Riesz means on homogeneous trees

Let $\mathbb{T}$ be a homogeneous tree. We prove that if $f\in L^{p}(X)$, $1\leq p\leq 2$, then the Riesz means $S_{R}^{z}\left( f\right) $ converge to $f$ almost everywhere as $R\rightarrow \infty $, whenever $\operatorname{Re}z>0 $.

A homogeneous tree T of degree Q+1, Q ≥ 2, is an infinite connected graph with no loops, in which every vertex is adjacent to Q + 1 other vertices. We shall identify T with its set of vertices V. T carries a natural distance d and a natural measure µ. Specifically, d(x, y) is the number of edges of the shortest path joining x to y and µ is the counting measure. For the counting measure, the volume of any sphere S(x, n) in T is given by |S(x, n)| = 1, if n = 0 (Q + 1)Q n−1 , if n ∈ N, and the Lebesgue spaces, associated with µ, have norms defined by L p (T) Let us fix a base point x 0 and set |x| = d(x, x 0 ). Functions depending only on |x| are called radial. If E(T) is a function space on T, we will denote by E(T) # the subspace of radial elements in E(T).
Let G be the group of isometries of T, and suppose that G acts transitively on T. If x 0 ∈ T is a fixed vertex, then the orbit Gx 0 is all of T. Therefore T may be identified through the map g → gx 0 with the quotient G/K, where K = {g ∈ G : gx 0 = x 0 }, [9, p. 46]. This means that every function on T may be lifted to a function on G, by definingf (g) = f (gx 0 ). The functionf has the property that f (gk) =f (g), for every k ∈ K, and conversely a K-right-invariant function on G may be identified with a function on T. Thus, we shall identify functions defined on T with right-K-invariant functions on G and radial functions with K-bi-invariant functions on G.
Let G be a locally compact group and K be a compact subgroup of G; then the pair (G, K) is called a Gelfand pair if the space C c (K\G/K) of complex continuous K-bi-invariant functions with compact support is a commutative algebra with the convolution product (the space C c (K\G/K) is always an algebra with the convolution product). Under the hypothesis that K acts transitively on the boundary of T, (G, K) is a Gelfand pair. Also, the transitive action of K on the boundary of T is also a necessary condition for (G, K) to be a Gelfand pair, [9, p. 47]. In fact, the following holds true, [22].
Proposition. For every finite subset F of V, denote by Aut F (T) the group of automorphisms g of T such that g(x) = x for all x ∈ F . Then G is equipped with a topology of locally compact totally discontinuous group such that the subgroups Aut F (T) form a fundamental system of neighborhoods of the identity in G.

Moreover a subgroup H is maximal, open, compact in G if and only if H is the stabiliser of a point x in V.
We normalize the Haar measure on G in such a way that K has a unit mass. Then This allows us to define the convolution of two functions on T by If f 2 is radial, then (2) rewrites In this short note, we deal with the Riesz means S z R , Re z > 0, R > 0, on T, which are defined as convolution operators (see Section 1 for more details). We prove the following result.
Let us recall that the Riesz means were first treated by E.M.Stein in [19], where he proved that if f ∈ L p ([0, 1] n ), n ≥ 1, p ∈ (1, 2], then Since then, many authors have studied Riesz means in various geometric contexts, as euclidean spaces, compact manifolds, Lie groups of polynomial volume growth, graphs and discrete groups of polynomial volume growth, Riemannian manifolds of nonnegative curvature, compact semisimple Lie groups and noncompact symmetric spaces. See [1,3,4,11,10,12,14,16,18,19,21]. Note that the multiplier m z R (λ) does not extend holomorphically to any strip containing the real line. So, by [7, Theorem 1.2], the Riesz means operator is not bounded on L p (T) if p = 2 and consequently the norm summability problem on L p (T), p = 2, is ill posed. Note also that in the case of homogeneous trees, there is no restriction on the size of Re z > 0, contrarily to the Euclidean case and even to the hyperbolic space case, due to absence of local obstructions.

Preliminaries
In this section we present the tools we need for the proof of our results. For details, see for example [7,8]. Let M be the mean operator Then, the Laplacian L on T is defined by The spherical function ϕ λ of index λ ∈ C is the unique radial eigenfunction of the operator L, which is associated with the eigenvalue and which is normalized by ϕ λ (0) = 1. Set τ = 2π log Q . Then, Note that ϕ λ is periodic with period τ .
The spherical Fourier transform of a radial function f on T is defined by The following inversion formula holds: where c is the meromorphic function We have the following Plancherel theorem [7]: the spherical Fourier transform extends to an isometry of L 2 (T) # onto L 2 ([−τ /2, τ /2], dλ |c(λ)| 2 ) and (11) Note also that the spherical Fourier transform is written as a composition H = F • A of the euclidean Fourier transform F and the Abel transform A [8].
Recall that the kernel κ z R of the Riesz means operator is given by the inverse spherical transform of the multiplier m z R : . For that, we shall make use of the following inversion formulas, [8]: 2. Estimates of the kernel κ z R Lemma 2. If Re z > 0, then . Then, using (13), (12) and (4), we obtain the following explicit expression of κ z R : sin λ sin(λ(n + 2k + 1))dλ Remark. Note that the same kernel estimate holds for every operator m R (L) with a uniformly bounded multiplier |m R (λ)| ≤ c.

Proof of Theorem 1
For the proof of Theorem 1, we need to introduce the maximal function associated with Riesz means: The proof will be given in steps.
First, proceeding as in [12], we have that Proof. We include the proof for the sake of completeness. Let h R , R > 0, be the heat kernel on T, i.e. let the heat semigroup be [20, Chapter III, MAXIMAL THEOREM], the heat maximal operator f → H * f := sup R>0 |f * h R | is bounded on L 2 (T). Thus, it suffices to prove the boundedness of (S z − H) * . Using the Mellin transform for the functions (1 − t/R) z + and e −tR , R > 0, [2], and the spectral theorem for L, we have where |c(z, s)| ≤ c(z)(1 + |s|) −(Re z+1) , [12], thus the integral above converges. Since L 2 (T) is a complete Banach lattice, from [6], we can since we have L is L 2 (T)→L 2 (T) ≤ 1 by the spectral theorem and the fact that L is self-adjoint on L 2 (T) [7, p.4271].