Grothendieck’s inequality in {\mathcal{S}}

# Grothendieck’s inequality in the noncommutative Schwartz space

Rupert H. Levene School of Mathematics and Statistics
University College Dublin
Belfield
Dublin 4
Ireland
and  Krzysztof Piszczek Faculty of Mathematics and Comp. Sci.
A. Mickiewicz University in Poznań
Umultowska 87
61-614 Poznań
Poland
###### Abstract.

In the spirit of Grothendieck’s famous inequality from the theory of Banach spaces, we study a sequence of inequalities for the noncommutative Schwartz space, a Fréchet algebra of smooth operators. These hold in non-optimal form by a simple nuclearity argument. We obtain optimal versions and reformulate the inequalities in several different ways.

###### Key words and phrases:
noncommutative Grothendieck inequality, noncommutative Schwartz space, m-convex Fréchet algebra, bilinear form, state.
###### 2010 Mathematics Subject Classification:
Primary: 47A30, 47A07, 47L10. Secondary: 47A63.
The research of the second-named author has been supported by the National Center of Science, Poland, grant no. UMO-2013/10/A/ST1/00091.
The authors are grateful to the anonymous referee, whose comments greatly improved the presentation of this paper.

## 1. Introduction

The noncommutative Schwartz space  is a weakly amenable m-convex Fréchet algebra whose properties have been investigated in several recent papers, see e.g. [2, 3, 13, 14]. It is not difficult to see that as a Fréchet space,  is nuclear. From this, we can easily deduce the following analogue of Grothendieck’s inequality, which we call Grothendieck’s inequality in : there exists a constant so that for any continuous bilinear form and any , there exists such that for every and any , we have

 (1) ∣∣m∑j=1u(xj,yj)∣∣⩽K∥u∥∗n∥(xj)∥RCk∥(yj)∥RCk

The norms on the right hand side arise naturally from the definition of , as explained in Section 2 below. Our goal in this note is to show that in fact always suffices, and that this is best possible.

This appears to be the first result concerning Grothendieck’s inequality in the category of Fréchet algebras; to the best of our knowledge, all previous results along these lines concern Banach spaces (including C-algebras, general Banach algebras and operator spaces). For Fréchet algebras, Grothendieck’s inequality seems to have a specific flavour. Every Fréchet space (and a fortiori, every Fréchet algebra) which appears naturally in analysis is nuclear, meaning that all tensor product topologies are equal. Since Grothendieck’s inequality can be understood as the equivalence of two tensor products, it seems that we can take inequality (1) for granted. The interesting question that remains is then optimality.

This paper is divided into four sections. In the remainder of this section, we recall a C-algebraic version of Grothendieck’s inequality due to Haagerup, and then review the definition and the basic properties of  which we require. In Section 2 we explain how nuclearity gives Grothendieck’s inequality in , and we estimate the constants and . Section 3 then settles the optimality question for  via a matricial construction. We conclude with a short section containing several reformulations of the inequality.

### 1.1. Grothendieck’s inequality

Pisier’s survey article  is a comprehensive reference for Grothendieck’s inequality. This presents many equivalent formulations and applications of this famous result, and recounts its evolution from ‘commutative’  to ‘noncommutative’. Of these reformulations and extensions, Haagerup’s noncommutative version most closely resembles (1), and we state it here for the convenience of the reader.

###### Theorem 1 (, [12, Theorem 7.1]).

Let  and  be C-algebras. For any bounded bilinear form  and any finite sequence in , we have

 ∣∣∑u(xj,yj)∣∣⩽2∥u∥∥(xj)∥RC∥(yj)∥RC

where .

### 1.2. The noncommutative Schwartz space

Let

 s={ξ=(ξj)j∈N∈CN:|ξ|n:=(+∞∑j=1|ξj|2j2n)12<+∞for alln∈N}

denote the so-called space of rapidly decreasing sequences. This space becomes Fréchet when endowed with the sequence of norms defined above. The basis of zero neighbourhoods of is defined by . The topological dual  of is the so-called space of slowly increasing sequences, namely

 {η=(ηj)j∈N∈CN:|η|′n:=(+∞∑j=1|ηj|2j−2n)12<+∞for somen∈N}

where the duality pairing is given by for , .

The noncommutative Schwartz space is the Fréchet space of all continuous linear operators from into , endowed with the topology of uniform convergence on bounded sets. The formal identity map is a continuous embedding and defines a product on by for . There is also a natural involution on given by for , . With these operations, becomes an m-convex Fréchet -algebra. The inclusion map is continuous, and in fact it is a spectrum-preserving -homomorphism . Moreover [13, Proposition 3], an element is positive (i.e., for some ), if and only if the spectrum of is contained in , or equivalently for all . On the other hand, by [3, Cor. 2.4] and [4, Theorems 8.2, 8.3], the topology of cannot be given by a sequence of C-norms. This causes some technical inconvenience (e.g. there is no bounded approximate identity in ) meaning we cannot apply C-algebraic techniques directly.

## 2. The inequality

Let be a non-decreasing sequence of norms which gives the topology of . For a continuous bilinear form, we write

 ∥u∥∗n:=sup{|u(x,y)|:x,y∈Un}

where ; similarly, for a functional , we write

 ∥ϕ∥∗n:=sup{|ϕ(x)|:x∈Un}.

Following Pisier [11, p. 316], for and , we write

 ∥(xj)∥RCk=max{∥∥m∑j=1x∗jxj∥∥12k,∥∥m∑j=1xjx∗j∥∥12k}.

Relative to our choice of norms , we have now defined each term in our hoped-for inequality (1). We will now reformulate it using tensor products.

For C-algebras, such a reformulation is standard. Indeed, by [6, Theorem 1.1] (formulated along the lines of [8, Theorem 2.1]), Haagerup’s noncommutative Grothendieck inequality entails the existence of a such that for any C-algebras and in the algebraic tensor product , we have where is the projective tensor norm and is the absolute Haagerup tensor norm [8, p. 164] on , given by

 ∥z∥ah=inf∥∥m∑j=1|xj|2∥∥12∥∥m∑j=1|yj|2∥∥12.

Here for an element of a C-algebra, and the infimum is taken over all representations where .

We proceed similarly for . For , let and consider the sequence of absolute Haagerup tensor norms on the algebraic tensor product given by

 ∥z∥ah,n:=inf∥∥m∑j=1|xj|2∥∥12n∥∥m∑j=1|yj|2∥∥12n

where the infimum runs over all ways to represent in . As usual, we write for the sequence of projective tensor norms on .

Just as in the C-algebra case, inequality (1) will follow once we show that the sequences of projective and absolute Haagerup tensor norms are equivalent on . In fact, the equivalence of these norms follows immediately from the nuclearity of  (see [9, Theorem 28.15] and [7, Ch. 21, §2, Theorem 1] for details). On the other hand, the optimal values of and  (depending on  and our choice of norms ) for which (1) hold are not given by such general considerations. These optimal parameters will be denoted by and .

Henceforth, we focus only on the sequence of norms where

 ∥x∥n:=sup{|xξ|n:ξ∈U∘n},n∈N, x∈S

and . In other words, is the norm of , considered as a Hilbert space operator from to . This sequence does indeed induce the topology of . In this context, we will estimate  and compute the exact values of .

We start with the following result, which can be compared with [13, Lemma 8]. To fix some useful notation, for we define an infinite diagonal matrix which we consider as an isometry and simultaneously as an isometry .

###### Proposition 2.

Let . We have

1. for every positive ;

2. for every self-adjoint ; and

3. for every .

Moreover, inequalities (ii) and (iii) are sharp.

###### Proof.

(i) Observe that . Furthermore, since  is positive, is positive and we have

 ∥x∥n =∥dnxdn∥B(ℓ2)=sup{⟨xdnξ,dnξ⟩:|ξ|ℓ2⩽1} =sup{⟨xξ,ξ⟩:|ξ|′n⩽1}.

 ∥x2∥2n=∥d2nx2d2n∥B(ℓ2)=∥d2nx∥2B(ℓ2),

and by [1, Proposition II.1.4.2],

 ∥x∥n=∥dnxdn∥B(ℓ2)=ν(dnxdn)=ν(d2nx)⩽∥d2nx∥B(ℓ2),

where denotes the spectral radius. This gives the desired inequality.

(iii) Since , any is also a Hilbert space operator, and the block-matrix operator is positive in (see e.g. [10, p. 117]). Equivalently,

 (2) |⟨xξ,η⟩|2⩽⟨(xx∗)1/2η,η⟩⟨(x∗x)1/2ξ,ξ⟩∀ξ,η∈ℓ2.

For , let us write where is the identity matrix. Now fix and choose . Then for all and (2) gives

 |⟨pmxpmξ,η⟩|2⩽⟨pm(xx∗)1/2pmη,η⟩⟨pm(x∗x)1/2pmξ,ξ⟩.

Since is an approximate identity in (see [13, Proposition 2]), we obtain

 |⟨xξ,η⟩|2⩽⟨(xx∗)1/2η,η⟩⟨(x∗x)1/2ξ,ξ⟩.

Taking the supremum over all in the unit ball of we get

 ∥x∥2n⩽∥(xx∗)1/2∥n∥(x∗x)1/2∥n.

Applying (ii) to the positive operators and we conclude that .

For sharpness, observe that if  is a diagonal rank one matrix unit then we have equality in both (ii) and (iii). ∎

###### Proposition 3.

For any and , we have

 m∑j=1∥xj∥n∥yj∥n⩽π26∥∥m∑j=1x∗jxj∥∥142n+1∥∥m∑j=1xjx∗j∥∥142n+1∥∥m∑j=1y∗jyj∥∥142n+1∥∥m∑j=1yjy∗j∥∥142n+1.
###### Proof.

Let and let . We claim that

 m∑k=1∥x∗kxk∥p⩽C∥∥m∑k=1x∗kxk∥∥p+1.

By the Cauchy–Schwarz inequality and Proposition 2(iii) this will then imply the desired inequality. To establish the claim, let and let us write for the standard basis vectors in . We have

 m∑k=1⟨x∗kxkξk,ξk⟩=m∑k=1+∞∑i,j=1⟨x∗kxkej,ei⟩(ij)p¯¯¯¯¯ξkii−pξkjj−p.

Applying the Cauchy–Schwarz inequality to summation over gives

 m∑k=1⟨x∗kxkξk,ξk⟩⩽m∑k=1(+∞∑i,j=1|⟨x∗kxkej,ei⟩|2(ij)2p)12.

Since is positive for any , and for positive operators we have (where ), this implies that

 m∑k=1⟨x∗kxkξk,ξk⟩ ⩽+∞∑j=1⟨(m∑k=1x∗kxk)jpej,jpej⟩ ⩽C∥∥m∑k=1x∗kxk∥∥p+1.

By Proposition 2(i), for any there are with

 m∑k=1∥x∗kxk∥p

Taking the infimum over yields the claim. ∎

As a straightforward consequence of Proposition 3, we obtain:

###### Theorem 4 (Grothendieck’s inequality in S).

There is a constant such that for any and . Moreover, every continuous bilinear form satisfies inequality (1) with , for any and any . In particular, taking where , we obtain

 (3) m∑j=1|ϕ(xj)|2⩽K(∥ϕ∥∗n∥(xj)∥RC2n+1)2
###### Remark.

This shows that . On the other hand, it is easy to show that . Indeed, if not, then (3) would hold with replaced by some . Take , define and by . Then for we get and . On the other hand, is equivalent (up to a constant) to . Therefore (3) takes the form for some constant (independent of ). Letting tend to infinity, we obtain , a contradiction. Hence .

## 3. Optimality

We will now show that . For this, we will use the tensor product formulation, noting that

 κ(n)=min{k∈N:sup{∥z∥π,n∥z∥ah,k:z∈S⊗S,z≠0}<∞}.

Recall the diagonal operator defined on page 2 above. Since every is an operator on via the canonical inclusions , it is clear that if , then and are both operators on . This leads to the following observation.

###### Proposition 5.

If , then

 ∥z∥π,n=∥∥k∑j=1dnxjdn⊗dnyjdn∥∥π.
###### Proof.

Write

 Δnz=k∑j=1dnxjdn⊗dnyjdn∈B(ℓ2)⊗B(ℓ2).

If and for some , then and

 ∥z∥π,n⩽m∑l=1∥d−1nald−1n∥n∥d−1nbld−1n∥n=m∑l=1∥al∥∥bl∥<∥Δnz∥π+ε.

This gives . The reverse inequality is proved similarly. ∎

We also need the following well-known fact.

###### Proposition 6.

If is a Hilbert space and , then

 ∥∥m∑j=1xj⊗x∗j∥∥h=∥∥m∑j=1xjx∗j∥∥.
###### Proof.

By [15, Theorem 4.3], the Haagerup norm on the left hand side is equal to the completely bounded norm of the map on  given by , which is completely positive, so attains its completely bounded norm at the identity operator. ∎

###### Theorem 7.

For every , we have .

###### Proof.

By Theorem 4, it only remains to show that . Choose sufficiently large that for all . This inequality ensures that for every , if we define

 i1=2k,ik+1=⌊k14n2k⌋,ij=ik+1+j−(k+1), 2⩽j⩽k,

then . Denote by the standard matrix units, and for , consider the self-adjoint operators

Let

 zk:=k+1∑j=2xj⊗xj.

Since and , by Propositions 5 and 6 we obtain

 ∥zk∥π,n =∥∥k+1∑j=2dnxjdn⊗dnxjdn∥∥π ⩾∥∥k+1∑j=2dnxjdn⊗dnxjdn∥∥h=∥∥k+1∑j=2(dnxjdn)2∥∥=i2n1k+1∑j=2i2nj.

On the other hand,

 |xj|2=x2j=ei1,i1+eij,ijandk+1∑j=2d2nx2jd2n=i4n1kei1,i1+k+1∑j=2i4njeij,ij.

Therefore

 ∥zk∥ah,2n ⩽i4n1k+i4nk+1.

Hence

 ∥zk∥π,n∥zk∥ah,2n>i2n1∑k+1j=2i2nji4n1k+i4nk+1>i−2n1i2n21+k−1i−4n1i4nk+1→∞ as k→∞,

by our choice of . So as required. ∎

## 4. Reformulations of the inequality

Here we give several different ways of stating our inequality; in each case, an analogous result may be found in . The methods here are fairly standard, so full proofs are often omitted. Throughout, we write .

### 4.1. Grothendieck’s inequality with states

Given , let be given by , . We call an element of the closed convex hull of an -state on . Note that by Proposition 2(i), for any positive element we have , where is the set of all -states on . The next result may be deduced from Theorem 4 by closely following the Hahn–Banach Separation argument of [12, §23].

###### Theorem 8.

For any continuous bilinear form and , there are -states on with

for all .

### 4.2. ‘Little’ Grothendieck inequality

As a consequence we obtain the following ‘little’ Grothendieck inequality in . Recall that if is a linear map between Fréchet spaces, then .

###### Theorem 9.

For any Fréchet-Hilbert space , if are continuous linear maps, and , then

 ∣∣m∑j=1⟨u1(xj),u2(yj)⟩k∣∣⩽K∥u1∥n,k∥u2∥n,k∥(xj)∥RC2n+1∥(yj)∥RC2n+1.

Equivalently, for any there are -states such that for all we have

 |⟨u1(x),u2(y)⟩k|⩽K∥u1∥n,k∥u2∥n,k×(ϕ1(x∗x)+ϕ2(xx∗))12(ψ1(y∗y)+ψ2(yy∗))12.
###### Proof.

Apply Theorems 4 and 8 to for . ∎

Using the same argument as in the proof of Theorem 8 we can obtain an equivalent version of the ‘little’ Grothendieck inequality.

###### Theorem 10.

For any Fréchet-Hilbert space , if is a continuous linear map and , then there exist -states on such that for all we have

 ∥ux∥k⩽√K∥u∥n,k(ϕ1(x∗x)+ϕ2(xx∗))12.

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