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STRONGLY EXPOSED POINTS IN THE UNIT BALL
OF TRACE-CLASS OPERATORS
KOUROSH NOUROUZI
Received 23 April 2001 and in revised form 22 September 2001
A theorem of Arazy shows that every extreme point of the unit ball of trace-class operators is strongly exposed. We give this result a simpler and direct proof here.
2000 Mathematics Subject Classification: 46A55, 47B20.
The main purpose of this paper is to give a direct proof of an interesting special case of a far-reaching theorem of Arazy [1,2].
Arazy studied the extreme, exposed, and strongly exposed points in the unit balls of symmetrically normed ideals (of operators) acting on a separable Hilbert space, and he obtained very useful and complete results in [2] on the extremal structure of these operator balls. Arazy’s study of symmetrically normed ideals is, on the one hand, quite general. On the other hand, the ideal of trace-class operators is, for a variety of reasons, perhaps the most interesting of the symmetrically normed ideals. Because of the importance of the trace-class operators, and in the light of sustained interest in exposed points amongst Banach space theorists, we thought it is worthwhile to find a fairly direct proof of Arazy’s theorem in this particular case. Our proof, presented herein, is straightforward in the sense that it relies essentially only on fundamental properties of singular values, as explained in the monograph of Gohberg and Kre˘ın [3].
Theorem 1(Arazy). LetS1 denote the Banach space of the trace-class operators acting on an infinite-dimensional separable complex Hilbert space, and assume that
x∈BallS1. The following statements are equivalent: (a) xhas rank1andtr(x∗x)=1;
(b) xis an extreme point ofBallS1; (c) xis a strongly exposed point ofBallS1.
The equivalence of (a) and (b) seems to have first been determined by Holub in [5]. Before moving to the proof, the relevant definitions are reviewed below.
An elementωin a convex setCin a complex Banach spaceXis anextremepoint of Cif the equationω=tω1+(1−t)ω2, fort∈(0,1)⊂Randω1,ω2∈C, is satisfied only withω1=ω2=ω. A pointω∈C isstrongly exposed if there is a continuous linear functionf:X→Csuch that
(i) Ref (γ) <Ref (ω), for allγ∈C\{ω},
(ii) Ref (γk)→Ref (ω), for a sequence,{γk}k⊂C, only ifγk→ω.
For any Banach spaceX, let (1) BallX= {x∈X:x ≤1},
(2) ext BallX= {x∈BallX:xis an extreme point of BallX}.
It is an elementary consequence of the Cauchy-Schwarz inequality that the extreme points of BallHin any Hilbert spaceH(of any dimension) are strongly exposed. In con-trast, ifB(H)denotes the algebra of a bounded operator acting on a Hilbert spaceH, then the extreme points of BallB(H)are exposed if and only ifHis separable, and they are strongly exposed if and only ifHis finite dimensional. (These results were proved by Grza´slewicz [4].) Arazy’s work provides a complete analysis of the situation concerning strongly exposed points in the unit balls of symmetrically normed ideals. Henceforth,Hwill denote a separable, infinite-dimensional Hilbert space;B(H)is theC∗-algebra of (bounded) operators acting onH; andK(H)denotes the ideal of compact operators acting onH. Ifx∈B(H), thenxhas a polar decompositionx=
w|x|, where|x|is the (unique) positive square root ofx∗xand wherewis the (unique) partial isometry whose initial space is the closure of the range of|x|and whose final space is the closure of the range ofx.
Forx∈K(H), thesingular values ofx are the elementssk(x)of the decreasing sequence{sk(x)}k∈Z+ of nonnegative real numberssk(x)=λk(x∗x), where
λkx∗x=minmaxxξ2:ξ∈L⊥,ξ =1:L⊂H,dimL=k−1. (1)
Let · denote the operator norm on B(H), namely, x =sup{xξ :ξ ∈H,
ξ =1}. Ifx∈K(H), then, by [3, page 29],
s1(x)= x;
sn(x)=minz−x: rankz≤n−1, forn≥2. (2)
Thetrace classis the ideal setS1ofB(H), defined byS1= {x∈K(H):
nsn(x) <
∞}, is an ideal ofB(H), and the function·1:S1→R+0, given by
x1= ∞
n=1
sn(x), x∈S1, (3)
is a norm onS1under whichS1is a Banach space.
For everyx∈S1, nxφn,φn, where{φn}n∈Z+ is an orthonormal basis ofH is absolutely convergent. This defines a linear functional onS1called thetrace
tr(x)=
∞
n=1
xφn,φn , x∈S1. (4)
It is well known that the definition of the trace is independent of the choice of or-thonormal basis.
Ifx∈S1, then letxbe the vector in1whosenth component issn(x). It is clear from the definition ofS1thatx∈BallS1if and only ifx∈Ball1.
Proof of Arazy’s theorem. We show that (b)⇒(a)⇒(c)⇒(b).
Ifx=w|x|is the polar decomposition ofx, where|x|has the spectral decomposition
|x|ξ=nsn(x)ξ,φnφn, forξ∈H. Leta,b∈B(H)be defined so that the action of aandbon eachξ∈His
ν
n=1
αnξ,φn wφn, ν
n=1
βnξ,φn wφn, (5)
whereαnandβndenote thenth components of the vectorsα,β∈1andν∈Z+∪{∞}. Thus,x=(1/2)a+(1/2)b. Becausea1≤ w(
n|αn|)≤1 and, likewise,b1≤1, x is an average of two elements (namely,aand b) from the unit ball ofS1. Hence, a=b=x.
The projectionw∗whas the range(Span{φn}ν
n=1)−, and thereforew∗wφn=φn, for alln. Thus, for allξ∈H,
|x|ξ=w∗xξ=w∗aξ= ν n=1
αnξ,φn φn, (6)
which means thatαn=sn(x)for everyn. Similarly,βn=sn(x). Hence,α=β=x, which proves thatx∈ext Ball1.
Now leten∈1be the vector with the real number 1 in positionnand zero in all other positions. Because ext Ball1= {−en,en:n∈Z+},x∈ext BallS1if and only ifx has exactly one nonzero singular value, namelys1(x), ands1(x)=1. In other words, x2=tr(x∗x)=rankx=1, completing the proof that (b)⇒(a).
To prove that (a)⇒(c), letxbe a rank-1 operator of norm 1. From the polar decom-positionx=w|x|ofx, there are unit vectorsφ1,ψ1∈H such thatxξ= ξ,φ1ψ1, for everyξ∈H, wherewφ1=ψ1andww∗is the projection onto Span{ψ1}.
Setp = |x|. Becausex is a rank-1 operator of norm 1,p is a rank-1 projection whose range is spanned byφ1. Extend the singleton set{φ1}to an orthonormal basis {φn}n∈Z+ofH. Thus, the trace of everyz∈S1is given by tr(z)=
nzφn,φn. Define a linear functionalfonS1by
f (y)=trpw∗yp, ∀y∈S
1. (7)
Thenf (x)=tr(pw∗w|x|p)=tr(p)=1, becausepis a rank-1 projection. So, Ref (x)= f (x)=1. Moreover, ify∈BallS1, then
f (y)=w∗yφ
1,φ1 ≤w∗y =s1(y)≤ ∞
n=1
sn(y)= y1≤1. (8)
Thus,fis a support functional for BallS1.
Assume that{yk}k⊂BallS1and limkRef (yk)=Ref (x)=f (x)=1, that is,
lim k Re
w∗ykφ1,φ1 =1. (9)
Then, by (8), the sequence{yk}khas the property that, for alln≥2,
lim k s1
yk=lim k
yk=limyk1=1, lim k sn
Furthermore,
ykφ1−xφ12
=ykφ12−2 Reykφ1,xφ1 +xφ12
=ykφ12−2 Reykφ1,wφ1 +ψ12
≤yk2
φ12−2 Re
w∗ykφ
1,φ1 +ψ12 ≤21−Rew∗ykφ
1,φ1 .
(11)
Thus, by (9),
ykφ1−xφ1 →0. (12)
Becausexis a rank-1 operator, the inequalities in [3, page 29] are
sn+1yk≤snyk−x≤sn−1yk, forn≥2. (13)
Hence, for everyk,
yk−x1=s1
yk−x+s2yk−x+ ∞
n=3
snyk−x
≤2s1
yk−x+
∞
n=2 snyk
=2yk−x+yk1−s1yk
≤2yk−x+1−s1
yk.
(14)
Therefore, to prove that limkyk = x in S1 it is sufficient, by (10), to prove that yk−x →0.
The singular value s2(yk) measures the distance inB(H) from yk to the set of operators whose rank is at most 1. Thus, by (10), there is a sequence{zk} ∈S1 of operators such that each zk is zero or rank-1 and yk−zk →0 as k→ ∞. From
yk−x ≤ yk−zk+zk−xwe see that it is now enough to prove thatzk−x →0. Becausezkφ1−xφ1 ≤ zk−ykφ1 + ykφ1−xφ1, we have that zkφ1− xφ1 →0. Hence, there existsN∈Z+such thatzkφ1≠0, for allk≥N. In all cases for whichzkφ1≠0, the vectorzkφ1spans the range ofzkand, therefore, there exist vectorsηk∈Hsuch that, for everyξ∈H,
zkξ=ξ,ηk zkφ1. (15)
Formula (15) also holds for allkfor whichzkφ1=0 by simply choosingηk=0. Ifk≥Nthenzkφ1= φ1,ηkzkφ1; that is,
φ1,ηk =1, ∀k≥N. (16)
Also,
Therefore, the sequence{ηk}kis bounded and
lim
k→∞ηk=limk≥N zk
zkφ1=1. (18)
For anyξ∈H, zk−x
ξ=ξ,ηk zkφ1−
ξ,φ1 xφ1 =ξ,ηk zkφ1−
ξ,ηk xφ1+
ξ,ηk xφ1−
ξ,φ1 xφ1 ≤ξ,ηk zkφ1−xφ1+xφ1ξ,ηk−φ1
≤ ξηkzkφ1−xφ1+ηk−φ1.
(19)
Therefore, we havezk−x →0 if we can prove that ηk−φ1 →0. But this is so, because
ηk−φ12
=ηk2−2 Reηk,φ1 +φ1 2
=ηk2−2+1=ηk2−1. (20)
Thus, from (18) we conclude thatηk−φ1 →0. Hence, (8)⇒(10).
The proof of (c)⇒(b) is a standard argument in convexity theory, which is, therefore, omitted here.
Acknowledgments. The author wishes to express his debt to Professor Douglas
R. Farenick for much more than just conversations related to the contents of this paper. This work is supported in part by the Ministry of Higher Education of Iran.
References
[1] J. Arazy,Basic sequences, embeddings, and the uniqueness of the symmetric structure in unitary matrix spaces, J. Funct. Anal.40(1981), no. 3, 302–340.
[2] ,On the geometry of the unit ball of unitary matrix spaces, Integral Equations Op-erator Theory4(1981), no. 2, 151–171.
[3] I. C. Gohberg and M. G. Kre˘ın,Introduction to the Theory of Linear Nonselfadjoint Oper-ators, translated from the Russian by A. Feinstein. Translations of Mathematical Monographs, vol. 18, American Mathematical Society, Rhode Island, 1969. [4] R. Grza´slewicz,Exposed points of the unit ball ofᏸ(H), Math. Z.193(1986), no. 4, 595–596. [5] J. R. Holub,On the metric geometry of ideals of operators on Hilbert space, Math. Ann.201
(1973), 157–163.
