The Case of Inner Product Spaces

The case of inner product spaces, in which we may provide a simpler condition of equality, is of interest in applications [6].

Theorem 6.4 (Dragomir, 2004). Let (X,k·k)be an inner product space over the real or complex number field K, e_k, x_i ∈ H\ {0}, k ∈ {1, . . . , m}, i ∈ {1, . . . , n}.IfM_ik ≥0fori∈ {1, . . . , n},{1, . . . , n}such that

(6.18) kxik −Rehxi, eki ≤Mik

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for each i∈ {1, . . . , n}, k ∈ {1, . . . , m},then we have the inequality

(6.19)

i=1

kx_ik ≤

1 m

k=1

e_k

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik.

The case of equality holds in (6.19) if and only if (6.20)

i=1

kx_ik ≥ 1 m

k=1 n

i=1

M_ik

and (6.21)

i=1

x_i = m Pn

i=1kx_ik −_m¹ Pm k=1

i=1M_ik kPm

k=1ekk²

k=1

e_k.

Proof. As in the proof of Theorem6.3, we have

(6.22)

i=1

kx_ik ≤Re

* 1 m

k=1

e_k,

i=1

x_i +

+ 1 m

k=1 n

i=1

M_ik,

andPm

k=1e_k6= 0.

On utilising the Schwarz inequality in the inner product space(H;h·,·i)for

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i=1x_i,Pm

k=1e_k,we have

i=1

x_i

k=1

e_k

≥

* _n X

i=1

x_i,

k=1

e_k +

(6.23)

≥

* _n X

i=1

x_i,

k=1

e_k +

≥Re

* _n X

i=1

x_i,

k=1

e_k +

By (6.22) and (6.23) we deduce (6.19).

Taking the norm in (6.21) and using (6.20), we have

i=1

x_i

= m Pn

i=1kx_ik −_m¹ Pm k=1

i=1M_ik kPm

k=1e_kk ,

showing that the equality holds in (6.19).

Conversely, if the case of equality holds in (6.19), then it must hold in all the inequalities used to prove (6.19). Therefore we have

(6.24) kx_ik= Rehx_i, e_ki+M_ik for each i∈ {1, . . . , n}, k ∈ {1, . . . , m},

(6.25)

i=1

x_i

k=1

e_k

* _n X

i=1

x_i,

k=1

e_k +

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and

(6.26) Im

* _n X

i=1

x_i,

k=1

e_k +

= 0.

From (6.24), on summing overiandk,we get

(6.27) Re

* _n X

i=1

x_i,

k=1

e_k +

i=1

kx_ik −

k=1 n

i=1

M_ik.

On the other hand, by the use of the identity (3.22), the relation (6.25) holds if and only if

i=1

x_i = hPn

i=1xi,Pm k=1eki kPm

k=1e_kk²

k=1

e_k,

giving, from (6.26) and (6.27), that

i=1

x_i = mPn

i=1kx_ik −Pm k=1

Pn i=1M_ik kPm

k=1e_kk²

k=1

e_k.

If the inequality holds in (6.19), then obviously (6.20) is valid, and the theorem is proved.

Remark 5. If in the above theorem the vectors {e_k}_k=1,m are assumed to be orthogonal, then (6.19) becomes:

(6.28)

i=1

kx_ik ≤ 1 m

k=1

ke_kk²

!¹₂

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik.

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Moreover, if{e_k}_k=1,mis an orthonormal family, then (6.28) becomes

(6.29)

i=1

kx_ik ≤

√m m

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik,

which has been obtained in [12].

Before we provide some natural consequences of Theorem 6.4, we need some preliminary results concerning another reverse of Schwarz’s inequality in inner product spaces (see for instance [4, p. 27]).

Lemma 6.5 (Dragomir, 2004). Let(X,k·k)be an inner product space over the real or complex number field Kandx, a ∈ H, r > 0.Ifkx−ak ≤r, then we have the inequality

(6.30) kxk kak −Rehx, ai ≤ 1

2r². The case of equality holds in (6.30) if and only if

(6.31) kx−ak=r and kxk=kak.

Proof. The conditionkx−ak ≤ris clearly equivalent to (6.32) kxk²+kak² ≤2 Rehx, ai+r². Since

(6.33) 2kxk kak ≤ kxk²+kak²,

with equality if and only if kxk =kak,hence by (6.32) and (6.33) we deduce (6.30).

The case of equality is obvious.

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Utilising the above lemma we may state the following corollary of Theorem 6.4[6].

Corollary 6.6. Let (H;h·,·i), e_k, x_i be as in Theorem 6.4. If r_ik > 0, i ∈ {1, . . . , n}, k ∈ {1, . . . , m}such that

(6.34) kx_i−e_kk ≤r_ik for each i∈ {1, . . . , n} andk ∈ {1, . . . , m}, then we have the inequality

(6.35)

i=1

kx_ik ≤

1 m

k=1

e_k

i=1

x_i

+ 1 2m

k=1 n

i=1

r_ik².

The equality holds in (6.35) if and only if

i=1

kx_ik ≥ 1 2m

k=1 n

i=1

r²_ik

and n

i=1

x_i = m Pn

i=1kxik − _2m¹ Pm k=1

Pn i=1r_ik² kPm

k=1e_kk²

k=1

e_k.

The following lemma may provide another sufficient condition for (6.18) to hold (see also [4, p. 28]).

Lemma 6.7 (Dragomir, 2004). Let (H;h·,·i)be an inner product space over the real or complex number fieldKandx, y ∈H, M ≥m >0.If either

(6.36) RehM y−x, x−myi ≥0

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or, equivalently, (6.37)

x− m+M 2 y

≤ 1

2(M −m)kyk, holds, then

(6.38) kxk kyk −Rehx, yi ≤ 1

4· (M −m)² m+M kyk².

The case of equality holds in (6.38) if and only if the equality case is realised in (6.36) and

kxk= M +m 2 kyk.

The proof is obvious by Lemma6.5fora= ^M^+m₂ yandr= ¹₂ (M −m)kyk. Finally, the following corollary of Theorem6.4may be stated [6].

Corollary 6.8. Assume that (H,h·,·i), e_k, x_i are as in Theorem6.4. IfM_ik ≥ m_ik >0satisfy the condition

RehM_ke_k−x_i, x_i−µ_ke_ki ≥0 for each i∈ {1, . . . , n}andk ∈ {1, . . . , m},then

i=1

kx_ik ≤

1 m

k=1

e_k

i=1

x_i

+ 1 4m

k=1 n

i=1

(M_ik−m_ik)² Mik+mik

ke_kk².

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7. Other Additive Reverses for m Functionals

A different approach in obtaining other additive reverses for the generalised triangle inequality is incorporated in the following new result:

Theorem 7.1. Let(X,k·k)be a normed linear space over the real or complex number field K. AssumeFk, k ∈ {1, . . . , m},are bounded linear functionals on the normed linear space X and x_i ∈ X, i ∈ {1, . . . , n}, M_ik ≥ 0, i ∈ {1, . . . , n}, k ∈ {1, . . . , m}are such that

(7.1) kx_ik −ReF_k(x_i)≤M_ik for each i∈ {1, . . . , n} and k∈ {1, . . . , m}.

(i) Ifc∞is defined by (c∞), then we have the inequality

(7.2)

i=1

kx_ik ≤c∞

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik.

(ii) Ifc_pis defined by (c_p) forp≥1,then we have the inequality:

(7.3)

i=1

kx_ik ≤ 1 m¹^p

c_p

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik.

Proof. (i) Since

1≤k≤mmax kF_k(x)k ≤c∞kxk for anyx∈X,

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then we have

k=1

F_k

i=1

x_i

≤m max

1≤k≤m

F_k

i=1

x_i

≤mc_∞

i=1

x_i .

Using (6.16), we may state that 1

k=1

ReFk n

i=1

≤c∞

i=1

which, together with (6.15) imply the desired inequality (7.2).

(ii) Using the fact that, obviously

k=1

|F_k(x)|^p

!¹_p

≤c_pkxk for anyx∈X, then, by Hölder’s inequality forp >1,¹_p +¹_q = 1,we have

k=1

Fk n

i=1

≤m¹^q

k=1

Fk n

i=1

p!¹_p

≤c_pm¹^q

i=1

x_i ,

which, combined with (6.15) and (6.16) will give the desired inequality (7.3).

The casep= 1goes likewise and we omit the details.

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Remark 6. Since, obviouslyc∞≤ max

1≤k≤mkF_kk,then from (7.2) we have

(7.4)

i=1

kx_ik ≤ max

1≤k≤m{kF_kk} ·

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik.

Finally, sincec_p ≤(Pm

k=1kF_kk^p)¹^p, p≥1,hence by (7.3) we have

(7.5)

i=1

kxik ≤ Pm

k=1kF_kk^p m

_p¹

i=1

+ 1 m

k=1 n

i=1

Mik.

The following corollary for semi-inner products may be stated as well.

Corollary 7.2. Let(X,k·k)be a real or complex normed space and[·,·] :X× X →Ka semi-inner product generating the normk·k.Assumee_k, x_i ∈H and Mik ≥0, i∈ {1, . . . , n}, k ∈ {1, . . . , m}are such that

(7.6) kx_ik −Re [x_i, e_k]≤M_ik, for anyi∈ {1, . . . , n}, k ∈ {1, . . . , m}.

(i) If

d∞ := sup

x6=0

max1≤k≤n|[x, e_k]|

kxk ≤ max

1≤k≤nke_kk

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then

i=1

kx_ik ≤d∞

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik (7.7)

≤ max

1≤k≤nke_kk ·

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik

;

(ii) If

d_p := sup

x6=0

k=1|[x, e_k]|^p kxk^p

¹_p



≤

k=1

ke_kk^p

!¹_p

,

wherep≥1,then

i=1

kx_ik ≤ 1 m^p¹

d_p

i=1

x_i

+ 1 m

k=1 n

i=1

M_ik (7.8)

≤ Pm

k=1ke_kk^p m

¹_p

i=1

+ 1 m

k=1 n

i=1

Mik

! .

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8. Applications for Complex Numbers

LetCbe the field of complex numbers. Ifz = Rez+iImz,then by|·|_p :C→ [0,∞), p∈[1,∞]we define thep−modulus ofz as

|z|_p :=







max{|Rez|,|Imz|} if p=∞, (|Rez|^p+|Imz|^p)¹^p if p∈[1,∞), where|a|, a∈Ris the usual modulus of the real numbera.

Forp= 2,we recapture the usual modulus of a complex number, i.e.,

|z|₂ = q

|Rez|²+|Imz|² =|z|, z ∈C.

It is well known that

C,|·|_p

, p ∈ [1,∞] is a Banach space over the real number field R.

Consider the Banach space(C,|·|₁)andF :C→C,F (z) =azwitha∈C, a 6= 0.Obviously,F is linear onC. Forz 6= 0,we have

|F (z)|

|z|₁ = |a| |z|

|z|₁ = |a|

|Rez|²+|Imz|²

|Rez|+|Imz| ≤ |a|. Since, forz₀ = 1,we have|F (z₀)|=|a|and|z₀|₁ = 1,hence

kFk₁ := sup

z6=0

|F (z)|

|z|₁ =|a|,

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showing thatF is a bounded linear functional on(C,|·|₁)andkFk₁ =|a|. We can apply Theorem3.1 to state the following reverse of the generalised triangle inequality for complex numbers [5].

Proposition 8.1. Leta_k, x_j ∈ C,k ∈ {1, . . . , m}andj ∈ {1, . . . , n}.If there exist the constantsr_k ≥0, k ∈ {1, . . . , m}withPm

k=1r_k >0and (8.1) r_k[|Rex_j|+|Imx_j|]≤Rea_k·Rex_j −Ima_k·Imx_j for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.2)

j=1

[|Rex_j|+|Imx_j|]≤ |Pm k=1a_k| Pm

k=1r_k

j=1

Rex_j

j=1

Imx_j

# .

The case of equality holds in (8.2) if both Re

k=1

a_k

! Re

j=1

x_j

−Im

k=1

a_k

! Im

j=1

x_j

k=1

! _n X

j=1

[|Rexj|+|Imxj|]

k=1

a_k

j=1

Rex_j

j=1

Imx_j

# .

The proof follows by Theorem3.1applied for the Banach space(C,|·|₁)and F_k(z) = a_kz, k ∈ {1, . . . , m}on taking into account that:

k=1

F_k 1

k=1

a_k .

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Now, consider the Banach space (C,|·|_∞).If F(z) = dz,then for z 6= 0 we have

|F (z)|

|z|_∞ = |d| |z|

|z|_∞ = |d|

|Rez|² +|Imz|² max{|Rez|,|Imz|} ≤√

2|d|.

Since, forz₀ = 1 +i,we have|F (z₀)|=√

2|d|,|z₀|_∞= 1,hence kFk_∞:= sup

z6=0

|F (z)|

|z|_∞ =√ 2|d|,

showing thatF is a bounded linear functional on(C,|·|_∞)andkFk_∞=√ 2|d|. If we apply Theorem 3.1, then we can state the following reverse of the generalised triangle inequality for complex numbers [5].

Proposition 8.2. Letak, xj ∈ C,k ∈ {1, . . . , m}andj ∈ {1, . . . , n}.If there exist the constantsr_k ≥0, k ∈ {1, . . . , m}withPm

k=1r_k >0and r_kmax{|Rex_j|,|Imx_j|} ≤Rea_k·Rex_j −Ima_k·Imx_j for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.3)

j=1

max{|Rex_j|,|Imx_j|}

≤√

2·|Pm k=1a_k| Pm

k=1r_k max (

j=1

Rex_j ,

j=1

Imx_j

) .

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The case of equality holds in (8.3) if both Re

k=1

a_k

! Re

j=1

x_j

−Im

k=1

a_k

! Im

j=1

x_j

k=1

r_k

! _n X

j=1

max{|Rex_j|,|Imx_j|}

=√ 2

k=1

a_k

max (

j=1

Rex_j ,

j=1

Imx_j

) .

Finally, consider the Banach space

C,|·|_2p

withp≥1.

LetF :C→C,F(z) = cz.By Hölder’s inequality, we have

|F (z)|

|z|_2p = |c|

|Rez|²+|Imz|²

|Rez|^2p+|Imz|^2p_2p¹

≤2¹²⁻^2p¹ |c|.

Since, forz0 = 1 +iwe have|F (z0)|= 2¹² |c|,|z0|_2p = 2^2p¹ (p≥1),hence kFk_2p := sup

z6=0

|F (z)|

|z|_2p = 2¹²⁻^2p¹ |c|,

showing thatF is a bounded linear functional on

C,|·|_2p

, p≥1andkFk_2p = 2¹²⁻^2p¹ |c|.

If we apply Theorem3.1, then we can state the following proposition [5].

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Proposition 8.3. Leta_k, x_j ∈ C,k ∈ {1, . . . , m}andj ∈ {1, . . . , n}.If there exist the constantsrk ≥0, k ∈ {1, . . . , m}withPm

k=1rk >0and rk

|Rexj|^2p+|Imxj|^2p_2p¹

≤Reak·Rexj −Imak·Imxj

for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.4)

j=1

|Rex_j|^2p+|Imx_j|^2p_2p¹

≤2¹²⁻^2p¹ |Pm k=1a_k| Pm

k=1r_k





j=1

Rex_j

j=1

Imx_j

2p



1 2p

The case of equality holds in (8.4) if both:

k=1

a_k

! Re

j=1

x_j

−Im

k=1

a_k

! Im

j=1

x_j

k=1

! _n X

j=1

|Rexj|^2p+|Imxj|^2p_2p¹

= 2¹²⁻^2p¹

k=1

a_k





j=1

Rex_j

j=1

Imx_j

2p



1 2p

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Remark 7. If in the above proposition we choose p = 1, then we have the following reverse of the generalised triangle inequality for complex numbers

j=1

|x_j| ≤ |Pm k=1a_k| Pm

k=1r_k

j=1

x_j

providedx_j, a_k, j ∈ {1, . . . , n},k ∈ {1, . . . , m}satisfy the assumption rk|xj| ≤Reak·Rexj−Imak·Imxj

for each j ∈ {1, . . . , n}, k ∈ {1, . . . , m}.Here |·| is the usual modulus of a complex number andr_k>0, k ∈ {1, . . . , m}are given.

We can apply Theorem6.3 to state the following reverse of the generalised triangle inequality for complex numbers [6].

Proposition 8.4. Letak, xj ∈ C,k ∈ {1, . . . , m}andj ∈ {1, . . . , n}.If there exist the constantsM_jk ≥0, k ∈ {1, . . . , m}, j∈ {1, . . . , n}such that

(8.5) |Rex_j|+|Imx_j| ≤Rea_k·Rex_j −Ima_k·Imx_j +M_jk for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.6)

j=1

[|Rex_j|+|Imx_j|]

≤ 1 m

k=1

a_k

j=1

Rex_j

j=1

Imx_j

# + 1

k=1 n

j=1

M_jk.

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The proof follows by Theorem6.3applied for the Banach space(C,|·|₁)and Fk(z) = akz, k ∈ {1, . . . , m}on taking into account that:

k=1

F_k 1

k=1

a_k .

If we apply Theorem6.3 for the Banach space(C,|·|_∞), then we can state the following reverse of the generalised triangle inequality for complex numbers [6].

Proposition 8.5. Leta_k, x_j ∈ C,k ∈ {1, . . . , m}andj ∈ {1, . . . , n}.If there exist the constantsMjk ≥0, k ∈ {1, . . . , m}, j∈ {1, . . . , n}such that

max{|Rex_j|,|Imx_j|} ≤Rea_k·Rex_j−Ima_k·Imx_j+M_jk for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.7)

j=1

max{|Rexj|,|Imxj|}

≤

√2 m

k=1

a_k

max (

j=1

Rex_j ,

j=1

Imx_j

) + 1

k=1 n

j=1

M_jk.

Finally, if we apply Theorem6.3, for the Banach space

C,|·|_2p

withp≥1, then we can state the following proposition [6].

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Proposition 8.6. Leta_k, x_j, M_jk be as in Proposition8.5. If

|Rex_j|^2p+|Imx_j|^2p_2p¹

≤Rea_k·Rex_j−Ima_k·Imx_j +M_jk for eachj ∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(8.8)

j=1

|Rex_j|^2p+|Imx_j|^2p_2p¹

≤ 2¹²⁻^2p¹ m

k=1

a_k





j=1

Rex_j

j=1

Imx_j

2p



1 2p

+ 1 m

k=1 n

j=1

M_jk.

wherep≥1.

Remark 8. If in the above proposition we choose p = 1, then we have the following reverse of the generalised triangle inequality for complex numbers

j=1

|x_j| ≤

1 m

k=1

a_k

j=1

x_j

+ 1 m

k=1 n

j=1

M_jk

providedx_j, a_k, j ∈ {1, . . . , n},k ∈ {1, . . . , m}satisfy the assumption

|x_j| ≤Rea_k·Rex_j −Ima_k·Imx_j+M_jk

for each j ∈ {1, . . . , n}, k ∈ {1, . . . , m}.Here |·| is the usual modulus of a complex number andMjk >0, j ∈ {1, . . . , n},k ∈ {1, . . . , m}are given.

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9. Karamata Type Inequalities in Hilbert Spaces

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 (9.1)

Z b a

f(x)dx

≤ Z b

|f(x)|dx,

plays a fundamental role in Mathematical Analysis and its applications.

It appears, see [20, p. 492], that the first reverse inequality for (9.1) was obtained by J. Karamata in his book from 1949, [14]. It can be stated as

(9.2) cosθ

Z b a

|f(x)|dx≤

Z b a

f(x)dx provided

−θ ≤argf(x)≤θ, x∈[a, b]

for givenθ ∈ 0,^π₂ .

This result has recently been extended by the author for the case of Bochner integrable functions with values in a Hilbert spaceH (see also [10]):

Theorem 9.1 (Dragomir, 2004). If f ∈ L([a, b] ;H) (this means that f : [a, b]→His strongly measurable on[a, b]and the Lebesgue integralRb

a kf(t)kdt is finite), then

(9.3)

Z b a

kf(t)kdt≤K

Z b a

f(t)dt ,

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provided thatf satisfies the condition

(9.4) kf(t)k ≤KRehf(t), ei for a.e. t∈[a, b], wheree∈H,kek= 1andK ≥1are given.

The case of equality holds in (9.4) if and only if (9.5)

Z b a

f(t)dt = 1 K

Z b a

kf(t)kdt

As some natural consequences of the above results, we have noticed in [10]

that, ifρ∈[0,1)andf ∈L([a, b] ;H)are such that (9.6) kf(t)−ek ≤ρ for a.e. t∈[a, b], then

(9.7) p

1−ρ² Z b

kf(t)kdt ≤

Z b a

f(t)dt with equality if and only if

Z b a

f(t)dt=p 1−ρ²

Z b a

kf(t)kdt

·e.

Also, foreas above and ifM ≥m >0, f ∈L([a, b] ;H)such that either (9.8) RehM e−f(t), f(t)−mei ≥0

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or, equivalently, (9.9)

f(t)− M +m 2 e

≤ 1

2(M −m) for a.e. t ∈[a, b],then

(9.10)

Z b a

kf(t)kdt ≤ M+m 2√

Z b a

f(t)dt ,

with equality if and only if Z b

f(t)dt = 2√ mM M +m

Z b a

kf(t)kdt

·e.

The main aim of the following sections is to extend the integral inequalities mentioned above for the case of Banach spaces. Applications for Hilbert spaces and for complex-valued functions are given as well.

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10. Multiplicative Reverses of the Continuous Triangle Inequality

10.1. The Case of One Functional

Let (X,k·k) be a Banach space over the real or complex number field. Then one has the following reverse of the continuous triangle inequality [11].

Theorem 10.1 (Dragomir, 2004). Let F be a continuous linear functional of unit norm onX.Suppose that the functionf : [a, b]→Xis Bochner integrable on[a, b]and there exists ar≥0such that

(10.1) rkf(t)k ≤ReF [f(t)] for a.e. t∈[a, b]. Then

(10.2) r

Z b a

kf(t)kdt ≤

Z b a

f(t)dt ,

where equality holds in (10.2) if and only if both

(10.3) F

Z b a

f(t)dt

=r Z b

kf(t)kdt

and

(10.4) F

Z b a

f(t)dt

Z b a

f(t)dt .

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Proof. Since the norm ofF is one, then

|F(x)| ≤ kxk for any x∈X.

Applying this inequality for the vectorRb

af(t)dt,we get

Z b a

f(t)dt

≥ F

Z b a

f(t)dt

(10.5)

≥

ReF Z b

f(t)dt

Z b a

ReF (f(t))dt .

Now, by integration of (10.1), we obtain (10.6)

Z b a

ReF (f(t))dt ≥r Z b

kf(t)kdt,

and by (10.5) and (10.6) we deduce the desired inequality (10.2).

Obviously, if (10.3) and (10.4) hold true, then the equality case holds in (10.2).

Conversely, if the case of equality holds in (10.2), then it must hold in all the inequalities used before in proving this inequality. Therefore, we must have (10.7) rkf(t)k= ReF(f(t)) for a.e. t∈[a, b],

(10.8) ImF

Z b a

f(t)dt

= 0

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and (10.9)

Z b a

f(t)dt

= ReF Z b

f(t)dt

Integrating (10.7) on[a, b],we get

(10.10) r

Z b a

kf(t)kdt= ReF Z b

f(t)dt

On utilising (10.10) and (10.8), we deduce (10.3) while (10.9) and (10.10) would imply (10.4), and the theorem is proved.

Corollary 10.2. Let (X,k·k)be a Banach space,[·,·] : X×X → Ra semi-inner product generating the norm k·kande ∈ X, kek = 1.Suppose that the functionf : [a, b]→X is Bochner integrable on[a, b]and there exists ar ≥ 0 such that

(10.11) rkf(t)k ≤Re [f(t), e] for a.e. t∈[a, b]. Then

(10.12) r

Z b a

kf(t)kdt ≤

Z b a

f(t)dt where equality holds in (10.12) if and only if both (10.13)

Z b a

f(t)dt, e

=r Z b

kf(t)kdt

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and (10.14)

Z b a

f(t)dt, e

Z b a

f(t)dt .

The proof follows from Theorem 10.1 for the continuous linear functional F (x) = [x, e], x∈X,and we omit the details.

The following corollary of Theorem10.1may be stated [8].

Corollary 10.3. Let (X,k·k) be a strictly convex Banach space, [·,·] : X × X → Ka semi-inner product generating the normk·kande ∈X,kek = 1.If f : [a, b]→Xis Bochner integrable on[a, b]and there exists ar≥0such that (10.11) holds true, then (10.12) is valid. The case of equality holds in (10.12) if and only if

(10.15)

Z b a

f(t)dt=r Z b

kf(t)kdt

Proof. If (10.15) holds true, then, obviously

Z b a

f(t)dt

=r Z b

kf(t)kdt

kek=r Z b

kf(t)kdt,

which is the equality case in (10.12).

Conversely, if the equality holds in (10.12), then, by Corollary10.2, we must have (10.13) and (10.14). Utilising Theorem4.2, by (10.14) we can conclude that there exists aµ >0such that

(10.16)

Z b a

f(t)dt =µe.

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Replacing this in (10.13), we get µkek² =r

Z b a

kf(t)kdt,

giving

(10.17) µ=r

Z b a

kf(t)kdt.

Utilising (10.16) and (10.17) we deduce (10.15) and the proof is completed.

In document JJ II (Pldal 39-65)