N) if and only if

    ∑_i=0^N d_i(x_i,y_i) = 1.

Here d_i stands for the edge-path distance on X_i. Observe that every X_i can be isometrically embedded into X by simply fixing the other coordinates. We will prove an extension of Theorem <ref> to products. For n=(n_1,...,n_N)∈^N and m=(m_1,...,m_N)∈^N, we denote by m∧ n the element of ^N given by

    (m∧ n)_i=min{m_i,n_i},  ∀ i∈{1,...,N}.

We will also write m≤ n if m_i≤ n_i for all i. Fix 2≤ q_1,...,q_N<∞ and set X=𝒯_q_1×⋯×𝒯_q_N. Now we consider N families of polynomials (P_n^(q_i))_n∈ (i=1,...,N) defined as in (<ref>).








Let φ:→, and let X be as above. The following are equivalent:


  a) The function φ belongs to ℛ_+(X).

  b) The operator B=(B_n,m)_m,n∈^N on ℓ_2(^N), given by

    B_m,n=∑_l≤ m∧ n1/q_1^l_1⋯ q_N^l_N∑_k=0^NNk(-1)^k φ(|m|+|n|-2|l|+2k),

is of trace class and positive. Moreover, the following limits

    l_0   =lim_n→∞φ(2n),     l_1   =lim_n→∞φ(2n+1)

 exist, and

    |l_1|≤ l_0.


  c) There exists a positive Borel measure μ on [-1,1]^N such that

    φ(n_1+⋯ +n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N),

for all (n_1,...,n_N)∈^N.



Observe that Theorem <ref> (for q<∞) is a particular case of Theorem <ref>. The latter could also be stated in a more general form, allowing some of the q_i to be ∞, but this case does not really give new information, as we will explain now.

Let X be a connected graph. For x,y∈ X, we define the interval between x and y as the set

    I(x,y)={u∈ X  :  d(x,y)=d(x,u)+d(u,y)}.

We say that  X is median if 

    |I(x,y)∩ I(y,z)∩ I(z,x)|=1,  ∀ x,y,z∈ X.

The class of median graphs is very large, and it contains trees and finite products of trees. It was proven by Chepoi <cit.> that a graph is median if and only if it is the 1-skeleton of a CAT(0) cube complex. The following result is essentially a consequence of the fact that the edge-path distance on such graphs is a conditionally negative kernel (see <cit.>).


Let X be a median graph. Then

    ℛ_+(𝒯_∞)⊂ℛ_+(X).



Therefore, if a median graph X contains an isometric copy of 𝒯_∞, then ℛ_+(X)=ℛ_+(𝒯_∞), and we already know what this set looks like, thanks to Theorem <ref>. In particular,

    ℛ_+(𝒯_∞)= ℛ_+(𝒯_∞^N),

for all N≥ 1. Here 𝒯_∞^N stands for the direct product of N copies of 𝒯_∞. So again, this is a very different situation than the one of radial Schur multipliers. Indeed, the following is a consequence of <cit.> and <cit.>.


Let N and q be natural numbers with N≥ 1 and 2≤ q<∞. Then

    ℛ(𝒯_q^N)=ℛ(𝒯_∞^N),

and

    ℛ(𝒯_∞^N+1)⊊ℛ(𝒯_∞^N).



Again, since there is an isometric inclusion 𝒯_q^N↪𝒯_q'^N' for q'≥ q and N'≥ N, we know that

    ℛ_+(𝒯_q'^N')⊂ℛ_+(𝒯_q^N).

It seems natural to ask whether some of these inclusions are actually equalities. We do not have a complete answer to this question, but Theorem <ref> allows us to conclude the following.


For all 2≤ q<∞, we have

    ℛ_+(𝒯_q×𝒯_q)⊊ℛ_+(𝒯_q).



We make one final remark concerning this question. The equivalence (a)⇔(c) in Theorem <ref> associates, to each element of ℛ_+(X), a Borel measure on [-1,1]^N, but not conversely. Indeed, the left-hand side of (<ref>) depends only on the sum n_1+⋯ +n_N, and this imposes severe restrictions on the measure μ. To illustrate this fact, suppose that φ:→ satisfies

    φ(n_1+n_2)=∫_[-1,1]^2t_1^n_1t_2^n_2  dμ(t_1,t_2),  ∀(n_1,n_2)∈^2,

for some positive measure μ on [-1,1]^2. Then, by computing the integral of t_1^n_1t_2^n_2(t_1-t_2)^2, one can show that μ must be supported on the diagonal of [-1,1]^2. This is actually another way of proving that ℛ_+(𝒯_∞^2)=ℛ_+(𝒯_∞). This suggests that a more thorough analysis of the identity (<ref>) could lead to a better understanding of the set ℛ_+(𝒯_q^N) and the inclusions (<ref>).



 §.§ Organisation of the paper

In Section <ref>, we state some general results concerning completely positive multipliers and trace-class operators. Section <ref> is devoted to the equivalence (a)⇔(b) in both Theorems <ref> and <ref>. In Section <ref>, we study a variant of the Hamburger moment problem and prove the implication (b)⇒(c). We complete the proof of the main results in Section <ref>. Corollaries <ref> and <ref> are proven in Section <ref>. Finally, in Section <ref>, we discuss median graphs and prove Proposition <ref>.





§ PRELIMINARIES


We present here some general facts on completely positive multipliers and positive trace-class operators, which will be crucial in our proofs.



 §.§ Completely positive multipliers


Let A and B be C^*-algebras. A linear map u:A→ B is said to be completely bounded if

    u_cb=sup_n≥ 1id_n⊗ u:M_n(A)→ M_n(B) <∞.

It is completely positive if id_n⊗ u is positive for each n. The following fact is essential for our purposes. See e.g. <cit.> for a proof.


Let A and B be unital C^*-algebras and let u:A→ B be a unital linear map. Then u is completely positive if and only if

    u_cb=1.



Every Schur multiplier ψ:X× X→ defines a completely bounded map M_ψ:ℬ(X)→ℬ(X), and the cb norm coincides with the usual operator norm. See <cit.> for a proof. We shall write

    ψ_cb=M_ψ_cb=M_ψ.


The following characterisation of positive definite radial kernels will be one of our main tools. It is a direct consequence of Propositions <ref> and <ref>.


Let X be a connected graph and let ψ:X× X→ be a radial Schur multiplier defined by a function φ:→. Then ψ is positive definite if and only if

    ψ_cb=φ(0).





 §.§ Positive trace-class operators


Let ℋ be a separable Hilbert space. We denote by S_1(ℋ) the space of trace-class operators on ℋ, and by S_1(ℋ)_+ the subset of positive elements of S_1(ℋ). See <cit.> for details on trace-class operators. The following fact is well known. We include its proof for the sake of completeness.


Let A∈ S_1(ℋ). Then A is positive if and only if Tr(A)=A_S_1.


If A is positive, then A_S_1= Tr(A) by the definition of the S_1-norm. Suppose now that A_S_1= Tr(A). By the singular value decomposition of compact operators (see e.g. <cit.>), A can be written as

    A=∑_n∈λ_n u_n⊙ v_n,

where (λ_n) is a sequence of non-negative numbers converging to 0, (u_n) and (v_n) are two orthonormal families in ℋ, and u_n⊙ v_n is the rank 1 operator given by (u_n⊙ v_n)f=⟨ f,v_n⟩ u_n. Moreover, if A=U|A| is the polar decomposition of A, then

    |A|=∑_n∈λ_n v_n⊙ v_n,

and u_n=Uv_n. We have

    A_S_1=Tr(|A|)=∑λ_n,

and

    Tr(A) = ∑_n∈⟨ Av_n,v_n⟩ = ∑_n∈⟨λ_nu_n,v_n⟩ 
    = ∑_n∈λ_n⟨ Uv_n,v_n⟩.

Hence

    ∑_n∈λ_n=∑_n∈λ_n⟨ Uv_n,v_n⟩.

Since λ_n≥ 0 and |⟨ Uv_n,v_n⟩|≤ 1, we must have ⟨ Uv_n,v_n⟩=1 for all n, which implies that Uv_n=v_n. Therefore u_n=v_n and A=|A|.








§ PRODUCTS OF TREES AND MULTI-RADIAL KERNELS


In this section we deal with the equivalence (a)⇔(b) in Theorems <ref> and <ref>. We will work in the slightly more general context of multi-radial kernels. The main reason is that this will make computations on products a little easier, but it may also be of independent interest.



 §.§ Homogeneous trees

We begin by recalling some facts and definitions from <cit.>. Let 𝒯_q be the (q+1)-homogeneous tree (2≤ q<∞). We fix an infinite geodesic ray ω_0:→𝒯_q. For each x∈𝒯_q there exists a unique geodesic ray ω_x:→𝒯_q which eventually flows with ω_0, meaning that |ω_xΔω_0|<∞, where we view ω_0 and ω_x as subsets of 𝒯_q. Using these rays, we can construct a family of vectors (δ_x')_x∈𝒯_q in ℓ_2(𝒯_q) such that

    ⟨δ_x',δ_y'⟩ = 
    1    if  x=y,
    
    -1/q-1   if  x≠ y  and ω_x(1)=ω_y(1),
    
    0    otherwise.

See the discussion in <cit.> after Remark 2.4. Now let S denote the forward shift operator on ℓ_2(), and define

    S_m,n = (1-1/q)^-1(S^m(S^*)^n-1/q S^*S^m(S^*)^nS),

for all m,n∈.

<cit.>
Let x,y∈𝒯_q. Let m and n be the smallest numbers such that ω_x(m)∈ω_y and ω_y(n)∈ω_x (this implies ω_x(m)=ω_y(n) and d(x,y)=m+n). Then

    (S_m,n)_ij=⟨δ_ω_y(j)',δ_ω_x(i)'⟩,

for all i,j∈.




 §.§ Multi-radial kernels


Let X=X_1×⋯× X_N be a product of N connected graphs. We say that ϕ:X× X→ is a multi-radial function if there exists ϕ̃:^N→ such that

    ϕ(x,y)=ϕ̃(d(x_1,y_1),...,d(x_N,y_N)),  ∀ x,y∈ X,

where x=(x_i)_i=1^N, y=(y_i)_i=1^N, and d_i is the edge-path distance on X_i. It readily follows that a radial function ϕ:X× X→, given by ϕ(x,y)=φ(d(x,y)), is multi-radial with

    ϕ̃(n_1,...,n_N)=φ(n_1+⋯ +n_N),  ∀ (n_1,...,n_N)∈^N.

We recall some facts and notations from <cit.>. Put [N]={1,...,N}. For each I⊂[N], we define χ^I∈{0,1}^N by

    χ^I_i= 1    if  i∈ I 
    
    0    if  i∉ I.

Consider the space ℓ_2(^N)=ℓ_2()⊗⋯⊗ℓ_2() and denote by S_i the forward shift operator on the i-th coordinate. Namely,

    S_1(δ_n_1⊗δ_n_2⊗⋯⊗δ_n_N) = (δ_n_1+1⊗δ_n_2⊗⋯⊗δ_n_N),

and so forth. For each i∈{1,...,N}, define S_i,m,n∈ℬ(ℓ_2(^N)) like in (<ref>), and put

    S(m,n)=∏_i=1^N S_i,m_i,n_i,  ∀ m,n∈^N.




 §.§ Products of homogeneous trees

A characterisation of multi-radial multipliers on products of trees was given in <cit.>, together with some estimates on the norms. We shall revisit the proof of this result and obtain an exact formula for the norm in the homogeneous case. This will generalise <cit.>. For details on Schur multipliers, see <cit.>.

From now on, fix 2≤ q_1,...,q_N<∞ and let 𝒯_q_i denote the (q_i+1)-homogeneous tree (i=1,...,N). We will focus on the graph

    X=𝒯_q_1×⋯×𝒯_q_N.

Consider, for each i=1,...,N, the operator τ_i:ℬ(ℓ_2(^N))→ℬ(ℓ_2(^N)) defined by

    τ_i(T)=S_iTS_i^*.

Since τ_i is isometric on S_1(ℓ_2(^N)) (see the proof of <cit.>), the formula

    ∏_i=1^N(1-1/q_i)^-1(I-τ_i/q_i)

defines an isomorphism of S_1(ℓ_2(^N)). Observe that this is a product of a commuting family of operators, hence there is no ambiguity in this definition.



Let ϕ̃:^N → be a function such that the limits

    l_0   =lim_|n|→∞
     |n| evenϕ̃(n)     l_1   =lim_|n|→∞
     |n| oddϕ̃(n)

exist. Then ϕ̃ defines a multi-radial Schur multiplier on X if and only if the operator T=(T_n,m)_m,n∈^N given by

    T_m,n=∑_I⊂[N](-1)^|I|ϕ̃(m+n+2χ^I),  ∀ m,n∈^N

is an element of ∈ S_1(ℓ_2(^N)). Moreover, in that case, the associated multiplier ϕ satisfies

    ϕ_cb=∏_i=1^N(1-1/q_i)(I-τ_i/q_i)^-1T_S_1 + |c_+|+|c_-|,

where c_±=l_0± l_1.


The characterisation of such multipliers was already proven in <cit.>. Hence we only need to show that the identity (<ref>) holds. Observe that one of the inequalities is somehow implicit in <cit.>. By <cit.>, there exists a trace-class operator T' such that

    ϕ̃(m+n)=c_++(-1)^|m|+|n|c_-+Tr(S(m,n)T'),  ∀ m,n∈,

and

    T' _S_1 + |c_+|+|c_-| ≤ϕ_cb.

Then it is shown in (the proof of) <cit.> that

    T'=∏_i=1^N(1-1/q_i)(I-τ_i/q_i)^-1T.

In order to prove the other inequality, we shall use the characterisation of Schur multipliers as scalar products on a Hilbert space (see <cit.>). Let T' be as in (<ref>). Since T' is of trace class, there exist sequences (ξ^(k))_k≥ 0, (η^(k))_k≥ 0 in ℓ_2(^N) such that

    T'=∑_k≥ 0ξ^(k)⊙η^(k),

and

    T'_S_1=∑_k≥ 0ξ^(k)_2η^(k)_2.

See the discussion right after <cit.>. This implies that

    T_m,n'=∑_k≥ 0ξ^(k)_m η^(k)_n,  ∀ m,n∈^N.

Now, for each k∈, define functions P_k,Q_k:X→ℓ_2(^N) by

    P_k(x)   =∑_l_1,...,l_N≥ 0ξ^(k)_(l_1,...,l_N)δ_ω_x_1(l_1)'⊗⋯⊗δ_ω_x_N(l_N)',
    
    Q_k(y)   =∑_j_1,...,j_N≥ 0η^(k)_(j_1,...,j_N)δ_ω_y_1(j_1)'⊗⋯⊗δ_ω_y_N(j_N)',

for all x=(x_1,...,x_N)∈ X and y=(y_1,...,y_N)∈ X. Observe that

    P_k(x)^2=∑_l_1,...,l_N≥ 0ξ^(k)_(l_1,...,l_N)^2=ξ^(k)^2,  ∀ x∈ X.

Similarly, Q_k(y)=η^(k) for all y∈ X. Hence

    ∑_k≥ 0P_k_∞Q_k_∞ = ∑_k≥ 0ξ^(k)η^(k) = T'_S_1.

Now fix x,y∈ X and observe that

    ∑_k≥ 0⟨ P_k(x),Q_k(y)⟩
       = ∑_k≥ 0∑_l_1,...,l_N=0
     j_1,...,j_N=0^∞(∏_i=1^N⟨δ_ω_x_i(l_i),δ_ω_y_i(j_i)⟩)ξ^(k)_(l_1,...,l_N)η^(k)_(j_1,...,j_N)
       = ∑_l_1,...,l_N=0
     j_1,...,j_N=0^∞(∏_i=1^N⟨δ_ω_x_i(l_i),δ_ω_y_i(j_i)⟩)T_(l_1,...,l_N),(j_1,...,j_N)'.

Then, by Lemma <ref>, there exist m_1,...,m_N and n_1,...,n_N in  such that d(x_i,y_i)=m_i+n_i and

    ⟨δ_ω_x(l_i)',δ_ω_y(j_i)'⟩=(S_i,m_i,n_i)_j_il_i,

for all l_i,j_i∈ (i=1,...,N). Hence, using (<ref>), the identity (<ref>) can be rewritten as

    ∑_k≥ 0⟨ P_k(x),Q_k(y)⟩   = ∑_l,j∈^N S(m,n)_j,lT_l,j'
       = Tr(S(m,n)T'),

which, by (<ref>), gives

    ϕ(x,y)=ϕ̃(m+n)=c_++(-1)^|m|+|n|c_-+∑_k≥ 0⟨ P_k(x),Q_k(y)⟩.

Observe that the function (m,n)↦(-1)^|m|+|n| defines a Schur multiplier on X of norm 1 (see e.g. <cit.>). Therefore

    ϕ_cb   ≤ |c_+|+|c_-|+∑_k≥ 0P_k_∞Q_k_∞
       =|c_+|+|c_-|+ T'_S_1.



Now we will use this lemma to characterise positive definite multi-radial kernels on X. Observe that a more general form of Proposition <ref> still holds in this context. Namely, ϕ is a positive definite multi-radial kernel if and only if

    ϕ_cb=ϕ̃(0,...,0).



Let ϕ̃:^N → be a function such that the limits

    l_0   =lim_|n|→∞
     |n| evenϕ̃(n)     l_1   =lim_|n|→∞
     |n| oddϕ̃(n)

exist, and let T be as in (<ref>). Then ϕ̃ defines a positive definite multi-radial kernel on X if and only if the operator

    T'=∏_i=1^N(1-1/q_i)(I-τ_i/q_i)^-1T

is an element of ∈ S_1(ℓ_2(^N))_+, and

    |l_1|≤ l_0.



We know from Lemma <ref> that ϕ̃ defines a multi-radial Schur multiplier ϕ on X if and only if T (equivalently T') is of trace class. And in that case

    ϕ_cb=T'_S_1+|c_+|+|c_-|,

where c_±=l_0± l_1. On the other hand, we know that ϕ is positive definite if and only if ϕ_cb=ϕ̃(0,...,0), or equivalently,

    ϕ̃(0,...,0) = T'_S_1 + |c_+| + |c_-|.

Moreover, by (<ref>),

    ϕ̃(0,...,0) =c_++c_-+Tr(T').

This means that ϕ is positive definite if and only if

    Tr(T')+c_++c_-= T'_S_1 + |c_+| + |c_-|.

By Lemma <ref>, this is equivalent to T' ∈ S_1(ℓ_2(^N))_+ and c_+,c_-≥ 0. Finally observe that the latter condition can be rewritten as |l_1|≤ l_0.


Now we go back to radial kernels. Recall that for m,n∈^N, we denote by m∧ n the element of ^N given by

    (m∧ n)_i=min{m_i,n_i},  ∀ i∈{1,...,N}.

By m≤ n we mean m_i≤ n_i for all i.


A function φ:→ belongs to ℛ_+(X) if and only if the following two conditions hold


  a) The operator B=(B_n,m)_m,n∈^N given by

    B_m,n=∑_l≤ m∧ n1/q_1^l_1⋯ q_N^l_N∑_k=0^NNk(-1)^k φ(|m|+|n|-2|l|+2k),

is of trace class and positive.

  b) The following limits

    l_0   =lim_n→∞φ(2n),     l_1   =lim_n→∞φ(2n+1)

 exist, and

    |l_1|≤ l_0.




We will apply Corollary <ref> to the function ϕ̃:^N→ given by

    ϕ̃(n)=φ(|n|),  ∀ n∈^N.

For this purpose, we need to show that the limits (<ref>) exist. If we assume (b), then this is part of the hypotheses. If we suppose that φ∈ℛ_+(X), then by restriction, φ∈ℛ_+(𝒯_q_1), and the limits l_0,l_1 exist by <cit.>. So ϕ̃ satisfies the hypotheses of Corollary <ref>. We conclude that φ∈ℛ_+(X) if and only if T' belongs to S_1(ℓ_2(^N))_+ and |l_1|≤ l_0. Observe that

    T_m,n   =∑_I⊂[N](-1)^|I|ϕ̃(m+n+2χ^I)
       = ∑_I⊂[N](-1)^|I|φ(|m|+|n|+2|I|)
       = ∑_k=0^NNk(-1)^kφ(|m|+|n|+2k),

for all m,n∈^N. Moreover,

    T'=α∑_l∈^N1/q_1^l_1⋯ q_N^l_NS_1^l_1⋯ S_N^l_NT(S_N^*)^l_N⋯(S_1^*)^l_1,

where α=∏_i=1^N(1-1/q_i). Hence

    T_m,n'    =α∑_l∈^N1/q_1^l_1⋯ q_N^l_N⟨ T(S_N^*)^l_N⋯(S_1^*)^l_1δ_n, (S_N^*)^l_N⋯(S_1^*)^l_1δ_m⟩
       =α∑_l≤ m∧ n1/q_1^l_1⋯ q_N^l_N⟨ Tδ_n-l, δ_m-l⟩
       =α∑_l≤ m∧ n1/q_1^l_1⋯ q_N^l_N∑_k=0^NNk(-1)^kφ(|m|+|n|-2|l|+2k),

for all m,n∈^N. This shows that T'=α B, which finishes the proof.







 §.§ The homogeneous tree of infinite degree 𝒯_∞


Corollary <ref> deals with homogeneous trees of finite degree. As we discussed in the introduction, an important role is played by the homogeneous tree of infinite degree 𝒯_∞. Recall that ℛ_+(𝒯_∞) stands for the set of functions on  defining positive definite radial kernels on 𝒯_∞.


Let φ:→, and let H be the Hankel operator given by

    H_i,j=φ(i+j)-φ(i+j+2).

Then φ belongs to ℛ_+(𝒯_∞) if and only if H belongs to S_1(ℓ_2())_+ and

    |lim_n→∞φ(2n+1)|≤lim_n→∞φ(2n).

In particular, those limits exist.


It was proven in <cit.> that φ belongs to ℛ(𝒯_∞) if and only if H belongs to S_1(ℓ_2()), and in that case the associated multiplier ψ satisfies

    ψ_cb=H_S_1+|c_+|+|c_-|,

where

    c_±=1/2lim_n→∞φ(2n)±1/2lim_n→∞φ(2n+1).

Since Tr(H)=φ(0)-c_+-c_-, the same arguments as in the proof of Corollary <ref> show that φ∈ℛ_+(𝒯_∞) if and only if

    Tr(H)+c_++c_-=H_S_1+|c_+|+|c_-|,

and this is equivalent to H∈ S_1(ℓ_2())_+ and c_+,c_-≥ 0, which proves the result.





§ A VARIATION OF THE HAMBURGER MOMENT PROBLEM


This section is devoted to the implication (b)⇒(c) in Theorems <ref> and <ref>, whose proof is inspired by the Hamburger moment problem.

Given a sequence (a_n) of real numbers, the moment problem asks whether there exists a positive measure μ on  such that

    a_n=∫_-∞^∞ s^n dμ(s),

for all n∈. Hamburger proved <cit.> that this happens if and only if the Hankel matrix H=(a_i+j)_i,j∈ is positive definite. Observe that one of the implications is quite simple. If we assume that (a_n) is the sequence of moments of a positive measure μ, then for every finite sequence x=(x_i)_i=0^m in ,

    ∑_i,j=0^m H_ijx_ix_j    = ∑_i,j=0^m a_i+jx_ix_j
       =∫_-∞^∞∑_i,j=0^m s^i+jx_ix_j dμ(s)
       =∫_-∞^∞|∑_i=0^m s^ix_i|^2 dμ(s),

which proves that H is positive definite. The converse is more delicate. We refer the reader to <cit.> for a full proof of it, which relies on the spectral theory of unbounded self-adjoint operators on Hilbert spaces. Since we deal only with trace-class operators here, our proofs will become simpler.





 §.§ The moment problem for trace-class operators


We begin with the proof of (b)⇒(c) in Theorem <ref> for q=∞. The properties of the measures in Hamburger's theorem for bounded, compact and trace-class Hankel operators are well known. See <cit.> and <cit.> for more information. Nevertheless, we shall give a complete proof of the construction of the measure in our setting, as it will illustrate the main ideas that we will generalise for finite values of q.


Let φ:→ be a function such that the Hankel operator

    H=(φ(i+j)-φ(i+j+2))_i,j∈

belongs to S_1(ℓ_2())_+, and

    |lim_n→∞φ(2n+1)|≤lim_n→∞φ(2n).

Then there exists a positive Borel measure μ on [-1,1] such that

    φ(n)=∫_-1^1 s^n dμ(s),

for all n∈.



The existence of the limits in (<ref>) is a consequence of the fact that H belongs to S_1(ℓ_2()) (see <cit.>). The real hypothesis here is that the inequality (<ref>) holds.



Let c_00() be the space of finitely supported sequences in . Since H is positive, we can define a scalar product on c_00() by

    ⟨ f,g⟩_ℋ=⟨ Hf,g⟩_ℓ_2.

Let N be the subspace consisting of all f∈ c_00() such that ⟨ f,f⟩_ℋ=0 (it is possible that N={0}). We define ℋ as the completion of c_00()/N for the scalar product ⟨ · , · ⟩_ℋ. Let S be the forward shift operator on c_00(). Since H is a Hankel matrix, we have

    ⟨ S f,g⟩_ℋ   = ⟨ HSf,g⟩_ℓ_2
       = ⟨ Hf,Sg⟩_ℓ_2
       =⟨ f,Sg⟩_ℋ,

for all f,g∈ c_00(). This tells us two things. It shows that S is symmetric, and that it passes to the quotient. Indeed, by the Cauchy–Schwarz inequality,

    ⟨ Sf, Sf⟩_ℋ   = ⟨ f, S^2f⟩_ℋ
       ≤⟨ f, f⟩_ℋ^1/2⟨ S^2f, S^2f⟩_ℋ^1/2,

for all f∈ c_00(). We shall prove that it extends to a bounded self-adjoint operator on ℋ. For this purpose, fix f∈ c_00() with f_ℋ=1 and define the following sequence,

    a_n=S^nf_ℋ^2=⟨ S^2nf,f⟩_ℋ,  ∀ n∈.

Again, by the Cauchy–Schwarz inequality,

    a_n≤S^2nf_ℋf_ℋ = (a_2n)^1/2.

So by induction,

    S f_ℋ^2=a_1≤(a_2^n)^1/2^n,  ∀ n∈.

Now observe that

    a_n    =⟨ S^2nf,Hf⟩_ℓ_2
       ≤S_ℬ(ℓ_2)^2nH_ℬ(ℓ_2)f_ℓ_2^2
       ≤H_ℬ(ℓ_2)f_ℓ_2^2.

Thus the sequence (a_n) is bounded, which by (<ref>) implies that

    S f_ℋ^2≤lim inf_n→∞(a_2^n)^1/2^n≤ 1.

This shows that S extends to a bounded self-adjoint operator Ψ on ℋ with Ψ_ℬ(ℋ)≤ 1. Let E_Ψ denote its spectral measure (for a very nice introduction to von Neumann's spectral theorem, we refer the reader to Attal's lecture notes <cit.>, which are available online). We can define a positive measure μ on [-1,1] by

    μ(A)=⟨ E_Ψ(A)δ_0,δ_0⟩_ℋ,

for every Borel set A⊂[-1,1]. Hence

    ∫_-1^1 s^n dμ(s)    = ⟨Ψ^nδ_0,δ_0⟩_ℋ
       = ⟨ HS^nδ_0,δ_0⟩_ℓ_2
       = φ(n)-φ(n+2),

for all n∈. On the other hand, since H is of trace class, we know that

    ∑_n≥ 0(φ(2n)-φ(2n+2)) <∞,

and this implies that μ({-1,1})=0 and

    ∫_(-1,1)1/1-s^2 dμ(s) = ∫_(-1,1)∑_n≥ 0 s^2n dμ(s) < ∞.

Let ν be the finite measure on (-1,1) given by

    dν(s)=(1-s^2)^-1dμ(s).

Then

    φ(n)-φ(n+2)=∫_(-1,1) s^n dν(s) - ∫_(-1,1) s^n+2 dν(s).

Recall the notation

    c_±=1/2lim_n→∞φ(2n)±1/2lim_n→∞φ(2n+1).

By the dominated convergence theorem, for all n∈,

    φ(n)-(c_++(-1)^nc_-)    =∑_j≥ 0φ(n+2j)-φ(n+2j+2)
       =∫_(-1,1) s^n dν(s) - lim_j→∞∫_(-1,1) s^n+2j+2 dν(s)
       =∫_(-1,1) s^n dν(s).

Since by hypothesis c_+,c_-≥ 0, we can define a new positive measure, on the closed interval [-1,1] now, by

    ν̃=c_+δ_1+c_-δ_-1+ν.

This measure satisfies

    ∫_[-1,1] s^n dν̃(s)=φ(n),  ∀ n∈.

This finishes the proof.



In the general moment problem, the forward shift operator does not necessarily extend to a bounded operator on ℋ. It is only a densely defined symmetric operator, and the proof uses the existence of a self-adjoint extension on a larger space. In our setting, we can avoid those technical issues thanks to the fact that H is bounded.





 §.§ Construction of a measure for 2≤ q < ∞


Now we will adapt the previous arguments to the case when the operator B from Theorem <ref>(b) is positive. If q is finite, this is not a Hankel matrix, so the forward shift operator S is not symmetric on the Hilbert space defined by it. We shall need to consider a perturbation of S that will allow us to adapt the proofs to this new setting.
 
We recall the definition of the family of polynomials (P_n^(q)). Fix 2≤ q<∞ and define

    P_0^(q)(x)=1,    P_1^(q)(x)=x,

and

    P_n+2^(q)(x)=(1+1q)xP_n+1^(q)(x)-1qP_n^(q)(x),  ∀ n≥ 0.

The following lemma is classical and it can be found in the several references given in the introduction. A short proof of it can be given using the results in <cit.>.


For all s∈[-1,1], the function φ:→, given by φ(n)=P_n^(q)(s), belongs to ℛ_+(𝒯_q).


By <cit.>, the Schur norm of the associated multiplier on 𝒯_q is 1. Since φ(0)=1, Proposition <ref> implies that this multiplier is positive definite.


Our goal is to prove the following.


Let φ:→. Assume that the operator B=(B_i,j)_i,j∈ given by

    B_i,j =∑_k=0^min{i,j}1/q^k(φ(i+j-2k)-φ(i+j-2k+2))

is of trace class and positive, and

    |lim_n→∞φ(2n+1)|≤lim_n→∞φ(2n).

Then there exists a positive Borel measure μ on [-1,1] such that

    φ(n)=∫_-1^1P_n^(q)(s) dμ(s),

for all n∈.


For this purpose, we shall need to study another family of polynomials, which is defined in similar fashion. We define (Q_n^(q))_n∈ by

    Q_0^(q)(x)=1,    Q_1^(q)(x)=(1+1q)x,

and

    Q_n+2^(q)(x)=(1+1q)xQ_n+1^(q)(x)-1qQ_n^(q)(x),  ∀ n≥ 0.

We begin by establishing a relation between (Q_n^(q)) and (P_n^(q)).


For all n∈,

    P_n^(q)(x)-P_n+2^(q)(x)= (1+1q)(1-x^2)Q_n^(q)(x).



We proceed by induction. First,

    P_0^(q)(x)-P_2^(q)(x)    = 1-(1+1q)x^2+1q
       = (1+1q)(1-x^2),

and

    P_1^(q)(x)-P_3^(q)(x)    = x - (1+1q)xP_2^(q)(x) + 1qP_1^(q)(x)
       = x - (1+1q)x[(1+1q)x^2-1q] + 1qx
       = (1+1q)x( 1 - (1+1q)x^2 + 1q)
       = Q_1^(q)(x)(1+1q)(1-x^2).

This proves (<ref>) for n=0,1. Now assume that it holds for n and n-1 with n≥ 1. Then

    P_n+1^(q)(x)-P_n+3^(q)(x)    = (1+1q)xP_n^(q)(x) - 1qP_n-1^(q)(x)
            - (1+1q)xP_n+2^(q)(x) + 1qP_n+1^(q)(x)
       = (1+1q)x(P_n^(q)(x)-P_n+2^(q)(x))
            - 1q(P_n-1^(q)(x)-P_n+1^(q)(x))
       = (1+1q)^2x(1-x^2)Q_n^(q)(x)
            - 1q(1+1q)(1-x^2)Q_n-1^(q)(x)
       = (1+1q)(1-x^2)((1+1q)xQ_n^(q)(x)- 1qQ_n-1^(q)(x))
       = (1+1q)(1-x^2)Q_n+1^(q)(x).

This finishes the proof.




Under the hypotheses of Lemma <ref>, there exists a positive Borel measure μ on [-1,1] such that

    φ(n)-φ(n+2)=∫_-1^1 Q_n^(q)(s) dμ(s),  ∀ n∈.



We define a scalar product on c_00() by

    ⟨ f,g⟩_ℋ=⟨ Bf,g⟩_ℓ_2,

and let ℋ be the completion (after taking quotient if necessary) of c_00() for this scalar product. Observe that

    B   =(1-1q)(I-1qτ)^-1H
       =(1-1q)∑_k≥ 01q^kS^kH(S^*)^k,

where τ(A)=SAS^* and H=(φ(i+j)-φ(i+j+2))_i,j∈. Recall that the operator (I-1qτ) is an isomorphism of S_1(ℓ_2()). Define

    Ψ f=(1+1q)^-1(S+1qS^*)f,  ∀ f∈ c_00(),

where S^* stands for the adjoint of S in ℓ_2() (not in ℋ). We claim that Ψ is a symmetric operator. Indeed, since H is a Hankel matrix, we have

    HS=S^* H.

Now observe that (<ref>) can be expressed as

    (1-1q) H = B - 1qSBS^*,

so (<ref>) can be written as

    BS-1qSB=S^*B-1qBS^*,

or equivalently

    B(S+1qS^*)=(S+1qS^*)^*B.

This shows that

    ⟨Ψ f,g⟩_ℋ=⟨ f,Ψ g⟩_ℋ,  ∀ f,g∈ c_00().

By the same arguments as in the proof of Lemma <ref>, Ψ extends to a bounded self-adjoint operator on ℋ with Ψ_ℋ≤ 1. Let E_Ψ denote its spectral measure. Again, we can define a positive measure μ on [-1,1] by

    μ(A)=⟨ E_Ψ(A)δ_0,δ_0⟩_ℋ,

for every Borel set A⊂[-1,1]. Now we claim that

    Q_n^(q)(Ψ)δ_0=S^nδ_0,  ∀ n∈.

Indeed, for n=0,1, this is a short computation. Then, assuming that this holds for n-1 and n, we get

    Q_n+1^(q)(Ψ)δ_0    =(1+1q)Ψ Q_n^(q)(Ψ)δ_0-1qQ_n-1^(q)(Ψ)δ_0
       = (S+1qS^*)S^nδ_0-1qS^n-1δ_0
       = S^n+1δ_0,

and the claim follows by induction. Using this, we obtain

    ∫_-1^1 Q_n^(q)(s) dμ(s) = ⟨ Q_n^(q)(Ψ)δ_0,δ_0⟩_ℋ = ⟨ BS^nδ_0,δ_0⟩_ℓ_2,

which implies by (<ref>) that

    (1-1q)^-1∫_-1^1 Q_n^(q)(s) dμ(s)    = ∑_k≥ 01q^k⟨ H(S^*)^kS^nδ_0,(S^*)^kδ_0⟩_ℓ_2
       = ⟨ HS^nδ_0,δ_0⟩_ℓ_2
       = φ(n)-φ(n+2).

Thus, the measure (1-1q)^-1μ satisfies (<ref>).


In order to prove that the measure obtained in Lemma <ref> does not have atoms in {-1,1}, we need to establish some properties of the polynomials Q_n^(q).


The polynomial functions Q_n^(q) satisfy the following:


  a) If n is even (resp. odd), then Q_n^(q) is an even (resp. odd) function. 

  b) For all n∈,

    Q_n^(q)(1)=∑_k=0^n1/q^k.




Both assertions follow by induction. For the first one, simply observe that

    Q_n+2^(q)(-s) = -(1+1q)sQ_n+1^(q)(-s)-1qQ_n^(q)(-s),

and that Q_0^(q) and Q_1^(q) are even and odd respectively. The second one is given by the identity

    Q_n+2^(q)(1) = (1+1q)Q_n+1^(q)(1)-1qQ_n^(q)(1).



We shall also need the following properties of the family (P_n^(q)).


The family of polynomials (P_n^(q)) satisfy the following.


  a) For all s∈[-1,1] and all n∈,

    |P_n^(q)(s)|≤ 1.


  b) For all n∈, P_n^(q)(1)=1 and P_n^(q)(-1)=(-1)^n.

  c) For all s∈(-1,1),

    lim_n→∞P_n^(q)(s)=0.




Let s∈[-1,1], and let ψ be the radial kernel on 𝒯_q given by

    ψ(x,y)=P_d(x,y)^(q)(s),  ∀ x,y∈𝒯_q.

By Lemma <ref>, ψ is positive definite. Then, for all n∈,

    |P_n^(q)(s)|≤ψ_cb=1.

This proves (a). Property (b) follows by induction. To prove (c), define

    l_0(s)=lim_n→∞P_2n^(q)(s),        l_1(s)=lim_n→∞P_2n+1^(q)(s).

By Corollary <ref>,

    |l_1(s)|≤ l_0(s).

On the other hand, we know that the sequence (P_n^(q)(s)) satisfies

    P_2n+2^(q)(s)+1/qP_2n^(q)(s)=(1+1/q)sP_2n+1^(q)(s),

so taking the limit n→∞, we get

    l_0(s)=sl_1(s).

If |s|<1, this can only hold if l_0(s)=l_1(s)=0.



By Lemma <ref>, we know that there exists a positive Borel measure μ on [-1,1] such that

    φ(n)-φ(n+2)=∫_[-1,1] Q_n^(q)(s) dμ(s),  ∀ n∈.

Moreover, using Lemma <ref>, we get

    φ(n)-φ(n+2)    =∫_[-1,1] Q_n^(q)(s) dμ(s)
       =Q_n^(q)(-1)μ({-1})+Q_n^(q)(1)μ({1})
            +∫_(-1,1) P_n^(q)(s)-P_n+2^(q)(s) dν(s),

where ν is the positive measure on (-1,1) given by

    dν(s)=(1+1q)^-1(1-s^2)^-1dμ(s).

This looks very similar to (<ref>), but first we need to prove that μ({-1,1})=0 and that P_n^(q)∈ L_1(ν), so we can write (<ref>) as a sum of two integrals. Define f_n:(-1,1)→ by

    f_n(s)=1-P_2n^(q)(s),  ∀ s∈(-1,1).

By Lemma <ref>, for all s∈(-1,1),

    f_n(s)≥ 0,   ∀ n∈,

and

    lim_n→∞ f_n(s) = 1.

Therefore, by Fatou's lemma, together with (<ref>),

    ∫_(-1,1) 1 dν(s)    ≤lim inf_n→∞∫_(-1,1) f_n dν(s)
       = lim inf_n→∞∑_k=0^n-1∫_(-1,1) P_2k^(q)(s)-P_2k+2^(q)(s) dν(s)
       = lim inf_n→∞∑_k=0^n-1φ(2k)-φ(2k+2)-Q_2k^(q)(1)μ({-1,1})
       =φ(0)-lim_n→∞φ(2n)-∑_k≥ 0Q_2k^(q)(1)μ({-1,1}).

Here we used the fact that Q_2k^(q)(-1)=Q_2k^(q)(1), given by Lemma <ref>(a). Moreover, by Lemma <ref>(b), we must have μ({-1,1})=0, or else the last sum would be infinite. Observe that this also shows that ν is a finite measure, from which we deduce that (<ref>) can be written as

    φ(n)-φ(n+2)=∫_(-1,1) P_n^(q)(s) dν(s) - ∫_(-1,1) P_n+2^(q)(s) dν(s).

As always, we define

    c_±=1/2lim_n→∞φ(2n)±1/2lim_n→∞φ(2n+1).

By the dominated convergence theorem, together with Lemma <ref>, for all n∈,

    φ(n)   -(c_++(-1)^nc_-)
       =lim_k→∞∑_j=0^kφ(n+2j)-φ(n+2j+2)
       =lim_k→∞∑_j=0^k∫_(-1,1) P_n+2j^(q)(s) dν(s) - ∫_(-1,1) P_n+2j+2^(q)(s) dν(s)
       =∫_(-1,1) P_n^(q)(s) dν(s) - lim_k→∞∫_(-1,1) P_n+2k+2^(q)(s) dν(s)
       =∫_(-1,1) P_n^(q)(s) dν(s).

Since c_+,c_-≥ 0, we can define a new positive measure on [-1,1] by

    ν̃=c_+δ_1+c_-δ_-1+ν,

and

    ∫_[-1,1] P_n^(q)(s) dν̃(s)=φ(n),  ∀ n∈.







 §.§ The multi-radial case

Now we will prove a multi-radial version of Lemma <ref> for products of trees. As before, we fix 2≤ q_1,...,q_N<∞ and define

    X=𝒯_q_1×⋯×𝒯_q_N,

where 𝒯_q_i is the (q_i+1)-homogeneous tree (i=1,...,N). Recall that, given a function ϕ̃:^N →, we can define the operators T and T' by

    T_m,n=∑_I⊂[N](-1)^|I|ϕ̃(m+n+2χ^I),  ∀ m,n∈^N,

and

    T'= ∏_i=1^N(1-1/q_i)(I-τ_i/q_i)^-1T.

Now we consider N families of polynomials (P_n^(q_i))_n∈ (i=1,...,N) defined as before:

    P_0^(q_i)(x)=1,    P_1^(q_i)(x)=x,


    P_n+2^(q_i)(x)=(1+1q_i)xP_n+1^(q_i)(x)-1q_iP_n^(q_i)(x),  ∀ n≥ 0.



Let ϕ̃:^N → be a function such that the limits

    l_0   =lim_|n|→∞
     |n| evenϕ̃(n)     l_1   =lim_|n|→∞
     |n| oddϕ̃(n)

exist. Assume that T' belongs to ∈ S_1(ℓ_2(^N))_+ and that |l_1|≤ l_0. Then there exists a positive Borel measure μ on [-1,1]^N such that

    ϕ̃(n_1,...,n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N),

for all (n_1,...,n_N)∈^N.


The proof follows the same ideas as the one of Lemma <ref>, hence we shall only sketch it. We construct a Hilbert space ℋ with scalar product given by

    ⟨ f,g⟩_ℋ=⟨ T'f,g⟩_ℓ_2(^N),   ∀ f,g∈ c_00(^N).

We denote by Ψ_1,...,Ψ_N the family of densely defined operators on ℋ given by

    Ψ_i=(1+1q_i)^-1(S_i+1q_iS_i^*),

where S_i stands for the forward shift operator on the i-th coordinate. Then Ψ_1,...,Ψ_N is a commuting family of operators on ℋ, and by the same arguments as in the proof of Lemma <ref>, they are bounded, self-adjoint and of norm at most 1. Then the spectral theorem for commuting families of self-adjoint operators allows us to construct a measure μ on [-1,1]^N by

    μ(A_1×⋯× A_N)=⟨ E_Ψ_1(A_1)⋯ E_Ψ_N(A_1)δ_(0,...,0),δ_(0,...,0)⟩_ℋ,

where E_Ψ_i is the spectral measure of the operator Ψ_i. Defining a new family of polynomials (Q_n^(i))_n∈ (i=1,...,N) like in (<ref>), one sees that

    Q_n^(q_i)(Ψ_i)δ_(0,...,0)=S_i^nδ_(0,...,0),

for every n∈. Therefore

    ∫_[-1,1]^N Q_n_1^(q_1)(s_1)    ⋯ Q_n_N^(q_N)(s_N) dμ(s_1,...,s_N) 
       = ⟨ Q_n_1^(q_1)(Ψ_1)⋯ Q_n_N^(q_N)(Ψ_N)δ_(0,...,0),δ_(0,...,0)⟩_ℋ
       = ⟨ T'δ_(n_1,...,n_N),δ_(0,...,0)⟩_ℓ_2(^N).

Since

    T'= ∏_i=1^N(1-1/q_i)∑_k_1,...,k_N≥ 01/q^k_1⋯ q^k_NS_1^k_1⋯ S_N^k_NT(S_N^*)^k_N⋯(S_1^*)^k_1,

rescaling the measure μ by the factor ∏_i=1^N(1-1/q_i)^-1, we get

    ∫_[-1,1]^N Q_n_1^(q_1)(s_1) ⋯ Q_n_N^(q_N)(s_N)    dμ(s_1,...,s_N) 
       = ⟨ Tδ_(n_1,...,n_N),δ_(0,...,0)⟩_ℓ_2(^N)
       =∑_I⊂[N](-1)^|I|ϕ̃((n_1,...,n_N)+2χ^I).

In particular, by Lemma <ref>,

    ∑_I⊂[N](-1)^|I|   ϕ̃((n_1,0,...,0)+2χ^I)
       = ∫_{-1,1}×[-1,1]^N-1 Q_n_1^(q_1)(s_1) dμ(s_1,...,s_N)
          + ∫_(-1,1)×[-1,1]^N-1 P_n_1^(q_1)(s_1)-P_n_1+2^(q_1)(s_1) dν_1(s_1,...,s_N),

where ν_1 is given by

    dν_1(s_1,...,s_N)=(1+1q_1)^-1(1-s_1^2)^-1dμ(s,...,s_N).

Again, using Fatou's lemma, one shows that

    ∫_(-1,1)×[-1,1]^N-1  1 dν_1(s_1,...,s_N)
       ≤lim inf_n→∞∑_k=0^n-1∫_(-1,1)×[-1,1]^N-1 P_2k^(q_1)(s_1)-P_2k+2^(q_1)(s_1) dν_1(s_1,...,s_N)
       = lim inf_n→∞∑_k=0^n-1∑_I⊂[N](-1)^|I|ϕ̃((2k,0,...,0)+2χ^I)
               -∑_k≥ 0Q_2k(1)μ({-1,1}×[-1,1]^N-1).

Observe that

    ∑_I⊂[N](-1)^|I|ϕ̃   ((2k,0,...,0)+2χ^I)
       =∑_J⊂[N]∖{1}(-1)^|J|ϕ̃((2k,0,...,0)+2χ^J)
           -∑_J⊂[N]∖{1}(-1)^|J|ϕ̃((2k+2,0,...,0)+2χ^J),

which gives us a telescoping sum in (<ref>), and thus

    ν_1((-1,1)×[-1,1]^N-1)    ≤∑_J⊂[N]∖{1}(-1)^|J|ϕ̃((0,...,0)+2χ^J) - l_0
               -∑_k≥ 0Q_2k(1)μ({-1,1}×[-1,1]^N-1).

This shows that ν_1 is finite and that

    μ({-1,1}×[-1,1]^N-1)=0.

Repeating this argument for the other coordinates, one shows that

    μ(∂[-1,1]^N)=0,

where ∂[-1,1]^N stands for the boundary of [-1,1]^N in ^N. Furthermore,

    ∑_I⊂[N](-1)^|I|ϕ̃((n_1,...,n_N)+2χ^I)
       =∫_(-1,1)^N∏_i=1^N(P_n_i^(q_i)(s_i)-P_n_i+2^(q_i)(s_i))dν(s_1,...,s_N),

where ν is the finite measure on (-1,1)^N given by

    dν(s_1,...,s_N)=(∏_i=1^N(1+1q_i)^-1(1-s_i^2)^-1)dμ(s,...,s_N).

Hence the expression

    ∑_k_1,...,k_N≥ 0∑_I⊂[N](-1)^|I|ϕ̃((n_1+2k_1,...,n_N+2k_N)+2χ^I)

can be computed from N telescoping sums, from which we obtain

    ϕ̃(n_1,...,n_N)-   (c_++(-1)^n_1+⋯ +n_Nc_-)
               =∫_(-1,1)^N P_n_1^(q_1)(s_1)⋯ P_n_N^(q_N)(s_N) dν(s_1,...,s_N),

where c_±=l_0± l_1≥ 0. Therefore

    ϕ̃(n_1,...,n_N)=∫_[-1,1]^N P_n_1^(q_1)(s_1)⋯ P_n_N^(q_N)(s_N) dν̃(s_1,...,s_N),

where

    ν̃=c_+δ_(1,...,1)+c_-δ_(-1,...,-1)+ν.




Let φ:→ be a function such that the operator B=(B_n,m)_m,n∈^N given by

    B_m,n=∑_l≤ m∧ n1/q_1^l_1⋯ q_N^l_N∑_k=0^NNk(-1)^k φ(|m|+|n|-2|l|+2k)

belongs to S_1(ℓ_2(^N))_+, and such that the following limits

    l_0   =lim_n→∞φ(2n),     l_1   =lim_n→∞φ(2n+1)

exist and satisfy

    |l_1|≤ l_0.

Then there exists a positive Borel measure μ on [-1,1]^N such that

    φ(n_1+⋯ +n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N),

for all (n_1,...,n_N)∈^N.


Simply apply Lemma <ref> to the function

    ϕ̃(n_1,...,n_N)=φ(n_1+⋯+n_N).







§ PROOF OF THE MAIN RESULTS


Sections <ref> and <ref> were devoted to the implications (a)⇔(b)⇒(c) of Theorems <ref> and <ref>. The remaining implication essentially states that the average of positive definite kernels over a positive measure is again positive definite. Like in the previous sections, we will state the result in the multi-radial context first. Fix 2≤ q_1,...,q_N<∞ and define

    X=𝒯_q_1×⋯×𝒯_q_N,

where 𝒯_q_i is the (q_i+1)-homogeneous tree, and let (P_n^(q_i))_n be the family of polynomials defined in (<ref>) (i=1,...,N).


Let μ be a positive and finite Borel measure on [-1,1]^N, and let ϕ̃:^N→ be given by

    ϕ̃(n_1,...,n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N).

Then ϕ̃:^N→ defines a positive definite multi-radial kernel on X.


Since μ is positive, it is sufficient to prove that, for each (t_1,...,t_N)∈[-1,1]^N, the multi-radial kernel given by

    (n_1,...,n_N)⟼ P_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)

is positive definite. For each i∈{1,...,N}, let ψ_i:𝒯_q_i×𝒯_q_i→ be given by

    ψ_i(x_i,y_i)=P_d(x_i,y_i)^(q_i)(t_i),  ∀ x_i,y_i∈𝒯_q_i.

We know from Lemma <ref> that ψ_i is positive definite on 𝒯_q_i. So by <cit.>, there is a Hilbert space ℋ_i and a map ξ_i:𝒯_q_i→ℋ_i such that

    ψ_i(x_i,y_i)=⟨ξ_i(x_i),ξ_i(y_i)⟩,  ∀ x_i,y_i∈𝒯_q_i.

Now define

    ℋ=ℋ_1⊗⋯⊗ℋ_N,

and ξ:X→ℋ by

    ξ(x)=ξ_1(x_1)⊗⋯⊗ξ_N(x_N),

for all x=(x_1,...,x_N)∈ X. Then

    ⟨ξ(x),ξ(y)⟩   = ⟨ξ_1(x_1),ξ_1(y_1)⟩⋯⟨ξ_N(x_N),ξ_N(y_N)⟩
       = ψ_1(x_1,y_1)⋯ψ_N(x_N,y_N),

for all x,y∈ X. Again by <cit.>, this shows that the kernel

    (x,y)⟼ψ_1(x_1,y_1)⋯ψ_N(x_N,y_N)

is positive definite on X, which is exactly what we wanted to prove.


The following corollary is a direct consequence of Lemma <ref>.


Let φ:→, and suppose that there is a positive Borel measure μ on [-1,1]^N such that

    φ(n_1+⋯+n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N),

for all n_1,...,n_N∈. Then φ belongs to ℛ_+(X).


Now we are ready to give the proofs of the main results of this article.


The equivalence (a)⇔(b) is given by Corollary <ref>. Thanks to Corollary <ref>, we have (b)⇒(c). Finally, (c)⇒(a) is given by Corollary <ref>.



For 2≤ q<∞, this is a particular case of Theorem <ref>. For q=∞, (a)⇔(b) and (b)⇒(c) are given by Lemmas <ref> and <ref> respectively. The implication (c)⇒(a) was proven in <cit.>. Another way of seeing it is by noticing that 𝒯_∞ is a median graph and applying Lemma <ref>.


Observe that our methods also allow us to obtain a similar result for multi-radial kernels. Take X=𝒯_q_1×⋯×𝒯_q_N and (P_n^(q_i)) as above.


Let ϕ̃:^N →, and assume that the limits

    l_0   =lim_|n|→∞
     |n| evenϕ̃(n)     l_1   =lim_|n|→∞
     |n| oddϕ̃(n)

exist. Then the following are equivalent:


  a) The function ϕ̃ defines a multi-radial positive definite kernel on X.

  b) The operator T' defined in (<ref>) belongs to S_1(ℓ_2(^N))_+, and |l_1|≤ l_0.

  c) There exists a positive Borel measure μ on [-1,1]^N such that

    ϕ̃(n_1,...,n_N)=∫_[-1,1]^NP_n_1^(q_1)(t_1)⋯ P_n_N^(q_N)(t_N)  dμ(t_1,...,t_N),

for all (n_1,...,n_N)∈^N.



This is a consequence of Corollary <ref>, Lemma <ref> and Lemma <ref>.



In contrast to Theorem <ref>, we need to include the existence of the limits l_0 and l_1 in the hypotheses of Lemma <ref> for the three statements to be equivalent. Indeed, take N=2 and φ_i∈ℛ_+(𝒯_q_i) (i=1,2). Then the function ϕ̃(n_1,n_2)=φ_1(n_1)φ_2(n_2) satisfies (a) and (c), but the limits l_0,l_1 do not necessarily exist, so (b) does not really make sense. On the other hand, if f_1,f_2:→ are any functions, then for ϕ̃(n_1,n_2)=f_1(n_1)+f_2(n_2), the operator T' in (b) is 0. So, in particular, it belongs to S_1(ℓ_2(^N))_+, but ϕ̃ does not even define a Schur multiplier on 𝒯_q_1×𝒯_q_2, unless some additional conditions are imposed to the functions f_1,f_2.





§ A CONCRETE EXAMPLE

Now we turn to the proof of Corollaries <ref> and <ref>. Fix 2≤ q< ∞. We will focus on the function φ:→ given by

    φ(n)=(-1/q)^n/2    n even,
    
    0     n odd,

with the convention (-1/q)^0=1. Observe that

    φ(n)=P_n^(q)(0),  ∀ n∈,

where (P_n^(q)) is the family of polynomials defined in (<ref>). So by Lemma <ref>, we know that φ∈ℛ_+(𝒯_q).


The function φ given by (<ref>) does not belong to ℛ_+(𝒯_q+1).


It suffices to prove, by Theorem <ref>, that the operator

    B=∑_n≥ 01/(q+1)^nS^nH(S^*)^n

is not positive. Here H=(φ(i+j)-φ(i+j+2))_ij, and S is the forward shift operator. We shall prove something stronger: For every ε>0, the operator

    A_ε=∑_n≥ 01/(q+ε)^nS^nH(S^*)^n

is not positive. Observe that

    φ(2n)-φ(2n+2)=(1+1q)(-1q)^n,

for all n∈. Hence,

    ⟨ A_εδ_1,δ_1⟩   = ∑_n≥ 01(q+ε)^n⟨ H(S^*)^nδ_1,(S^*)^nδ_1⟩
       = (1+1q)(-1q) + 1q+ε(1+1q)
       = (1+1q)(1q+ε-1q),

and this is strictly negative.


We conclude that φ belongs to ℛ_+(𝒯_q)∖ℛ_+(𝒯_q+1), which proves Corollary <ref>.


The function φ given by (<ref>) does not belong to ℛ_+(𝒯_q×𝒯_q).


Observe that

    φ(2k)-2φ(2k+2)+φ(2k+4)=(1+2/q+1/q^2)(-1/q)^k,  ∀ k∈.

Hence, by Theorem <ref>, we only need to check that the operator B=(B_m,n)_m,n∈^2 given by

    B_m,n=∑_l≤ m∧ n1/q^|l|(-1/q)^|m|+|n|/2-|l|   if  |m|+|n|  is even,
    
    0     otherwise,

is not positive. For this purpose, we shall prove that

    ⟨ B(δ_(0,1)+δ_(1,0)), δ_(0,1)+δ_(1,0)⟩ < 0.

Indeed, first observe that for |m|+|n| even,

    B_m,n   =(-1/q)^|m|+|n|/2∑_l≤ m∧ n(-1)^|l|
       =(-1/q)^|m|+|n|/21/4(1+(-1)^m_1∧ n_1)(1+(-1)^m_2∧ n_2).

Thus

    ⟨ Bδ_(0,1), δ_(0,1)⟩ = ⟨ Bδ_(1,0), δ_(1,0)⟩ = 0

and

    ⟨ Bδ_(0,1), δ_(1,0)⟩ = ⟨ Bδ_(1,0), δ_(0,1)⟩ = -1/q.

This shows that B is not positive.


Therefore φ belongs to ℛ_+(𝒯_q)∖ℛ_+(𝒯_q×𝒯_q), which proves Corollary <ref>.




§ MEDIAN GRAPHS


This section is devoted to the proof of Proposition <ref>. Recall that a connected graph X is median if 

    |I(x,y)∩ I(y,z)∩ I(z,x)|=1,  ∀ x,y,z∈ X,

where

    I(x,y)={u∈ X  :  d(x,y)=d(x,u)+d(u,y)}.

Chepoi proved the following (see <cit.>).


Median graphs are exactly the 1-skeletons of CAT(0) cube complexes.


For details on CAT(0) cube complexes, we refer the reader to <cit.> and the references therein. Thanks to this identification, we may talk about the dimension of a median graph. We say that a median graph X is of dimension N if there exists an isometric embedding from the N-dimensional cube {0,1}^N into X, but there is not such an embedding from {0,1}^N+1. A 1-dimensional median graph is a tree.

A hyperplane in a median graph is an equivalence class of edges under the equivalence relation generated by

    {x,y}∼{u,v}  if  {x,y,v,u} is a square.

If H is a hyperplane and {x,y}∈ H, we say that H separates x from y, and that {x,y} crosses H. More generally, we say that a path crosses H if one of its edges does. Sageev characterised the distance on a median graph in terms of hyperplanes (see <cit.>).


Let x,y be two vertices in a median graph X, and let γ be a geodesic joining x and y. Then γ crosses every hyperplane separating x from y, and it does so only once. Moreover, γ does not cross any other hyperplane.


This says that the distance between two points is exactly the number of hyperplanes separating them.

Niblo and Reeves <cit.> proved that the distance on a median graph is a conditionally negative kernel. Let Y be a nonempty set. We say that a symmetric function ψ:Y× Y→ is conditionally negative if ψ(y,y)=0 for every y∈ Y, and for every finite sequence y_1,...,y_n∈ Y and every λ_1,...,λ_n∈ such that ∑λ_i=0, we have

    ∑_i,j=0^nλ_iλ_jψ(y_i,y_j)≤ 0.

It can be shown that this is equivalent to the fact that there exists a Hilbert space ℋ and a function b:Y→ℋ such that

    ψ(x,y)=b(x)-b(y)^2,  ∀ x,y∈ Y.



Let X be a median graph. Then the edge-path distance on X is a conditionally negative kernel.


An alternative proof of this fact can be given using the language of spaces with walls. See e.g. <cit.>. We reproduce the proof in our setting for the reader's convenience.


By Theorem <ref>, the distance between two vertices in X is exactly the number of hyperplanes separating them. Let 𝒳 be the set of hyperplanes of X, and fix x_0∈ X. For each x∈ X, let S_x be the set of hyperplanes separating x and x_0. Define b:X→ℓ_2(𝒳) by

    b(x)=∑_h∈ S_xδ_h,  ∀ x∈ X.

Then

    b(x)-b(y)^2=|S_xΔ S_y|,  ∀ x,y∈ X.

Fix x,y∈ X and let m be the median point between x,y and x_0. Then Theorem <ref> implies that

    |S_xΔ S_y|=d(x,m)+d(y,m)=d(x,y).

We conclude that

    d(x,y)=b(x)-b(y)^2,  ∀ x,y∈ X,

which implies that d is conditionally negative.


By Schoenberg's theorem (see e.g. <cit.>), Lemma <ref> implies that, for every s∈[0,1], the function

    (x,y)∈ X ⟼ s^d(x,y)

is positive definite. Moreover, it is not hard to see that on every bipartite graph the function (x,y)↦(-1)^d(x,y) is positive definite (see for example <cit.>). Hence, by composition, the function (<ref>) is positive definite for every s∈[-1,1].


Let X be a median graph and let μ be a finite positive Borel measure on [-1,1]. Then the function φ:→ given by

    φ(n)=∫_-1^1s^n dμ(s),  ∀ n∈,

belongs to ℛ_+(X).


Let x_1,...,x_n be a finite sequence of elements of X. Let z_1,...,z_n∈. Then

    ∑_i,j=1^nφ(d(x_i,x_j))z_iz_j = ∫_-1^1∑_i,j=1^ns^d(x_i,x_j)z_iz_j dμ(s),

and this is non-negative since it is the integral of a non-negative function with respect to a positive measure. This shows that φ∈ℛ_+(X).



Let X be a median graph and let φ∈ℛ_+(𝒯_∞). By Theorem <ref>, φ is of the form (<ref>) (recall that P_n^(∞)(s)=s^n). By Lemma <ref>, this implies that φ∈ℛ_+(X).





 §.§ Acknowledgements

I thank Gilles Pisier for suggesting this problem to me. I am deeply grateful to Adam Skalski for encouraging me to write this paper, for many interesting discussions, and for his careful reading of several different versions of these notes. I also thank Mikael de la Salle for fruitful discussions.

This work was supported by the ANR project GAMME (ANR-14-CE25-0004).


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