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In this note we show that, for any power bounded elementaofA,θ=σ(A, A∗)−limn→∞(1+a2 )n exists,θ is an idempotent and θa=aθ=θ

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Dedicated to Professor Serban Stratila on the occasion of his 70th birthday

A NOTE ON POWER BOUNDED ELEMENTS OF VON NEUMANN ALGEBRAS

A. ULGER

LetA be a von Neumann algebra with predualAand the unit element 1. An elementaofAis said to be power bounded ifsupn≥0||an||<∞. In this note we show that, for any power bounded elementaofA,θ=σ(A, A)−limn→∞(1+a2 )n exists,θ is an idempotent and θa==θ.

AMS 2010 Subject Classication: 46L10, 47B99.

Key words: Von Neumann algebras, power bounded elements.

1. INTRODUCTION

Let A is a von Neumann algebra with the unit element 1. By the weak topology ofAwe mean the topologyσ(A, A), whereAis the (unique) predual of A. In this note we prove that, for any power bounded element aof A, the weak limit of the sequence((1+a2 )n)n≥0 always exists, it is an idempotent, say θ, and is such that θa = aθ = θ. An important problem here (see Example 5 below) is to characterize the power bounded elements aof A for which the weak−limn→∞(1+a2 )n=θis dierent from zero. We do not know the answer of this question.

2. THE MAIN RESULT

Let A be a von Neumann algebra and 1 be its unit element. As stated above, our aim is to prove that, for any power bounded element a of A, the sequence ((1+a2 )n)n≥0 weak converges to some idempotent element ofA.

The main ingredient of the proofs below is the fact that the multiplication operator ϕ : A×A → A, ϕ(a, b) = ab, is weak-weak continuous in each variable when the other variable is kept xed.

We need two preliminary results, Lemma 1 and Corollary 3 below. In connection with Lemma 1, we recall the following well-known result due to Katznelson and Tzafriri [4]:

REV. ROUMAINE MATH. PURES APPL. 59 (2014), 2, 303306

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304 A. Ulger 2

For a power bounded linear operator T on a Banach space X,

||Tn+1−Tn|| →0, as n→ ∞, i σ(T)∩Γ⊆ {1}.

Here σ(T)is the spectrum ofT and Γis the unit circle. We shall not use this result; we cited it here because it explains when the hypothesis of the next lemma is satised.

Lemma 1. Let a be a power bounded element of Asuch that ||an+1−an||

→ 0, as n→ 0. Then weak−limn→∞an =θ exists, θ is an idempotent and θa=aθ=θ.

Proof. To prove that the sequence(an)n≥0 converges in the weak topol- ogy of A, since this sequence is bounded, it is enough to prove that it has only one weak cluster point. For a contradiction, let θ1 and θ2 be two weak cluster points of the sequence (an)n≥0. Then this sequence has two subnets (ani)i∈I and (amj)j∈J that converge weak, respectively, to θ1 and θ2. Since

||an+1−an|| →0, asn→0,

limi ||ani+1−ani||= 0 and lim

j ||amj+1−amj||= 0 too.

So, since the norm of A is weak lower semi-continuous, from the rst limit, we get

||aθ1−θ1|| ≤lim inf

i ||aani−ani||= lim inf

i ||ani+1−ani||= 0.

Hence, aθ11. In the preceding lines writing ani+1 as ani+1=ania, we getθ1a=θ1 too. Instead of the net(ani)i∈I, if we use the other net(amj)j∈J

we get that aθ22 and θ2a=θ2. Iterating, for instance equality aθ11, we get aniθ1 = θ1. Passing to the limit in the weak topology of A, we see thatθ1 is an idempotent. Similarly we see thatθ2 is also an idempotent. Now iterating the equalitiesaθ11 andθ1a=θ1, we get that

amjθ11 and θ1amj1.

Passing in these equalities to the limit in the weak topology ofA, we get

(1) θ2θ11θ21.

This time iterating the equalities aθ2 = θ2 and θ2a = θ2, we get the equalities

aniθ22 and θ2ani2. Passing to the limits in the weak topology, we get

(2) θ1θ22θ12.

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3 A Note On Power Bounded Elements of von Neumann Algebras 305

From (1) and (2), we see that θ1 = θ2 so that the bounded sequence (an)n≥0has only one weak cluster point, sayθ. Thus,θ=weak−limn→∞an exists, aθ=θa=θand θ is an idempotent.

The second result we need is the following result, known as Ishikawa Iteration Theorem. This result can be found in Ishikawa's paper ([3], p. 69) and also in the book ([2], Lemma 9.4) by K. Goebel and Kirk,

Theorem 2. LetX be a Banach space and T :X →X a (not necessarily linear) nonexpansive mapping (i.e., for x, y∈X, ||T(x)−T(y)|| ≤ ||x−y||).

Then, for 0< λ <1 andx∈X,

n→∞lim ||(λI+ (1−λ)T)n+1(x)−(λI+ (1−λ)T)n(x)||= 0.

IfT :X→Xis a bounded linear operator then, in the preceding theorem, instead of assuming that T is nonexpansive it is enough to assume that T is power bounded. Indeed, if T :X → X is a power bounded linear operator on X, for x ∈ X, |x| = Supn≥0||Tn(x)|| is a norm on X equivalent to the original norm of X. For this normT is nonexpansive, so that, the norms being equivalent, the following corollary holds.

Corollary 3. LetX be a Banach space andT :X →Xa power bounded linear operator. Then, for any x∈X,

n→∞lim ||(I+T

2 )n+1(x)−(I+T

2 )n(x)||= 0.

At this point we would like to include the following remarks of the referee.

The referee has kindly informed us that, from ([1], Lemma 2.1), one can get the inequality ||(I+T2 )n+1−(I+T2 )n|| ≤ c.1n for some constant c > 0. From this inequality one gets of course that ||(I+T2 )n+1−(I+T2 )n|| →0. He has also indicated that this latter result can also be deduced from the above mentioned theorem of Katznelson and Tzafriri. Since for our purpose the above weaker result is sucient, we decided to keep the above corollary as it is.

The main result of this note is the following theorem.

Theorem 4. Let A be a von Neumann algebra. Then, for any power bounded element aof A, the limit θ=weak−limn→∞(1+a2 )n exists, θis an idempotent and θa=aθ =θ.

Proof. Leta∈Abe a power bounded element ofA. Toawe associate the linear operatorLa:A→A dened byLa(b) =ab. As Lna(b) = (La◦...◦La) = anb, we see that La is a power bounded linear operator on the Banach space A. Hence, by the preceding Corollary, for eachb∈A,

n→∞lim ||(I+La

2 )n+1(b)−(I+La

2 )n(b)||= 0.

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306 A. Ulger 4

For b= 1, the unit element ofA, the preceding limit becomes

(3) lim

n→∞||(1 +a

2 )n+1−(1 +a

2 )n||= 0.

Hence, by Lemma 1,θ=weak−limn→∞(1+a2 )exists,θis an idempotent and

θ(1 +a

2 ) = (1 +a 2 )θ=θ.

Equivalently, θa=aθ =θ. This completes the proof.

The next example explains one of the reasons why understanding the power bounded elementsaofAfor which theweak−limn→∞(1+a2 )is dierent from zero is important.

Example 5. LetHbe a separable Hilbert space andA=B(H)the opera- tor algebra of the bounded linear operators onH. The predual ofB(H)is the space of the trace class operators on H. By Theorem 4 above, for any power bounded T ∈B(H),

P =weak− lim

n→∞(I +T 2 )n

exists,P is a projection andP◦T =T◦P =P. So, in the case whereI 6=P 6= 0, the closed subspace M =P(H) ofH is proper and it is an invariant subspace for T.

REFERENCES

[1] S.R. Foguel and B. Weiss, On convex power seriesof a concervative Markov operator . Proc. Amer. Math. Soc. 38 (1973), 325330.

[2] K. Goebel and W.A. Kirk, Topics in metric xed point theory. In: Cambridge Studies in Advanced Mathematics 28, Cambridge Univ. Press, 1990.

[3] S. Ishikawa, Fixed points and iteration of a nonexpansive mapping in a Banach space.

Proc. Amer. Math. Soc. 59 (1976), 6571.

[4] Y. Katznelson and L. Tzafriri, On power bounded operators. J. Funct. Anal. 68 (1986), 313328.

Received 12 September 2013 Koc University,

Department of Mathematics, Rumeli Feneri Yolu, 34450, Sariyer, Istanbul

Turkey

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