DIAGNÓSTICO DE LA BOMBA VERTICAL

By Rademacher’s theorem, every subset of a Euclidean space with positive Lebesgue measure is trivially seen to be a universal Fr´echet set. The converse however is false in any Euclidean space of dimension at least two. The original proof of this statement is due to Preiss - see [23] - and in this section we shall revisit his example.

We first recall the elementary fact that any Gδ subset of a complete metric

space is a complete topological space.

Lemma 4.5. If (Y,∆) is a complete metric space andO =∩_n≥1On, where the sets On are open, then the subspace topology on O is induced by a complete metric d where d(x, y) = ∆(x, y) + ∞ X n=1 1 2nmin 1 ∆(y, Oc n) − 1 ∆(x, Oc n) ,1 .

Proof. It is straightforward to check this formula defines a metric on O. That d

induces the subspace topology onO, considered as a subset of Y, follows from the inequality ∆(x, y)≤d(x, y) and the fact thatdis a continuous function onO×O, with respect to the metric ∆; the latter holds asdis a uniform limit of continuous functions.

Finally if (ym)m≥1 is a Cauchy sequence in O then as ∆(ym, yn)≤d(ym, yn)

and (Y,∆) is complete we have ∆(ym, y) → 0 for some y ∈ Y. Further, for each n≥1 the sequence 1 ∆(ym, Ocn) m≥1

is Cauchy; therefore it is a bounded sequence. It quickly follows that ∆(y, Oc n)>0

for alln≥1 so thaty∈O. Hencedis complete.

Theorem 4.6. Suppose thatY is a Banach space with separable dual, T is a dense subset of Y and S ⊆Y is a Gδ subset of Y containing every line segment joining two points ofT. Then S is a universal Fr´echet set in Y.

Proof. We let the space X=S with the topology inherited from Y; asS is Gδ, X

is a complete space. Let π:X → Y be the inclusion map; clearly π is continuous. Note that Y has an equivalent norm that is Fr´echet differentiable on Y \ {0} by Theorem 1.16. The conditions (4.1)-(4.3) of Theorem 4.2 can be satisfied for any

δ0 > 0 by taking γ to be an affine map and using the density of T. Therefore

π(X) =X=S is a universal Fr´echet set.

Instead of proving directly thatS may be taken to be Lebesgue null if Y =

Hausdorff dimension at most one for any Banach spaceY with separable dual. We first recall the definition of Hausdorff dimension.

Definition 4.7. If S is a metric space then we call a sequence(Ci)i≥1 of subsets of S a countable cover of S if S =∪i≥1Ci. We then define, for any non-negative real number d,

C_Hd(S) = infX

i≥1

diameter(Ci)d and then define the Hausdorff dimension

dH(S) = inf{d >0 such that CHd(S) = 0}.

We may consider any subsetS of a Banach spaceY as a metric space in its own right and apply this definition to calculate its Hausdorff dimension.

Lemma 4.8. If Y is a Banach space and L is a countable union of line segments in Y then there exists a Gδ subsetO of Y with L⊆O and such that the Hausdorff dimension ofO is less than or equal to one.

Proof. We first note that it is easily checked thatC_Hd is countably subadditive, when considered as a set function defined on subsets of a metric space, and that if l is a line segment of length at most one,d >0 and kis a positive integer then

C_Hd(B_1/k(l))≤k· 4 k d = 4d·k−(d−1), (4.14)

as we may cover B1/k(l) with k open balls of radius 2/k. Here, as usual, Br(S)

denotes the open r-neighbourhood ofS Br(S) :=

[

y∈S Br(y)

for any S⊆Y and r >0.

Now letLbe a countable union of line segments in a Banach space Y. One may write

L= [

m≥1

Lm

where eachLm is a line segment in Y of length at most one.

We let On= ∞ [ m=1 B1/2m+n(L_m)

and O = ∞ \ n=1 On.

Note that O is a Gδ subset of Y, containing L. We must verify that the

Hausdorff dimension ofO is no greater than one.

Letd >1. It suffices, by Definition 4.7, to show that

C_Hd(O) = 0.

But for eachn,

C_Hd(O)≤C_Hd(On) ≤ ∞ X m=1 C_Hd(B_1/2m+n(L_m)) ≤ ∞ X m=1 4d·2−(m+n)(d−1) = 4d·2 −(n+1)(d−1) 1−2−(d−1)

using the countable subadditivity of Cd

H, (4.14) and d > 1. The right hand side

tends to 0 asn→ ∞ so we deduce that

C_Hd(O) = 0 for alld >1, as required.

Corollary 4.9. Any Banach space Y with separable dual has a universal Fr´echet setS ⊆Y with Hausdorff dimension at most one.

Proof. This is immediate by combining Theorem 4.6 with Lemma 4.8.

We may remark that, for non-trivialY, the Hausdorff dimension of a such a setS must in fact be exactly one.

Lemma 4.10. If Y is a Banach space with Y 6= {0} and S ⊆ Y has Hausdorff dimension less than one then there exists a Lipschitz functiong:Y →_Rthat is not Fr´echet differentiable at any y∈S.

Proof. Let π ∈ Y∗ with kπk = 1. As π is Lipschitz and Hausdorff dimension is non-increasing under taking the Lipschitz image, the set

has Hausdorff dimension less than one. Henceπ(S) has Lebesgue measure zero. By Lemma 1.6 we can therefore find a Lipschitz function f: R → R that is nowhere

differentiable on π(S). We letg:Y →_Rbe given by

g=f◦π.

We note thatgis Lipschitz. Lete∈S(Y) withπe6= 0. For anyy∈S and t∈R, g(y+te)−g(y) =f(πy+ (πe)t)−f(πy).

Asf is not differentiable at πy∈π(S), we deduce that the directional derivative

g0(y, e)

does not exist for any y ∈ S; hence g: Y → _R is not Fr´echet differentiable at any

y∈S.

Finally, we use the elementary fact that any subset ofY =RM with Hausdorff

dimension strictly less thanM has Lebesgue measure zero to deduce the following, from Corollary 4.9.

Corollary 4.11. IfM ≥2thenRM has a universal Fr´echet set of Lebesgue measure zero.

See [23, Corollary 6.5], and the subsequent remark, for the original presen- tation of this result.

4.3

A compact null universal Fr´echet set in Euclidean