THE RADON NIKODYM PROPERTY AND LIPSCHITZ EMBEDDINGS

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1 THE RADON NIKODYM PROPERTY AND LIPSCHITZ EMBEDDINGS Motivation The idea here is simple. Suppose we have a Lipschitz homeomorphism f : X Y where X and Y are Banach spaces, namely c 1 x y f (x) f (y) c 2 x y. Now suppose that we can introduce some notion of differentiability so that f is differentiable with derivative T at some point a. Then we certainly have that c 1 x Tx c 2 x, namely, that a linear isomorphism exists. To make this precise we will need to define a suitable notion of differentiability. Of key note here is the Radon Nikodym property, so this is where we will start. 1 Some Calculus on Banach Spaces 1.1 The Radon Nikodym Property We say that a closed, convex, bounded set C E of a Banach space E has the Radon Nikodym Property if given a measurable space (Ω, B) with an E valued measure τ, and a scalar valued probability measure µ with τ(a)/µ(a) C for all A B (with µ(a) = 0m then there exists an f L 1 (µ, E) such that τ(a) = A f (w)dµ. We say that a space has the Radon Nikodym property when its unit ball has the Radon Nikodym property. We note that c 0 does not have the Radon Nikodym Property: define a measure τ(a) = ( A 1dt, A eit dt,... ). Such a measure on c 0 is absolutely continuous with respect to Lebesgue measure but there is no possible derivative in c 0. Theorem 1.1. Let E be a Banach space. Then E has the Radon Nikodym property just when every absolutely continuous function f : [0, 1] E is differentiable almost everywhere. Proof. Let f be such an absolutely continuous function. The variation of f, V(t) = sup{ f (s j ) f (s j 1 ) } where {s j } is some partition of [0, t] is an absolutely continuous scalar function, so we can represent V(t) = µ[0, t], where µ is some finite positive measure µ which is absolutely continuous with respect to Lebesgue measure. Define a vector measure τ by τ(a i, b i ) = f (b i ) f (a i ), extending by linearity on disjoint unions of intervals. If A is a general Borel set there is some sequence of sets G n of [0, 1] that are open and decrease to A. Since µ(g n ) exists, we see that µ(g m /G m ) 0, and thus τ(a) = lim τ(g n ) exists (as τ = µ.) Since E has the Radon Nikodym Property we can find some Bochner integrable g : [0, 1] E so that f (t) = f (0) + τ[0, t] = t g(x) dx + f (0). Such an f has derivative almost 0 everywhere by the obvious extension of the classical Lebesgue differentiation theorem. 1

2 We note that the converse is true here: namely, if there is some function f : [0, 1] E which is differentiable nowhere, then E fails the Radon Nikodym Property. This property is harder to prove, requiring a soujorn into the properties of Martingales, so we will not include it here. This gives us another way of thinking about the counter example given for c 0 : take the function f (t) = ( t 0 1 dt, t 0 sin t dt, t 0 sin 2t dt,... ) (this is in c 0 by direct computation). The only possible derivative here does not end up in c 0. We can also use this to prove that L 1 [0, 1] does not have the RNP 1 by considering the function f (t) = 1 [0,t]. This is clearly non differentiable (any derivative would converge to the delta functional: the delta functional is not in L 1.) Moreover, the Radon Nikodym property is preserved under isomorphisms. This is actually an d easy consequence, differentiation being a linear operator and all that: dt (F(g(t)) = F( d dt g(t)) if F is a linear isomorphism. (The fact that it is a linear isomorphism gives us the bounding required). 1.2 Gateaux Differentiability We say that a function f : E F between two Banach spaces is Gateaux differentiable at a point x if all its directional derivatives exist, and the resulting map is linear: namely if the function F(u+τψ) F(u) df(u, ψ) = lim τ 0 τ exists, and df(u, λψ + µφ) = λdf(u, ψ) + µdf(u, φ). Of particular note is when this limit exists uniformly in the unit sphere, we will then call the function Frechet Differentiable. First up, we want to be able to reduce to a countable case: Lemma 1.1. Suppose that f is a Lipschitz function between two open sets in Banach spaces E and F. If G E is a dense additive subgroup such that at x 0 the difference quotients exist and form an additive function then the function is Gateaux differentiable at x 0. Proof. The functions h t (u) = ( f (x 0 + tu) f (x 0 ))/t are equicontinuous with Lipschitz constant the same as the Lispchitz constant of f. Hence the existence of lim t 0 h t (u) on G implies the limit exists for all u G = E. Also the additivity of the limit on G implies the additivity of the limit on E. 1.3 Almost Everywhere on Banach Spaces We have another problem: what does almost all mean on a Banach space E? There are several possible definitions, however, it is worth noting that there is no obvious choice of definition of almost everywhere here. There is no natural choice, on R n we can view Lesbegue Measure as coming from the Riesz Representation theorem applied to C[ 1, 1] n : Lesbegue Measure is the unique measure extending Riemann Integration. However, in a Banach space, the unit sphere (the obvious analogue for [ 1, 1] n is never compact, so we are not free to apply Riesz. So, how do we clear this up? There are several ways we could potentially proceed. We will proceed by introducing Aronszajn null sets. The reader is cautioned that there are several other 1 We could do this in entirely the smae way as c 0, but this is nicer 2

3 useful ways of proceeding, namely, Gauss null sets, Cube null sets and Haar null sets are all valid choices. Definition: We say that a set A is Aronszajn null if it can be written as the union of Borel sets E n with the property that for any E n the intersection of E n with the line through x m has Lebesgue measure 0, for all x m in some separating set. We write A({y k }) = the family of Borel sets such that the intersection with the line through y k has Lesbegue measure 0, so equivalently, a set is Aronszajn null if it lies in A({x k }) for every separating set x k. 2 We note that the entire space isn t Aronszajn null (as the intersection of every line with the entire space is the line which has infinite measure) neither is any ball centred at zero, for basically the same reason (but with finite measure). It is worth noting that the concept of Aronszajn null is translation invariant: this either pops out of theorem 3.1 or can be obtained through several arguments relating Aronszajn null sets to other types of null sets. 3 Our goal now is to present a proof of the main theorem in this area: the set of points at which a Lipschitz function is not Gateaux differentiable is Aronszajn null. First of all, we show that the result is true for functions f : U F where U R n. 2 Functions from R n to F Theorem 2.1. Suppose F is a Banach space with the Radon Nikodym Property. Then, if U R n is an open subset, and f : U F is a Lipschitz function, then f is almost everywhere Gateaux differentiable. Proof. Note that we are free to take G some countably dense additive subgroup of R n4 and then apply the Lemma from the section on Gateaux Differentiability. First of all, note that the directional derivatives of f exist for each of the vectors in G, as F has the Radon Nikodym property. So our goal now is to show that these directional derivatives are almost everywhere additive functions on G. To prove this we will rely upon using convolutions and bump functions: take a bump function φ(x) in R n which is everywhere positive, compactly supported and has integral 1. Define φ k (x) = k n φ(kx). So, for g(x) = ( f φ)(x) (a C function) we have that D g (x)u = ( f D φ (x))u is a linear function of u R n for each x. On the other hand, Lesbegue s Dominated Convergence theorem gives us that D g (x)u = φ h u (x), where h u is the limit of the difference quotient (and a bounded measurable function). 5 Conssequently, we have that φ (h u+v h u h v ) = 0 for every u, v. 2 It is a technicality that we require every separating set x k, but an important one. 3 Of particular note is that a set is Aronszajn null if and only it is a null set with respect to every Gaussian measure (a measure on the space whose pushforward under any element of the dual space is a Gaussian measure). It is seen (quite easily) that the second concept is very translation invariant. 4 Probably Q n 5 Bounded because Lipschitz, this gives us our dominating function. 3

4 Finally, note that lim ϕ k p = p pointwise ae, for every bounded measurable function p mapping to R. Using the fact that the dual seperates, we can apply functionals and we are done. 3 A characterization of A({y k }) We need one final ingredient before we can show that the set of points is Aronszajn null, and this is a way of characterizing Aronszajn null sets a bit better. This will consist of a Lemma and a theorem. Lemma 3.1. Let F be a k dimensional subspace of E and let λ k denote the Lebesgue measure on F. Then the function f B (x) = λ k (F (B + x)) is Borel for every Borel subset B of E. Proof. We have that the set {x : f B (x) α} is certainly closed for every closed/bounded set B and every α, and that if B j increases or decreases to B, then f Bj (x) f B (x), and thus we are done. Theorem 3.2. Let F be an n-dimensional subspace of E and let y k be a basis for F. Let λ n be the Lebesgue measure on F and let A be a Borel subset of E such that λ n (F (A + x)) = 0 for every x E. Then A A({y k }). Proof. We prove this by induction on the dimension n of F. It is clear that this is true for n = 1, this is simply the definition of Aronszajn null sets. Let G be the span of {y 1,..., y n 1 } and assume that λ n is the product measure of λ n 1 on G and λ 1 on y n (suitable normalized). Let A n = {x A : λ 1 (A (x + Ry n )) = 0}. The set A n is Borel, and therefore, so is A = A/A n. This set has the property that for each line L parallel to y n either L A = or L A = L A and λ 1 (L A ) > 0. Fix some x E. We have that 0 = λ n ((A + x) F) λ n ((A + x) F), as we are shrinking the size of the set. This equals C λ 1((u + Ry n ) (A + x))dλ n 1, where C is the projection of F (A + x) onto G in the direction of y n. This is contained in the set B = G (A + x), and thus we have that 0 B λ 1((u + Ry n ) (A + x))dλ n 1. By construction, the integrand here is strictly positive, namely, λ n 1 (B) = 0 4 Main Proof Theorem 4.1. Suppose that E is a separable Banach space, F a Banach space with the Radon Nikodym Property. Let f be a Lispchitz function from an open set U E into F. Then the set of points at which f is not Gateaux differentiable is Aronszajn null. Proof. First up, we note that the the set of points at which f is Gateaux Differentiable is a Borel set: this is because we can use the hypothesis of separability to cut down to the countable case. A 4

5 point x is differentiable just when for all n N there is some m N and t, s 1 m we have that: f (x + tu) f (x) f (x + su) f (x) t s < 1/n. This condition is one that we can take a countable intersection of, namely, the set of differentiable points is Borel. 6 Take some sequence of linearly independent vectors whose linear span is dense in E. Let V n =span{x k : l n}. Let D n be the set of those x U for which f is differentiable in each direction in V n, and linear in its argument. The idea now is to consider V n E, this is roughly an n-dimensional space, so we want to use the result of section 2. For every y E the set (E/D n + y) V n is the subset of V n where the function f y (x) = f (x y) Vn is differentiable, and by Theorem 2.1, this set has measure 0. By Theorem 3.1 the set E/D n belongs to A({x k : k n}). It follows that E/ n D n A(x k ), eg, is Aronszajn null. The function is thus Gateuax differentiable on D n by theorem Now we are ready to use this to prove our main result: Theorem 4.2. Suppose f : E F is a Lipschitz embedding with x y f (x) f (y) K x y with E a separable Banach space and F a space with the Radon-Nikodym property then E is isomorphic to a subset of F. Proof. By the above, there is a Gateaux derivative for f at some point. By construction, the derivative D f is such that x D f K x, but this is an embedding into F as required. 6 We would have no hope of doing any analysis/finding the measure of anything if it were not Borel. 5