A Note on Simultaneous Polar and Cartesian Decomposition

F. Barthe^{∗}, M. Csörnyei^{†} and A. Naor^{‡}

November 6, 2001

Abstract

We study measures on ℝ^{n} which are product measures for the usual Cartesian product structure of ℝ^{n} as well as for the polar decomposition of ℝ^{n} induced by a convex body. For finite atomic measures and for absolutely continuous measures with density dμ/dx = e^{−V (x)}, where V is locally integrable, a complete characterization is presented.

1 Introduction

A subset K ⊂ ℝ^{n} is called a star-shaped body if it is star-shaped with respect to the origin, compact, has non-empty interior, and for every x ≠ 0 there is a unique r > 0 such that x/r ∈ ∂K. We denote this r by ∥x∥_{K} (∥ · ∥_{K} is the Minkowski functional of K). Note that ∥x∥_{K} is automatically continuous (If x_{n} tends to x ≠ 0, then for every subsequence x_{n_{k}} such that ∥x_{n_{k}}∥_{K} converges to r, the compactness ensures that x/r ∈ ∂K, so that r = ∥x∥_{K} by the uniqueness assumption). Any star-shaped body K ⊂ ℝ^{n} induces a polar product structure on ℝ^{n} \ {0} through the identification

x ↦ (∥x∥_{K}, \frac{x}{∥x∥_{K}} ) .

In this note we study the measures on ℝ^{n}, n ≥ 2 which are product measures with respect to the Cartesian coordinates, and the above polar decomposition.

In measure theoretic formulation, we will be interested in the measures μ on ℝ^{n} which are product measures with respect to the product structures ℝ^{n} = ℝ× ⋯×ℝ = ℝ^{+}·∂K. Here × is the usual Cartesian product and for R ⊂ ℝ^{+}, Ω ∈ ∂K, the polar product is by definition R · Ω = {rω; r ∈ R and ω ∈ Ω}. We adopt similar notation for product measures: ⊗ will be used for Cartesianproduct measures and ⊙ for polar-product measures. With this notation, we say that μ has a simultaneous product decomposition with respect to K if there are measures μ_{1}, … μ_{n} on ℝ such that μ = μ_{1} ⊗ ⋯ ⊗ μ_{n}, and there is a measure τ on ℝ^{+} and a measure ν on ∂K such that μ = τ ⊙ ν (in what follows, all measures are Borel). Notation like A^{k} or ∏_{i} A_{i} always refer to the Cartesian product.

∗CNRS-Université de Marne-la-Vallée, Equipe d’analyse et de Mathématiques appliquées, Cité Descartes, Champs-sur-Marne, 77454, Marne-la-Vallée, cedex 2, France. E-mail:

barthe@math.univ.mlv.fr. Partially supported by EPSRC grant GR/R37210.

†Department of Mathematics, University College London, Gower Street, London, WC1E 6BT, United Kingdom. E-mail: mari@math.ucl.ac.uk. Supported by the Hungarian National Foundation for Scientific Research, grant # F029768.

‡Institute of Mathematics, Hebrew University, Jerusalem 91904, Israel. E-mail:

naor@math.huji.ac.il. Supported in part by the Binational Science Foundation Israel-USA, the Clore Foundation and the EU grant HPMT-CT-2000-00037. This work is part of a Ph.D.

thesis being prepared under the supervision of Professor Joram Lindenstrauss.

For probability measures one can formulate the notion of simultaneous product decomposition as follows. A measure μ on ℝ^{n} has a simultaneous product decomposition with respect to K if and only if there are independent real valued random variables X_{1}, …, X_{n} such that if we denote X = (X_{1}, …, X_{n}) then μ(A) = P (X ∈ A) and X/∥X∥_{K} is independent of ∥X∥_{K}.

The standard Gaussian measure on ℝ^{n} is obviously a Cartesian product. A consequence of its rotation invariance is that it is also a polar-product measure for the usual polar structure induced by the Euclidean ball. Many characterizations of the Gaussian distribution have been obtained so far. The motivations for such characterizations arise from several directions. Maxwell proved that the Gaussian measure is the only rotation invariant product probability measure on ℝ^{3}, and deduced that this is the distribution of the velocities of gas particles. The classical Cramer and Bernstein characterizations of the Gaussian measure, as well as the numerous related results that appeared in the literature arose from various probabilistic and statistical motivations. We refer to the book [2] and the reference therein for a detailed account. The more modern characterization due to Carlen [3] arose from the need to characterize the equality case in a certain functional inequality.

To explain the motivation for the present paper, we begin by noting that the Gaussian density is in fact one member of a wider family of measures with simultaneous product decomposition, involving bodies other than the Euclidean ball. They will be easily introduced after setting notation. The cone measure on the boundary of K, denoted by μ_{K} is defined as:

μ_{K}(A) = vol([0, 1] · A).

This measure is natural when studying the polar decomposition of the Lebesgue measure with respect to K, i.e. for every integrable f : ℝ^{n} → ℝ, one has

∫_{ℝ^{n}} f(x) dx = ∫_{0}^{+∞} nr^{n−1} ∫_{∂K} f(rω) dμ_{K}(ω)dr.

For the particular case K = B_{p}^{n} = {x ∈ ℝ^{n}; ∥x∥_{p} ≤ 1}, where ∥x∥_{p} =

( ∑_{i=1}^{n} |x_{i}|^{p})^{1/p}, a fundamental result of Schechtman and Zinn [7] (see also Rachev and Rüschendorff [6]), gives a concrete representation of μ_{K}:

Theorem 1 Let g be a random variable with density e^{−|t|^{p}}/(2Γ(1+1/p)), t ∈ ℝ. If g_{1}, ..., g_{n} are i.i.d. copies of g, set:

S = (∑_{i=1}^{n} |g_{i}|^{p})^{1/p} ,

and consider the random vector:

Z = (\frac{g_{1}}{S} , ..., \frac{g_{n}}{S} ) ∈ ℝ^{n}.

Then the random vector Z is independent of S. Moreover, for every measurable A ⊂ ∂B_{p}^{n} we have:

\frac{μ_{B_{p}^{n}}(A)}{vol(B_{p}^{n})} = P (Z ∈ A) .

The independence of Z and S, that is the simultaneous product decomposition, turns out to be very useful for probabilistic as well as geometric purposes ([8],[5],[4],[1]). One might hope that such a statement holds true for other norms and other densities. The aim of this note is to show that the ℓ_{p}^{n} norm is in fact characterized by this property, although we will show that such an independence result holds for other measures. Section 2 is devoted to absolutely continuous measures. Section 3 presents a classifications for finite atomic measures, when K is convex. As the reader will see there are more examples. Some of them, however, are not interesting and we will discard them by suitable assumptions. For example: a constant random variable is independent of any other. This observation allows to produce several measures with simultaneous product decomposition. Any random variable X with values in the half-line {x; x_{1} > 0} works. Its law is clearly a Cartesian product measure, and X/∥X∥_{K} is constant regardless of K, so it is independent of ∥X∥_{K}. Similarly, if X has independent components and takes values in only one sphere r∂K it has a simultaneous product decomposition. If K is not assumed to be convex, many different examples may be produced: take two sets of positive numbers {x_{1}, …, x_{N}} and {y_{1}, …, y_{N}} and consider μ_{1} = ∑ δ_{x_{i}} and μ_{2} = ∑ δ_{y_{i}}. If one assumes that the numbers y_{i}/x_{j} are all different then the measure μ_{1} ⊗ μ_{2} is supported on points (x_{j}, y_{i}) all having different directions. So there are several origin-star-shaped bodies K such that μ_{1} ⊗ μ_{2} is supported on the boundary of K. For such K’s, μ_{1} ⊗ μ_{2} admits a polar decomposition.

Finally, if μ has simultaneous product decomposition, and ϵ_{1}, …, ϵ_{n} ∈ {−1, 1} then the restriction of μ to {x; x_{i}ϵ_{i} > 0} still has this property. This remark allows us to restrict the study to the positive ”octant” (0, +∞)^{n} (and one has to glue pieces together at the end).

2 Absolutely Continuous Measures

As in the classical characterizations of the Gaussian measure, the assumption that the measure is absolutely continuous reduces the characterization problem to a solution of a functional equation which holds almost everywhere (with respect to the Lebesgue measure). Unless we add some smoothness assumptions on the densities, the next step is to apply a smoothing procedure. Of course, after “guessing” the family of solutions of the equations, we must come up with a smoothing procedure which sends each member of this family to another member of the family. The classical Cramer and Bernstein characterizations use Fourier transform techniques (see [2]), while Carlen [3] applies the heat semigroup. The particular form of the equation we will derive will force us to use yet another smoothing procedure.

When μ is absolutely continuous the following (easy) characterization holds:

Lemma 2 Assume that μ is an absolutely continuous measure on ℝ^{n}, then it has a simultaneous product decomposition with respect to K if and only if there are locally integrable non-negative functions f_{1}, …, f_{n} defined on ℝ, g defined on ∂K (locally integrable with respect to μ_{K}) and h on (0, ∞) such that

\frac{dμ}{dx}(x) = ∏_{i=1}^{n} f_{i}(x_{i}) = g (\frac{x}{∥x∥_{K}} ) · h(∥x∥_{K}),

Lebesgue almost everywhere.

Proof: Assume that μ has a simultaneous product decomposition with respect to K. In the above notation, write μ = μ_{1} ⊗ ⋯ ⊗ μ_{n} = τ ⊙ ν. For every measurable B ⊂ ∂K:

ν(B) = μ(ℝ^{+} · B) = ∫_{R^{+}·B} \frac{dμ}{dx}(x)dx =

= ∫_{B} (∫_{0}^{∞} n · r^{n−1} \frac{dμ}{dx}(rω)dr) dμ_{K}(ω).

Similarly, for every measurable A ⊂ ℝ^{+}

τ(A) = μ(A · ∂K) = ∫_{A} n · (∫_{∂K} \frac{dμ}{dx}(rω)dμ_{K}(ω)) dr.

This shows that both τ and ν are absolutely continuous. Similarly μ_{1}, …, μ_{n} are absolutely continuous.

Now, for every measurable A ⊂ ℝ^{+}, B ⊂ ∂K, C_{1}, … C_{n} ⊂ ℝ:

μ(A · B) = τ(A)ν(B) = ∫_{0}^{∞} ∫_{∂K} \frac{dτ}{dr} (r) · \frac{dν}{dμ_{K}} (ω)drdμ_{K}(ω) =

= ∫_{A·B} \frac{1}{n · ∥x∥_{K}^{n−1}} · \frac{dτ}{dr} (∥x∥_{K}) · \frac{dν}{dμ_{K}} (\frac{x}{∥x∥_{K}} ) dx,

and

μ(C_{1} × … × C_{n}) = ∫_{C_{1}×...×C_{n}} ∏_{i=1}^{n} \frac{dμ_{i}}{dx_{i}} (x_{i})dx.

Since the product Borel σ-algebras on ℝ^{+}·∂K and ℝ×…×ℝ (n times) coincide, this shows that:

\frac{dμ}{dx}(x) = ∏_{i=1}^{n} \frac{dμ_{i}}{dx_{i}} (x_{i}) = \frac{1}{n · ∥x∥_{K}^{n−1}} · \frac{dτ}{dr} (∥x∥_{K}) · \frac{dν}{dμ_{K}} (\frac{x}{∥x∥_{K}} ) ,

Lebesgue almost everywhere. The reverse implication is even simpler. □

Fix some p > 0, b_{1}, … b_{n} > −1 and a_{1}, … a_{n} > 0. Let X_{1}, …, X_{n} be independent random variables, such that the density of X_{i} is:

\frac{pa_{i}^{(b_{i}+1)/p}}{2Γ (\frac{b_{i}+1}{p})} · |t|^{b_{i}}e^{−a_{i}|t|^{p}}.

Note that:

∏_{i=1}^{n} |x_{i}|^{b_{i}}e^{−a_{i}|x_{i}|^{p}} =

= (∑_{i=1}^{n} a_{i}|x_{i}|^{p})^{\frac{1}{p} ∑_{i=1}^{n} b_{i}} · e^{− ∑_{i=1}^{n} a_{i}|x_{i}|^{p}} ∏_{i=1}^{n} [\frac{|x_{i}|}{(∑_{j=1}^{n} a_{j}|x_{j}|^{p})^{1/p}} ]^{b_{i}} .

Hence, if we denote X = (X_{1}, …, X_{n}) then by Lemma 2, X/ (∑_{i=1}^{n} a_{i}|X_{i}|^{p})^{1/p} and (∑_{i=1}^{n} a_{i}|X_{i}|^{p})^{1/p} are independent. Moreover, if b_{1} = … = b_{n} = 0 and a_{1} = … = a_{n} = 1 then it follows from the proof of Lemma 2 that X/∥X∥_{p} generates the cone measure on the sphere of ℓ_{p}^{n}. We have therefore obtained a generalization of Theorem 1.

The main goal of this section is to prove that the above densities are the only way to obtain a measure with a simultaneous product decomposition with respect to a star shaped body K ⊂ ℝ^{n} (and that K must then be a weighted ℓ_{p}^{n} ball). In solving the functional equation of Lemma 2 we will require a smoothing procedure. Clearly, we require a way to smooth a function such that a function of the form c|t|^{b}e^{−a|t|^{p}} is transformed to a function of the same form. Let ψ : ℝ → ℝ be locally integrable. For any infinitely differentiable ρ : (0, ∞) → [0, ∞) which is compactly supported in (0, ∞) define:

(ρ ⋆ ψ)(x) = ∫_{0}^{∞} ρ(t)ψ (\frac{x}{t}) dt = ∫_{0}^{∞} xρ(sx)ψ (\frac{1}{s}) ds.

It is easy to verify that ρ ⋆ ψ is infinitely differentiable on (−∞, 0) ∪ (0, ∞). Fix some ϵ < 1/2 and let ρ_{ϵ} : (0, ∞) → [0, 1/(2ϵ)] be any infinitely differentiable function such that ρ_{ϵ}(t) = 1/(2ϵ) when |t − 1| ≤ ϵ and ρ_{ϵ}(t) = 0 when |t − 1| > ϵ + ϵ^{2}. Now, for x > 0 (and similarly when x < 0):

|(ρ_{ϵ} ⋆ ψ)(x) − \frac{1}{2xϵ} ∫_{x/(1+ϵ+ϵ^{2})}^{x/(1−ϵ−ϵ^{2})} ψ(u)du| =

= \frac{1}{x} ∫_{x/(1+ϵ+ϵ^{2})}^{x/(1−ϵ−ϵ^{2})} |\frac{1}{2ϵ} − (\frac{x}{u})^{2} ρ_{ϵ} (\frac{x}{u})| |ψ(u)|du ≤

≤ \frac{1}{x} ∫_{x/(1+ϵ)}^{x/(1−ϵ)} |ψ(u)|du + \frac{1}{2ϵx} [∫_{x/(1+ϵ+ϵ^{2})}^{x/(1+ϵ)} |ψ(u)|du + ∫_{x/(1−ϵ)}^{x/(1−ϵ−ϵ^{2})} |ψ(u)|du] ,

which, by the Lebesgue density theorem, implies that lim_{ϵ→0} ρ ⋆ ψ = ψ almost everywhere. Since, lim_{ϵ→0} ∫_{0}^{∞} ρ_{ϵ}(t)dt = 1, the same holds for β_{ϵ} = ρ_{ϵ}/ ∫_{0}^{∞} ρ_{ϵ}.

Since for every function of the form f(t) = c|t|^{b}e^{−a|t|^{p}}, the function exp(ρ_{ϵ} ⋆ (log f)) has the same form, the above smoothing procedure allows us to prove our main result. In what follows ε(x) ∈ {−1, 1} denotes the sign of x (any convention for the sign of zero will do).

Theorem 3 Let K ⊂ ℝ^{n} be a star shaped body. Assume that μ is an absolutely continuous probability measure on ℝ^{n} which has a simultaneous product decomposition with respect to K. Assume in addition that log (\frac{dμ}{dx}) is locally integrable. Then there is some p > 0 and there are b_{1}, …, b_{n} > −1 and r, a_{1}(1), a_{1}(−1), c_{1}(1), c_{1}(−1) …, a_{n}(1), a_{n}(−1), c_{n}(1), c_{n}(−1) > 0 such that:

K = {x ∈ ℝ^{n}; ∑_{i=1}^{n} a_{i}(ε(x_{i}))|x_{i}|^{p} ≤ r} ,

and

dμ(x) = ∏_{i=1}^{n} c_{i}(ε(x_{i}))|x_{i}|^{b_{i}}e^{−a_{i}(ε(x_{i}))|x_{i}|^{p}} dx_{i}.

Conversely, for K and μ as above, μ has simultaneous product decomposition with respect to K.

We will require the following elementary lemma:

Lemma 4 Fix α, α′ > 0. Let f : (0, ∞) → ℝ be a continuous function such that for every x > 0,

f(αx) = 2f(x)  and  f(α′x) = \frac{3}{2}f(x),

then for every x > 0, f(x) = f(1) x^{\frac{log 2}{log α}} (If α = 1, f(x) = 0 for all x).

Proof: We may assume that f is not identically zero. Then α ≠ α′ and α, α′ ≠ 1. Consider the set

P = {β > 0; there is c_{β} > 0 s.t. f(βx) = c_{β}f(x), for all x > 0}.

It is a multiplicative subgroup of (0, ∞). By classical results P is either dense in (0, ∞) or discrete. Assume first that it is dense. Fix some x_{0} such that f(x_{0}) ≠ 0. For any β ∈ P, c_{β} = f(βx_{0})/f(x_{0}). By continuity of f and by the density of P it follows that for every x, β > 0 one has

f(βx) = \frac{f(βx_{0})}{f(x_{0})}f(x).

So, if for some β, f(βx_{0}) = 0 then f is identically zero. Therefore, f does not vanish. We can chose x_{0} = 1 and setting g = f/f(1), we have that for every x, y > 0, g(xy) = g(x)g(y). It is well known that the continuity of g ensures that it is a power function.

To finish, let us note that P cannot be discrete. Indeed, if P is discrete, since it contains α and α′ ≠ α, it is of the form {T^{h}; h ∈ ℤ} for some positive T ≠ 1. So there are k, k′ ∈ ℤ \ {0} such that α = T^{k} and α′ = T^{k′}. Our hypothesis is that for all x > 0

2f(x) = f(αx) = f(T^{k}x) = c_{T}^{k} f(x),

\frac{3}{2}f(x) = f(α′x) = f(T^{k′}x) = c_{T}^{k′}f(x).

For an x such that f(x) ≠ 0 we get 2 = c_{T}^{k} and \frac{3}{2} = c_{T}^{k′}. It follows that 3^{k} = 2^{k+k′}. This is impossible because k ≠ 0. □ Proof of Theorem 3: Using the notation and the result of Lemma 2,

\frac{dμ}{dx}(x) = ∏_{i=1}^{n} f_{i}(x_{i}) = g (\frac{x}{∥x∥_{K}} ) · h(∥x∥_{K}).

For i = 1, …, n denote F_{i} = log f_{i}. Denote also G = log g and H = log h. Let φ : ℝ^{n} → ℝ be a compactly supported continuous function. For every t > 0:

∑_{i=1}^{n} ∫_{ℝ^{n}} φ(tx)F_{i}(x_{i})dx = ∫_{ℝ^{n}} φ(tx)G (\frac{x}{∥x∥_{K}} ) dx + ∫_{ℝ^{n}} φ(tx)H (∥x∥_{K}) dx.

Changing variables this translates to

∑_{i=1}^{n} ∫_{ℝ^{n}} φ(y)F_{i} (\frac{y_{i}}{t} ) dy = ∫_{ℝ^{n}} φ(y)G (\frac{y}{∥y∥_{K}} ) dy + ∫_{ℝ^{n}} φ(y)H (\frac{∥y∥_{K}}{t} ) dy.

Fix some ϵ > 0. Multiplying by β_{ϵ} and integrating, we get,

∑_{i=1}^{n} ∫_{ℝ^{n}} φ(y)(β_{ϵ}⋆F_{i})(y_{i})dy = ∫_{ℝ^{n}} φ(y)G (\frac{y}{∥y∥_{K}} ) dy+∫_{ℝ^{n}} φ(y)(β_{ϵ}⋆H)(∥y∥_{K})dy.

Denote ϕ_{i} = β_{ϵ} ⋆ F_{i} and η = β_{ϵ} ⋆ H. By the above identity for almost every y ∈ ℝ^{n}:

∑_{i=1}^{n} ϕ_{i}(y_{i}) = G (\frac{y}{∥y∥_{K}} ) + η(∥y∥_{K}).

Since ϕ_{i} and η are continuous on ℝ \ {0}, we can change G on a set of measure zero such that the latter identity holds for every y ∈ ℝ^{n} with non-zero coordinates.

Fix some y ∈ ℝ^{n} with non-zero coordinates. For every λ > 0, one has

∑_{i=1}^{n} ϕ_{i}(λy_{i}) = G (\frac{y}{∥y∥_{K}} ) + η(λ∥y∥_{K}).

Since both sides of the equation are differentiable in λ, taking derivatives at λ = 1 yields:

∑_{i=1}^{n} χ_{i}(y_{i}) = ζ(∥y∥_{K}),  (1)

where for simplicity, we write χ_{i}(t) = tϕ′_{i}(t) and ζ(t) = tη′(t). From this, we shall deduce that ζ, χ_{i} are power functions. This can be proved by differentiation along the boundary of K, under smoothness assumptions. Since we want to deal with general star shaped bodies, we present now another reasoning.

Note that lim_{t→0} χ_{i}(t) exists. Indeed since η is smooth on (0, ∞) and ∥ · ∥_{K} is continuous, the above equation for y^{t} = ∑_{j≠i} e_{j} + te_{i} gives

lim_{t→0} χ_{i}(t) = ζ (∥ ∑_{j≠i} e_{j}∥_{K} ) − ∑_{j≠i} χ_{i}(1).

Similarly, ζ may be extended by continuity at 0. Hence, (1) holds on ℝ^{n}. Applying (1) to λe_{i}, gives

χ_{i}(λ) = ζ(|λ| · ∥ε(λ)e_{i}∥_{K}) + γ_{i},

for some constant γ_{i}. Plugging this into (1) for y_{i} ≥ 0 we obtain an equation in ζ only:

∑_{i=1}^{n} ζ(y_{i}∥e_{i}∥_{K}) + γ_{i} = ζ(∥y∥_{K}).

Choosing y = λy_{1}e_{1}/∥e_{1}∥_{K} + λy_{2}e_{2}/∥e_{2}∥_{K}, with λ, y_{1}, y_{2} > 0, we get

ζ(λy_{1}) + ζ(λy_{2}) + ∑_{i=1}^{n} γ_{i} + (n − 2)ζ(0) = ζ(λ∥y_{1} \frac{e_{1}}{∥e_{1}∥_{K}} + y_{2} \frac{e_{2}}{∥e_{2}∥_{K}} ∥_{K}).

Differentiating in λ at λ = 1 and setting f(t) = tζ′(t), t > 0,

f(y_{1}) + f(y_{2}) = f(∥y_{1} \frac{e_{1}}{∥e_{1}∥_{K}} + y_{2} \frac{e_{2}}{∥e_{2}∥_{K}} ∥_{K}).  (2)

For y_{1} = y_{2} = t > 0, we obtain 2f(t) = f(αt), with α = ∥\frac{e_{1}}{∥e_{1}∥_{K}} + \frac{e_{2}}{∥e_{2}∥_{K}} ∥_{K}. Combining this relation with (2) gives

\frac{1}{2}f(y_{1}α) + f(y_{2}) = f(∥y_{1} \frac{e_{1}}{∥e_{1}∥_{K}} + y_{2} \frac{e_{2}}{∥e_{2}∥_{K}} ∥_{K}).

For y_{1} = t/α and y_{2} = t > 0 we get \frac{3}{2}f(t) = f(tα′), with α′ = ∥\frac{1}{α} \frac{e_{1}}{∥e_{1}∥_{K}} +

\frac{e_{2}}{∥e_{2}∥_{K}} ∥_{K}. Lemma 4 ensures that f(t) = f(1)t^{p} with p = log 2/ log α ≠ 0, and with the convention that p = 0 if α = 1 (in this case f(1) = 0). It follows that

ζ(t) = ζ′(1)t^{p}/p + ζ(0)

if p ≠ 0 and ζ(t) = ζ(0) otherwise. Integrating again, we get an expression for η = β_{ϵ} ∗ log h. Letting ϵ tend to zero shows that there are constants a, b, c ∈ ℝ such that for a.e. t > 0, h(t) = ct^{b}e^{−at^{p}} (this is valid even if p = 0).

Next we find an expression of the functions f_{i}. We start with the relation

tϕ′(t) = χ_{i}(t) = ζ(|t| · ∥ε(t)e_{i}∥_{K}) + γ_{i} = a_{i}(ε(t))|t|^{p} + b_{i},

for some constants a_{i}(1), a_{i}(−1), b_{i}. Thus for t ≠ 0, ϕ′(t) = a_{i}(ε(t))|t|^{p−1}ε(t) + b_{i}/t. Integrating (with different constants on (−∞, 0) and on (0, ∞)) and taking the limit ϵ → 0 as before we arrive at f_{i}(t) = c_{i}(ε(t))|t|^{b_{i}}e^{−a_{i}|t|^{p}}, for almost every t and for some constants c_{i}(1), c_{i}(−1). For f_{i} to have finite integral, p has to be non-zero. Our initial equation reads as: for a.e. x,

∏_{i=1}^{n} c_{i}(ε(x_{i}))|x_{i}|^{b_{i}}e^{−a_{i}|x_{i}|^{p}} = c∥x∥_{K}^{b}e^{−a∥x∥_{K}^{p}} g (\frac{x}{∥x∥_{K}} ) .

By continuity this holds on ℝ^{n} \ {x; ∏_{i=1}^{n} x_{i} = 0}. For such an x and λ > 0, the equation becomes

∏_{i=1}^{n} c_{i}(ε(x_{i}))λ^{∑_{i=1}^{n} b_{i}}e^{−λ^{p} ∑_{i=1}^{n} a_{i}(ε(x_{i}))|x_{i}|^{p}} ∏_{i=1}^{n} c_{i}|x_{i}|^{b_{i}}

= cλ^{b}∥x∥_{K}^{b}e^{−aλ^{p}∥x∥_{K}^{p}} g (\frac{x}{∥x∥_{K}} ) .

This clearly implies that a∥x∥_{K}^{p} = ∑_{i=1}^{n} a_{i}(ε(x_{i}))|x_{i}|^{p}. Since μ is a probability measure, necessarily a, a_{i}(1), a_{i}(−1) > 0 and b_{i} > −1. Thus K is determined. The boundedness of K forces p > 0. The proof is complete. □

We now pass to the case of μ being an infinite measure. In this case, every star-shaped body gives rise to a measure with a simultaneous product decomposition. Indeed, for every b_{1}, …, b_{n} > −1 in ℝ, and every star-shaped body K, the measure dμ(x) = ∏_{i=1}^{n} |x_{i}|^{b_{i}} dx admits such a decomposition, due to the identity:

∏_{i=1}^{n} |x_{i}|^{b_{i}} = ∏_{i=1}^{n} (\frac{|x_{i}|}{∥x∥_{K}} )^{b_{i}} · ∥x∥_{K}^{∑_{i=1}^{n} b_{i}}.

We can however prove that the above example is the only additional case. For simplicity we work with measures on (0, ∞)^{n}.

Theorem 5 Let K ⊂ ℝ^{n} be a star shaped body. Assume that μ is an absolutely continuous measure on (0, ∞)^{n} which has a simultaneous product decomposition with respect to K. Assume in addition that log (\frac{dμ}{dx}) is locally integrable. Then one of the following assertions holds:

1) There are p, r > 0, b_{1}, …, b_{n} ∈ ℝ and c ≥ 0 and a_{1}, …, a_{n} ≠ 0 all having the same sign such that:

K ∩ (0, ∞)^{n} = {x ∈ (0, ∞)^{n}; ∑_{i=1}^{n} |a_{i}| · |x_{i}|^{p} ≤ r} ,

and

dμ = c ∏_{i=1}^{n} (x_{i}^{b_{i}}e^{−a_{i}x_{i}^{p}}1_{{x_{i}>0}}dx_{i}) .

2) K is arbitrary and there are b_{1}, …, b_{n} ∈ ℝ and c > 0 such that

dμ = c ∏_{i=1}^{n} (x_{i}^{b_{i}}1_{{x_{i}>0}}dx_{i}) .

Conversely if K and μ satisfy 1) or 2) then μ has a simultaneous product decomposition with respect to K.

Proof: This result follows from the proof of Theorem 3. The writing is simpler since we work on (0, ∞)^{n}. We present the modifications. If in the argument p = 0, then f_{i}(t) = c_{i}t^{b_{i}} and we are done. If p ≠ 0 then the argument provides a, a_{1}, …, a_{n} such that whenever x_{i} > 0

a∥x∥_{K}^{p} = ∑_{i=1}^{n} a_{i}x_{i}^{p}.

If a = 0 then then a_{i} = 0 for all i’s, f_{i}(t) = c_{i}t^{b_{i}} and there is no constraint on K. If a ≠ 0 then the previous relation gives a_{i} = a∥e_{i}∥_{K}^{p}, so the a_{i}’s are not zero and have same sign. Since ∥x∥_{K} = (∑_{i=1}^{n} \frac{a_{i}}{a} |x_{i}|^{p})^{1/p}, the set K ∩ (0, ∞)^{n} is a weighted ℓ_{p}^{n}-ball. By boundedness p > 0. As before f_{i}(t) = c_{i}t^{b_{i}}e^{−a_{i}t^{p}}. This ends the proof. □

3 Atomic Measures

In this section we focus on finite atomic measures ∑_{P ∈S} α_{P} δ_{P}, where δ_{P} is the Dirac measure at P and S ⊂ ℝ^{n} is countable. For convenience, we write μ(P) for μ({P }). We also restrict to convex sets K. The following result deals with measures which are not supported on a sphere. Measures which concentrate on a sphere, when K is convex are much easier to classify and we leave this to the reader (one of the μ_{i}’s has to be a Dirac mass).

Theorem 6 Assume that K ⊂ ℝ^{n} is convex, symmetric and contains the origin in its interior, and that μ is a finite (and non-zero) atomic measure on (0, ∞)^{n}, which admits a simultaneous polar and Cartesian decomposition with respect to K: μ = μ_{1} ⊗⋯⊗μ_{n} = τ ⊙ν. Assume in addition that τ is not a Dirac measure. Then the following assertions hold:

a) There are λ_{1}, …, λ_{n} > 0 such that K ∩ [0, ∞)^{n} = ∏_{i=1}^{n}[0, λ_{i}].

b) There are c, r, α_{1}, … α_{n} > 0, 0 < q < 1 and D = ∏_{i=1}^{n} {rλ_{i}q^{k}; k ∈ ℕ} such that :

μ(x) = \{ \begin{matrix}c ∏_{i=1}^{n} x_{i}^{α_{i}} & if x ∈ D \\ 0 & if x ∉ D \end{matrix}

Conversely, if K and μ satisfy a) and b) then μ has a simultaneous product decomposition with respect to K.

As a matter of illustration, we check that conditions a) and b) ensure simultaneous Cartesian and polar product decompositions. Let x ∈ (0, ∞)^{n}. Since the set D is a Cartesian product, it is clear that μ defined in b) is a Cartesianproduct measure. Next, write x = ρω with ρ > 0 and ω ∈ ∂K then

μ(x) = μ(ρω) = c ρ^{∑_{i=1}^{n} α_{i}} ∏_{i=1}^{n} ω_{i}^{α_{i}}δ_{D}(ρω),

so we just need to check that D is a polar product set in order to show that the latter is the product of a quantity depending only on ρ times a quantity depending on ω. But this is easy: if x ∈ D then for each i, one has x_{i} = rλ_{i}q^{k_{i}}. By a), ∥x∥_{K} = r q^{min_{i} k_{i}} ∈ T = {rq^{k}; k ∈ ℕ} and

\frac{x}{∥x∥_{K}} = (λ_{i}q^{k_{i}−min_{j} k_{j}})_{i=1}^{n} ∈ Ω = \{(λ_{i}q^{h_{i}})_{i=1}^{n}; h_{i} ≥ 0 and ∏_{i} h_{i} = 0\}.

This shows that D ⊂ T · Ω. The converse inclusion is easily checked. Hence μ has simultaneous product decomposition.

The rest of this section is devoted to the proof of the necessary condition in Theorem 6. From now on we assume that μ and K satisfy the assumption of the theorem. We begin with some notation. If λ is a measure on a measurable space (Ω, Σ), let M_{λ} = {x ∈ Ω; λ(x) = sup_{ω∈Ω} λ(ω)}. Clearly, if λ is a finite measure then |M_{λ}| < ∞. We also put

M_{λ}^{2} = {x ∈ Ω; λ(x) = sup_{ω∈Ω\M_{λ}} λ(ω)} .

When λ is countably supported we define supp(λ) = {ω ∈ Ω; λ(ω) > 0}.

Returning to the setting of Theorem 6, we clearly have that supp(μ) = supp(μ_{1}) × ⋯ × supp(μ_{n}) = supp(τ) · supp(ν) and M_{μ} = M_{μ_{1}} × ⋯ × M_{μ_{n}} = M_{τ} · M_{ν}.

The next lemma will be used for a measure other than μ of the main theorem. This is why the fact that τ in the product decomposition is not Dirac is specifically stated as an hypothesis there.

Lemma 7 Define the slope function s_{1}(x) = \frac{x_{1}}{∥(x_{2},...,x_{n})∥_{2}} . If τ is not a Dirac measure, then s_{1} does not attain its minimum on supp(μ).

Proof: Assume that μ(x) > 0. Then μ_{1}(x_{1}), … μ_{n}(x_{n}) > 0, ν(x/∥x∥_{K}) > 0 and τ(∥x∥_{K}) > 0. Since τ is not a Dirac measure, there is r > 0, r ≠ ∥x∥_{K} such that τ(r) > 0. Let y = \frac{r}{∥x∥_{K}} x. Then μ(y) = τ(r)ν(x/∥x∥_{K}) > 0, so that

for every i = 1, …, n, μ_{i} (\frac{r}{∥x∥_{K}} x_{i}) > 0. Therefore for

u = (\frac{r}{∥x∥_{K}} x_{1}, x_{2}, …, x_{n}) ,

v = (x_{1}, \frac{r}{∥x∥_{K}} x_{2}, \frac{r}{∥x∥_{K}} x_{3}, …, \frac{r}{∥x∥_{K}} x_{n}) ,

μ(u) > 0 and μ(v) > 0. But s_{1}(u) = \frac{r}{∥x∥_{K}} s_{1}(x), s_{1}(v) = \frac{∥x∥_{K}}{r} s_{1}(x) and either

\frac{r}{∥x∥_{K}} < 1 or \frac{∥x∥_{K}}{r} < 1. □

Corollary 8 Under the assumptions of Theorem 6, |supp(μ)| = ∞ and |M_{τ} | = 1.

Proof: The first assertion is obvious, and the second assertion follows since μ|_{M_{μ}} has a simultaneous product decomposition. Indeed μ|_{M_{μ}} has also a finite support, therefore s_{1} attains its minimum on it. It follows that the radial measure τ|_{M_{μ}} has to be a Dirac measure. □

Put M_{τ} = {r}.

Lemma 9 supp(μ) ⊂ {x; ∥x∥_{K} ≤ r}.

Proof: If supp(μ) ⊄ {x; ∥x∥_{K} ≤ r} then since τ is a finite measure there is R > r such that τ(R) > 0 is maximal on (r, ∞). For every i = 1, …, n let M_{i} = max M_{μ_{i}} > 0. Now, x = (M_{1}, M_{2}, …, M_{n}) ∈ M_{μ}, so that ∥x∥_{K} = r. Put y = \frac{R}{r}x. Clearly μ(y) = τ(R)ν(x/∥x∥_{K}) is maximal on the set \{x; ∥x∥_{K} > r\}. For every i = 1, …, n define

x^{i} = (M_{1}, …, M_{i−1}, \frac{R}{r}M_{i}, M_{i+1}, …, M_{n}) .

Note that for every j = 1, …, n, M_{j} = max M_{μ_{j}} < \frac{R}{r}M_{j} = y_{j} so that y_{j} ∉ M_{μ_{j}}. It follows that for every i = 1, …, n, μ(x^{i}) > μ(y), so that ∥x^{i}∥_{K} ≤ r. Now, using the convexity of K we have

r = ∥x∥_{K} = ∥\frac{1}{n − 1 + \frac{R}{r}} ∑_{i=1}^{n} x^{i}∥_{K} ≤

≤ \frac{1}{n − 1 + \frac{R}{r}} ∑_{i=1}^{n} ∥x^{i}∥_{K} ≤ \frac{nr}{n − 1 + \frac{R}{r}} < r,

which is a contradiction. □

Since τ isn’t Dirac, there is some r′ ∈ M_{τ}^{2}. By Lemma 9, r′ < r.

Lemma 10 For every i = 1, …, n, inf supp(μ_{i}) = 0.

Proof: As in Lemma 7, we will study the function:

s_{i}(x) = \frac{x_{i}}{∥ ∑_{j≠i} x_{j}e_{j}∥_{2}} .

Since Lemma 9 implies in particular that supp(μ) is bounded, our claim will follow once we show that inf_{x∈supp(μ)} s_{i}(x) = 0. Let σ_{i} = inf_{x∈supp(μ)} s_{i}(x) and assume that σ_{i} > 0. For every ϵ > 0 there is x ∈ supp(μ) such that s_{i}(x) ≤ (1+ϵ)σ_{i}. From the proof of Lemma 7 it follows that for every ρ ∈ supp(τ) there are u, v ∈ supp(τ) such that s_{i}(u) = \frac{ρ}{∥x∥_{K}} s_{i}(x) and s_{i}(v) = \frac{∥x∥_{K}}{ρ} s_{i}(x). Hence,

min \{\frac{ρ}{∥x∥_{K}} , \frac{∥x∥_{K}}{ρ} \} ≥ \frac{1}{1+ϵ}. If ∥x∥_{K} = r take ρ = r′. Otherwise, ∥x∥_{K} ≤ r′, in which case take ρ = r. In both cases we get that r ≤ (1 + ϵ)r′, which is a contradiction when ϵ is small enough. □

In what follows we will continue to use the notation M_{i} = max M_{μ_{i}}, and we will also put m_{i} = min M_{μ_{i}}. Let x = (M_{1}, …, M_{n}), x′ = (m_{1}, …, m_{n}).

Corollary 11 For every J ⊂ {1, …, n}:

∥∑_{i∈J} M_{i}e_{i}∥_{K} ≤ r.

Proof: By Lemma 10 for every ϵ > 0 and i = 1, …, n there is z_{i} ∈ supp(μ_{i}) with z_{i} < ϵ. Now:

∑_{i∈J} M_{i}e_{i} + ∑_{i∉J} z_{i}e_{i} ∈ supp(μ),

so that by Lemma 9 we get:

∥∑_{i∈J} M_{i}e_{i} + ∑_{i∉J} z_{i}e_{i}∥_{K} ≤ r.

The result follows by taking ϵ → 0. □

Corollary 12 ∏_{i=1}^{n}[0, M_{i}] ⊂ rK.

Lemma 13 Let J be a non-empty subset of {1, …, n}. Then:

∥∑_{i∈J} M_{i}e_{i} + ∑_{i∉J} \frac{r′}{r} m_{i}e_{i}∥_{K} = r.

Proof: Denote y = \frac{r′}{r} x′. Since x′ ∈ M_{μ}, ∥x′∥_{K} = r. Now, μ(y) = τ(r′)ν(x′/∥x′∥_{K}), and because r′ ∈ M_{τ}^{2} and x′/∥x′∥_{K} ∈ M_{ν} we deduce that μ(y) is maximal on the set {x; ∥x∥_{K} ≠ r}. But since r′ < r, for any j = 1, …, n one has y_{j} = \frac{r′}{r} m_{j} < m_{j} = min M_{μ_{j}}, so that y_{j} ∉ M_{μ_{j}}. It follows that since J ≠ ∅,

μ (∑_{i∈J} M_{i}e_{i} + ∑_{i∉J} \frac{r′}{r} m_{i}e_{i}) = (∏_{i∈J} μ_{i}(M_{i})) (∏_{i∉J} μ_{i}(y_{i})) > ∏_{i=1}^{n} μ_{i}(y_{i}) = μ(y),

so that ∥∑_{i∈J} M_{i}e_{i} + ∑_{i∉J} \frac{r′}{r} m_{i}e_{i}∥_{K} = r. □

We can now prove the first part of Theorem 6:

Proposition 14 K ∩ [0, ∞)^{n} = ∏_{i=1}^{n} [0, \frac{M_{i}}{r} ].

Proof: We set Q = ∏_{i=1}^{n}[0, M_{i}]. Let 1 ≤ i ≤ n. Since 0 < m_{i}r′/r < M_{i} the point

P_{i} = M_{i}e_{i} + ∑_{j≠i} \frac{r′}{r} m_{j}e_{j}

lies in the interior of the facet Q ∩ {x; x_{i} = M_{i}} of Q. It is also a boundary point of rK by Lemma 13. As guaranteed by Corollary 12, Q ⊂ rK, so that any supporting hyperplane of rK at P_{i} is a supporting hyperplane of Q at this point. Therefore at P_{i} the convex set rK admits {x; x_{i} = M_{i}} as a (unique) supporting hyperplane. It follows that rK ⊂ {x; x_{i} ≤ M_{i}}. This is true for every 1 ≤ i ≤ n and the proof is complete. □

We now pass to the proof of the final assertion of Theorem 6. We have proved that there are real numbers t_{i} = r/M_{i} > 0, i = 1 … n such that for every x ∈ [0, ∞)^{n}, ∥x∥_{K} = max_{1≤i≤n} t_{i}x_{i}. Moreover for every x ∈ supp(μ), ∥x∥_{K} ≤ r and μ(r/t_{1}, …, r/t_{n}) > 0.

Lemma 15 If p, q ∈ supp(τ) and q > p then \frac{pr}{q} , \frac{pq}{r} ∈ supp(τ).

Proof: Since μ (\frac{r}{t_{1}} , …, \frac{r}{t_{n}} ) > 0, ν (\frac{1}{t_{1}} , …, \frac{1}{t_{n}} ) > 0. Now,

0 < τ(p)ν (\frac{1}{t_{1}} , …, \frac{1}{t_{n}} ) = μ (\frac{p}{t_{1}} , …, \frac{p}{t_{n}} ) = ∏_{i=1}^{n} μ_{i} (\frac{p}{t_{i}} ) ,

so that for every i = 1, …, n, \frac{p}{t_{i}} ∈ supp(μ_{i}). Similarly, \frac{q}{t_{i}} ∈ supp(μ_{i}). Hence, since p < q:

0 < μ (\frac{p}{t_{1}} , \frac{q}{t_{2}} , …, \frac{q}{t_{n}} ) = τ(q)ν (\frac{p}{qt_{1}} , \frac{1}{t_{2}} , …, \frac{1}{t_{n}} ) ,

so that ν (\frac{p}{qt_{1}} , \frac{1}{t_{2}} , …, \frac{1}{t_{n}} ) > 0. Now, using again the fact that p < q we have:

μ_{1} (\frac{pr}{qt_{1}} ) = \frac{μ (\frac{pr}{qt_{1}} , \frac{r}{t_{2}} , …, \frac{r}{t_{n}} )}{∏_{i=2}^{n} μ_{i} (\frac{r}{t_{i}} )} =

= \frac{τ(r)ν (\frac{p}{qt_{1}} , \frac{1}{t_{2}} , …, \frac{1}{t_{n}} )}{∏_{i=2}^{n} μ_{i} (\frac{r}{t_{i}} )} > 0.

This shows that \frac{pr}{qt_{1}} ∈ supp(μ_{1}). Similarly, for every i, \frac{pr}{qt_{i}} ∈ supp(μ_{i}). Hence,

0 < ∏_{i=1}^{n} μ_{i} (\frac{pr}{qt_{i}} ) = μ (\frac{pr}{qt_{1}} , …, \frac{pr}{qt_{n}} ) = τ (\frac{pr}{q} ) ν (\frac{1}{t_{1}} , …, \frac{1}{t_{n}} ) .

This shows that \frac{pr}{q} ∈ supp(τ). Now, since the remark preceding Lemma 15 implies that p ≤ r,

μ_{1} (\frac{pq}{rt_{1}} ) ∏_{i=2}^{n} μ_{i} (\frac{q}{t_{i}} ) = μ (\frac{pq}{rt_{1}} , \frac{q}{t_{2}} , …, \frac{q}{t_{n}} ) = τ(q)ν (\frac{p}{rt_{1}} , \frac{1}{t_{2}} , …, \frac{1}{t_{n}} ) =

= \frac{τ(q)μ (\frac{p}{t_{1}} , \frac{r}{t_{2}} , …, \frac{r}{t_{n}} )}{τ(r)} = \frac{τ(q)}{τ(r)}μ_{1} (\frac{p}{t_{1}} ) ∏_{i=2}^{n} μ_{i} (\frac{r}{t_{i}} ) > 0.

Hence, \frac{pq}{rt_{1}} ∈ supp(μ_{1}). Similarly, for every i, \frac{pq}{rt_{i}} ∈ supp(μ_{i}), so that:

0 < μ (\frac{pq}{rt_{1}} , …, \frac{pq}{rt_{n}} ) = τ (\frac{pq}{r} ) ν (\frac{1}{t_{1}} , …, \frac{1}{t_{n}} ) ,

which shows that \frac{pq}{r} ∈ supp(τ). □

Lemma 16 For every i = 1, …, n, t_{i}supp(μ_{i}) = supp(τ).

Proof: In the proof of Lemma 15 we have seen that supp(τ) ⊂ t_{i}supp(μ_{i}). We show the other inclusion. First, note that inf supp(τ) = 0. Indeed for every ϵ > 0, Lemma 10 ensures the existence of x_{i} ∈ supp(μ_{i}) such that x_{i} < ϵ/t_{i}. Now,

0 < ∏_{i=1}^{n} μ_{i}(x_{i}) = τ ( max_{1≤i≤n} t_{i}x_{i}) ν (\frac{x}{∥x∥_{K}} ) .

So inf supp(τ) ≤ ϵ.

Take any p ∈ supp(μ_{i}). There is some q ∈ supp(τ) such that q < t_{i}p. By the proof of Lemma 15, for every j, q/t_{j} ∈ supp(μ_{j}), so that:

0 < μ_{i}(p) ∏_{j≠i} μ_{j} (\frac{q}{t_{j}} ) = μ (pe_{i} + ∑_{j≠i} \frac{q}{t_{j}} e_{j}) = τ(t_{i}p)ν (\frac{1}{t_{i}} e_{i} + ∑_{j≠i} \frac{q}{t_{i}t_{j}p}e_{j}) ,

so that t_{i}p ∈ supp(τ). □

Lemma 17 For every i = 1, …, n and for every p, q ∈ supp(μ_{i}),

μ_{i} (\frac{t_{i}pq}{r} ) = \frac{μ_{i}(p)μ_{i}(q)}{μ_{i} (\frac{r}{t_{i}} )}.

Proof: Note that since t_{i}p ∈ supp(τ), t_{i}p ≤ r. Hence, using the fact that \frac{t_{i}q}{t_{j}} ∈ supp(μ_{j}) we have:

0 < μ_{i} (\frac{t_{i}pq}{r} ) ∏_{j≠i} μ_{j} (\frac{t_{i}q}{t_{j}} ) = μ (\frac{t_{i}pq}{r} e_{i} + ∑_{j≠i} \frac{t_{i}q}{t_{j}} e_{j}) =

= τ (max \{\frac{t_{i}^{2}pq}{r} , t_{i}q\}) ν (\frac{1}{max \{\frac{t_{i}^{2}pq}{r} , t_{i}q\}} (\frac{t_{i}pq}{r} e_{i} + ∑_{j≠i} \frac{t_{i}q}{t_{j}} e_{j})) =

= τ (t_{i}q) ν (\frac{p}{r}e_{i} + ∑_{j≠i} \frac{1}{t_{j}} e_{j}) = τ(t_{i}q)\frac{μ (pe_{i} + ∑_{j≠i} \frac{r}{t_{j}} e_{j})}{τ(r)} =

= \frac{τ(t_{i}q)}{τ(r)}μ_{i}(p) ∏_{j≠i} μ_{j} (\frac{r}{t_{j}} ) = μ_{i}(p)\frac{μ (qe_{i} + ∑_{j≠i} \frac{t_{i}q}{t_{j}} e_{j})}{ν (∑_{j=1}^{n} \frac{1}{t_{j}} e_{j})} · \frac{μ (∑_{j=1}^{n} \frac{r}{t_{j}} e_{j})}{μ_{i} (\frac{r}{t_{i}} ) τ(r)} =

= \frac{μ_{i}(p)μ_{i}(q)}{μ_{i} (\frac{r}{t_{i}} )} ∏_{j≠i} μ_{j} (\frac{t_{i}q}{t_{j}} ) .

□

Lemma 18 Assume that A ⊂ (0, 1], A ≠ {1}, A ≠ ∅, has the property that xy and x/y are in A whenever x, y ∈ A and x ≤ y. Let f : A → (0, ∞) be a function such that if x, y ∈ A then f(xy) = f(x)f(y) and:

∑_{a∈A} f(a) < ∞.

Then there are α > 0 and 0 < q < 1 such that f(a) = a^{α} and A = {q^{n}}_{n=0}^{∞}.

Proof: For any a ∈ A \ {1} and n ∈ ℕ, a^{n} ∈ A and f(a^{n}) = f(a)^{n}. Since ∑_{n=1}^{∞} f(a^{n}) < ∞, f(a) < 1. Now, if a, b ∈ A \ \{1\} and \frac{a^{n}}{b^{m}} < 1, \frac{a^{n}}{b^{m}} ∈ A, so that 1 > f (\frac{a^{n}}{b^{m}} ) = \frac{f(a)^{n}}{f(b)^{m}}. We have shown that for every n, m ∈ ℕ and a, b ∈ A \ \{1\}:

\frac{n}{m} > \frac{log b}{log a} ⟹ \frac{n}{m} > \frac{log f(b)}{log f(a)}.

Hence, \frac{log b}{log a} ≥ \frac{log f(b)}{log f(a)} for every a, b ∈ A \ \{1\}. By symmetry, there is α ∈ ℝ

such that for every a ∈ A, \frac{log f(a)}{log a} = α. This proves the first assertion (α > 0 since f(a) < 1).

Put B = {− log a; a ∈ A}. Clearly:

a, b ∈ B ⟹ a + b, |a − b| ∈ B.

Since f(a) = a^{α} and ∑_{a∈A} f(a) < ∞, for every x > 0 there are only finitely many a ∈ A with a ≥ x. In other words, for every x > 0 there are only finitely many b ∈ B with b ≤ x. In particular, if we let p = inf B \ {0} then p > 0 and p ∈ B. Now, for every n = 0, 1, 2, …, np ∈ B. We claim that B = {0, p, 2p, 3p, …}. Indeed, if x ∈ B \ {0, p, 2p, 3p, …} then there is an integer n such that 0 < |x − np| < p. But, |x − np| ∈ B, and this contradicts the definition of p. Finally, for q = e^{−p}, A = {1, q, q^{2}, …}. □

Remark: All the assumptions in Lemma 18 are necessary. Apart from the trivial examples such as A = (0, 1] and A = (0, 1] ∩ ℚ we would like to point out the more interesting example A = {2^{n}3^{m}; m, n ∈ ℤ and 2^{n}3^{m} ≤ 1}, f(2^{n}3^{m}) = 2^{αm}3^{βm} where α and β are distinct real numbers (of course in this case the condition ∑_{a∈A} f(a) < ∞ is not satisfied).

Proof of Theorem 6: Assertion a) is given by Proposition 14. To prove b) fix some 1 ≤ i ≤ n and define: A = \frac{t_{i}}{r} supp(μ_{i}). By Lemma 16 and Lemma 9, A = \frac{1}{r}supp(τ) ⊂ \frac{1}{r}(0, r] = (0, 1]. Additionally, if x, y ∈ A, x ≤ y, then there are p, q ∈ supp(τ), p ≤ q, such that x = \frac{p}{r}, y = \frac{q}{r}. Lemma 15 implies that \frac{x}{y} = \frac{1}{r} · \frac{pr}{q} ∈ \frac{1}{r}supp(τ) = A. Similarly, xy ∈ A. Define f_{i} : A → ℝ by

f_{i}(x) = \frac{μ_{i} (\frac{r}{t_{i}} x)}{μ_{i} (\frac{r}{t_{i}} )}.

Clearly for every x ∈ A, f_{i}(x) > 0 and since μ_{i} is a finite measure, ∑_{a∈A} f_{i}(a) < ∞. An application of Lemma 17 gives that for every x, y ∈ A, f_{i}(xy) = f_{i}(x)f_{i}(y). Now, Lemma 18 implies that there are α_{i} > 0 and 0 < q < 1 such that f_{i}(a) = a^{α_{i}} and A = {q^{n}}_{n=0}^{∞}. So, supp(τ) = {rq^{k}; k ∈ ℕ} and by Lemma 16, one gets supp(μ_{i}) = {rq^{k}/t_{i}; k ∈ ℕ}. Moreover, μ_{i}(rq^{k}/t_{i}) = μ_{i}(r/t_{i})q^{kα_{i}}. This concludes the proof of the theorem. □

4 Concluding Remarks

In this sections we list some remarks and open problems that arise from the results of the previous two sections.

1) There are examples when K is allowed to be unbounded (of course in this case it is no longer a body). Indeed the “unit ball” of ℓ_{p}^{n} for non-positive p gives such a decomposition with f_{i}(t) = |t|^{b_{i}} exp(−|t|^{p}).

2) Theorem 3 does not cover the case of the uniform measure on B_{∞}^{n} = [−1, 1]^{n}, which clearly has simultaneously the Cartesian and the polar decomposition with respect to K = B_{∞}^{n}. It is the natural limit case of the examples with the densities exp(−|t|^{p}). Under strong conditions on the density and its support, results can be obtained which encompass measures supported on the cube. It would be very nice to get rid of the conditions. It seems that one of the necessary steps would be to understand the structure of sets in ℝ^{n} which are products with respect to the Cartesian structure and for the polar structure generated by a convex set K. This is a problem of independent interest.

3) The classification of simultaneous product measures, without additional hypothesis, is a very challenging problem. Note that our results may be used. Indeed if μ has simultaneous product decomposition, then its absolutely continuous part has it too. Similarly, if a singular measure has the property, then its atomic part has it too, so Theorem 6 applies. The main obstacle seems to be dealing with singular continuous measures.

Acknowledgments: Part of this work was carried out while the first and lastnamed authors were visiting University College London. They are grateful to their UCL hosts, professors K. Ball and D. Preiss, for their invitation.

References

[1] F. Barthe and A. Naor. Hyperplane projections of the unit ball of ℓ_{p}^{n}. To appear in Discrete Comput. Geom.

[2] W. Bryc. The normal distribution. Characterizations with applications. Lecture Notes in Statistics, 100. Springer Verlag, New York, 1995.

[3] E. A. Carlen. Superadditivity of Fisher’s information and logarithmic Sobolev inequalities. J. Funct. Anal., 101(1):194–211, 1991.

[4] A. Naor. The Surface Measure and Cone Measure on the Sphere of ℓ_{p}^{n}. Preprint, 2001.

[5] A. Naor and D. Romik. Projecting the surface measure of the sphere of ℓ_{p}^{n}. Preprint, 2001.

[6] S. T. Rachev and L. Rüschendorf. Approximate independence of distributions on spheres and their stability properties. Ann. Probab., 19(3):1311– 1337, 1991.

[7] G. Schechtman and J. Zinn. On the volume of intersection of two L_{p}^{n} balls. Proc. Amer. Math. Soc, 110(1):217–224, 1990.

[8] G. Schechtman and J. Zinn. Concentration on the ℓ_{p}^{n} ball. Geometric aspects of functional analysis, Lecture Notes in Math., 1745, Springer Verlag, Berlin, 245-256, 2000.