Cauchy-Schwartz inequality revisited

Who? Dragoljub J. Keckic

From? University of Belgrade, Faculty of Mathematics

When? May 17th 2018, 15:00

Recall that...

...thisinequality

n∑

k=1

akbk

≤

n∑

k=1

|ak |21/2 n

∑

k=1

|bk |21/2

...and thisinequality

∫

X

f (t)g(t)dμ(t)

≤∫

X

|f (t)|2 dμ(t)1/2 ∫

X

|g(t)|2 dμ(t)1/2

...as well asmany others...

...are specialcases of

| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.

Recall that...

...thisinequality

n∑

k=1

akbk

≤

n∑

k=1

|ak |21/2 n

∑

k=1

|bk |21/2

...and thisinequality

∫

X

f (t)g(t)dμ(t)

≤∫

X

|f (t)|2 dμ(t)1/2 ∫

X

|g(t)|2 dμ(t)1/2

...as well asmany others...

...are specialcases of

| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.

Recall that...

...thisinequality

n∑

k=1

akbk

≤

n∑

k=1

|ak |21/2 n

∑

k=1

|bk |21/2

...and thisinequality

∫

X

f (t)g(t)dμ(t)

≤∫

X

|f (t)|2 dμ(t)1/2 ∫

X

|g(t)|2 dμ(t)1/2

...as well asmany others...

...are specialcases of

| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.

Recall that...

...thisinequality

n∑

k=1

akbk

≤

n∑

k=1

|ak |21/2 n

∑

k=1

|bk |21/2

...and thisinequality

∫

X

f (t)g(t)dμ(t)

≤∫

X

|f (t)|2 dμ(t)1/2 ∫

X

|g(t)|2 dμ(t)1/2

...as well asmany others...

...are specialcases of

| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.

Similarly

CS inequalityfor Hilbert

modules

| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2 leads to many inequalities, e.g.

1 Jocic, Proc. Amer. Math. Soc. (1998)2 Jocic, J London Math. Soc. (1999)3 Jocic, J Funct. Anal. (2005)4 Jocic, Krtinic, Moslehian, Math. Inequal. Appl. (2013)5 Jocic, Milosevic, Linear Algebra Appl. (2016)6 Milosevic, Adv. Oper. Theory (2016)7 Jocic, Milosevic, Ðuric, Filomat (2017)8 Jocic, Krtinic, Lazarevic, Melentijevic, Milosevic, Complex

Anal. Oper. Theory (2018)

Similarly

CS inequalityfor Hilbert

modules

| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2 leads to many inequalities, e.g.

1 Jocic, Proc. Amer. Math. Soc. (1998)2 Jocic, J London Math. Soc. (1999)3 Jocic, J Funct. Anal. (2005)4 Jocic, Krtinic, Moslehian, Math. Inequal. Appl. (2013)5 Jocic, Milosevic, Linear Algebra Appl. (2016)6 Milosevic, Adv. Oper. Theory (2016)7 Jocic, Milosevic, Ðuric, Filomat (2017)8 Jocic, Krtinic, Lazarevic, Melentijevic, Milosevic, Complex

Anal. Oper. Theory (2018)

Basic definitions

C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗

satisfying ‖a∗a‖ = ‖a‖2.

HilbertC∗-module

over A

(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:

1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.

The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.

More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)

Basic definitions

C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗

satisfying ‖a∗a‖ = ‖a‖2.

HilbertC∗-module

over A

(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:

1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.

The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.

More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)

Basic definitions

C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗

satisfying ‖a∗a‖ = ‖a‖2.

HilbertC∗-module

over A

(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:

1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.

The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.

More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)

Basic definitions

C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗

satisfying ‖a∗a‖ = ‖a‖2.

HilbertC∗-module

over A

(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:

1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.

The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.

More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)

Basic definitions

C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗

satisfying ‖a∗a‖ = ‖a‖2.

HilbertC∗-module

over A

(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:

1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.

The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.

More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)

Cauchy-Schwartz inequality

Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)

Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.

Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave

‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)

Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.

Cauchy-Schwartz inequality

Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)

Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.

Given by Paschke (1973). In books at very beginning.

Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave

‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)

Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.

Cauchy-Schwartz inequality

Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)

Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.

Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave

‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)

Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.

Cauchy-Schwartz inequality

Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)

Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.

Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave

‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)

Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.

Cauchy-Schwartz inequality

Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)

Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.

Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave

‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)

Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.

Examples

StandardHilbert module

l2(A) = x =(ξn)+∞n=1 |

∑+∞n=1 ξ

∗nξn converges in the norm of A

... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞

n=1 ξ∗nξn converges weakly

⟨x , y ⟩ =∑+∞

n=1 ξ∗nηn.

L2(Ω;A) If A has a predual then At ∈ L2 iff∫

ΩA∗tAt dμ(t) < +∞ – the

weak integral⟨A,B⟩ =

∫

ΩA∗tBt dμ(t).

Examples

StandardHilbert module

l2(A) = x =(ξn)+∞n=1 |

∑+∞n=1 ξ

∗nξn converges in the norm of A

... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞

n=1 ξ∗nξn converges weakly

⟨x , y ⟩ =∑+∞

n=1 ξ∗nηn.

L2(Ω;A) If A has a predual then At ∈ L2 iff∫

ΩA∗tAt dμ(t) < +∞ – the

weak integral⟨A,B⟩ =

∫

ΩA∗tBt dμ(t).

Examples

StandardHilbert module

l2(A) = x =(ξn)+∞n=1 |

∑+∞n=1 ξ

∗nξn converges in the norm of A

... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞

n=1 ξ∗nξn converges weakly

⟨x , y ⟩ =∑+∞

n=1 ξ∗nηn.

L2(Ω;A) If A has a predual then At ∈ L2 iff∫

ΩA∗tAt dμ(t) < +∞ – the

weak integral⟨A,B⟩ =

∫

ΩA∗tBt dμ(t).

Examples

StandardHilbert module

l2(A) = x =(ξn)+∞n=1 |

∑+∞n=1 ξ

∗nξn converges in the norm of A

... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞

n=1 ξ∗nξn converges weakly

⟨x , y ⟩ =∑+∞

n=1 ξ∗nηn.

L2(Ω;A) If A has a predual then At ∈ L2 iff∫

ΩA∗tAt dμ(t) < +∞ – the

weak integral

⟨A,B⟩ =∫

ΩA∗tBt dμ(t).

Examples

StandardHilbert module

l2(A) = x =(ξn)+∞n=1 |

∑+∞n=1 ξ

∗nξn converges in the norm of A

... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞

n=1 ξ∗nξn converges weakly

⟨x , y ⟩ =∑+∞

n=1 ξ∗nηn.

L2(Ω;A) If A has a predual then At ∈ L2 iff∫

ΩA∗tAt dμ(t) < +∞ – the

weak integral⟨A,B⟩ =

∫

ΩA∗tBt dμ(t).

Main idea

Elementaryoperators

X 7→∑

k AkXBk – (Lummer, Rosenblum 1959) – variousframeworks

Consider X as an operator, and Ak , Bk as variables!Then

∑

k A∗kXBk = ⟨A,XB⟩!

I.p.t.i.transformer

X 7→∫

ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩

Basicinequalities

Inequality (1) becomes

∑

kA∗kXBk

≤

∑

kA∗kAk

1/2

‖X‖

∑

kB∗kBk

1/2

∫

Ω

A∗tXBt

≤

∫

Ω

A∗tAt

1/2

‖X‖

∫

Ω

B∗tBt

1/2

.

Main idea

Elementaryoperators

X 7→∑

k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then

∑

k A∗kXBk = ⟨A,XB⟩!

I.p.t.i.transformer

X 7→∫

ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩

Basicinequalities

Inequality (1) becomes

∑

kA∗kXBk

≤

∑

kA∗kAk

1/2

‖X‖

∑

kB∗kBk

1/2

∫

Ω

A∗tXBt

≤

∫

Ω

A∗tAt

1/2

‖X‖

∫

Ω

B∗tBt

1/2

.

Main idea

Elementaryoperators

X 7→∑

k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then

∑

k A∗kXBk = ⟨A,XB⟩!

I.p.t.i.transformer

X 7→∫

ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩

Basicinequalities

Inequality (1) becomes

∑

kA∗kXBk

≤

∑

kA∗kAk

1/2

‖X‖

∑

kB∗kBk

1/2

∫

Ω

A∗tXBt

≤

∫

Ω

A∗tAt

1/2

‖X‖

∫

Ω

B∗tBt

1/2

.

Main idea

Elementaryoperators

X 7→∑

k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then

∑

k A∗kXBk = ⟨A,XB⟩!

I.p.t.i.transformer

X 7→∫

ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩

Basicinequalities

Inequality (1) becomes

∑

kA∗kXBk

≤

∑

kA∗kAk

1/2

‖X‖

∑

kB∗kBk

1/2

∫

Ω

A∗tXBt

≤

∫

Ω

A∗tAt

1/2

‖X‖

∫

Ω

B∗tBt

1/2

.

Main idea

Elementaryoperators

X 7→∑

k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then

∑

k A∗kXBk = ⟨A,XB⟩!

I.p.t.i.transformer

X 7→∫

ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩

Basicinequalities

Inequality (1) becomes

∑

kA∗kXBk

≤

∑

kA∗kAk

1/2

‖X‖

∑

kB∗kBk

1/2

∫

Ω

A∗tXBt

≤

∫

Ω

A∗tAt

1/2

‖X‖

∫

Ω

B∗tBt

1/2

.

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Other norms

SemifiniteW∗-algebra

A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.

p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.

More norms onB(H)

B(H) is a semifinite W∗-algebra with the usual tr.

Ky Fan norms ‖T‖(k) =∑k

j=1 sj (T ) =∑k

j=1 λj (T∗T )1/2.

Unitarilyinvariant norm

Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .

Ky Fan dom.property

(1951)

|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).

Interpolations

Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).

Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.

Complexinterpolation

betweenLp(A;τ)

S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M

1−1/p0 M

1/p1 .

Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).

Corollary The corresponding Riesz-Thorin theorem.

Interpolations

Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).

Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.

Complexinterpolation

betweenLp(A;τ)

S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M

1−1/p0 M

1/p1 .

Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).

Corollary The corresponding Riesz-Thorin theorem.

Interpolations

Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).

Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.

Complexinterpolation

betweenLp(A;τ)

S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M

1−1/p0 M

1/p1 .

Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).

Corollary The corresponding Riesz-Thorin theorem.

Interpolations

Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).

Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.

Complexinterpolation

betweenLp(A;τ)

S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M

1−1/p0 M

1/p1 .

Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).

Corollary The corresponding Riesz-Thorin theorem.

Interpolations

Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).

Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.

Complexinterpolation

betweenLp(A;τ)

S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M

1−1/p0 M

1/p1 .

Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).

Corollary The corresponding Riesz-Thorin theorem.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Modular conjugation

l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n

, ... ) have thefollowing properties:

1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).

Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.

Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.

Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.

Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =

¬

x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶

etc.

Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.

Inequalities in ‖ · ‖1

Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.

Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).

Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then

‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1

|τ(b ⟨x , ay ⟩)| =

= sp‖b‖=1

|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.

In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.

Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬

⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶

‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1

Inequalities in ‖ · ‖1

Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.

Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).

Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then

‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1

|τ(b ⟨x , ay ⟩)| =

= sp‖b‖=1

|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.

In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.

Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬

⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶

‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1

Inequalities in ‖ · ‖1

Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.

Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).

Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then

‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1

|τ(b ⟨x , ay ⟩)| =

= sp‖b‖=1

|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.

In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.

Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬

⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶

‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1

Inequalities in ‖ · ‖1

Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.

Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).

Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then

‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1

|τ(b ⟨x , ay ⟩)| =

= sp‖b‖=1

|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.

In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.

Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬

⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶

‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1

Inequalities in ‖ · ‖1

Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.

Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).

Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then

‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1

|τ(b ⟨x , ay ⟩)| =

= sp‖b‖=1

|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.

In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.

Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬

⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶

‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1

Inequalities in |||·|||

Proposition x and y normal implies

|||⟨x , ay ⟩||| ≤

⟨x , x⟩1/2 a ⟨y , y ⟩1/2

.

Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.

Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.

Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.

Inequalities in |||·|||

Proposition x and y normal implies

|||⟨x , ay ⟩||| ≤

⟨x , x⟩1/2 a ⟨y , y ⟩1/2

.

Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.

Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.

Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.

Inequalities in |||·|||

Proposition x and y normal implies

|||⟨x , ay ⟩||| ≤

⟨x , x⟩1/2 a ⟨y , y ⟩1/2

.

Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.

Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.

Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.

Inequalities in |||·|||

Proposition x and y normal implies

|||⟨x , ay ⟩||| ≤

⟨x , x⟩1/2 a ⟨y , y ⟩1/2

.

Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.

Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.

Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.

Inequalities in ‖ · ‖p

Idea Since ‖¬

x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶

‖ ≤ ‖a‖ and

‖¬

⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶

‖1 ≤ ‖a‖1, consider

z 7→¬

⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶

and apply complex interpolation.

Proposition If 1/ r + 1/q = 2/p then

‖ ⟨x , ay ⟩ ‖p ≤

⟨x , x⟩q−1 x , x1/2q

a

⟨y , y ⟩r−1 y , y1/2r

p.

Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.

Inequalities in ‖ · ‖p

Idea Since ‖¬

x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶

‖ ≤ ‖a‖ and

‖¬

⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶

‖1 ≤ ‖a‖1, consider

z 7→¬

⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶

and apply complex interpolation.

Proposition If 1/ r + 1/q = 2/p then

‖ ⟨x , ay ⟩ ‖p ≤

⟨x , x⟩q−1 x , x1/2q

a

⟨y , y ⟩r−1 y , y1/2r

p.

Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.

Inequalities in ‖ · ‖p

Idea Since ‖¬

x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶

‖ ≤ ‖a‖ and

‖¬

⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶

‖1 ≤ ‖a‖1, consider

z 7→¬

⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶

and apply complex interpolation.

Proposition If 1/ r + 1/q = 2/p then

‖ ⟨x , ay ⟩ ‖p ≤

⟨x , x⟩q−1 x , x1/2q

a

⟨y , y ⟩r−1 y , y1/2r

p.

Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.

Aczel type inequalities

Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2

Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑

n=0

| ⟨x , y ⟩ |n ≤+∞∑

n=0

‖x‖n‖y‖n ≤

+∞∑

n=0

‖x‖2n1/2 +∞

∑

n=0

‖y‖2n1/2

= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.

Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies

(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2

≤ |||a− ⟨x , ay ⟩|||

Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =

x⊗n, ay⊗n

.Put b = (I − T )−1a. Then:

Aczel type inequalities

Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2

Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑

n=0

| ⟨x , y ⟩ |n ≤+∞∑

n=0

‖x‖n‖y‖n ≤

+∞∑

n=0

‖x‖2n1/2 +∞

∑

n=0

‖y‖2n1/2

= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.

Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies

(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2

≤ |||a− ⟨x , ay ⟩|||

Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =

x⊗n, ay⊗n

.Put b = (I − T )−1a. Then:

Aczel type inequalities

Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2

Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑

n=0

| ⟨x , y ⟩ |n ≤+∞∑

n=0

‖x‖n‖y‖n ≤

+∞∑

n=0

‖x‖2n1/2 +∞

∑

n=0

‖y‖2n1/2

= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.

Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies

(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2

≤ |||a− ⟨x , ay ⟩|||

Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =

x⊗n, ay⊗n

.Put b = (I − T )−1a. Then:

Aczel type inequalities

Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2

Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑

n=0

| ⟨x , y ⟩ |n ≤+∞∑

n=0

‖x‖n‖y‖n ≤

+∞∑

n=0

‖x‖2n1/2 +∞

∑

n=0

‖y‖2n1/2

= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.

Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies

(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2

≤ |||a− ⟨x , ay ⟩|||

Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =

x⊗n, ay⊗n

.Put b = (I − T )−1a. Then:

Aczel type inequalities – second frame

Proof –continuation |||b||| =

∑

k≥0T ka

=

∑

k≥0

x⊗k , ay⊗k

=

=

·

k≥0x⊗k , a

∑

k≥0y⊗k

¶

≤

≤

·

k≥0x⊗k ,

∑

k≥0x⊗k

¶12 a¬∑

k≥0y⊗k ,

∑

k≥0y⊗k

¶12

.

However∑

n≥0 x⊗n,

∑

n≥0 x⊗n

=∑

n≥0

x⊗n, x⊗n

=∑

n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.

Hence

(I − T )−1a

≤

(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2

. Fewsubstitution, normality and we are done.

Aczel type inequalities – second frame

Proof –continuation |||b||| =

∑

k≥0T ka

=

∑

k≥0

x⊗k , ay⊗k

=

=

·

k≥0x⊗k , a

∑

k≥0y⊗k

¶

≤

≤

·

k≥0x⊗k ,

∑

k≥0x⊗k

¶12 a¬∑

k≥0y⊗k ,

∑

k≥0y⊗k

¶12

.

However∑

n≥0 x⊗n,

∑

n≥0 x⊗n

=∑

n≥0

x⊗n, x⊗n

=∑

n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.

Hence

(I − T )−1a

≤

(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2

. Fewsubstitution, normality and we are done.

Aczel type inequalities – second frame

Proof –continuation |||b||| =

∑

k≥0T ka

=

∑

k≥0

x⊗k , ay⊗k

=

=

·

k≥0x⊗k , a

∑

k≥0y⊗k

¶

≤

≤

·

k≥0x⊗k ,

∑

k≥0x⊗k

¶12 a¬∑

k≥0y⊗k ,

∑

k≥0y⊗k

¶12

.

However∑

n≥0 x⊗n,

∑

n≥0 x⊗n

=∑

n≥0

x⊗n, x⊗n

=∑

n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.

Hence

(I − T )−1a

≤

(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2

. Fewsubstitution, normality and we are done.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Aczel type inequalities – third frame

Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998

Morecorollaries

The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.

Idea Drop normality condition and use complex interpolation(obtained earlier)

Proposition ‖∆1−1/qx

a∆1−1/ry

‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where

∆z =¬

∑+∞n=0 z

⊗n,∑+∞

n=0 z⊗n¶−1/2

, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.

Proof Combine previous proof with complex interpolation.

Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.

Grüß inequality

Classic form

1

b − a

∫ b

a

fg −1

b − a

∫ b

a

f1

b − a

∫ b

a

g

≤

1

4(mx f −min f )(mx g −min g)

Hilbert spaceform

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1

(Observe LHS is a (semi)inner product.)

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the

ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.

(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)

Grüß inequality

Classic form

1

b − a

∫ b

a

fg −1

b − a

∫ b

a

f1

b − a

∫ b

a

g

≤

1

4(mx f −min f )(mx g −min g)

Hilbert spaceform

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1

(Observe LHS is a (semi)inner product.)

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the

ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.

(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)

Grüß inequality

Classic form

1

b − a

∫ b

a

fg −1

b − a

∫ b

a

f1

b − a

∫ b

a

g

≤

1

4(mx f −min f )(mx g −min g)

Hilbert spaceform

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1

(Observe LHS is a (semi)inner product.)

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the

ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.

(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)

Grüß inequality

Classic form

1

b − a

∫ b

a

fg −1

b − a

∫ b

a

f1

b − a

∫ b

a

g

≤

1

4(mx f −min f )(mx g −min g)

Hilbert spaceform

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1

(Observe LHS is a (semi)inner product.)

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the

ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.

(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)

Grüß inequality

Classic form

1

b − a

∫ b

a

fg −1

b − a

∫ b

a

f1

b − a

∫ b

a

g

≤

1

4(mx f −min f )(mx g −min g)

Hilbert spaceform

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1

(Observe LHS is a (semi)inner product.)

|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the

ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.

(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)

Grüß inequality – continuation

Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤

(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2

, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1

4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].

Proof Easy.

Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then

Φ(x , ay) =∫

Ω

x(t)∗ay(t)dμ(t) −∫

Ω

x(t)∗ dμ(t)∫

Ω

ay(t)dμ(t),

– MIA2013 (17 pages) main result.

Grüß inequality – continuation

Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤

(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2

, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1

4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].

Proof Easy.

Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then

Φ(x , ay) =∫

Ω

x(t)∗ay(t)dμ(t) −∫

Ω

x(t)∗ dμ(t)∫

Ω

ay(t)dμ(t),

– MIA2013 (17 pages) main result.

Grüß inequality – continuation

Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤

(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2

, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1

4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].

Proof Easy.

Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then

Φ(x , ay) =∫

Ω

x(t)∗ay(t)dμ(t) −∫

Ω

x(t)∗ dμ(t)∫

Ω

ay(t)dμ(t),

– MIA2013 (17 pages) main result.

Refinements

Remark Till now, we use only ‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. However, thereis more precise

| ⟨x , ay ⟩ | ≤ ‖x‖

y , a∗ay1/2

. (2)

Corollaries Apply (2) to M = An to prove some results of LAA 2016.Apply (2) to Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ to obtain mainresults of CAOT 2018.

Refinements

Remark Till now, we use only ‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. However, thereis more precise

| ⟨x , ay ⟩ | ≤ ‖x‖

y , a∗ay1/2

. (2)

Corollaries Apply (2) to M = An to prove some results of LAA 2016.Apply (2) to Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ to obtain mainresults of CAOT 2018.

Thanks for your attention

Completedetails

arXiv:1801.07953"The applications of Cauchy-Schwartz inequality for Hilbertmodules to elementary operators and i.p.t.i. transformers"submitted 2018. January 24th