The p^{\frac{1}{p}} in Pisier’s Factorization Theorem S.J. Montgomery-Smith

Abstract We show that the constants in Pisier’s factorization theorem for (p, 1)-summing operators from C(Ω) cannot be improved.

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A theorem of Pisier (see [3]) states the following. Theorem 1. Let T : C(Ω) → X be a bounded linear operator, where Ω is a compact Hausdorff topological space and X is a Banach space. Then the following are equivalent.

i) T is (p, 1)-summing. ii) There is a constant C < ∞ and a Radon probability measure μ on Ω such that for all

f ∈ C(Ω) we have

∥T f∥ ≤ C ∥f∥L_{1}(μ)^{\frac{1}{p}} ∥f∥_{∞}^{1− \frac{1}{p}} .  (1)

iii) There is a constant C < ∞ and a Radon probability measure μ on Ω such that for all

f ∈ C(Ω) we have

∥T f∥ ≤ C ∥f∥_{L_{p,1}(μ)} .  (2)

If we examine the proof of this theorem carefully, we can deduce the following. Let π_{p,1}(T) denote the (p, 1)-summing norm of T. Theorem 2. Let T : C(Ω) → X be a bounded linear operator as in Theorem 1 that is (p, 1)-summing. Then we can say the following.

i) The set of C and μ that satisfy (1) coincide with the set of C and μ that satisfy (2). ii) π_{p,1}(T) ≤ C for all C satisfying (1) or (2). iii) We can choose C = p^{\frac{1}{p}} π_{p,1}(T) in (1) and (2).

Part (i) of this theorem is straight forward to demonstrate (we merely note that one needs the inequality: ∥f∥_{p,1} ≤ ∥f∥1^{\frac{1}{p}} ∥f∥_{∞}^{1− \frac{1}{p}}). Part (ii) is also easy to verify. However, to show part (iii), the main part of Pisier’s theorem, is much harder. Furthermore, the methods do not indicate whether the p^{\frac{1}{p}} factor can be made smaller, or even completely removed (that is, replaced by 1).

The purpose of this paper is to show the rather surprising fact that the p^{\frac{1}{p}} factor cannot be reduced at all.

Theorem 3. Given ϵ > 0, there is an operator T : C(Ω) → X such that for any Radon probability measure μ on Ω, if C is the least number satisfying (1) or (2), then C ≥ p^{\frac{1}{p}} π_{p,1}(T) (1 − ϵ).

Construction. Let 1 ≤ S ≤ N be integers, and let Ω be the collection of S-subsets of {1, 2, …, N}. For each 1 ≤ n ≤ N, let Ω_{n} = { ω ∈ Ω : n ∈ ω}. We note the following facts for later on:

|Ω_{n}| = \binom{N − 1}{S − 1},  (3)
\frac{1}{N} ∑_{n=1}^{N} χ_{Ω_{n}} = \frac{S}{N} χ_{Ω}.  (4)

We give Ω the discrete topology, and define a norm ∥ · ∥_{∗} on C(Ω) by

∥f∥_{∗} = sup_{1≤n≤N} ∑_{ω∈Ω_{n}} |f(ω)| .

We let T be the canonical embedding

T : (C(Ω), ∥ · ∥_{∞}) → (C(Ω), ∥ · ∥_{∗}).

Lemma 4. For 1 ≤ p < ∞, the (p, 1)-summing norm of T may be estimated by

π_{p,1}(T) ≤ \frac{N^{S−1+ \frac{1}{p}}}{(S − 1)! (Sp − p + 1)^{\frac{1}{p}}} .

In order to show Lemma 4, we will need two more lemmas.

Lemma 5. If 1 ≤ p < ∞, and T : C(Ω) → X is a bounded linear operator, where X is a Banach space, then the (p, 1)-summing norm of T may be calculated by the formula

π_{p,1}(T) = sup \{(∑_{s=1}^{S} ∥T f_{s}∥^{p})^{\frac{1}{p}} \} ,

where the supremum is over all sequences f_{1}, f_{2}, …, f_{S} of disjoint elements of the unit ball of C(Ω).

Proof. See [2], Lemma 6 or [1], Proposition 14.4.

Lemma 6. If 1 ≤ n ≤ N, then

|Ω_{1} ∪ Ω_{2} ∪ … ∪ Ω_{n}| ≤ \frac{1}{(S − 1)!} ∑_{m=1}^{n}(N − m)^{S−1}.

Proof. A simple counting argument shows that

|Ω_{1} ∪ Ω_{2} ∪ … ∪ Ω_{m} \ Ω_{1} ∪ Ω_{2} ∪ … ∪ Ω_{m−1}| = \binom{N − m}{S − 1} ,

and this is bounded by (N − m)^{S−1}/(S − 1)!.

Proof of Lemma 4. By Lemma 5, it is easy to see that

π_{p,1}(T) = sup \{(∑_{n=1}^{N} ∥χ_{B_{n}}∥_{∗}^{p})^{\frac{1}{p}} \} = sup \{(∑_{n=1}^{N} |B_{n}|^{p})^{\frac{1}{p}} \} ,

where the supremum is over disjoint sets B_{1}, B_{2}, …, B_{N} ⊆ Ω such that B_{n} ⊆ Ω_{n} for each 1 ≤ n ≤ N. Since Ω_{1}, Ω_{2}, …, Ω_{N} interact with one another in a completely symmetric fashion, we may assume, without loss of generality, that |B_{1}| ≥ |B_{2}| ≥ … ≥ |B_{N}|.

Now

∑_{n=1}^{N} |B_{n}|^{p} = ∑_{n=1}^{N} ( ∑_{m=1}^{n} |B_{m}|) (|B_{n}|^{p−1} − |B_{n+1}|^{p−1}) .

(We take B_{N+1} = ∅.) Since |B_{n}|^{p−1} − |B_{n+1}|^{p−1} ≥ 0, and B_{1} ∪ B_{2} ∪ … ∪ B_{n} ⊆ Ω_{1} ∪ Ω_{2} ∪ … ∪ Ω_{n}, we have, by Lemma 6, that

∑_{n=1}^{N} |B_{n}|^{p} ≤ \frac{1}{(S − 1)!} ∑_{n=1}^{N} ( ∑_{m=1}^{n}(N − m)^{S−1}) (|B_{n}|^{p−1} − |B_{n+1}|^{p−1})

= \frac{1}{(S − 1)!} ∑_{n=1}^{N}(N − n)^{S−1} |B_{n}|^{p−1} .

Now, applying Hölder’s inequality and dividing, we deduce

(∑_{n=1}^{N} |B_{n}|^{p})^{\frac{1}{p}} ≤ \frac{1}{(S − 1)!} (∑_{n=1}^{N}(N − n)^{Sp−p})^{\frac{1}{p}} .

Finally, we estimate the last quantity by an integral, and derive

(∑_{n=1}^{N} |B_{n}|^{p})^{\frac{1}{p}} ≤ \frac{1}{(S − 1)!} (∫_{0}^{N} x^{Sp−p} dx)^{\frac{1}{p}}

≤ \frac{N^{S−1+ \frac{1}{p}}}{(S − 1)! (Sp − p + 1)^{\frac{1}{p}}} ,

as desired. Proof of Theorem 3. By the hypothesis on C, there is a probability measure μ on Ω such that inequality (1) holds. In particular, if we substitute f = χ_{Ω_{n}}, we deduce that

|Ω_{n}|^{p} = ∥χ_{Ω_{n}}∥_{∗}^{p} ≤ C^{p} ∫ χ_{Ω_{n}} dμ.

Hence

\frac{1}{N} ∑_{n=1}^{N} |Ω_{n}|^{p} ≤ C^{p} ∫ \frac{1}{N} ∑_{n=1}^{N} χ_{Ω_{n}} dμ,

and so by equalities (3) and (4) we have

C ≥ \frac{N^{\frac{1}{p}} \binom{N − 1}{S − 1}}{S^{\frac{1}{p}}}.

Thus, by Lemma 4, we deduce

C ≥ \frac{N^{\frac{1}{p}} \binom{N − 1}{S − 1}(S − 1)! (Sp − p + 1)^{\frac{1}{p}}}{S^{\frac{1}{p}} N^{S−1+ \frac{1}{p}}} π_{p,1}(T).

Choosing N much larger than S, we find that

C ≥ (\frac{Sp − p + 1}{S})^{\frac{1}{p}} π_{p,1}(T) (1 − \frac{1}{2}ϵ).

Finally, choosing S large, we have the desired result, that is, C ≥ p^{\frac{1}{p}} π_{p,1}(T) (1 − ϵ).

References

1. Jameson G.J.O.: Summing and Nuclear Norms in Banach Space Theory. London Math. Soc., Student Texts 8, 1987.

2. Maurey B.: Type et cotype dans les espaces munis de structures locales inconditionelles, Exposés 24–25. In: Seminaire Maurey-Schwartz 1973–74 (Ecole Polytechnique).

3. Pisier G.: Factorization of operators through L_{p∞} or L_{p1} and non-commut-ative generalizations. Math. Ann. 276 105–136 (1986).

S.J. Montgomery-Smith, Department of Mathematics, University of Missouri at Columbia, Columbia, Missouri 65211, U.S.A.