Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 46, 2005
REVERSES OF THE CONTINUOUS TRIANGLE INEQUALITY FOR BOCHNER
INTEGRAL OF VECTOR-VALUED FUNCTIONS IN HILBERT SPACES
S.S. DRAGOMIR
S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428, MCMC 8001
VIC, AUSTRALIA .
sever@csm.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 07 June, 2004; accepted 15 April, 2005
Communicated by R.N. Mohapatra
A BSTRACT. Some reverses of the continuous triangle inequality for Bochner integral of vectorvalued functions in Hilbert spaces are given. Applications for complex-valued functions are
provided as well.
Key words and phrases: Triangle inequality, Reverse inequality, Hilbert spaces, Bochner integral.
2000 Mathematics Subject Classification. 46C05, 26D15, 26D10.
1. I NTRODUCTION
Let f : [a, b] → K, K = C or R be a Lebesgue integrable function. The following inequality,
which is the continuous version of the triangle inequality,
Z b
Z b
(1.1)
f (x) dx ≤
|f (x)| dx,
a
a
plays a fundamental role in Mathematical Analysis and its applications.
It appears, see [4, p. 492], that the first reverse inequality for (1.1) was obtained by J. Karamata in his book from 1949, [2]. It can be stated as
Z b
Z b
(1.2)
cos θ
|f (x)| dx ≤ f (x) dx ,
a
a
provided
−θ ≤ arg f (x) ≤ θ, x ∈ [a, b]
for given θ ∈ 0,
π
2
.
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
115-04
2
S.S. D RAGOMIR
This integral inequality is the continuous version of a reverse inequality for the generalised
triangle inequality
n
n
X
X
|zi | ≤ zi ,
(1.3)
cos θ
i=1
i=1
provided
a − θ ≤ arg (zi ) ≤ a + θ, for i ∈ {1, . . . , n} ,
where a ∈ R and θ ∈ 0, π2 , which, as pointed out in [4, p. 492], was first discovered by M.
Petrovich in 1917, [5], and, subsequently rediscovered by other authors, including J. Karamata
[2, p. 300 – 301], H.S. Wilf [6], and in an equivalent form, by M. Marden [3].
The first to consider the problem in the more general case of Hilbert and Banach spaces, were
J.B. Diaz and F.T. Metcalf [1] who showed that, in an inner product space H over the real or
complex number field, the following reverse of the triangle inequality holds
n
n
X X
kxi k ≤ xi ,
(1.4)
r
i=1
i=1
Re hxi , ai
,
kxi k
i ∈ {1, . . . , n} ,
provided
0≤r≤
and a ∈ H is a unit vector, i.e., kak = 1. The case of equality holds in (1.4) if and only if
!
n
n
X
X
(1.5)
xi = r
kxi k a.
i=1
i=1
The main aim of this paper is to point out some reverses of the triangle inequality for Bochner
integrable functions f with values in Hilbert spaces and defined on a compact interval [a, b] ⊂
R. Applications for Lebesgue integrable complex-valued functions are provided as well.
2. R EVERSES FOR A U NIT V ECTOR
We recall that f ∈ L ([a, b] ; H) , the space of Bochner integrable functions with values in a
Hilbert space H, if and only if f : [a, b] → H is Bochner measurable on [a, b] and the Lebesgue
Rb
integral a kf (t)k dt is finite.
The following result holds:
Theorem 2.1. If f ∈ L ([a, b] ; H) is such that there exists a constant K ≥ 1 and a vector
e ∈ H, kek = 1 with
(2.1)
kf (t)k ≤ K Re hf (t) , ei for a.e. t ∈ [a, b] ,
then we have the inequality:
Z
(2.2)
a
b
Z b
kf (t)k dt ≤ K f (t) dt
.
a
The case of equality holds in (2.2) if and only if
Z b
Z b
1
f (t) dt =
(2.3)
kf (t)k dt e.
K
a
a
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R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
3
Proof. By the Schwarz inequality in inner product spaces, we have
Z b
Z b
(2.4)
f (t) dt = f (t) dt
kek
a
a
Z b
≥
f (t) dt, e a
Z b
f (t) dt, e ≥ Re
a
Z b
Z b
≥ Re
f (t) dt, e =
Re hf (t) , ei dt.
a
a
From the condition (2.1), on integrating over [a, b] , we deduce
Z b
Z
1 b
(2.5)
Re hf (t) , ei dt ≥
kf (t)k dt,
K a
a
and thus, on making use of (2.4) and (2.5), we obtain the desired inequality (2.2).
If (2.3) holds true, then, obviously
Z b
Z b
Z b
K
f (t) dt = kek
kf (t)k dt =
kf (t)k dt,
a
a
a
showing that (2.2) holds with equality.
If we assume that the equality holds in (2.2), then by the argument provided at the beginning
of our proof, we must have equality in each of the inequalities from (2.4) and (2.5).
Observe that in Schwarz’s inequality kxk kyk ≥ Re hx, yi , x, y ∈ H, the case of equality
holds if and only if there exists aR positive scalar µ such that x = µe. Therefore, equality holds
b
in the first inequality in (2.4) iff a f (t) dt = λe, with λ ≥ 0 .
If we
R b assume that a strict
R b inequality holds in (2.1) on a subset of nonzero Lebesgue measures,
then a kf (t)k dt < K a Re hf (t) , ei dt, and by (2.4) we deduce a strict inequality in (2.2),
which contradicts the assumption. Thus, we must have kf (t)k = K Re hf (t) , ei for a.e. t ∈
[a, b] .
If we integrate this equality, we deduce
Z b
Z b
kf (t)k dt = K
Re hf (t) , ei dt
a
a
Z b
f (t) dt, e
= K Re
a
= K Re hλe, ei = λK,
giving
1
λ=
K
and thus the equality (2.3) is necessary.
This completes the proof.
Z
b
kf (t)k dt,
a
A more appropriate result from an applications point of view is perhaps the following result.
Corollary 2.2. Let e be a unit vector in the Hilbert space (H; h·, ·i) , ρ ∈ (0, 1) and f ∈
L ([a, b] ; H) so that
(2.6)
kf (t) − ek ≤ ρ for a.e. t ∈ [a, b] .
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4
S.S. D RAGOMIR
Then we have the inequality
Z b
Z b
p
2
(2.7)
1−ρ
kf (t)k dt ≤ f (t) dt
,
a
a
with equality if and only if
Z b
Z b
p
2
(2.8)
f (t) dt = 1 − ρ
kf (t)k dt · e.
a
a
Proof. From (2.6), we have
kf (t)k2 − 2 Re hf (t) , ei + 1 ≤ ρ2 ,
giving
kf (t)k2 + 1 − ρ2 ≤ 2 Re hf (t) , ei
for a.e. t ∈ [a, b]
p.
Dividing by 1 − ρ2 > 0, we deduce
p
kf (t)k2
2 Re hf (t) , ei
p
(2.9)
+ 1 − ρ2 ≤ p
1 − ρ2
1 − ρ2
for a.e. t ∈ [a, b] .
On the other hand, by the elementary inequality
p
√
+ qα ≥ 2 pq, p, q ≥ 0, α > 0
α
we have
p
kf (t)k2
(2.10)
2 kf (t)k ≤ p
+ 1 − ρ2
1 − ρ2
for each t ∈ [a, b] .
Making use of (2.9) and (2.10), we deduce
1
kf (t)k ≤ p
1 − ρ2
for a.e. t ∈ [a, b] .
Applying Theorem 2.1 for K = √ 1
1−ρ2
Re hf (t) , ei
, we deduce the desired inequality (2.7).
In the same spirit, we also have the following corollary.
Corollary 2.3. Let e be a unit vector in H and M ≥ m > 0. If f ∈ L ([a, b] ; H) is such that
(2.11)
Re hM e − f (t) , f (t) − mei ≥ 0 for a.e. t ∈ [a, b] ,
or, equivalently,
M +m 1
(2.12)
e
≤ (M − m) for a.e. t ∈ [a, b] ,
f (t) −
2
2
then we have the inequality
√
Z b
Z
2 mM b
kf (t)k dt ≤ f (t) dt
(2.13)
,
M +m a
a
or, equivalently,
√
√ 2 Z
Z b
Z b
M− m b
.
(2.14)
(0 ≤)
kf (t)k dt − f
(t)
dt
f (t) dt ≤
M +m
a
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a
a
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R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
5
The equality holds in (2.13) (or in the second part of (2.14)) if and only if
√
Z b
Z b
2 mM
(2.15)
kf (t)k dt e.
f (t) dt =
M +m
a
a
Proof. Firstly, we remark that if x, z, Z ∈ H, then the following statements are equivalent
(i) Re hZ − x, x − zi ≥ 0
1
and
≤ kZ − zk .
(ii) x − Z+z
2
2
Using this fact, we may simply realise that (2.9) and (2.10) are equivalent.
Now, from (2.9), we obtain
kf (t)k2 + mM ≤ (M + m) Re hf (t) , ei
√
for a.e. t ∈ [a, b] . Dividing this inequality with mM > 0, we deduce the following inequality
that will be used in the sequel
kf (t)k2 √
M +m
√
+ mM ≤ √
Re hf (t) , ei
mM
mM
(2.16)
for a.e. t ∈ [a, b] .
On the other hand
kf (t)k2 √
2 kf (t)k ≤ √
+ mM ,
mM
(2.17)
for any t ∈ [a, b] .
Utilising (2.16) and (2.17), we may conclude with the following inequality
M +m
kf (t)k ≤ √
Re hf (t) , ei ,
2 mM
for a.e. t ∈ [a, b] .
Applying Theorem 2.1 for the constant K :=
m+M
√
2 mM
≥ 1, we deduce the desired result.
3. R EVERSES FOR O RTHONORMAL FAMILIES OF V ECTORS
The following result for orthonormal vectors in H holds.
Theorem 3.1. Let {e1 , . . . , en } be a family of orthonormal vectors in H, ki ≥ 0, i ∈ {1, . . . , n}
and f ∈ L ([a, b] ; H) such that
ki kf (t)k ≤ Re hf (t) , ei i
(3.1)
for each i ∈ {1, . . . , n} and for a.e. t ∈ [a, b] .
Then
! 21 Z
Z b
n
b
X
2
(3.2)
ki
kf (t)k dt ≤ f (t) dt
,
i=1
a
a
where equality holds if and only if
Z b
X
Z b
n
(3.3)
f (t) dt =
kf (t)k dt
ki ei .
a
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S.S. D RAGOMIR
Rb
Proof. By Bessel’s inequality applied for a f (t) dt and the orthonormal vectors {e1 , . . . , en } ,
we have
2
Z b
2
n Z b
X
(3.4)
f
(t)
dt
≥
f
(t)
dt,
e
i
a
≥
=
a
i=1
n X
Re
f (t) dt, ei
i=1
n Z b
X
i=1
2
b
Z
a
2
Re hf (t) , ei i dt .
a
Integrating (3.1), we get for each i ∈ {1, . . . , n}
Z b
Z b
0 ≤ ki
kf (t)k dt ≤
Re hf (t) , ei i dt,
a
a
implying
(3.5)
n Z
X
i=1
b
2
2 X
Z b
n
2
ki
kf (t)k dt .
Re hf (t) , ei i dt ≥
a
a
i=1
On making use of (3.4) and (3.5), we deduce
Z b
2
2
Z b
n
X
2
f (t) dt
ki
kf (t)k dt ,
≥
a
a
i=1
which is clearly equivalent to (3.2).
If (3.3) holds true, then
Z b
Z b
n
X
kf (t)k dt ki ei f (t) dt =
a
a
i=1
# 21
"X
Z b
n
kf (t)k dt
ki2 kei k2
=
a
n
X
=
i=1
ki2
! 12 Z
b
kf (t)k dt,
a
i=1
showing that (3.2) holds with equality.
Now, suppose that there is an i0 ∈ {1, . . . , n} for which
ki0 kf (t)k < Re hf (t) , ei0 i
on a subset of nonzero Lebesgue measures enclosed in [a, b]. Then obviously
Z b
Z b
ki0
kf (t)k dt <
Re hf (t) , ei0 i dt,
a
a
and using the argument given above, we deduce
! 21 Z
Z b
n
b
X
2
.
ki
kf (t)k dt < f
(t)
dt
a
i=1
a
Therefore, if the equality holds in (3.2), we must have
(3.6)
ki kf (t)k = Re hf (t) , ei i
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R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
7
for each i ∈ {1, . . . , n} and a.e. t ∈ [a, b] .
Also, if the equality holds in (3.2), then we must have equality in all inequalities (3.4), this
means that
Z b
n Z b
X
(3.7)
f (t) dt =
f (t) dt, ei ei
a
a
i=1
and
b
Z
(3.8)
Im
= 0 for each i ∈ {1, . . . , n} .
f (t) dt, ei
a
Using (3.6) and (3.8) in (3.7), we deduce
Z b
Z b
n
X
f (t) dt =
Re
f (t) dt, ei ei
a
=
i=1
n Z b
X
i=1
=
a
Re hf (t) , ei i ei dt
a
n Z
X
kf (t)k dt ki ei
a
i=1
Z
b
b
kf (t)k dt
=
a
n
X
ki ei ,
i=1
and the condition (3.3) is necessary.
This completes the proof.
The following two corollaries are of interest.
Corollary 3.2. Let {e1 , . . . , en } be a family of orthonormal vectors in H, ρi ∈ (0, 1) , i ∈
{1, . . . , n} and f ∈ L ([a, b] ; H) such that:
(3.9)
kf (t) − ei k ≤ ρi for i ∈ {1, . . . , n} and a.e. t ∈ [a, b] .
Then we have the inequality
n−
n
X
ρ2i
! 12 Z
a
i=1
b
Z b
,
kf (t)k dt ≤ f
(t)
dt
a
with equality if and only if
!
Z b
Z b
n
X
1
f (t) dt =
kf (t)k dt
1 − ρ2i 2 ei .
a
a
i=1
Proof. From the proof of Theorem 2.1, we know that (3.3) implies the inequality
q
1 − ρ2i kf (t)k ≤ Re hf (t) , ei i , i ∈ {1, . . . , n} , for a.e. t ∈ [a, b] .
p
Now, applying Theorem 3.1 for ki := 1 − ρ2i , i ∈ {1, . . . , n}, we deduce the desired result.
Corollary 3.3. Let {e1 , . . . , en } be a family of orthonormal vectors in H, Mi ≥ mi > 0,
i ∈ {1, . . . , n} and f ∈ L ([a, b] ; H) such that
(3.10)
Re hMi ei − f (t) , f (t) − mi ei i ≥ 0
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8
S.S. D RAGOMIR
or, equivalently,
1
M
+
m
i
i
f (t) −
≤ (Mi − mi )
e
i
2
2
for i ∈ {1, . . . , n} and a.e. t ∈ [a, b] . Then we have the reverse of the generalised triangle
inequality
" n
# 12 Z
Z b
b
X 4mi Mi
,
kf (t)k dt ≤ f
(t)
dt
2
(m + M )
i=1
i
a
i
with equality if and only if
Z b
Z
f (t) dt =
a
a
b
a
!
√
n
X
2 m i Mi
ei .
kf (t)k dt
m i + Mi
i=1
Proof. From the proof of Corollary 2.3, we know (3.10) implies that
√
2 m i Mi
kf (t)k ≤ Re hf (t) , ei i , i ∈ {1, . . . , n} and a.e. t ∈ [a, b] .
m i + Mi
Now, applying Theorem 3.1 for ki :=
√
2 mi Mi
,
mi +Mi
i ∈ {1, . . . , n} , we deduce the desired result.
4. A PPLICATIONS FOR C OMPLEX -VALUED F UNCTIONS
Let e = α+iβ (α, β ∈ R) be a complex number with the property that |e| = 1, i.e., α2 +β 2 =
1.
The following proposition holds.
Proposition 4.1. If f : [a, b] → C is a Lebesgue integrable function with the property that there
exists a constant K ≥ 1 such that
(4.1)
|f (t)| ≤ K [α Re f (t) + β Im f (t)]
for a.e. t ∈ [a, b] , where α, β ∈ R, α2 + β 2 = 1 are given, then we have the following reverse
of the continuous triangle inequality:
Z b
Z b
(4.2)
|f (t)| dt ≤ K f (t) dt .
a
a
The case of equality holds in (2.2) if and only if
Z b
Z b
1
f (t) dt =
(α + iβ)
|f (t)| dt.
K
a
a
The proof is obvious by Theorem 2.1, and we omit the details.
Remark 4.2. If in the above Proposition 4.1 we choose α = 1, β = 0, then the condition (4.1)
for Re f (t) > 0 is equivalent to
[Re f (t)]2 + [Im f (t)]2 ≤ K 2 [Re f (t)]2
or with the inequality:
|Im f (t)| √ 2
≤ K − 1.
Re f (t)
Now, if we assume that
(4.3)
|arg f (t)| ≤ θ,
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,
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R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
9
then, for Re f (t) > 0,
|tan [arg f (t)]| =
and if we choose K =
1
cos θ
|Im f (t)|
≤ tan θ,
Re f (t)
> 1, then
√
K 2 − 1 = tan θ,
and by Proposition 4.1, we deduce
b
Z
cos θ
(4.4)
a
Z b
|f (t)| dt ≤ f (t) dt ,
a
which is exactly the Karamata inequality (1.2) from the Introduction.
Obviously, the result from Proposition 4.1 is more comprehensive since for other values of
(α, β) ∈ R2 with α2 + β 2 = 1 we can get different sufficient conditions for the function f such
that the inequality (4.2) holds true.
A different sufficient condition in terms of complex disks is incorporated in the following
proposition.
Proposition 4.3. Let e = α + iβ with α2 + β 2 = 1, r ∈ (0, 1) and f : [a, b] → C be a Lebesgue
integrable function such that
f (t) ∈ D̄ (e, r) := {z ∈ C| |z − e| ≤ r} for a.e. t ∈ [a, b] .
(4.5)
Then we have the inequality
√
(4.6)
Z
1 − r2
a
b
Z b
|f (t)| dt ≤ f (t) dt .
a
The case of equality holds in (4.6) if and only if
Z b
Z b
√
2
f (t) dt = 1 − r (α + iβ)
|f (t)| dt.
a
a
The proof follows by Corollary 2.2 and we omit the details.
Finally, we may state the following proposition as well.
Proposition 4.4. Let e = α + iβ with α2 + β 2 = 1 and M ≥ m > 0. If f : [a, b] → C is such
that
h
i
(4.7)
Re (M e − f (t)) f (t) − me ≥ 0 for a.e. t ∈ [a, b] ,
or, equivalently,
1
M
+
m
f (t) −
e ≤ (M − m) for a.e. t ∈ [a, b] ,
2
2
(4.8)
then we have the inequality
(4.9)
√
Z b
Z
2 mM b
|f (t)| dt ≤ f (t) dt ,
M +m a
a
or, equivalently,
√
√ 2
Z
b
M− m
b
(0 ≤)
|f (t)| dt − f (t) dt ≤
M +m
a
a
Z
(4.10)
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Z b
.
f
(t)
dt
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S.S. D RAGOMIR
The equality holds in (4.9) (or in the second part of (4.10)) if and only if
√
Z b
Z b
2 mM
f (t) dt =
(α + iβ)
|f (t)| dt.
M +m
a
a
The proof follows by Corollary 2.3 and we omit the details.
Remark 4.5. Since
M e − f (t) = M α − Re f (t) + i [M β − Im f (t)] ,
f (t) − me = Re f (t) − mα − i [Im f (t) − mβ]
hence
(4.11)
h
i
Re (M e − f (t)) f (t) − me
= [M α − Re f (t)] [Re f (t) − mα] + [M β − Im f (t)] [Im f (t) − mβ] .
It is obvious that, if
mα ≤ Re f (t) ≤ M α for a.e. t ∈ [a, b] ,
(4.12)
and
mβ ≤ Im f (t) ≤ M β for a.e. t ∈ [a, b] ,
(4.13)
then, by (4.11),
h
i
Re (M e − f (t)) f (t) − me ≥ 0 for a.e. t ∈ [a, b] ,
and then either (4.9) or (4.12) hold true.
We observe that the conditions (4.12) and (4.13) are very easy to verify in practice and may be
useful in various applications where reverses of the continuous triangle inequality are required.
Remark 4.6. Similar results may be stated for functions f : [a, b] → Rn or f : [a, b] → H,
with H particular instances of Hilbert spaces of significance in applications, but we leave them
to the interested reader.
R EFERENCES
[1] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces,
Proceedings Amer. Math. Soc., 17(1) (1966), 88–97.
[2] J. KARAMATA, Teorija i Praksa Stieltjesova Integrala (Serbo-Croatian) (Stieltjes Integral, Theory
and Practice), SANU, Posebna izdanja, 154, Beograd, 1949.
[3] M. MARDEN, The Geometry of the Zeros of a Polynomial in a Complex Variable, Amer. Math. Soc.
Math. Surveys, 3, New York, 1949.
[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis,
Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
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