M. KHOSRAVI1, H. MAHYAR2 AND M. S. MOSLEHIAN3
Abstract. We prove several versions of reverse triangle inequality in HilbertC∗-modules based on some works of S. S. Dragomir. In particular, we show that if e1,· · · , em are
vectors in a Hilbert moduleX over aC∗-algebraA such thathei, eji= 0 (1≤i6=j ≤m)
andkeik= 1 (1≤i≤m), and alsork, ρk ∈R(1≤k≤m) andx1,· · ·, xn∈Xsatisfy
0≤r2kkxjk ≤Rehrkek, xji, 0≤ρ2kkxjk ≤Imhρkek, xji,
then
"m X
k=1
(rk2+ρ2k)
#12 n X
j=1 kxjk ≤
n X
j=1
xj
,
and the equality holds if and only if
n X
j=1
xj= n X
j=1 kxjk
m X
k=1
(rk+iρk)ek.
1. Introduction and preliminaries
The triangle inequality is one of the most fundamental inequalities in mathematics. Several
mathematician have been investigated its generalizations and reverses.
Petrovitch [15] in 1917, proved that for complex number z1,· · · , zn, we have
|
n
X
j=1
zj| ≥cosθ n
X
j=1
|zj|,
where 0< θ < π2 and α−θ < arg zj < α+θ (1≤j ≤n) for a real number α.
This inequality can be found also in Karamata’s book [8]. The first generalization of the
triangle inequality in Hilbert space was given by Diaz and Matcalf [4]. They proved that for
x1,· · · , xn in a Hilbert space H, if e be a unit vector H such that 0≤r ≤
Rehxj,ei
kxjk for some
r ∈R and each 1≤j ≤n, then
r n
X
j=1
kxjk ≤ k n
X
j=1
xjk,
2000 Mathematics Subject Classification. Primary 46L08; Secondary 15A39, 26D15, 46L05, 51M16.
Key words and phrases. Triangle inequality, Reverse inequality, HilbertC∗-module.
and the equality holds if and only if Pn
j=1xj =r
Pn
j=1kxjke.
Recently, a number of mathematicians have represented several refinements of the reverse
triangle inequality in Hilbert spaces and normed spaces. See [1, 2, 3, 6, 7, 9, 12, 14].
Our aim is to give some generalizations of results in Hilbert spaces to the framework of
Hilbert C∗-modules. For this purpose, first we recall some fundamental definitions in the theory of Hilbert modules. We also use the elementary C∗-algebra theory, in particular we utilize this property that if a ≤ b then a1/2 ≤ b1/2, where a, b are positive elements of a C∗-algebra A. We also repeatedly apply the following known relation:
1 2(aa
∗
+a∗a) = (Rea)2+ (Ima)2, ()
whereais an arbitrary element ofA. For details onC∗-algebra theory we referred the readers to [13].
Suppose that A is a C∗-algebra and X is a linear space which is an algebraic right A -module. The space X is called a pre-Hilbert A-module (or an inner product A-module) if
there exists an A-valued inner product h., .i:X×X→Awith the following properties: (i) hx, xi ≥0 and hx, xi= 0 if and only if x= 0
(ii) hx, λy+zi=λhx, yi+hx, zi
(iii) hx, yai=hx, yia (iv) hx, yi∗ =hy, xi
for all x, y, z ∈ X, a ∈ A, λ ∈ C. By (ii) and (iv), h., .i is conjogate linear in the first variable. Using the Cauchy–Schwartz inequality hy, xihx, yi ≤ khx, xikhy, yi [10, Page 5], it follows that kxk = khx, xik12 is a norm on X making it into a right normed module. The
pre-Hilbert module X is called a Hilbert A-module if it is complete with respect to this
norm. Notice that the inner structure of a C∗-algebra is essentially more complicated than complex numbers. For instance, the notations such as orthogonality and theorems such as
Riesz’ representation in the complex Hilbert space theory cannot simply be generalized or
transferred to the theory of Hilbert C∗-modules.
One may define an “A-valued norm” |.| by |x|= hx, xi1/2. Clearly, k |x| k=kxk for each
x ∈ X. It is known that |.| does not satisfy the triangle inequality in general. See [10, 11] for more information on Hilbert C∗-modules.
2. Main results
Theorem 2.1. Let A be a C∗-algebra with unit 1, let X be a Hilbert A-module and let
x1,· · · , xn ∈X. If there exist real numbers k1, k2 ≥0 with
0≤k1kxjk ≤Rehe, xji, 0≤k2kxjk ≤Imhe, xji,
for some e∈X with |e| ≤1 and all 1≤j ≤n, then
(k21+k22)12
n
X
j=1
kxjk ≤
n
X
j=1
xj
. (2.1)
Proof. Applying the Cauchy–Schwarz inequality, we get
|he, n
X
j=1
xji|2 ≤ kek2
n
X
j=1
xj
2
≤
n
X
j=1
xj
2
,
and
|h
n
X
j=1
xj, ei|2 ≤
n
X
j=1
xj
2
|e|2 ≤
n
X
j=1
xj
2
,
whence
Pn
j=1xj
2
≥ 1 2
|he,Pn
j=1xji|2+|h
Pn
j=1xj, ei|2
= 12he,Pn
j=1xji
∗he,Pn
j=1xji+h
Pn
j=1xj, ei ∗hPn
j=1xj, ei
= (Rehe,Pn
j=1xji)2+ (Imhe,
Pn
j=1xji)2 by ()
= (RePn
j=1he, xji)
2+ (ImPn
j=1he, xji) 2
≥k2 1(
Pn
j=1kxjk) 2+k2
2(
Pn
j=1kxjk) 2
= (k21+k22)(Pn
j=1kxjk) 2.
Using the same argument as in the proof of Theorem 2.1 one can obtain the following
result, where k1, k2 are hermitian elements of A.
Theorem 2.2. If the vectors x1,· · · , xn∈X satisfy the conditions
0≤k21kxjk2 ≤(Rehe, xji)2, 0≤k22kxjk2 ≤(Imhe, xji)2,
for some hermitian elements k1, k2 in A, some e ∈ X with |e| ≤ 1 and all 1 ≤ j ≤ n then the inequality 2.1 holds.
Corollary 2.3. LetXbe a Hilbert module over aC∗-algebraAwith unit 1and letf : [a, b]→
X be strongly measurable such that the Lebesgue integral Rabkf(t)kdt exist and be finite. If there exist self-adjoint elements a1, a2 in A with
a21kf(t)k2 ≤Rehf(t), ei2, a2 2kf(t)k
2 ≤Imhf(t), ei2 ( a.e. t∈[a, b]),
where e∈X with |e| ≤1, then
(a21+a22)12 Z b
a
kf(t)kdt≤ k Z b
a
f(t)dtk.
Now we prove a useful lemma which is frequently applied in the next theorems.
Lemma 2.4. Let X be a Hilbert A-module and let x, y ∈X. If |hx, yi|=kxkkyk, then
y= xhx, yi
kxk2 .
Proof. For x, y ∈X we have
0≤ y−
xhx,yi kxk2
2
=hy−xkhxx,yk2i, y−
xhx,yi kxk2 i
=hy, yi − 1
kxk2hy, xihx, yi+
1
kxk4hy, xihx, xihx, yi −
1
kxk2hy, xihx, yi ≤ |y|2− 1
kxk2|hx, yi|2 =|y|2−
1
kxk2kxk2kyk2
=|y|2− kyk2 ≤0,
whence
y−
xhx,yi kxk2
= 0. Hence y=
xhx,yi
kxk2 .
Using the Cauchy-Schwarz inequality, we have the following theorem for Hilbert modules,
which is similar to [1, Theorem 2.5].
Theorem 2.5. Let e1,· · ·, em be a family of vectors in a Hilbert C∗-module X such that
hei, eji= 0 (1≤i6=j ≤m)and keik= 1 (1≤i≤m). Suppose thatrk, ρk∈R (1≤k≤m)
and that the vectors x1,· · · , xn ∈X satisfy
0≤r2kkxjk ≤Rehrkek, xji, 0≤ρ2kkxjk ≤Imhρkek, xji,
Then
" m X
k=1
(r2k+ρ2k)
#12 n X
j=1
kxjk ≤
n
X
j=1
xj
and the equality holds if and only if
n
X
j=1
xj = n
X
j=1
kxjk m
X
k=1
(rk+iρk)ek. (2.3)
Proof. There is nothing to prove if Pm
k=1(r 2
k +ρ2k) = 0. Assume that
Pm
k=1(r 2
k+ρ2k) 6= 0. From the hypothesis we have
Pm
k=1(r 2
k+ρ2k)
2 Pn
j=1kxjk
2
≤ RehPm
k=1rkek,
Pn
j=1xji+ Imh
Pm
k=1ρkek,
Pn
j=1xji
2
= RehPn
j=1xj,
Pm
k=1(rk+iρk)eki
2
by Im(a) = Re(ia∗),Re(a∗) = Re(a) (a∈A)
≤ 1 2|h
Pn j=1xj,
Pm
k=1(rk+iρk)eki| 2
+|hPm
k=1(rk+iρk)ek,
Pn
j=1xji|
2 by ()
≤ 1 2k
Pn
j=1xjk 2|Pm
k=1(rk+iρk)ek| 2
+kPm
k=1(rk+iρk)ekk2|
Pn
j=1xj|2
≤ kPn
j=1xjk2k
Pm
k=1(rk+iρk)ekk
2
by|a| ≤ kak (a∈A) =kPn
j=1xjk
2khPm
k=1(rk+iρk)ek,
Pm
k=1(rk+iρk)ekik
=kPn
j=1xjk 2Pm
k=1|rk+iρk| 2kekk2
=kPn
j=1xjk2
Pm
k=1(r 2
k+ρ
2
k).
Hence
" m X
k=1
(rk2+ρ2k)
#
( n
X
j=1
kxjk)2 ≤
n X j=1 xj 2 .
By taking square roots the desired result follows.
Clearly we have equality in 2.2 if condition 2.3 holds. To see the converse, first note that if
equality holds in 2.2, then all inequalities in the above relations should be equality. Therefore
rk2kxjk=Rehrkek, xji, ρ2kkxjk=Imhρkek, xji,
Reh n X j=1 xj, m X k=1
(rk+iρk)eki=h
n X j=1 xj, m X k=1
(rk+iρk)eki,
and also
|h
m
X
k=1
(rk+iρk)ek, n
X
j=1
xji|=k n
X
j=1
xjkk m
X
k=1
From Lemma 2.4 and these equalities we have
Pn
j=1xj =
Pm
k=1(rk+iρk)ek
kPm
k=1(rk+iρk)ekk2h Pm
k=1(rk+iρk)ek,
Pn
j=1xji
=
Pm
k=1(rk+iρk)ek
Pm
k=1(r2k+ρ2k) Reh Pm
k=1(rk+iρk)ek,
Pn
j=1xji
=
Pm
k=1(rk+iρk)ek
Pm
k=1(r2k+ρ2k) Pm
k=1
Pn
j=1(r 2
kkxjk+ρ2kkxjk) =Pn
j=1kxjk
Pm
k=1(rk+iρk)ek,
which is the desired result.
From now on we assume that X is a right Hilbert module over a C∗-algebra A, which is an algebraic left A-module subject to
hx, ayi=ahx, yi (x, y ∈X, a∈A). (†)
For example if A is a C∗-algebra and Ibe a commutative right ideal of A, thenI is a right
Hilbert module over A and
hx, ayi=x∗(ay) =ax∗y =ahx, yi (x, y ∈I, a∈A).
The next theorem is a refinement of [6, Theorem 2.1]. To prove it we need the following
lemma.
Lemma 2.6. Let X be a Hilbert A-module and e1,· · · , en ∈ X be a family of vectors such that hei, eji= 0 (i6=j) and keik= 1. If x∈X, then
|x|2 ≥
n
X
k=1
|hek, xi|2 and |x|2 ≥
n
X
k=1
|hx, eki|2.
Proof. The first result follows from the following inequality:
0≤ |x−Pn
k=1ekhek, xi|2 =hx−
Pn
k=1ekhek, xi, x−
Pn
j=1ejhej, xii =hx, xi+Pn
k=1
Pn
j=1hek, xi∗hek, ejihej, xi −2
Pn
k=1|hek, xi|2 =hx, xi+Pn
k=1hek, xi ∗he
k, ekihek, xi −2
Pn
k=1|hek, xi| 2
≤ |x|2+Pn
k=1hek, xi ∗he
k, xi −2Pnk=1|hek, xi|2 =|x|2−Pn
k=1|hek, xi| 2.
By considering |x−Pn
k=1hek, xiek|
2, similarly, we have the second one.
Theorem 2.7. Let e1,· · · , em ∈ X be a family of vectors with hei, eji= 0 (1 ≤i 6=j ≤m) and keik= 1 (1≤i≤m). If the vectors x1,· · ·, xn ∈X satisfy the conditions
0≤rkkxjk ≤Rehek, xji, 0≤ρkkxjk ≤Imhek, xji (1≤j ≤n,1≤k ≤m), (2.4)
where rk, ρk ∈[0,∞) (1≤k ≤m), then
" m X
k=1
(r2k+ρ2k)
#12 n X
j=1
kxjk ≤
n
X
j=1
xj
. (2.5)
Proof. Applying the previous lemma for x=Pn
j=1xj, we obtain
Pn
j=1xj
2
≥ 1 2
Pm
k=1|hek,
Pn
j=1xji|2+
Pm
k=1|h
Pn
j=1xj, eki|2
=Pm
k=1 1 2
hek,
Pn
j=1xji∗hek,
Pn
j=1xji+h
Pn
j=1xj, eki∗h
Pn
j=1xj, eki
=Pm
k=1(Rehek,
Pn
j=1xji)2+ (Imhek,
Pn
j=1xji)2 by ()
=Pm
k=1(Re
Pn
j=1hek, xji)2+ (Im
Pn
j=1hek, xji)2
≥Pm
k=1(rk2(
Pn
j=1kxjk)2+ρ2k(
Pn
j=1kxjk)2) by (2.4)
=Pm
k=1(r 2
k+ρ2k)(
Pn
j=1kxjk) 2.
Proposition 2.8. In Theorem 2.7, if hek, eki = 1, then the equality holds in (2.5) if and
only if
n
X
j=1
xj = ( n
X
j=1
kxjk) m
X
k=1
(rk+iρk)ek. (2.6)
Proof. If (2.6) holds, then inequality in (2.5) turns trivially into equality.
Next, assume that equality holds in (2.5). Then two inequalities in the proof of Theorem
2.7 should be equality. Hence
|
n
X
j=1
xj|2 = m
X
k=1
|hek, n
X
j=1
xji|2 and | n
X
j=1
xj|2 = m
X
k=1
|h
n
X
j=1
xj, eki|2,
which is equivalent to
n
X
j=1
xj = m
X
k=1
n
X
j=1
ekhek, xji= m
X
k=1
n
X
j=1
hek, xjiek,
and also
So n
X
j=1
xj = m
X
k=1
n
X
j=1
ekhek, xji= m
X
k=1
n
X
j=1
ek(rk+iρk)kxjk= ( n
X
j=1
kxjk) m
X
k=1
(rk+iρk)ek.
There are some versions of additive reverse of triangle inequality. In [5], S. S. Dragomir
established the following theorem:
Theorem 2.9. Let {ek}mk=1 be a family of orthonormal vectors in Hilbert space H and
Mjk ≥0 (1≤i≤n,1≤k ≤m) such that
kxjk −Rehek, xji ≤Mjk,
for each 1≤i≤n and 1≤k ≤m. Then
n
X
j=1
kxjk ≤ 1
√
mk n
X
j=1
xjk+ 1 m
n
X
j=1
m
X
k=1
Mjk;
and the equality holds if and only if
n
X
j=1
kxjk ≥ 1 m
n
X
j=1
m
X
k=1
Mjk,
and
n
X
j=1
xj = n
X
j=1
kxjk − 1 m
n
X
j=1
m
X
k=1
Mjk
! m
X
k=1
ek.
Now we extend this theorem for Hilbert modules.
Theorem 2.10. Let {ek}mk=1 be a family of vectors in Hilbert A-module Xwith|ek| ≤1 (1≤ k ≤ m) and hei, eji = 0 (1 ≤ i =6 j ≤ m) and xj ∈ X(1 ≤ i ≤ n). If for some scalars Mjk ≥0 (1≤i≤n, 1≤k ≤m),
kxjk −Rehek, xji ≤Mjk (1≤i≤n,1≤k ≤m), (2.7)
then
n
X
j=1
kxjk ≤ 1
√
mk n
X
j=1
xjk+ 1 m
n
X
j=1
m
X
k=1
Mjk. (2.8)
Moreover, if |ek|= 1 (1≤k≤m), then the equality in (2.8) holds if and only if n
X
j=1
kxjk ≥ 1 m
n
X
j=1
m
X
k=1
and
n
X
j=1
xj = n
X
j=1
kxjk − 1 m n X j=1 m X k=1 Mjk ! m X k=1
ek. (2.10)
Proof. Taking the summation in (2.7) over ifrom 1 to n, we obtain n
X
j=1
kxjk ≤Rehek, n
X
j=1
xji+ n
X
j=1
Mjk,
for each k∈ {1,· · · , m}. Summing these inequalities over k from 1 to m, we deduce n
X
j=1
kxjk ≤ 1
mReh m X k=1 ek, n X j=1
xji+ 1 m m X k=1 n X j=1
Mjk. (2.11)
Using the Cauchy–Schwarz we obtain
RehPm
k=1ek,
Pn
j=1xji
2 ≤ 1
2(|h
Pm
k=1ek,
Pn
j=1xji|
2+|hPm
k=1ek,
Pn
j=1xji
∗|2) by ()
≤ 1 2(k
Pm
k=1ekk2|
Pn
j=1xj|2+|
Pm
k=1ek|2k
Pn
j=1xjk2)
≤ kPm
k=1ekk 2kPn
j=1xjk 2
≤mkPn
j=1xjk 2, (2.12) since k m X k=1
ekk2 =kh m
X
k=1
ek, m
X
k=1
ekik=k m X k=1 m X l=1
hek, elik=k m
X
k=1
|ek|2k ≤m .
Using (2.12) in (2.11), we deduce the desired inequality.
If (2.9) and (2.10) hold, then
1 √ m Pn j=1xj
= 1 √ m Pn
j=1kxjk − 1
m
Pn j=1
Pm k=1Mjk
kPm
k=1ekk
=Pn
j=1kxjk −m1
Pn
j=1
Pm
k=1Mjk, and then the equality in 2.8 holds true.
Conversely, if the equality holds in (2.8), then obviously (2.9) is valid and we have equalities
all over in the above proof. This means that
kxjk −Rehek, xji=Mjk,
Reh
m
X
k=1
ek, n
X
j=1
xji=h m
X
k=1
ek, n
X
j=1
xji,
and
|h
m
X
k=1
ek, n
X
j=1
xji|=k m
X
k=1
ekkk n
X
j=1
Now from Lemma 2.4 and these relations, we have
Pn
j=1xj =
Pm k=1ek
kPm k=1ekk2h
Pm k=1ek,
Pn j=1xji =
Pm k=1ek
m Reh
Pm
k=1ek,
Pn
j=1xji =
Pm k=1ek
m
Pm
k=1
Pn
j=1(kxjk −Mjk)
=Pn
j=1kxjk − m1
Pn
j=1
Pm
k=1Mjk
Pm
k=1ek.
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Maryam Khosravi: Department of Mathematics, Teacher Training University, Tahran,
Iran.
E-mail address: khosravi−[email protected]
Hakimeh Mahyar: Department of Mathematics, Teacher Training University, Tahran,
Iran.
E-mail address: [email protected]
Mohammad Sal Moslehian: Department of Mathematics, Ferdowsi University, P. O. Box
1159, Mashhad 91775, Iran;
Centre of Excellence in Analysis on Algebraic Structures (CEAAS), Ferdowsi University,
Iran.
