The Case of m Functionals. The following result may be stated [8]:

and by (11.13) and (11.14) we deduce (11.12).

Remark 11.4. IfX =H,(H;h·,·i)is a Hilbert space, then from Corollary 11.3 we deduce the additive reverse inequality obtained in [7]. For further similar results in Hilbert spaces, see [7]

and [9].

k=1

F_k Z b

f(t)dt

Z b a

f(t)dt

k=1

F_k

and m

k=1

F_k Z b

f(t)dt

= 0.

These imply that (11.17) and (11.18) hold true, and the theorem is completely proved.

Remark 11.6. If F_k, k ∈ {1, . . . , m} are of unit norm, then, from (11.16) we deduce the inequality

(11.21)

Z b a

kf(t)kdt ≤

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt,

which is obviously coarser than (11.16) but, perhaps more useful for applications.

The following new result may be stated as well:

Theorem 11.7. Let (X,k·k) be a Banach space over the real or complex number fieldKand F_k : X → K, k ∈ {1, . . . , m} continuous linear functionals on X. Assume also that f : [a, b]→Xis a Bochner integrable function on[a, b]andM_k : [a, b]→[0,∞), k ∈ {1, . . . , m}

are Lebesgue integrable functions such that

(11.22) kf(t)k −ReF_k[f(t)]≤M_k(t) for eachk ∈ {1, . . . , m}and a.e. t∈[a, b].

(i) Ifc_∞is defined by (c_∞), then we have the inequality (11.23)

Z b a

kf(t)kdt ≤c_∞

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt.

(ii) Ifc_p, p≥1,is defined by (c_p) , then we have the inequality Z b

kf(t)kdt≤ c_p m^1/p

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt.

The proof is similar to the ones from Theorem 7.1 and 11.5 and we omit the details.

The case of Hilbert spaces for Theorem 11.5, in which one may provide a simpler condition for equality, is of interest in applications [8].

Theorem 11.8 (Dragomir, 2004). Let (H,h·,·i) be a Hilbert space over the real or complex number fieldKande_k∈H, k ∈ {1, . . . , m}.Iff : [a, b]→His a Bochner integrable function on[a, b], f(t) 6= 0for a.e. t ∈[a, b]andM_k : [a, b]→ [0,∞), k ∈ {1, . . . , m}is a Lebesgue integrable function such that

(11.24) kf(t)k −Rehf(t), eki ≤Mk(t) for eachk ∈ {1, . . . , m}and for a.e. t∈[a, b],then

(11.25)

Z b a

kf(t)kdt≤

1 m

k=1

e_k

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt.

The case of equality holds in (11.25) if and only if (11.26)

Z b a

kf(t)kdt ≥ 1 m

k=1

Z b a

M_k(t)dt

and (11.27)

Z b a

f(t)dt=

a kf(t)kdt−_m¹ Pm k=1

aM_k(t)dt kPm

k=1e_kk²

k=1

ek. Proof. As in the proof of Theorem 11.5, we have

(11.28)

Z b a

kf(t)kdt ≤Re

*1 m

k=1

ek, Z b

f(t)dt +

+ 1 m

k=1

Z b a

Mk(t)dt andPm

k=1e_k 6= 0.

On utilising Schwarz’s inequality in Hilbert space(H,h·,·i)forRb

a f(t)dt andPm

k=1e_k,we have

Z b a

f(t)dt

k=1

e_k

≥

*Z b

f(t)dt,

k=1

e_k +

(11.29)

≥

*Z b

f(t)dt,

k=1

e_k +

≥Re

*Z b

f(t)dt,

k=1

e_k +

. By (11.28) and (11.29), we deduce (11.25).

Taking the norm on (11.27) and using (11.26), we have

Z b a

f(t)dt

= m

akf(t)kdt− _m¹ Pm k=1

a Mk(t)dt kPm

k=1ekk ,

showing that the equality holds in (11.25).

Conversely, if the equality case holds in (11.25), then it must hold in all the inequalities used to prove (11.25). Therefore we have

(11.30) kf(t)k= Rehf(t), e_ki+M_k(t) for eachk∈ {1, . . . , m}and for a.e.t ∈[a, b],

(11.31)

Z b a

f(t)dt

k=1

e_k

*Z b a

f(t)dt,

k=1

e_k +

and

(11.32) Im

*Z b a

f(t)dt,

k=1

e_k +

= 0.

From (11.30) on integrating on[a, b]and summing overk,we get

(11.33) Re

*Z b a

f(t)dt,

k=1

e_k +

=m Z b

kf(t)kdt−

k=1

Z b a

M_k(t)dt.

On the other hand, by the use of the identity (3.22), the relation (11.31) holds if and only if Z b

f(t)dt = DRb

a f(t)dt,Pm k=1e_kE kPm

k=1e_kk²

k=1

e_k, giving, from (11.32) and (11.33), that (11.27) holds true.

If the equality holds in (11.25), then obviously (11.26) is valid and the theorem is proved.

Remark 11.9. If in the above theorem, the vectors{e_k}k∈{1,...,m}are assumed to be orthogonal, then (11.25) becomes

(11.34)

Z b a

kf(t)kdt ≤ 1 m

k=1

ke_kk²

!¹₂

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt.

Moreover, if{e_k}k∈{1,...,m}is an orthonormal family, then (11.34) becomes (11.35)

Z b a

kf(t)kdt ≤ 1

√m

Z b a

f(t)dt

+ 1 m

k=1

Z b a

M_k(t)dt which has been obtained in [4].

The following corollaries are of interest.

Corollary 11.10. Let (H;h·,·i), e_k, k ∈ {1, . . . , m} and f be as in Theorem 11.8. If r_k : [a, b]→[0,∞), k ∈ {1, . . . , m}are such thatr_k ∈L²[a, b], k ∈ {1, . . . , m}and

(11.36) kf(t)−e_kk ≤r_k(t),

for eachk ∈ {1, . . . , m}and a.e. t∈[a, b], then

(11.37)

Z b a

kf(t)kdt≤

1 m

k=1

e_k

Z b a

f(t)dt

+ 1 2m

k=1

Z b a

r_k²(t)dt.

The case of equality holds in (11.37) if and only if Z b

kf(t)kdt ≥ 1 2m

k=1

Z b a

r_k²(t)dt

and

Z b a

f(t)dt=

a kf(t)kdt− _2m¹ Pm k=1

a r²_k(t)dt kPm

k=1ekk²

k=1

e_k. Finally, the following corollary may be stated.

Corollary 11.11. Let(H;h·,·i),e_k, k ∈ {1, . . . , m}andf be as in Theorem 11.8. IfM_k, µ_k : [a, b]→Rare such thatM_k≥µ_k >0a.e. on[a, b], ^(M_M^k^−µ^k⁾²

k+µk ∈L[a, b]and RehM_k(t)e_k−f(t), f(t)−µ_k(t)e_ki ≥0 for eachk ∈ {1, . . . , m}and for a.e. t∈[a, b],then

Z b a

kf(t)kdt ≤

1 m

k=1

e_k

Z b a

f(t)dt

+ 1 4m

k=1

ke_kk² Z b

[M_k(t)−µ_k(t)]² M_k(t) +µ_k(t) dt.

12. APPLICATIONS FORCOMPLEX-VALUEDFUNCTIONS

We now give some examples of inequalities for complex-valued functions that are Lebesgue integrable on using the general result obtained in Section 10.

Consider the Banach space (C,|·|₁)over the real field RandF : C→C, F (z) = ez with e=α+iβand|e|² =α²+β² = 1, thenF is linear onC. Forz 6= 0,we have

|F (z)|

|z|₁ = |e| |z|

|z|₁ = q

|Rez|²+|Imz|²

|Rez|+|Imz| ≤1.

Since, forz₀ = 1,we have|F (z₀)|= 1and|z₀|₁ = 1,hence kFk₁ := sup

z6=0

|F (z)|

|z|₁ = 1, showing thatF is a bounded linear functional on(C,|·|₁).

Therefore we can apply Theorem 10.1 to state the following result for complex-valued func-tions.

Proposition 12.1. Letα, β ∈ R withα² +β² = 1, f : [a, b] → Cbe a Lebesgue integrable function on[a, b]andr≥0such that

(12.1) r[|Ref(t)|+|Imf(t)|]≤αRef(t)−βImf(t) for a.e. t∈[a, b].Then

(12.2) r Z b

|Ref(t)|dt+ Z b

|Imf(t)|dt

≤

Z b a

Ref(t)dt

Z b a

Imf(t)dt . The equality holds in (12.2) if and only if both

α Z b

Ref(t)dt−β Z b

Imf(t)dt=r Z b

|Ref(t)|dt+ Z b

|Imf(t)|dt

and

α Z b

Ref(t)dt−β Z b

Imf(t)dt =

Z b a

Ref(t)dt

Z b a

Imf(t)dt . Now, consider the Banach space(C,|·|_∞).IfF (z) = dzwithd=γ+iδand|d|=

√ 2 2 ,i.e., γ²+δ² = ¹₂,thenF is linear onC. Forz 6= 0we have

|F (z)|

|z|_∞ = |d| |z|

|z|_∞ =

√2

2 · q

|Rez|²+|Imz|² max{|Rez|,|Imz|} ≤1.

Since, forz₀ = 1 +i,we have|F (z₀)|= 1,|z₀|_∞= 1,hence kFk_∞ := sup

z6=0

|F (z)|

|z|_∞ = 1,

showing thatF is a bounded linear functional of unit norm on(C,|·|_∞).

Therefore, we can apply Theorem 10.1, to state the following result for complex-valued functions.

Proposition 12.2. Let γ, δ ∈ R withγ² +δ² = ¹₂, f : [a, b] → C be a Lebesgue integrable function on[a, b]andr≥0such that

rmax{|Ref(t)|,|Imf(t)|} ≤γRef(t)−δImf(t) for a.e. t∈[a, b].Then

(12.3) r Z b

max{|Ref(t)|,|Imf(t)|}dt≤max

Z b a

Ref(t)dt ,

Z b a

Imf(t)dt

. The equality holds in (12.3) if and only if both

γ Z b

Ref(t)dt−δ Z b

Imf(t)dt=r Z b

max{|Ref(t)|,|Imf(t)|}dt

and γ

Z b a

Ref(t)dt−δ Z b

Imf(t)dt = max

Z b a

Ref(t)dt ,

Z b a

Imf(t)dt

. Now, consider the Banach space

C,|·|_2p

with p ≥ 1.Let F : C→C, F(z) = cz with

|c|= 2^2p¹⁻¹² (p≥1).Obviously,F is linear and by Hölder’s inequality

|F (z)|

|z|_2p = 2^2p¹⁻¹² q

|Rez|²+|Imz|²

|Rez|^2p+|Imz|^2p_2p¹

≤1.

Since, forz0 = 1 +iwe have|F (z0)|= 2¹^p,|z0|_2p = 2^2p¹ (p≥1),hence kFk_2p := sup

z6=0

|F (z)|

|z|_2p = 1, showing thatF is a bounded linear functional of unit norm on

C,|·|_2p

,(p≥1). Therefore on using Theorem 10.1, we may state the following result.

Proposition 12.3. Letϕ, φ∈Rwithϕ²+φ² = 2^2p¹ ⁻¹² (p≥1), f : [a, b]→ Cbe a Lebesgue integrable function on[a, b]andr≥0such that

|Ref(t)|^2p+|Imf(t)|^2p_2p¹

≤ϕRef(t)−φImf(t) for a.e. t∈[a, b],then

(12.4) r Z b

|Ref(t)|^2p+|Imf(t)|^2p_2p¹ dt

≤

Z b a

Ref(t)dt

Z b a

Imf(t)dt

2p#_2p¹

, (p≥1)

where equality holds in (12.4) if and only if both ϕ

Z b a

Ref(t)dt−φ Z b

Imf(t)dt =r Z b

|Ref(t)|^2p+|Imf(t)|^2p_2p¹ dt and

ϕ Z b

Ref(t)dt−φ Z b

Imf(t)dt =

Z b a

Ref(t)dt

Z b a

Imf(t)dt

2p#_2p¹ . Remark 12.4. Ifp= 1above, and

r|f(t)| ≤ϕRef(t)−ψImf(t) for a.e. t∈[a, b],

providedϕ,ψ ∈Randϕ²+ψ² = 1, r≥ 0,then we have a reverse of the classical continuous triangle inequality for modulus:

r Z b

|f(t)|dt≤

Z b a

f(t)dt ,

with equality iff ϕ

Z b a

Ref(t)dt−ψ Z b

Imf(t)dt =r Z b

|f(t)|dt

and

ϕ Z b

Ref(t)dt−ψ Z b

Imf(t)dt =

Z b a

f(t)dt .

If we apply Theorem 11.1, then, in a similar manner we can prove the following result for complex-valued functions.

Proposition 12.5. Let α, β ∈ R with α² +β² = 1, f, k : [a, b] → C Lebesgue integrable functions such that

|Ref(t)|+|Imf(t)| ≤αRef(t)−βImf(t) +k(t) for a.e. t∈[a, b].Then

(0≤) Z b

|Ref(t)|dt+ Z b

|Imf(t)|dt−

Z b a

Ref(t)dt

Z b a

Imf(t)dt

≤ Z b

k(t)dt.

Applying Theorem 11.1, for(C,|·|_∞)we may state:

Proposition 12.6. Let γ, δ ∈ R with γ² + δ² = ¹₂, f, k : [a, b] → C Lebesgue integrable functions on[a, b]such that

max{|Ref(t)|,|Imf(t)|} ≤γRef(t)−δImf(t) +k(t) for a.e. t∈[a, b].Then

(0≤) Z b

max{|Ref(t)|,|Imf(t)|}dt−max

Z b a

Ref(t)dt ,

Z b a

Imf(t)dt

≤ Z b

k(t)dt.

Finally, utilising Theorem 11.1, for

C,|·|_2p

withp≥1,we may state that:

Proposition 12.7. Letϕ, φ∈Rwithϕ²+φ² = 2^2p¹⁻¹² (p≥1), f, k : [a, b]→Cbe Lebesgue integrable functions such that

|Ref(t)|^2p+|Imf(t)|^2p_2p¹

≤ϕRef(t)−φImf(t) +k(t) for a.e. t∈[a, b].Then

(0≤) Z b

|Ref(t)|^2p +|Imf(t)|^2p_2p¹ dt

−

Z b a

Ref(t)dt

Z b a

Imf(t)dt

2p#_2p¹

≤ Z b

k(t)dt.

Remark 12.8. Ifp= 1in the above proposition, then, from

|f(t)| ≤ϕRef(t)−ψImf(t) +k(t) for a.e. t∈[a, b],

provided ϕ, ψ ∈ Rand ϕ² +ψ² = 1,we have the additive reverse of the classical continuous triangle inequality

(0≤) Z b

|f(t)|dt−

Z b a

f(t)dt

≤ Z b

k(t)dt.

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In document (1)http://jipam.vu.edu.au/ Volume 6, Issue 5, Article 129, 2005 REVERSES OF THE TRIANGLE INEQUALITY IN BANACH SPACES S.S (Pldal 38-46)