Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 1, Article 1, 2002
NORM INEQUALITIES IN STAR ALGEBRAS
A.K. GAUR
M ATH D EPARTMENT
D UQUESNE U NIVERSITY
P ITTSBURGH , PA 15282, U.S.A.
Gaur@mathcs.duq.edu
Received —–; accepted 21 June, 2001.
Communicated by S.P. Singh
A BSTRACT. A norm inequality is proved for elements of a star algebra so that the algebra is
noncommutative. In particular, a relation between maximal and minimal extensions of regular
norm on a C ∗ -algebra is established.
Key words and phrases: W ∗ -algebras, C ∗ -algebras, Self-adjoint elements (operators), Maximal and minimal extensions of a
regular norm, Furuta type inequalities.
2000 Mathematics Subject Classification. 46H05, 46J10, 47A50.
Let H be a Hilbert space and B(H) be the algebra of all bounded linear operators on H. A
subset of B(H) is a W ∗ -algebra on H if X is a C ∗ -algebra which is closed in the weak operator
topology, see [1]. Also, a W ∗ -algebra is a C ∗ -subalgebra of B(H) which is weakly closed. In
particular, a W ∗ -algebra is an algebra of operators. We note that a C ∗ -algebra acting on H is
commutative if and only if zero is the only nilpotent element of the algebra.
Let X be a W ∗ -algebra and XSA be the set of self-adjoint elements of X, that is if T ∈
S (X) =⇒ T = T ∗ , where T ∗ is the adjoint of T . Here we prove the following theorem.
Theorem 1. A unital W ∗ -algebra X of operators is noncommutative if ∀A, B ∈ XSA ,
kAk = 1 = kBk =⇒ kA + Bk > 1 + kABk .
Proof. Since X is noncommutative, there exists an operator T in X such that T 2 = 0. Suppose
X1 is the range of T and X2 is the orthogonal complement of X1 . Then H = X1 ⊕ X2 . Let S
be an operator with kSk = 1. Then
0 0
0 S
∗
.
T =
and T =
0 0
S∗ 0
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
The author is grateful to the referee for valuable suggestions.
051-01
2
A.K. G AUR
We use these representations for T and T ∗ to define the operators A and B as follows: For
σ1 > 0, σ2 > 0 and σ1 + σ2 = 1, we have
SS ∗ 0
A =
= T T ∗,
0 0
σ1 SS ∗ σ2 S
= σ1 (T T ∗ + T ∗ T ) + σ2 (T + T ∗ ) .
B =
σ2 S ∗ σ1 S ∗ S
Clearly, A = A∗ , B = B ∗ and A, B ∈ X. Next, we consider the following two cases.
Case 1. Let σ1 = σ2 =
1
2
and
1
1 SS ∗ S
= (T T ∗ + T ∗ T + T + T ∗ ) .
B=
∗
∗
S
S S
2
2
It is not difficult to see that kAk = 1 and kBk ≤ 1. To obtain kBk ≥ 1, let kSk = 1 then
∃an ∈ X 3: kan k = 1. Also,
kSS ∗ an − an k2 = kSS ∗ an k2 − 2 kS ∗ an k2 + kan k2
≤ 2 kan k2 − kS ∗ an k2
and if kS ∗ an k → 1 then SS ∗ an − an → 0. Further, (Bb − b) → 0, where b = (an + S ∗ an ) and
hence, kBk ≥ 1, which concludes that kBk = 1.
Let
1 SS ∗ S
AB =
and
S ∗ S ∗S
2
∗
S
1
SS ∗ + SS2
2
A+B =
= (T T ∗ + T ∗ T + T + T ∗ ) .
S∗
S∗S
2
2
2
Choose an as above and bn =
S ∗ an
,
(2µ−1)
where µ > 1. Let
1
1
1 2
µ1 = σ1 + + σ2 +
2
4
so that it satisfies the equation
(µ1 − σ1 − 1) (µ1 − σ1 ) = σ22 .
Then
[(A + B) (an + bn ) − µ (an + bn )] → 0
and kA + Bk ≥ µ > 1. If we choose σ1 and σ2 so that
1
1
1
1 2
σ1 + + σ2 +
> 1 + σ12 + σ22 2 ,
2
4
then we have
kA + Bk > 1 + kABk .
For example, it is sufficient to take σ1 = 23 and σ2 = 13 . We note that σ1 > σ2 . If σ1 < σ2 then
√
the above inequality fails. Since µ > 1 + 2σ1 the proof in this case is complete.
Case 2. Let σ1 6= σ2 . Then kABk ≤ a0 , (by mimicking the proof of Case 1), where
q 2
2
2
a0 = sup σ1 kak + σ2 kbk : kak + kbk = 1 = σ1 + σ22 .
J. Inequal. Pure and Appl. Math., 3(1) Art. 1, 2002
http://jipam.vu.edu.au/
N ORM I NEQUALITIES IN S TAR A LGEBRAS
3
Let bn (µ1 − σ1 ) = σ2 S ∗ an , where µ1 depends on σ1 and σ2p
. Then kA + Bk ≥ µ1 and one can
have the following form of µ1 , that is, 2µ1 = (2σ1 + 1) + 1 + σ12 . Hence, µ1 > 1 + a0 and
this concludes the proof of the theorem.
Remark 2. If X is commutative then for A, B ∈ XSA with kAk = 1 = kBk, we have 0 ≤
I − B − A + AB, where I is the identity operator. Thus kA + Bk ≤ 1 + kABk.
Let 0 < p, q, r be real numbers such that q (2r + 1) ≥ (2r + p) and q ≥ 1. If two bounded
1
linear operators A, B ∈ B (H) on a Hilbert space H satisfy 0 ≤ B ≤ A then (B r Ap B r ) q ≥
p
2r
B q B q . This inequality is called the Furuta inequality and can be found in [3]. Recently, Kotaro
and others in [6] have extended this inequality in a unital hermitian Banach ∗-algebras with
continuous involution. We give a slightly different version of these inequalities in the following
corollary.
Corollary 3. Suppose that XC ∗ is a C ∗ -algebra acting on H. Let λ, µ and σ be three real
numbers with σ > 0, λ > 0. Then there exists operators T1 , T2 and T3 in X 3: λT1 + µT2 +
σT3 ≥ 0 ⇐⇒ λσ ≥ µ2 .
Proof. We recall that an operator O ∈ B (H) is positive if hOh, hi ≥ 0 for every vector h.
Using the techniques of Theorem 1, the following operators belong to XC ∗ . That is,
√
∗S
SS ∗ 0
0
SS
√
T1 =
,
T2 =
, and
0 0
S ∗ SS ∗
0
0 0
T3 =
0 S ∗S
are in XC ∗ . In this case we have
λT1 + µT2 + σT3 =
√
∗
∗S
λSS
µ
SS
√
.
µS ∗ SS ∗
σS ∗ S
Let λT1 + µT2 + σT3 = Λ. Then we observe that the determinant of Λ is zero if λσ = µ2 . If
ε < 0 and h ∈ H, then we have
k(Λ − ε) hk2 = kΛhk2 − 2ε hΛh, hi + ε2 khk2 ≥ −2ε hΛh, hi + ε2 khk2 ≥ +ε2 khk2 .
Thus ε ∈
/ SPap (Λ) , the approximate point spectrum of Λ. This means that (Λ − ε) is left
invertible. Since (Λ − ε) is hermitian, it must also be right invertible. That is, ε ∈
/ SP (Λ) and
so Λ ≥ 0 ⇐⇒ λσ ≥ µ2 .
Alternatively, for a ∈ X1 and b ∈ X2 , we have
2 r
√
∗
SS µ2
a + λ −
kS ∗ ak2 .
hΛ (a + b) , (a + b)i = σSb + µ
σ σ
Since a ∈ X1 , therefore
√
r
SS ∗
µ2
a → 0 =⇒ Λ ≥ 0 ⇐⇒ λ −
≥ 0.
σ
σ
Hence the proof of the corollary is complete.
∃bn ∈ X2 3:
σ (Sbn ) + µ
Remark 4. By reducing the matrix Λ into a product of three matrices the above corollary can
also be proved. That is, Λ = L∗ DL, where
µ√ ∗ √ ∗
I W
L=
and W =
SS S S S.
0 I
σ
J. Inequal. Pure and Appl. Math., 3(1) Art. 1, 2002
http://jipam.vu.edu.au/
4
A.K. G AUR
By the partial commutation relation, we have
√
√
SS ∗ S = S S ∗ S
and hence
 
D=
λ−
µ2
σ
SS ∗
0


.
0
σS ∗ S
Under the above assumption about σ and S ∗ S, the Sylvester type test applies. That is, Λ is
2
positive (semi definite) if and only if σ > 0 and λ − µσ ≥ 0.
Let A, B > 0 be invertible operators on H. In this case a Furuta type inequality is obtainable
by replacing 1 with 0 in the original Furuta inequality in Remark 2. In fact, we have
2r
A2r ≥ (Ar B p Ar ) (2r+p) .
Also, if A ≥ B ≥ 0, 3: A > 0, then for each α ∈ [0, 1] and p ≥ 1 we have
(1+r−α)
n r −α
s r o [(p−α)s+r]
α −α
2
2
2
2
A A B A
A
≤ A(1+r−α) for s ≥ 1 and r ≥ α.
For more details, see [3]. The following examples give an application of these inequalities in
case of C ∗ -algebras.
Example 0.1. Let X be a commutative C ∗ -algebra acting on H. If we take
A = 6T1 + 0T2 + 3T3 and
B = 3T1 + 2T2 + T3
then
A − B = 3T1 − 2T2 + T3
and by the Corollary 3 we have A − B ≥ 0. We further note that A2 is not greater than or equal
to B 2 , since for b ∈ X2 ,
2
A − B 2 b, b + (S ∗ S)2 b, b = 0.
Example 0.2. Let
A = 2T1 , B = T1 + T2 + T3 and
C = 4T1 + T2 + T3 .
Then A ≥ 0 and B + C ≥ 0. Further, by the Corollary 3, we have B + C − A ≥ 0. Let Ψ ≤ B
and Φ ≤ C, where A = Ψ + Φ. Then Ψ ≤ A. Hence, from Corollary 3, for a ∈ X1 and b ∈ X2 ,
we have
√
2
∗
h(T1 + T2 + T3 ) (a + b) , (a + b)i = SS a + Sb .
√
Also, bn ∈ X2 =⇒ Ψ = 0, because Sbn + SS ∗ a → 0. Thus
A = Φ = 2T1 ≤ 4T1 + T2 + T3 .
Example 0.3. Let
1
T1 , B = T1 + T2 + T3 and
3
C = 4T1 + 2T2 + T3 .
A =
Then A ≥ 0 and B +C ≥ 0. Next, by Corollary 3, we get B +C −A ≥ 0. Now by Example 0.2
it follows that A = Φ = 31 T1 ≤ C. This contradicts Corollary 3, since for (C − A) , λσ < µ2 .
J. Inequal. Pure and Appl. Math., 3(1) Art. 1, 2002
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N ORM I NEQUALITIES IN S TAR A LGEBRAS
5
Remark 5. The algebra norm k·k on a non-unital Banach algebra J can be extended to an
algebra norm on the unitization J+ = Ce + J, (where e is the unit in the algebra) in many ways.
In particular, the following two norms,
l1 − norm = k(λ) e + ak1 = |λ| + kak
and the operator norm,
k(λ) e + akOP = sup {k(λ) b + abk , k(λ) b + bak ; b ∈ J, kbk ≤ 1} , λ ∈ C, a ∈ J
are the maximal and the minimal extensions of the original norm respectively, if it is a regular
norm, that is,
kak = sup {kabk , kbak ; b ∈ J, kbk ≤ 1} ,
+
see [4]. The unitization J is complete under both k·k1 and k·kOP , so by the two norm lemma,
[2, II, 2.5] these two norms are equivalent. If J is a C ∗ -algebra, a ∈ J is self-adjoint, and λ is
complex, then k(λ) e + ak1 ≤ 3 k(λ) e + akOP . So far the constant 3 is the best possible in this
case.
Recently, in [5], we extended the above result to locally m-convex algebras. Now we prove
the following corollary.
Corollary 6. If is a commutative non-unital C ∗ -algebra A, B ∈ YSA with kAk = 1 = kBk,
and λ is complex, then
k(λI + Ψ)k1 ≤ λ0 + 3 kΦkOP ,
where λ0 > 0, Ψ = A + B, and Φ = AB.
Proof. Since
k(λI + A + B)k1 ≤ 3 k(λI + A + B)kOP ,
we have
1
k(λI + A + B)k1 ≤ |λ| + k(A + B)kOP =⇒ k(λI + Ψ)k1 ≤ λ0 + 3 kΦkOP ,
3
whereλ0 = 3 (|λ| + 1) > 0. This concludes the proof of the corollary.
Remark 7. k(λi + Ψ)kOP ≤
λ0
3
+ kΦkOP .
R EFERENCES
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[2] N. DUNFORD AND J.T. SCHWARTZ , Linear Operators, Part I, Interscience, New York, (1958).
[3] T. FURUTA, M. HASHIMOTO AND M. ITO, Equivalence relation between generalized Furuta inequality and related operator functions, Sci. Math. I (electronic), 2 (1998), 257–259.
[4] A.K. GAUR AND Z.V. KOVARIK, Norms on unitizations of Banach algebras, Proc. Amer. Soc.,
117(1) (1993), 111–113.
[5] A.K. GAUR, Maximal and minimal semi-norms and g-bounded operators, IJNSNS, 2 (2001), 283–
288.
[6] T. KOTARO AND U. ATSUSHI, The Furuta inequality in Banach *-algebras, Proc. Amer. Math. Soc.
128(6) (2000), 1691–1695.
J. Inequal. Pure and Appl. Math., 3(1) Art. 1, 2002
http://jipam.vu.edu.au/