ON PISIER’S CONSTRUCTION

JOHN E. McCARTHY

Abstract. A classical proof of Pisier’s construction of a polynomially bounded operator not similar to a contraction is given.

0. Introduction

In [6], G. Pisier gave a very ingenious example of an operator that is polynomially bounded but not similar to a contraction, answering a question posed by Halmos in [3]. His proof uses martingales. The object of this expository note is to describe Pisier’s example without using techniques of probability theory. We remark that this has also been done by S. Kisliakov and by K. Davidson and V. Paulsen.

Let H be a separable, infinite dimensional Hilbert space. Let H^{2}(H) denote the space of H-valued analytic functions F on the disk for which

∥F ∥^{2} = lim_{r→1} \frac{1}{2π} ∫ ∥F (re^{iθ})∥_{H}^{2}dθ = \frac{1}{2π} ∫ ∥F (e^{iθ})∥_{H}^{2}dθ < ∞

Let K^{2}(H) denote the corresponding space of conjugate-analytic functions (not required to vanish at the origin). By passing to boundary values on the circle 𝕋, both of these spaces can be thought of as subspaces of L^{2}(H), the square-summable functions from 𝕋 to H. Let Q denote the orthogonal projection from L^{2}(H) onto K^{2}(H).

Let {C_{k}}_{k=1}^{∞} be a fixed set of operators on H satisfying the canonical anti-commutation relations, i.e.

C_{i}C_{j} + C_{j}C_{i} = 0
C_{i}^{∗}C_{j} + C_{j}C_{i}^{∗} = δ_{ij}  (0.1)

It is easy to construct n such operators on a 2^{n} dimensional Hilbert space. For example, let

D = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) E = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )

Date: March 11 1996.

1991 Mathematics Subject Classification. 47B38, 31A20.

The author was partially supported by the National Science Foundation grant DMS 9296099.

Work done at MSRI is supported in part by NSF grant DMS-9022140.

On the n-fold (or infinite) tensor product of ℂ^{2} with itself, one can define

C_{1} = D ⊗ I ⊗ I ⊗ ⋯ ⊗ I
C_{2} = E ⊗ D ⊗ I ⊗ ⋯ ⊗ I
C_{3} = I ⊗ E ⊗ D ⊗ ⋯ ⊗ I
…
C_{n} = I ⊗ I ⊗ I ⊗ ⋯ ⊗ E ⊗ D

Let Φ(z) be the B(H)-valued conjugate analytic function

Φ(z) = ∑_{k=1}^{∞} 2^{−k}z̄^{2^{k}−1}C_{k}

and let Γ_{Φ} : H^{2}(H) → K^{2}(H) be the vectorial Hankel operator

Γ_{Φ}F = Q(Φ(F)).

Let S : H^{2}(H) → H^{2}(H) be the operator of multiplication by z, and X : K^{2}(H) → K^{2}(H) be multiplication by z followed by the projection Q. Finally, let T be the operator on K^{2}(H) ⊕ H^{2}(H) given by

T = ( \begin{matrix}X & Γ_{Φ} \\ 0 & S \end{matrix} )  (0.1)

Pisier proved that T is polynomially bounded but not similar to a contraction. Notice that, if p is a polynomial and M_{p} denotes multiplication by p, then

p(T) = ( \begin{matrix}p(X) & Γ_{Φ}M_{p′} \\ 0 & p(S) \end{matrix} )  (0.2)

(this can be checked for p(z) = z^{n}, and then follows for general p by linearity).

1. T is not similar to a contraction

First we need a lemma.

Lemma 1.1. Let α_{1}, α_{2}, … be complex numbers. Then

∥ ∑ α_{i}C_{i}∥^{2} = ∑ |α_{i}|^{2}

Proof: Let R = ∑ α_{i}C_{i}. Notice that, by (0.1),

R^{∗}R + RR^{∗} = (∑ ᾱ_{i}C_{i}^{∗})(∑ α_{j}C_{j}) + (∑ α_{j}C_{j})(∑ ᾱ_{i}C_{i}^{∗})
= ∑_{i,j} ᾱ_{i}α_{j}(C_{i}^{∗}C_{j} + C_{j}C_{i}^{∗})
= ∑ |α_{i}|^{2}

and

R^{2} = \frac{1}{2}(∑_{i,j} α_{i}α_{j}C_{i}C_{j} + ∑_{i,j} α_{i}α_{j}C_{j}C_{i})
= 0

Therefore

∥R∥^{4} = ∥R^{∗}RR^{∗}R∥
= ∥R^{∗}(R^{∗}R + RR^{∗})R∥
= ∥R^{∗}(∑ |α_{i}|^{2})R∥
= ∑ |α_{i}|^{2}∥R∥^{2}

So ∥R∥^{2} = ∑ |α_{i}|^{2}, as required. □

An operator R is called completely polynomially bounded if there exists a constant c such that for every matrix valued (or operator valued) polynomial P, the inequality ∥P (R)∥ ≤ c∥P ∥ holds, where ∥P ∥ is the supremum of the operator norms of P (z) as z ranges over the unit disk. If R were similar to a contraction, then it would be completely polynomially bounded. Reason: suppose R = A^{−1}BA, and B is a contraction. Let P (z) = ∑ z^{n}D_{n} be an operator valued polynomial. Then

P (R) = ∑ R^{n} ⊗ D_{n} = (A^{−1} ⊗ I)P (B)(A ⊗ I)

By the operator-valued version of von Neumann’s inequality (which can be proved using the Sz.-Nagy dilation theorem just as the scalar one), ∥P (B)∥ ≤ ∥P ∥, so ∥P (R)∥ ≤ ∥A^{−1}∥∥A∥∥P ∥. The converse of this statement - completely polynomially bounded implies similar to a contraction - is true, but we shall not use this fact [5].

We shall prove that T is not completely polynomially bounded, and therefore cannot be similar to a contraction. Fix a positive integer n. Let

P (z) = ∑_{k=1}^{n} z^{2^{k}}C_{k}

By Lemma (1.1), we have ∥P ∥ = \sqrt{n}. Let us calculate the norm of

P (T) = ∑_{k=1}^{n} T^{2^{k}} ⊗ C_{k}

With respect to the 2-by-2 decomposition of T in (0.1), this also has a 2-by-2 decomposition, with the (1, 2) entry being

R = Γ_{Φ} ∑_{k=1}^{n} 2^{k}M_{z^{2^{k}−1}} ⊗ C_{k} = Q ∑_{k=1}^{n} 2^{k}(∑_{j=1}^{∞} 2^{−j}z̄^{2^{j}−1}z^{2^{k}−1}C_{j}) ⊗ C_{k}.

The constant term in R is then ∑_{k=1}^{n} C_{k} ⊗ C_{k}, and it will suffice to prove that the norm of this is larger than c\sqrt{n} for any fixed c.

Let {e_{i}}_{i=1}^{∞} be an orthonormal basis for H, and assume that the C_{k} are constructed as in the previous section, so in particular ⟨C_{k}e_{i}, e_{j}⟩ is always real (this assumption is to avoid introducing an antilinear isometry of H). Then

∥ ∑_{k=1}^{n} C_{k} ⊗ C_{k}∥ ≥ \frac{1}{2^{n}} ⟨∑_{k=1}^{n} C_{k} ⊗ C_{k}(∑_{i=1}^{2^{n}} e_{i} ⊗ e_{i}), (∑_{j=1}^{2^{n}} e_{j} ⊗ e_{j})⟩
= \frac{1}{2^{n}} ∑_{i,j,k}⟨C_{k}e_{i}, e_{j}⟩ ⟨C_{k}e_{i}, e_{j}⟩ = \frac{1}{2^{n}} ∑_{i,j,k}⟨C_{k}e_{i}, e_{j}⟩ ⟨e_{j}, C_{k}e_{i}⟩

= \frac{1}{2^{n}} ∑_{k=1}^{n} ∑_{i=1}^{2^{n}}⟨C_{k}e_{i}, C_{k}e_{i}⟩

= \frac{1}{2^{n}} ∑_{k=1}^{n} 2^{n−1} = \frac{n}{2}

Therefore ∥p(T)∥/∥p∥ ≥ \sqrt{n}/2 and is not bounded, so T is not completely polynomially bounded, and hence is not similar to a contraction.

2. T is polynomially bounded

To prove that T is polynomially bounded, by (0.2) we must prove that there exists a constant c such that ∥Γ_{Φ}M_{p′}∥ ≤ c∥p∥_{∞} for all polynomials p. Now,

∥Γ_{Φ}M_{p′}∥ = sup{| ⟨Γ_{Φ}p′F, G⟩ | : ∥F ∥_{H^{2}(H)} = 1, ∥G∥_{K^{2}(H)} = 1}

Let

F (z) = ∑_{n=0}^{∞} z^{n}ξ_{n} , G(z) = ∑_{m=0}^{∞} z̄^{m}η_{m}

where

∑_{n=0}^{∞} ∥ξ_{n}∥_{H}^{2} = 1 = ∑_{m=0}^{∞} ∥η_{m}∥_{H}^{2}.

Let σ denote normalized Lebesgue measure on 𝕋. Then

⟨Γ_{Φ}p′F, G⟩ = ∫_{𝕋} dσ(z)p′(z) ∑_{k=1}^{∞} 2^{−k}z̄^{2^{k}−1} ∑_{n,m=0}^{∞} z^{n+m}⟨C_{k}ξ_{n}, η_{m}⟩.

By (1.1) the map v : H → B(H) which sends e_{i} to C_{i} is an isometry. Therefore

⟨C_{k}ξ_{n}, η_{m}⟩ = tr[C_{k}(ξ_{n} ⊗ η_{m})]
= (v(e_{k}), ξ_{n} ⊗ η_{m})_{tr}
= (e_{k}, v^{t}(ξ_{n} ⊗ η_{m}))_{tr}

where (., .)_{tr} is the linear pairing between B(H) and the trace class operators, given by (A, B)_{tr} = tr(AB). Therefore

⟨Γ_{Φ}p′F, G⟩ = ∫_{𝕋} dσ(z)p′(z) ∑_{k=1}^{∞} 2^{−k}z̄^{2^{k}−1}(v(e_{k}), [∑_{n=0}^{∞} z^{n}ξ_{n}] ⊗ [ ∑_{m=0}^{∞} z^{m}η_{m}])_{tr} (2.1)

Now, [∑_{n=0}^{∞} z^{n}ξ_{n}] ⊗ [∑_{m=0}^{∞} z^{m}η_{m}] is in H^{1}(H ⊗ H), and moreover

∥[∑_{n=0}^{∞} z^{n}ξ_{n}] ⊗ [ ∑_{m=0}^{∞} z^{m}η_{m}]∥_{H^{1}(H⊗H)} = ∫_{𝕋} dσ(z)∥[∑_{n=0}^{∞} z^{n}ξ_{n}] ⊗ [ ∑_{m=0}^{∞} z^{m}η_{m}]∥_{H⊗H}
= ∫_{𝕋} dσ(z)∥ ∑_{n=0}^{∞} z^{n}ξ_{n}∥_{H}∥ ∑_{m=0}^{∞} z^{m}η_{m}∥_{H}
≤ [∫_{𝕋} dσ(z)∥ ∑_{n=0}^{∞} z^{n}ξ_{n}∥_{H}^{2}]^{1/2}[∫_{𝕋} dσ(z)∥ ∑_{m=0}^{∞} z^{m}η_{m}∥_{H}^{2}]^{1/2}
= ∥F ∥_{H^{2}(H)}∥G∥_{K^{2}(H)}
= 1

Combining this with (2.1), we get that

∥Γ_{Φ}M_{p′}∥ ≤ ∥Q ∑_{k=1}^{∞} 2^{−k}p′(z)z̄^{2^{k}−1}e_{k}∥_{H^{1}(H)^{∗}} ∥ ∑_{n,m=0}^{∞} z^{n+m}v^{t}(ξ_{n} ⊗ η_{m})∥_{H^{1}(H)}

The dual of H^{1}(H) is BMO(H) [1]. Therefore we are done if we can prove the following:

∥Q ∑_{k=1}^{∞} 2^{−k}p′(z)z̄^{2^{k}−1}e_{k}∥_{BMO(H)} ≤ c∥p∥_{∞}  (2.2)

We do this in section 3.

3. The key estimate

Let

K_{k}(e^{iθ}) := 2^{−k} ∑_{n=−2^{k}}^{2^{k}}(2^{k} − |n|)e^{inθ}

be the Fejér kernel, and let

T_{k}f(e^{iθ}) = f ∗ K_{k}(e^{iθ}) = \frac{1}{2π} ∫_{𝕋} f(e^{iϕ})K_{k}(e^{i(θ−ϕ)})dϕ

Then what we want to estimate in (2.2) can be written

Q ∑_{k=1}^{∞} 2^{−k}p′(z)z̄^{2^{k}−1}e_{k} = Q ∑_{k=1}^{∞} T_{k}p(z)z̄^{2^{k}}e_{k},  (3.1)

where we write T_{k}p(z)z̄^{2^{k}} to mean T_{k}[p(z)z̄^{2^{k}}], i.e.

[T_{k}p(z)z̄^{2^{k}}](e^{iθ}) = \frac{1}{2π} ∫_{𝕋} p(e^{iϕ})e^{−i2^{k}ϕ}K_{k}(e^{i(θ−ϕ)})dϕ

As was observed in this context by S. Kisliakov, the norm in the dual of H^{1}(H) depends only on the conjugate analytic part of the function; so one can remove the Q in (3.1), and it is sufficient to show

∥ ∑_{k=1}^{∞} T_{k}p(z)z̄^{2^{k}}e_{k}∥_{BMO(H)} ≤ c∥p∥_{∞}  (3.2)

Now

∥ ∑_{k=1}^{∞} T_{k}p(z)z̄^{2^{k}}e_{k}∥_{BMO(H)}^{2} = [sup_{I} \frac{1}{|I|} ∫_{I} [∑_{k=1}^{∞} |T_{k}p(z)z̄^{2^{k}} − MV_{k}|^{2}]^{1/2} dθ]^{2}

≤ sup_{I} ∑_{k=1}^{∞} \frac{1}{|I|} ∫_{I} |T_{k}p(z)z̄^{2^{k}} − MV_{k}|^{2}dθ  (3.3)

where I ranges over the arcs of the circle, and MV_{k} denotes the mean value, over I, of the preceding expression.

Let us fix some arc I; we shall estimate (3.3) by breaking the sum into two pieces, k small and k large. We shall let c denote a universal constant, independent of I, k and p, that may vary from one line to the next.

Case (a): 2^{−k} ≥ |I|.

We shall estimate the size of the derivative of T_{k}p(z)z̄^{2^{k}}. Note that for any function f,

|\frac{d}{dθ}T_{k}f(e^{iθ})| ≤ ∥f∥_{∞}∥\frac{d}{dθ}K_{k}(e^{iθ})∥_{1}

The Fejér kernel can be expressed in closed form as

K_{k}(e^{iθ}) = \frac{1}{2^{k}} (\frac{sin(2^{k−1}θ)}{sin(θ/2)})^{2}  (3.3)

(see e.g. [4]), and one can estimate the L^{1} norm of the derivative by summing the variation over intervals of length π/2^{k}, to get c2^{k}. Therefore

∥\frac{d}{dθ}T_{k}f∥_{i} ≤ c2^{k}∥f∥_{∞}

Now let the end-points of I be e^{ia} and e^{ib}. Then for any function f

\frac{1}{|I|} ∫_{I} |T_{k}f − MV |^{2}dθ ≤ \frac{1}{|I|} ∫_{a}^{b} |T_{k}f(e^{iθ}) − T_{k}f(e^{ia})|^{2}dθ
≤ \frac{1}{|I|} ∫_{a}^{b} ∥\frac{d}{dθ}T_{k}f∥_{∞}(θ − a)^{2}dθ
≤ c2^{2k}|I|^{2}∥f∥_{∞}^{2}

Letting f be z̄^{2^{k}}p(z), and summing over k such that 2^{k} is less than or equal to 1/|I|, we get

∑_{2^{−k}≥|I|} \frac{1}{|I|} ∫_{I} |T_{k}p(z)z̄^{2^{k}} − MV_{k}|^{2}dθ ≤ c∥p∥_{∞}^{2}  (3.4)

Case (b): 2^{−k} ≤ |I|.

For large k, we can drop the mean value term. Let 3I denote the interval concentric with I of three times the length. Write p = p_{1} + p_{2}, where p_{1} is supported on 3I, and p_{2} is supported on 𝕋 \ 3I.

Case (i): p_{1}.

Note that

∑_{k=1}^{∞} ∫_{𝕋} |T_{k}z̄^{2^{k}}p_{1}|^{2} = ∑_{k=1}^{∞}\{∑_{n=0}^{2^{k}} |p̂_{1}(n)|^{2} \frac{n^{2}}{2^{2k}} + ∑_{n=2^{k}+1}^{2^{k+1}} |p̂_{1}(n)|^{2} \frac{(2^{k+1} − n)^{2}}{2^{2k}} \}

Interchanging the order of summation, one gets that the sum is less than or equal to

\frac{7}{3} ∑_{n=0}^{∞} |p̂_{1}(n)|^{2} ≤ \frac{7}{3}∥p_{1}∥_{2}^{2} ≤ 7|I| ∥p∥_{∞}^{2}

Therefore

∑_{k=1}^{∞} \frac{1}{|I|} ∫_{I} |T_{k}p_{1}(z)z̄^{2^{k}} − MV_{k}|^{2}dθ ≤ c∥p∥_{∞}^{2}  (3.5)

Case (ii): p_{2}.

We have

∑_{2^{−k}≤|I|} \frac{1}{|I|} ∫_{I} |T_{k}z̄^{2^{k}}p_{2}|^{2}dθ = ∑_{2^{−k}≤|I|} \frac{1}{|I|} ∫_{I} dθ |∫_{𝕋\3I} dϕ e^{−i2^{k}ϕ}p(e^{iϕ})K_{k}(e^{i(θ−ϕ)})|^{2}

≤ ∑_{2^{−k}≤|I|} \frac{1}{|I|} ∫_{I} dθ∥p∥_{∞}^{2} |∫_{|I|/2}^{2π−|I|/2} K_{k}(e^{iϕ})dϕ|^{2}

From (3.3) we get that

∫_{|I|/2}^{2π−|I|/2} K_{k}(e^{iϕ})dϕ ≤ c2^{−k} \frac{1}{|I|}.

Therefore

∑_{2^{−k}≤|I|} \frac{1}{|I|} ∫_{I} |T_{k}z̄^{2^{k}}p_{2}|^{2}dθ ≤ ∑_{2^{−k}≤|I|} ∥p∥_{∞}^{2}2^{−2k} \frac{1}{|I|^{2}}
≤ c∥p∥_{∞}^{2}  (3.6)

Combining (3.4), (3.5) and (3.6) we get

sup_{I} \frac{1}{|I|} ∑_{k=1}^{∞} ∫_{I} |T_{k}p(z)z̄^{2^{k}} − MV_{k}|^{2}dθ ≤ c∥p∥_{∞}^{2}

as desired.

4. Finite dimensional estimates

Let R be any operator on an n-dimensional Hilbert space. There are two quantities associated with it (which may be infinite):

∥R∥_{s.b.} := inf{∥A∥∥A^{−1}∥ : ∥ARA^{−1}∥ ≤ 1}
∥R∥_{p.b.} := sup{∥p(R)∥ : p a polynomial and ∥p∥_{∞} ≤ 1}

Adopt the convention that if R is not similar to any contraction, then ∥R∥_{s.b.} = ∞. By von Neumann’s inequality, one always has ∥R∥_{s.b.} ≤ ∥R∥_{p.b.}. In [2], J. Bourgain proved that, for any matrix R on an n-dimensional Hilbert space,

∥R∥_{s.b.} ≤ ∥R∥_{p.b.}^{4} log n

One of the consequences of Pisier’s result, as he points out, is that there is a lower bound:

Corollary 4.1 (Pisier). There is a constant δ > 0 such that, for all K > 1, and all n ≥ 1, there exists an n-by-n matrix R with ∥R∥_{p.b.} ≤ K and ∥R∥_{s.b.} ≥ δ(K − 1)\sqrt{log n}.

Proof: It is sufficient to prove for n ≥ 4 and a power of two; let n = 2^{m+1}. Let H_{m} = H^{2}(l_{2^{m}}^{2}) ⊖ z^{n}H^{2}(l_{2^{m}}^{2}), and K_{m} = K^{2}(l_{2^{m}}^{2}) ⊖ z̄^{n}K^{2}(l_{2^{m}}^{2}). Let C_{1}, …, C_{m} satisfy the canonical anti-commutation relations on l_{2^{m}}^{2}, let Φ_{m}(z) = ∑_{k=1}^{m} 2^{−k}z̄^{2^{k}−1}C_{k} and let Γ_{Φ_{m}} : H_{m} → K_{m} be Φ_{m} followed by projection. Let S_{m} and X_{m} be truncated versions of the forward and backward shifts, and on K_{m} ⊕ H_{m}, let

T_{m} = ( \begin{matrix}X_{m} & εΓ_{Φ_{m}} \\ 0 & S_{m} \end{matrix} )

where ε > 0 will be chosen later. Note that T is an n-by-n matrix. Then for any polynomial p,

p(T_{m}) = ( \begin{matrix}p(X_{m}) & εΓ_{Φ_{m}}p′(S_{m}) \\ 0 & p(S_{m}) \end{matrix} )

Just as above, one gets that

∥T_{m}∥_{s.b.} ≥ ε\sqrt{m}/2

and ∥Γ_{Φ_{m}}p′(S_{m})∥ ≤ c∥p∥_{∞} where c is the constant in 2.2, and so

∥T_{m}∥_{p.b.} ≤ εc + 1.

Now let ε = (K − 1)/c, and let δ = 1/4c. Then T_{m} works.

□

Acknowledgements

The author would like to thank M. Christ and A. Arias for valuable discussions.

References

[1] J. Bourgain. Vector-valued singular integrals and the H^{1}−BMO duality. In J.-A. Chao and W.A. Woyczyński, editors, Probability Theory and Harmonic Analysis, pages 1–17. Marcel-Dekker, New York, 1986.

[2] J. Bourgain. On the similarity problem for polynomially bounded operators on Hilbert space. Israel J. of Math., 54:227–241, 1986.

[3] P. R. Halmos. Ten problems in Hilbert space. Bull. Amer. Math. Soc., 76:887–933, 1970.

[4] Henry Helson. Harmonic Analysis. Addison-Wesley, Reading MA, 1983.

[5] V. Paulsen. Every completely polynomially bounded operator is similar to a contraction. J. Funct. Anal., 55:1–17, 1984.

[6] G. Pisier. A polynomially bounded operator on Hilbert space which is not similar to a contraction. Preprint.

John E. McCarthy, MSRI and Department of Mathematics, Washington University, St. Louis, MO 63130