Quantizing Analysis

8/9/2019 Quantizing Analysis

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Quantization of AnalysisKelvin Lui

12/11/2014

Department of MathematicsColby College

2015


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Definition 8. A ring is a datum (R, +, ×, 0, 1) where R is a set, 1, 0 ∈ R and +, ×are binary operations on R defined such that

1. R is an abelian group under + with identity element 0;

2. The binary operator × is associative and 1 × x = x × 1 = x ∀x ∈ R;

3. Multiplication distributes over addition:

x × (y + z ) = (x × y) + (x × z )

(x + y) × z = (x × z ) + (y × z ) ∀x,y,z ∈ R

If the operation × is commutative, x × y = y × x ∀x, y ∈ R, we say R is acommutative ring. In addition, if 1 = 0 then the ring has only one element and isreferred to as the ”zero ring”. An example of a ring would be the integers Z or the

integers modulo n Z/nZ , n ∈ Z. If R is a ring, we can construct a new ring R[t] of polynomials in t with coefficients in R.

Definition 9. If R is a ring, a subset S ⊆ R is said to be a subring if it inherits the structure of a ring from R, therefore 0, 1 ∈ S and S is closed under the addition and multiplication operations in R; (S, +) is a subgroup of (R, +).

An example would be the integers Z are a subring of Q.

Lemma 2.1. Subring Criterion Let R be a ring and S ⊆ R, then S is a sunning if and only if 1 ∈ S and ∀s1, s2 ∈ S

we have s1s2, s1 − s2 ∈ S .

Proof. (⇒) If we assume that S is a subring the proof is trivial.(⇐) Note S = {0} because 1 ∈ S . If s1s2, s1 − s2 ∈ S, ∀s1, s2 ∈ S that implies

that S is an additive subgroup by the subgroup test. It is also easy to see that 0 ∈ S and that the other conditions for a subring hold.

Definition 10. A map f : R → S between rings R and S is said to be a (ring)homomorphism if

1. f (1R) = 1S

2. f (r1 +R r2) = f (r1) +S f (r2)

3. f (r1.Rr2) = f (r1).S f (r2)


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The image of a ring homomorphism f : R → S is a subring of S . If the image isall of S we say f is surjective, and if f : R → S is an isomorphism then ∃g : S toRsuch that f ◦ g = idS and g ◦ f = idR.

We can also look at polynomials with coefficients in a ring. If R is a ring then the

set R[t] of sequences (an)n∈N, an ∈ R where ∃N ∈ N such that an = 0, ∀n ≥ N . Wecan denote elements of R[t] as

N n=0

antn. (2.1.2)

We will define addition and multiplication to be

N

n=0ant

n +M

n=0bnt

n =

max{N,M }

n=0(an + bn)t

n (2.1.3a)

N n=0

antn ×

M n=0

bntn =

N +M n=0

nk=0

akbn−k

tn (2.1.3b)

and with this we see that R[t] is a ring. The set of all sequences (an)n∈N forms aring as well with the same definitions of addition and multiplication. It is called thering of formal power series in t and denoted as R[[t]]. We view elements of R[[t]] as”infinite sums” ,

∞n+0 ant

n.

Definition 11. If R is a ring, then an element a ∈ R\{0} is a zero-divisor if there is some b ∈ R\{0} such that a.b = 0.

Definition 12. Let f : R → S be a ring homomorphism. The kernel of f is

ker (F ) = {r ∈ R : f (r) = 0}, (2.1.4)

and the image of f is

im (f ) = {s ∈ S : ∃r ∈ R, f (r) = 2}. (2.1.5)

The image of a homomorphism is a subring of the target ring. The kernel, howeveris not a subring of the target ring. The kernel is closed under addition and multi-

plication and since 0.x = 0, ∀x, it obeys a stronger kind of closure with respect tomultiplication. If x ∈ ker(f ) and ∀r ∈ R then f (x.r) = f (x).f (r) = 0.f (r) = 0 sothat x.r ∈ ker(f ).

Definition 13. Let R be a ring. A subset I ⊆ R is called an ideal if it is a subgroupof (R, +) and moreover for any a ∈ I and r ∈ R then a.r ∈ I .

Lemma 2.2. If f : R → S is a homomorphism, then ker (f ) is an ideal.


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C 1 + C 2 = {c1 + c2 : c1 ∈ C i, i = 1, 2} (2.2.1)

C 1 + C 2 is a subset of R and we claim it to be a single coset of I . Definitely, it

is a union of cosets, since c1 + c2 ∼ r, for some r ∈ R we have r = (c1 + c2) + i forsome i ∈ I and so r = (c1 + i) + c2 ∈ C 1 + C 2; the sum is an element of the unionof cosets. To show that it is a single coset, that addition is well-defined, fix elementsdi ∈ C i, i = 1, 2. Then if ci ∈ C i we have ci = di + ki, ki ∈ I , hence

c1 + c2 + (d1 + k1) + (d2 + k2) = (d1 + d2) + (k1 + k2) ∈ (d1 + d2) + I. (2.2.2)

To give R/I a ring structure, we need to define a multiplication operation. Giventwo cosets C 1, C 2 define:

C 1 · C 2 = {c1 · c2 : ci ∈ C i, i = 1, 2}. (2.2.3)We also claim that the set C 1 · C 2 is contained in a single coset of R/I , that is

C 1 · C 2 + I is a single I -coset. Suppose the ci, di are given as above, then

c1 · c2 = (d1 + k1) · (d2 + k2) = d1 · d2 + (d1 · k2 + k1 · 22 + k1 · k2) ∈I

∈ d1 · d2 + I (2.2.4)

We get a binary operation on R/I by setting:

C 1 × C 2 = C 1 · C 2 + I (2.2.5)

Theorem 2.4. The datum (R/I, +, ×, 0 + I , i + I ) defines a ring structure on R/I and moreover the map 1 : R → R/I given by q (r) = r + I is a surjective ring homomorphism.

Proof. It is clear that R/I is an abelian group under + with an identity element0. We then check to see that the binary operation × is associative and that themultiplication distributes over addition:

C 1 × (C 2 × C 3) = (a1 + I ) × ((a2 + I ) × (a3 + I ))

= (a1 + I )((a1 · a2 + I )) = (a1)(a2 · a3) + I = (a1 · a2)a3 + I = (a1 · a2 + I ) × (a3 + I )

= ((a1 + I ) × (a2 + I )) × (a3 + I )

= (C 1 × C 2) × C 3

∀C 1, C 2, C 3 ∈ R/I,a1, a2, a3 ∈ R (2.2.6)


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a simple adaptation with the multiplicative identity will show associativity of 1R/O × C 1 = C 1 = C 1 × 1R/O, where 1R/O = 1 + I .

C 1 × (C 2 + C 3) = (a1 + I ) × ((a2 + I ) + (a3 + I ))= (a1 + I ) × ((a2 + a3) + I ) = a1(a2 + a3) + I

= (a1a2 + a1a + 3) + I

= (a1a2 + I ) + (a1a3 + I )

= C 1 × C 2 + C 1 × C 3

∀C 1, C 2, C 3 ∈ R/I,a1, a2, a3 ∈ R (2.2.7)

A similar process can be done to show (C 1 + C 2) × C 3 = (C 1 × C 3) + (C 2 × C 3).Finally the map q : R → R/I is surjective and that it is a homomorphism can

been seen immediately from the definitions.

The map q : R → R/I is called the quotient homomorphism.

Corollary 2.4.1. Any ideal is the kernel of a ring homomorphism.

Proof. If I is an ideal and q : R → R/I is the quotient map then clearly ker(q ) = I .

Next, we will look at comparisons between ideals in a quotient ring with ideals inthe original ring.

Lemma 2.5. Let R be a ring, I an ideal in R and q : R → R/I the quotient map.If J is an ideal then q (J ) is an ideal in R/I , and if K is an ideal in R/I then q −1(K ) = {r ∈ R : q (r) ∈ K } is an ideal in R which contains I . Moreover, these correspondences give a bijection between the ideals in R/I and the ideals in R which contain I .

Proof. Let q : R → R/I denote the quotient homomorphism. If J is an ideal inR we need to check that q(J) is an ideal in R/I . q is a homomorphism so it is asubgroup of R/I under addition, we need to check the multiplicative property. Anyelement of q (J ) has the form q ( j), ∀ j ∈ J . From Theorem 2.4 we know that themap q is surjective and so for any element of R/I has the form q (r), ∀r ∈ R. Since

q (r) · q ( j) = q (r · j) and r · j ∈ J j then q (J ) is an ideal in R/I .If K ⊆ R/I is an ideal then q −1(K ) is an ideal of R. Let t1, t2 ∈ q −

1(K ) theq (t1 + t2) = q (t1) + q (t2) ∈ K because K is an ideal, then t1 + t2 ∈ q −1(K ). If t ∈ q −1(K ) and r ∈ R then q (rṫ) = q (r) · q (t) ∈ K because K is an ideal of R/I .Then this means that r · t ∈ q −1(K ) and therefore q −1(K ) is an ideal in R.

Note that q (q −1(K )) = K , true for surjective map q of sets. If J is an ideal of Rthen we claim that the composition q −1(q (J )) is the ideal J + I . Assuming that if


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I ⊆ J then q −1(q (J )) = J , hence q is bijective when restricted to ideals containing I which is required. Now we need to check that q −1(q (J )) + J + I . Let x ∈ q −1(q (J ))then q (x) = q ( j) for some j ∈ J and so q (x − j ) = 0, then x − j ∈ I , becauseq (J ) ∈ R/I . However, x = j + (x − j) ∈ J + I so that q −1(q (J )) ⊆ J + I . The reverse

inclusion is clear because q ( j + i) = ( j + i) + I = j + I, ∀ j ∈ J, i ∈ I .

Theorem 2.6. First Isomorphism theorem Let f : R → S be a homomorphism of rings, and let I = ker (f ). Then f induces

an isomorphism f̄ : R/I → im (f ) given by:

f̄ (r + I ) = f (r) (2.2.8)

Proof. Note that if r − s ∈ I then f (r − s) = 0 and therefore f (r) = f (s), f take asingle value on each coset r + I so f̄ is well-defined. Clearly from the definition of the

ring structure on R/I it is a ring homomorphism. It is clearly surjective, if s ∈ im(f )then s = f (r) for some r ∈ R thus s = f̄ (r + I ). To see that f̄ is injective it is enoughthat ker( f̄ ) = 0. If f̄ (R + I ) = 0 ⇒ f (r) = 0 hence r ∈ I so that r + I = I .

Let R be a ring and A, B be subsets of R. Let A + B = {a + b : a ∈ A, b ∈ B}.

Theorem 2.7. Second Isomorphism theorem Let R be a ring and A a subring of R, B an ideal of R. Then A +B is a subring of

R and the restriction of the quotient map q : R → R/B to A induces an isomorphism

A/A ∩ B ∼= (A + B)/B (2.2.9)Proof. First let us check that A + B is a subring of R. It is clear that 1 ∈ A + Bsince 1 ∈ A and for 1 ∈ A + B, 1 = 1 + 0. Then using the two other conditions of thesubring criterion

(a1 + b1) × (a2 + b2) = a1a2 + b1a2 + a1b2 + b1b2 ∈B

∈ A + B (2.2.10a)

(a1 + b1) + (a2 + b2) = (a1 + a2) + (b1 + b2) ∈ A + B (2.2.10b)

It is easy to see that A ∩ B is an ideal in A and that B is an ideal in A + B, sotherefore the two quotients A/A ∩ B and (A + B)/B exist. We let q : R → R/Bbe the quotient map. The restriction of q to p : A → R/B, whose image is clearly(A + B)/B, so by the first isomorphism theorem it is enough to check that the kernelof p is in A ∩ B. It is clear: if a ∈ A has p(a) = 0 then a + B = 0 + B so that a ∈ Band so a ∈ A ∩ B.


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Theorem 2.8. Universal property of Quotients If f : R → S is a homomorphism and I ⊆ ker (f ) is an ideal, then there is a

unique homomorphism f̄ : R/I → S such that f = f̄ ◦ q where q : R → R/I is the quotient map.

Proof. Note that since q is surjective the requirement that f̄ (q (r)) = f (r) uniquelydetermines f̄ if it exists. But if r1 − r2 ∈ I then since I ⊆ ker(f ), we have 0 =f (r1 − r2) = f (r1) − f (r2) hence f (r1) = f (r2). Follows that f is constant on the I -cosets and so it induces a unique map f̄ : R/I → S such that f̄ ◦q = f . It is immediatefrom the definitions of the ring structure on R/I that f̄ is a homomorphism.

Theorem 2.9. Third Isomorphism theorem If I ⊆ J are ideals, then J/I = { j + I : j ∈ J } is an ideal in R/I and

(R/I )/(J/I ) ∼= R/J (2.2.11)

Proof. Let q I : R → R/I and let q J : R → R/J be the two quotient maps. By theuniversal property for the quotient q J : R toR/J/ applied to the homomorphism q I wesee that there is a homomorphism q̄ I : R/J → R/I induced by the map q I : R → R/I and q̄ ◦ q J = q I . Clearly q̄ I is surjective since q I , and if q̄ (r + j) = 0 then sincer + J = q J (r) so that q̄ I (r + J ) = q I (r) we have r ∈ I so that ker(q̄ I ) = J/I and theresult follows from the first homomorphism theorem.

2.3 Modules

Definition 14. Let R be a ring with identity 1R. A module over R is an abelian group (M, +) together with a multiplication action of R on M satisfying:

1R.m = m, ∀m ∈ M (2.3.1a)

(r1.r2)m = r1(r2.m), ∀r1, r2 ∈ R, m ∈ M (2.3.1b)

(r1 + r2)m = r1.m + r2.m, ∀r1, r2 ∈ R, m ∈ M (2.3.1c)

r.(m1 + m2) = r.m1 + r.m2, ∀r ∈ R, m1, m2 ∈ M (2.3.1d)

A concrete sense of a module M over a ring R, could be quickly understood if one

were to let R be a field, then the module would simply be a vector space over R. Amodule over a field are essentially vector spaces over that field.

Definition 15. If M is an R-module, a subset N ⊆ M is called a submodule if it is an abelian subgroup of M and whenever r ∈ R and n ∈ N then r.n ∈ N .

One extension from linear algebra is that of linear independence.


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Definition 16. If M is a module over R, we say a set S ⊆ M is linear independent if whenever we have an equation r1s1 + r2s2 + ... + rksk = 0 for ri ∈ R, s1 ∈ S 1 ≤ i ≤ kwe have r1 = r2 = ... = rk = 0. We can say that a set S is a basis for a module M if and only if it is linearly independent and it spans M . Any module which has a basis

is called a free module.

On the other hand there is a notion of a spanning set.

Definition 17. A spanning set for an R-module is a subset whose finite R-linear combinations fill up the module. These always exist, since the entire module is a spanning set.

To continue on we assume that all rings R are integral domains. As for vectorspaces, modules have a natural notion of a quotient module. An essential aspect of the quotient module is the condition that N be a submodule so that the multiplicationon M induced a module structure on M/N .

Definition 18. If N is a submodule of M , then in particular it is a subgroup of an abelian group, so we can form the quotient M/N . The module M/N is called the quotient module of M by N . We define multiplication on the quote module by:

r.(m + N ) = r.m + N, rinR,m + N ∈ M/N (2.3.2)

The multiplication well defined because if m1 + N = m2 + N we then have m1 −m2 ∈ N , same as quotient rings, and so r.(m1 − m2) ∈ N ⇒ r.m1 + N = r.m2 + N .Similar to rings, we can define a natural analogue of linear maps.

Definition 19. If M 1, M 2 are modules, we say that φ : M 1 → M 2 is a modulehomomorphism if:

φ(m1 + m2) = φ(m1) + φ(m2), ∀m1, m2 ∈ M 1 (2.3.3a)

φ(r.m) = r.φ(m), ∀r ∈ R, m ∈ M (2.3.3b)

Lemma 2.10. Submodule correspondence Let M be an R-module and N a submodule. Let q : M → M/N be the quotient

map. If S is a submodule of M then q (s) is a submodule of M/N , while if T is a submodule of M/N then q −1(T ) is a submodule of M . Moreover, the map T → q −1(T )gives an injective map from submodules of M/N to the submodules of M which contain N , thus submodules of M/N correspond bijectively to submodules of M which contain N .

Proof. We can check from the definitions that q (S ) and q −1(T ) are submodules of N and M respectively. As the proof for both are the same, we will only prove it forq (S ).

Let S be a submodule of M and q : M → M/N be the quotient map. Lets1, s2 ∈ S then q (s1), q (s2) ∈ M/N . Since S is a submodule then s1 + s2 ∈ S ⇒


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q (s1 + s2) = q (s1) + q (s2) ∈ M/N . Finally, if r ∈ R then r.s ∈ S , by definition of submodule, and φ(r.s) = r.φ(s) ∈ M/N . Therefore q (S ) is a submodule of M/N .

Now, let T be any subset of M/N . We have that q (q −1(T )) = T is surjectivebecause q is surjective. Also because know that q −1(T ) is always a submodule in M ,

the map of S → q (S ) is a surjective map from submodules in M to submodules inM/N and that T → q −1(T ) is an injective map, furthermore since q (N ) = 0 ∈ T for any submodule T of M/N we have N ⊆ q −1(T ) so that the image of the mapT → q −1(T ) consists of submodules of M which contain N . We only now need tocheck that the submodules of M of the form q −1(T ) are precisely these submodules.

Suppose that S is an arbitrary submodule of M and consider q −1(q (S )). Bydefinition

q −1(q (S )) = {m ∈ M : q (m) ∈ q (S )}

= {m ∈ M : ∃s ∈ S m + N = s + N }= {m ∈ M : m ∈ s + N } (2.3.4)

The right hand side is simply writing the submodule S + N , S/N . However, if N ⊆ S then we have S + N = S and therefore q −1(q (S )) = S and any submodule S which contains N is indeed the preimage of a submodule M/N .

Theorem 2.11. Universal property of quotients Suppose that φ : M → N is a homomorphism of R-modules, and S is a submodule

of M with S ⊂ ker (φ). Then there is a unique homomorphism φ̄ : M/S → N such

that φ = φ̄◦q where q : M toM/S is the quotient homomorphism, that is the following diagram commutes:

M N

M/S

φ

q φ̄

Moreover, ker (φ̄) is the submodule ker (φ)/S = {m + S : m ∈ ker (φ)}.

Proof. q is surjective, so the formula φ̄(q (m)) = φ(m) determines uniquely the valuesof φ̄, thus φ̄ is unique if it exists. If m−m ∈ S then because by assumption S ⊆ ker(φ)it follows that 0 = φ(m − m) = φ(m) − φ(m) since φ is a homomorphism. Thereforeφ is constant on the S -cosets. This induces a map φ̄(m + S ) = φ(m). φ̄ is then ahomomorphism because it follows directly from the module structure on the quotientM/S , and by definition φ = φ̄ ◦ q .


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φ̄

(m1 + S ) + (m2 + S )

= φ̄

m1 + m2 + S

= φ(m1 + m2)

= φ(m1) + φ(m2)

= φ̄(m1 + S ) + φ̄(m2 + S ) (2.3.5a)

φ̄

r.(m + S )

= φ̄(r.m + S )

= φ(r.m) = r.φ(m)

= r.φ̄(m + S ) (2.3.5b)

Note that φ̄(m + S ) = φ(m) = 0 if and only if m ∈ ker(φ), therefore the kernel of φ̄ is m + S ∈ ker(φ)/S .

As we have done above, we will now look at the isomorphism theorems for modules.First we let M be an R-module.

Theorem 2.12. First Isomorphism theorem If φ : M → N is a homomorphism then φ induces an isomorphism φ̄ : M/ker (φ) →

im (φ).

Proof. Let K = ker(φ). Using the universal property of quotients to K , then we haveker(φ̄) = ker(φ)/ker(φ) = 0 it follows that φ̄ is injective and therefore induces anisomorphism onto its image which from the equation φ̄ ◦ q = φ = I’m(φ).

Theorem 2.13. Second isomorphism theorem If M is an R-module and N 1, N 2 are submodules of M then

(N 1 + N 2)/N 2 ∼= N 1/N 1 ∩ N 2 (2.3.6)

Proof. First, let q : M → M/N 2 be the quotient map, which restricts p from N 1 toM/N 2, and whose image is clearly (N 1 + N 2)/N 2. Therefore by the first isomorphismtheorem it is only necessary to check that the kernel of p is N 1 ∩ N 2. If n ∈ N 1 has

p(n) = 0 then n + N 2 = 0 + n2 so that m ∈ N 2 and so n ∈ N 1 ∩ N 2.

Theorem 2.14. Third isomorphism theorem Suppose that N 1 ⊆ N 2 are submodules of M . Then we have

(M/N 1)/(N 2/N 1) ∼= M/N 2 (2.3.7)


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Proof. We first let q i : M → M/N i for i = 1, 2. Using the universal property of quotients for q 1 we see that there is a homomorphism q̄ 2 : M/N 1 → M/N 2 inducedby the map q 2 : M → M/N 2, with the kernel being ker(q 2/N 1) = N 2/N 1 and q̄ 2 ◦q 1 = q 2. Therefore q̄ 2 is surjective because q 2 is and the result follows from the first

isomorphism theorem.

These proofs mainly followed that of the proofs for the isomorphism theorems’ of rings.

3 Tensors and Tensor Products

3.1 Dual Spaces

The study of dual spaces is essential to understanding the notion of a tensor product.To formalize our understanding of the dual space we will look at real-valued functionsdefined on a real vector space T , f : T → R. The set of all such functions is thengiven the structure

(f + g)(v) = f (v) + (v), ∀v ∈ T (3.1.1a)

(αf )(v) = α(f (v)) ∀v ∈ T (3.1.1b)

0(v) = 0 ∀v ∈ T (3.1.1c)

(−f )(v) = −(f (v)) ∀v ∈ T, (3.1.1d)

which defines a vector space. Furthermore, we will restrict the space of all real-valued functions to only those that are linear

f (αu + β v) = αf (u) + βf (v) ∀α, β ∈ R, ∀u, v ∈ T. (3.1.2)

The real-valued linear functions on a real vector space are also called linear func-tionals and the space constitutes a vector space called the dual of T , denoted byT ∗.

For simplicity we will point out the two types of vectors, those in T an those inT ∗, and distinguish them from one another. For the basis vectors of T ∗ we will usesuperscripts and the components of vectors we will use subscripts, therefore if {ea}

is a basis of T ∗, then a vector λ ∈ T ∗ has the unique expression λ = a λaea.In a natural way, we will now prove that the that the dual space of T has the

same dimensionality as T .

Lemma 3.1. The dual space,T ∗, of T has the same dimensionality as T .


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Proof. First, let T be a N -dimenionsal vector space and let {ea} be the given basisof T . We define {ea} to be real-valued functions which maps λ ∈ T into the realnumber λa, which is the ath component relative to the basis vectors of T , {ea}. Theyare given to be

ea(λ) = λa λ ∈ T (3.1.3a)

ea(eb) = δ ab (3.1.3b)

This clearly gives N real-valued functions which satisfy Eq(3.1.3b). Now we willshow that they are linear and they constitute a basis for T ∗.

The linearity of the real-valued functions is shown by the restriction

ea(αλ + βµ) = αλa + βµa ∀α, β ∈ R ∀λ, µ ∈ T (3.1.4)

To prove that {ea

} is a basis for any given ν ∈ T ∗

we can define N real numbers,denoted ν a by ν (ea) = ν a. Then

ν (λ) = ν (λaea) = λaν (ea)

= λaν a

= ν aea(λ) ∀λ ∈ T (2.1.5)

Therefore for any ν ∈ T ∗ we have ν = ν aea, showing that {ea} stands T ∗. Theindependence of the basis vectors comes from Eq(3.1.3b). Thus dim(T ∗) = dim(T ).

Theorem 3.2. Let F be a linear functional on a normed space (E , · ). The following statements are then equivalent to each other:

1. F is continuous;

2. F is continuous at 0;

3. sup {|F (x)| : x ∈ E, x ≤ 1} < ∞

F is continuous if and only if it is bounded on the unit ball of E .

Proof. From statement 1 the proof of statement two is trivial.Assuming statement 2, we know ∀ > 0 ∃δ > 0 such that x < δ ⇒ |F (x)| < .

Take = 1 then there exists a δ > 0 such that x < δ ⇒ |F (x)| < 1. Therefore∀x ∈ E such that x ≤ 1 we have δx

2 < δ ⇒ |F ( δx

2 )|


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F

(x − y)

x − y

≤ M

|F (x) − F (y)| ≤ M x − y; (2.1.6)

line 2 follows from linearity of the linear functional. Let δ ≤ M

then it is clearthat F is continuous.

Lemma 3.3. Let E ∗ be the set of all continuous linear functionals on the normed space (E, · ) and F ∈ E ∗. Then F = supx∈E,x≤1|F (x)| is a norm on E

∗.

Proof. We simply need to check if the norm satisfies the following three conditions:

1. F > 0 if F = 0;

2. λF = |λ|F ∀λ ∈ C ∀F ∈ E ∗

3. F + F ≤ F + F ∀F, F ∈ E ∗

Clearly conditions 1 and 2 are easily shown by definition of the norm. Condition3 is satisfied by the triangle inequality.

Theorem 3.4. The set E ∗ of all continuous linear functionals on the normed space (E, · ) is itself a Banach space with respect to pointless algebraic operations and

norm

F = supx∈E,x≤1F (x) (3.1.7)

Proof. E ∗ is a vector space over the same field as E . From Theorem 2.3, we knowthat F is a real number and that it is finite. Lemma 2.3 also tells us that the givennorm is a norm on E ∗. We will now show that E is complete.

Let (F n) be a Cauchy sequence in E ∗ so that F n − F m → 0 as n, m → ∞.

Therefore ∀x ∈ E ⇒ |F n(x) − |F m| → 0 as n, m → ∞. We see that this is Cauchysequence of scalars; we denote the limit of this scalar sequence by F (x). We mustnow show that F ∈ E ∗ and that F n → F with respect to the norm on E ∗.

Let > 0 and pick n0 ∈ N such that m, n ≥ n0 ⇒ F m − F n < ∀x ∈ E such that x ≤ 1 and m, n ≥ n0,

|F m(x) − F n(x)| < (3.1.8)

Letting m → ∞, ∀n ≥ n0 and ∀x x ≤ 1,

|F (x) − F n(x)| ≤ (3.1.9)


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Since ∀x ∈ E x ≤ 1 implies F − F 0 and F n0 are bounded, then F is alsobounded on the unit ball and by Theorem 2.3 it is continuous. Thus F ∈ E ∗.Whenever n ≥ n0 ⇒ F − F n ≤ implies that F n → F ∈ E ∗ and so E ∗ is complete.

Theorem 3.5. For any vector x in a normed space E and any continuous linear functional F on E ,

|F (x)| ≤ F x. (3.1.10)

Proof. Suppose that x ∈ E x = 0. Then xx

is a unit vector and therefore

F

x

x

≤ F

|F (x)|x ≤ F (2.1.11)

Lemma 3.6. Let V be an inner product space. ∀x,y,z ∈ V , if (x, z ) = (y, z ) ∀z ∈V ⇒ x = y.

Proof. If (x, z ) = (y, z ) then

0 = (x, z ) + (−1)(y, z )

= (x, z ) + (−y, z )(x − y, z ) (2.1.12)

If this is true for all z ∈ V then z = x − y ⇒ x − y = 0.

Lemma 3.7. Let M be a linear subspace of an H and let x ∈ H . Then x ∈ M ⊥ if and only if

x − y ≥ x ∀y ∈ M (3.1.13)

Proof. (⇒) If x ∈ M ⊥

then, ∀y ∈ M x and y are orthogonal and Pythagoras’ theoremgives

x − y2 = x2 + y2 ≤ x2 (3.1.14)

(⇐) Suppose that Eq(3.1.13) holds. ∀y ∈ M and ∀λ ∈ C, then λy ∈ M and

x − λy2 ≥ x2. (3.1.15)


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Rewriting this as an inner product we have

(x − λy,x − λy) ≥ x2

x2 − λ̄(x, y) − λ(x, y)− + |λ|2y2 ≥ x2

−2Reλ̄(x, y) + |λ|2y2 ≥ 0. (2.1.16)

The last line comes from ∀z ∈ C ⇒ z + z̄ = 2Re{z }. Eq(2.1.16) holds for anyλ ∈ C, it also holds for λ = tz where t > 0 and z ∈ C, where |z | = 1, thenz̄ (x, y) = |(x, y)|. Thus

−2t|(x, y)| + t2y2 ≥ 0

(x, y)| ≤ 1

2

ty2. (2.1.17)

Letting t → ∞ the result shows (x, y) = 0.

Lemma 3.8. Let M be a closed linear subspace of a Hilbert space H and let x ∈ H .There exists y ∈ M z ∈ M ⊥ such that x = y + z .

Proof. Take y ∈ M and let it be the closest point to x in M , by definition

x − y ≤ x − m ∀m ∈ M. (3.1.18)

Let z = x − y then x = y + z . ∀m ∈ M ⇒ y + m ∈ M and so

z = x − y ≤ x − (y + m)

≤ z − m ∀m ∈ M. (2.1.19)

By Lemma 2.7, z ∈ M ⊥.

Theorem 3.9. Riesz-Frechet Theorem Let H be a Hilbert space and let F be a continuous linear functional on H . There

exists a unique y ∈ H such that

F (x) = (x, y) ∀x ∈ H. (3.1.20)

Furthermore y = F .


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Proof. y is unique because

(x, y) = F (x) = (x, y) ∀x ∈ H (3.1.21)

implies y = y

by Lemma 2.6.A trivial case of F is when F is the zero operator and we take y = 0.Let M be the kernel of the linear functional F ,

M = KerF = {x ∈ H : F (x) = 0}, (3.1.22)

and M is a proper closed subspace of H . From Lemma 2.8, we know that H =M ⊕ M ⊥ and therefore M ⊥ = {0}. Let z ∈ M ⊥ z = 0. We can scale z with a scalarsuch that F (z ) = 1. Then pick a z ∈ M ⊥ such that for any x ∈ H

x = (x − F (x)z ) + F (x)z (3.1.23)

Because H is a direct some of M and it’s orthogonal complement M ⊥, the firstterm on the right hand expression is an element of M and the second term is anelement of M ⊥. Taking an inner product of both sides with z ,

(x, z ) = (F (x)z, z ) = F (x)z 2 ∀x ∈ H, (3.1.24)

since z ⊥ M . If we let y = zz2

we have

(x, y) = F (x) (3.1.25)

If x ≤ 1 then by Cauchy-Schwarz,

|F (x)| = |(x, y)| ≤ xy ≤ y (3.1.26)

Let x = yy

, which is a unit vector, and therefore

F ≥ |F (x)| = |F (y)|

y =

|(y, y)|

y = y. (3.1.27)

Therefore F = y.

3.2 Tensor Product

Tensor products came around for vector spaces due to its inherent need in physicsand engineering. The following description of tensor products is of vector spaces.

Let R be a commutative ring and M and N be R-modules. We assume thatthe rings R will have a multiplicative identity and that the modules are to be unital1 · m = m, ∀m ∈ M . The product operation M ⊗R N , is calle the tensor product.However, we will begin with the tensor product of vector spaces first.


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Let V and W be vector spaces over a field K , and let {ei} and {f j} be a basisfor V and W respectively. Then the tensor product of these two vector spaces isV ⊗K W defined to be the K -vector space with a basis of formal symbols ei ⊗ f j;simply claim that these new symbols are linearly dependent by definition. Then the

tensors or elements of the K -vector space are written in formal sums with the formi,j cijei ⊗ f j with cij ∈ K . Furthermore, ∀v ∈ V, w ∈ W we define v ⊗ w to be the

element of V ⊗K W given by writing v and w in terms of the original bases of bothV and W then expanding as if the ⊗ is a non commutative product. For example wecould let V = W = R2 Then we would have a 4 dimensional space R2 ⊗R R

2 with 4basis vectors: e1 ⊗ e1, e1 ⊗ e2, e2 ⊗ e1, e2 ⊗ e2. If we let v = e1 + 2e2 and w = −3e1 + e2,then

v ⊗ w = (e1 + 2e2) ⊗ (−3e1 + e2) = −3e1 ⊗ e1 + e1 ⊗ e2 − 6e2 ⊗ e1 + 2e3 ⊗ e3 (3.2.1)

If one were to pick another basis of R2 for the tensor product, we realize that v ⊗whas a meaning in the space R2 ⊗R R2 independent of the chosen basis. Looking backat a more generalized version, with modules, but the properties of the product ⊗ areto be satisfied in general, and not just on a basis: for R-modules M and N , theirtensor product M ⊗R N is an R-modules spanned, by a spanning set, by all symbolsm ⊗ n, m ∈ M, n ∈ N . The symbols satisfy these distributive laws:

(m + m) ⊗ n = m ⊗ n + m ⊗ n (3.2.2a)

m ⊗ (n + n) = m ⊗ n + m ⊗ n (3.2.2b)

r(m ⊗ n) = (rm) ⊗ n = m ⊗ (rn) (3.2.2c)

∀r ∈ Rm,m ∈ M,n,n ∈ N

In the space of M ⊗R N there are usually more elements than products m ⊗ n.The general element of M ⊗R N , similar to linear algebra, is called a tensor and is anR-linear combination

r1(m1 ⊗ n1) + r2(m2 ⊗ n2) + ... + rk(mk + nk), (3.2.3a)

k ≥ 1, ri ∈ R, mi ∈ M, ni ∈ N

Since we know from above that ri(mi ⊗ ni) = (rimi) ⊗ ni then let rimi be renamed just as mi. Therefore the above linear combination can be written as a sum

ki=1

mi ⊗ ni = m1 ⊗ n1 + ... + mk ⊗ nk. (3.2.4)

The idea of two sums equal in M ⊗RN is not trivial in terms of the given descriptionof a tensor product. However, there is one case: let M and N be free R-modules with


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bases {ei} and {f j} respectively. The resulting space M ⊗rN is then also a free modulewith basis {ei ⊗ f j}. An element is then equal to another when the coefficients of thefinite sum

i,j cijei ⊗ f j are the same.

In a general case, where both M and N don’t have bases, equality in M ⊗R N

relies on using a universal mapping property of the tensor product: M ⊗R N is theuniversal object that turns bilinear maps on M × N into linear maps. Before wecontinue to define the tensor product rigorously, we will first define a bilinear map.

Definition 20. A function B : M × N → P , where M, N , and P are R-modules, is called bilinear when it is linear in each argument with the other one fixed:

B(m1 + m2, n) = B(m1, n) + B(m2, n) (3.2.5a)

B(rm,n) = rB(m, n) (3.2.5b)

B(m, n1 + n2) + B(m, n1) + B(m, n2) (3.2.5c)

B(m,rn) = rB(m, n) (3.2.5d)∀r ∈ R,m,m1, m2 ∈ M,n,n1, n2 ∈ N

This idea of bilinearity can be extended further to multilinearity.Now, any bilinear map M × N → P to an R- module P can be composed with a

linear map P → Q to induce a map M × N → Q that is bilinear.

M × N P

Q

bilinear

bilinear linear

The construction of the tensor product of M and N will be the solution to auniversal mapping problem: find an R-module T and bilinear map b : M × N → T such that every bilinear map on M × N is the composite of the bilinear map b and aunique linear map out of T . By definition then,

Definition 21. The tensor product M ⊗RN is an R-module equipped with a bilinear

map M × N ⊗−→ M ⊗R N such that for any bilinear map M × N

B−→ P there is a linear

map M ⊗R N L−→ P making the following diagram commute:

M × N M ⊗R N

P

⊗

B L


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Despite the definition M ⊗R N is not explicitly given but instead gives a universalmapping property involving it. Note that maps out of M × N are bilinear and thoseout of M ⊗R N are linear; bilinear maps out of M × N turn into linear maps out of M ⊗R N . We can show that any of the two constructions are the same.

Let T, T be R-modules, and b : M × N → T and b : M × N → T satisfy theuniversal mapping property of the tensor product of M and N . From the definition,the universality of b the map b factors uniquely through T : a unique linear mapf : T → T . Furthermore, the universality of b induces a unique factorization in Bthrough T : a unique linear map f : T → T . Combining the two together we have:

T

M × N T

T

b

b

b

f

f

We also notice that there is an unique mapping from f ◦ f : T → T . This isalso because of the universality of the map b to the R-module T . We can let thecomposition map of f ◦ f = idT be the identity map. Similarly we have f ◦ f = idT .Therefore both T and T are isomorphic R-modules by f and also f ◦ b = b, whichmeans f identifies b with b. Any two tensor products of M and N can be identified

with each other in a unique way compatible with the distinguished bilinear maps tothem from M × N .

Theorem 3.10. The tensor product of M and N exists.

Proof. The tensor product of M and N , M ⊗R N will be the quotient module of afree R-module. For any set S , we will write the free R-module on S as F R(S ) =

s∈S Rδ s, where δ s is the term with 1 in the s-position and 0 everywhere else. ie. if s = s ∈ S ⇒ rδ s − rδ s has r in the s-position, −r in the s-position, 0 elsewhere.Let S = M × N :

F R(M × N ) = (m,n)∈M ×N

Rδ (m,n). (3.2.6)

Let D be a submodule of F R(M × N ) spanned by all the elements of:

δ (m+m,n) − δ (m,n) − δ (m,n), δ (m,n+n) − δ (m,n) − δ (m,n), δ (rm,n) − δ (m,rn)

rδ (m,n) − δ (rm,n), rδ (m,n) − δ (m,rn). (3.2.7a)


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The tensor product is then the quotient module by D:

M ⊗R N := F R(M × N )/D. (3.2.8)

We write the coset δ (m,n)

+D in M ⊗R

N as m⊗

n. Remembering that the elementsof D are equivalent to 0 in F R(M × N )/D gives us

δ (m+m,n) ≡ δ (m,n) + δ (m,n)

(m + m) ⊗ n = m ⊗ n + m ⊗ n ∈ M ⊗R N. (3.2.9a)

The other relations of a bilinear function also hold. This shows that the function⊗ : M × N → M ⊗R N is bilinear. Now, we need to show that all bilinear maps out of M × N factor uniquely through the bilinear map ⊗. Suppose P is an R-module andB : M × N → P is a bilinear map. Treat M × N as a set, so B is just a function onthis set, the universal mapping property of free modules extends B from a function

M × NtoP to a linear function : F R(M × N ) → P with (δ (m,n) = B(m, n). Wewant to show that make sense as a function on M ⊗R N ; show that D ⊆ ker()From the linearity of B ,

B(m + m, n) = B(m, n) + B(m, n), B(m, n + n) = B(m, n) + B(m, n)

rB(m, n) = B(rm,n) = B(m,rbn) (3.2.10a)

therefore

(δ (m+m,n)) = (δ (m,n)) + (δ (m,n)), (δ (m,n+n)) = (δ (m,n)) + (δ (m,n))

r(δ (m,n)) = (δ (rm,n)) = (δ (m,rn)). (3.2.11a)

Since is linear, these conditions must hold as well:

(δ (m+m,n)) = (δ (m,n) + δ (m,n)), (δ (m,n+n)) = (δ (m,n) + δ (m,n)),

(rδ (m,n)) = (δ (rm,n)) = (δ (m,rn)). (3.2.12a)

Therefore the kernel of contains all the generators of the submodule D, so will induce a linear map L : F R(M × N )/D → P where L(δ (m,n) + D) = (δ (m,n)) =B(m, n). Since F R(M × N )/D = M ⊗R N and δ (m,n) + D = m ⊗ n there for the

diagram below commutes as such:

M × N M ⊗R N

P

⊗

B L


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This diagram shows that every linear map B out of M × N comes from a linearmap L out of M ⊗R N such that L(m ⊗ n) = B(m, n), ∀m ∈ M, n ∈ N . Now weshow that the linear map L is the only one that makes the diagram commute.

The definition of M ⊗R N as a quotient of the free module F R(M × N ) tells us

that every element of F R(M × N ) is a finite sum

r1δ (m1,n1) + r2δ (m1,n2) + · · · + rk2δ (mk,nk). (3.2.13)

The reduction map F R(M × N ) → F R(M × N )/D = M ⊗R N is linear, so everyelement of M ⊗R N is a finite sum of the form

r1(m1 ⊗ n1) + r2(m1 ⊗ n2) + · · · + rk2(mk ⊗ nk). (3.2.14)

Therefore this means the elements of m⊗n in M ⊗RN span it as a R-module. Thelinear maps out of M ⊗R N are determined by the values of all the elements m ⊗ n,

thus there is at most one linear map L : M ⊗R N → P such that m ⊗ n → B(m, n).

Definition 22. Tensors in M ⊗R N that have the form m ⊗ n are called elementary tensors.

Similar to linear algebra elements of the the tensor product, or tensors, are gen-erally linear combinations of elementary tensors.

We will now prove two simple theorems that run parallel with our idea of linearalgebra.

Theorem 3.11. Le M and N be R-modules with respective spanning sets {xi}i∈I and

{y j} j∈J . The tensor product M ⊗R N is spanned linearly by the elementary tensors xi ⊗ y j.

Proof. Since an elementary tensor in M ⊗R N has the form m ⊗ n, we write m andn as spanned by the sets:

m =i

aixi, n = j

b jy j (3.2.15)

where ai = b j = 0 for all but finitely many i and j . Since ⊗ is bilinear we have

m ⊗ n = i aixi ⊗ j b jy j = i,j aib jxi ⊗ y j (3.2.16)which is a linear combination of the tensors xi ⊗ y j. Since every tensor is a sum

of elementary tensors, the xi ⊗ y j ’s span M ⊗R N as an R-module.

Theorem 3.12. In M ⊗R N , m ⊗ 0 = 0 × n = 0.


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Proof. Since m ⊗ n is additive in n with m fixed, we have

m ⊗ 0 = m ⊗ (0 + 0) = m ⊗ 0 + m ⊗ 0 ⇒ m ⊗ 0 = 0. (3.2.17)

The same argument is applied to 0 ⊗ n.

3.3 Tensor products and the Hilbert-Schmidt Class

Now we will look at specific examples of tensor products limited to Hilbert spaces.In defining the (Hilbert) tensor product H of two Hilbert spaces H1, H2, the

approach we take will utilize and emphasize what we did above, the ”universal”property of the tensor product. In terms of operators, rather than mappings, TheHilbert space H is characterized, up to isomorphism, by the existence of a bilinearmapping p : H1×H2 → H. It has the following property that each ”suitable” bilinear

mapping L from H1 × H2 into a Hilbert space K has a unique factorization L = T p,with T being a bounded linear operator from H into K.

Before the formal construction of the theory, we need to understand the intuitiveaspects of it. When x1 ∈ H1, x2 ∈ H2, we want to view the element p(x1, x2) ∈ H asa ”product” x1 ⊗ x2. The linear combinations of such products form an everywhere-dense subspace of H. The bilinearity of p implies that these products satisfy certainlinear relations:

(x1 + y1) ⊗ (x2 + y2) − x1 ⊗ x2 − x1 ⊗ y2 − y1 ⊗ x2 − y1 ⊗ y2 = 0

x1, y1 ∈ H1, x2, y2 ∈ H2 (3.3.1a)

All the linear relations satisfied by product vectors can be achieved by use of thebilinearity of p. The inner product on H satisfies the conditions:

x1 ⊗ x2, y1 ⊗ y2 = x1, y1x2, y2 (3.3.2a)

x1 ⊗ x22 = x1 ⊗ x2, x1 ⊗ x2

= x1, x1x2, x2 = x12x2

2

x1 ⊗ x2 = x1x2. (3.3.2b)

Our construction of the Hilbert space H, the elements of H are complex-valuedfunctions defined on the product H1 × H2 and conjugate-linear in both variables. If v1 ∈ H1, v2 ∈ H2, v1 ⊗ v2 is the function that assigns the value v1, x1v2, x2 to theelement (x1, x2) ∈ H1 × H2.

Now, suppose that H1,..., Hn are Hilbert spaces and φ is a mapping from thecartoon product of all these Hilbert spaces into the scalar field C. φ is called a


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bounded multilinear functional on H1 × · · · × Hn if φ is linear in each of its variables,assuming that the other variables remain fixed, and there is a real number c such that

|φ(x1,...,xn)| ≤ cx1 · · · xn, x1 ∈ H1,...,xn ∈ Hn. (3.3.3)

If this is so then the lest such constant c is denoted by the norm of the mappingφ. Then φ is a continuous mapping from H1 × · · · × Hn → C relative to the producttor the norm topologies on the Hilbert spaces.

Theorem 3.13. Suppose that H1,..., Hn are Hilbert spaces and φ is a bounded mul-tilinear functional on H1 × · · · × Hn.

(i) The sum

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 (3.3.4)

has the same finite or infinite value for all orthonormal bases Y 1 of H1,..., Y n of Hn.

(ii) If K1,..., Kn are Hilbert spaces, Am ∈ B (Hm, Km), (m = 1,...,n), ψ is a bounded multilinear functional on K1 × · · · Kn, and

φ(x1,...,xn) = ψ(A1x1,...,Anxn)

x1 ∈ H1,...,xn ∈ Hn (3.3.5)

then

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 ≤ A1

2 · · · An2z1∈Z 1

· · ·zn∈Z n

|φ(z 1,...,z n)|2, (3.3.6)

when Y m and Z m are orthonormal bases of Hm and Km, respectively where m =1,...,n.

Proof. To prove (i) it is sufficient to show that

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 ≤

z1∈Z 1

· · ·zn∈Z n

|φ(z 1,...,z n)|2, (3.3.7)

, whenever Y m, Z m are orthonormal bases of Hm, m = 1,...,n. Note that sinceY m, Z m are orthonormal bases we can represent the basis vectors in one in terms of alinear combination of the other.

y1 =z1∈Z 1

y1, z 1z 1,...,yn =

zn∈Z n

yn, z nz n (3.3.8)

Therefore,


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y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 =

y1∈Y 1

· · ·yn∈Y n

|φ(z1∈Z 1

y1, z 1z 1,...,

zn∈Z n

yn, z nz n)|2

= y1∈Y 1

· · · yn∈Y n

| z1∈Z 1

· · · zn∈Z n

y1, z 1 · · · yn, z nφ(z 1,...,z n)|2

≤z1∈Z 1

· · ·zn∈Z n

z 12 · · · z n

2|φ(z 1,...,z n)|2

=z1∈Z 1

· · ·zn∈Z n

|φ(z 1,...,z n)|2

(3.3.9a)

The third line is arrived by noticing that we can rewrite the z m ∈ Z m basis interms of ym ∈ Y m

z 1 =y1∈Y 1

z 1, y1y1,...,z n =yn∈Y n

z n, ynyn. (3.3.10)

Also note that by definition ym, z m = z m, ym and that

z m2 =

ym∈Y m

z m, ymym,

ym∈Y m

z m, ymym

=

ym∈Y m

ym, z mym, ym, z mym

= ym∈Y m

|ym, z m|2ym2

=

ym∈Y m

|ym, z m|2 (3.3.11a)

Using the same argument and exchanging the orthonormal bases we can showequality.

Then for the proof of (ii), we suppose that 1 ≤ m ≤ n we choose and fix vectorsy1 ∈ Y 1,...,ym−1 ∈ Y m−1, z m+1 ∈ Z m+1,...,z n ∈ Z m. The mapping

z → ψ(A1y1,...,Am−1ym−1, z , z m+1,...,z n) : Km → C (3.3.12)is a bounded linear functional on Km, so ∃w ∈ Km such that

ψ(A1y1,...,Am−1ym−1, z , z m+1,...,z n) = z, w, z ∈ Km (3.3.13)

From Parseval’s equation


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ym∈Y M

|ψ(A1y1,...,Am−1ym−1, z , z m+1,...,z n)|2

= ym∈Y M

|Amym, w|2 = ym∈Y M

|ym, A∗mw|2

= A∗mw2 ≤ Am

2w2 = Am2

zm∈Z m

|z m, w|2

= Am2

zm∈Z m

|ψ(A1y1,...,Am−1ym−1, z , z m+1,...,z n)|2. (3.3.14a)

A further summation now yields

y1∈Y 1

· · · ym∈Y m

zm+1∈Z m+1

· · · zn∈Z n

|ψ(A1y1,...,A

mym

, z m+1

,...,z n

)|2

≤ Am2y1∈Y 1

· · ·

ym−1∈Y m−1

zm∈Z m

· · ·

zn∈Z n

|ψ(A1y1,...,Am−1ym−1, z m,...,z n)|2

(3.3.15)

Therefore

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 =

y1∈Y 1

· · ·yn∈Y n

|ψ(A1y1,...,Anyn)|2

≤ An2 y1∈Y 1

· · · yn−1∈Y n−1

zn∈Z n

|ψ(A1y1,...,An−1yn−1, z n)|2

≤ An−12An

2

×y1∈Y 1

· · ·

yn−2∈Y n−2

zn−1∈Z n−1

zn∈Z n

|ψ(A1y1,...,An−2yn−2, z n−1, z n)|2

≤ · · · ≤ A12 · · · An

2z1∈Z 1

· · ·zn∈Z n

|ψ(z 1,...,z n)|2 (3.3.16)

With H1,..., Hn Hilbert spaces, a mapping φ : H1 × · · · × Hn → C is called aHilbert-Schmidt functional on H1 × · · · × Hn if it si a bounded multilinear functional,and the sum of Eq(3.3.4) is finite for any choice of the orthonormal bases Y 1 in H1,..., Y n in Hn.

Theorem 3.14. If H1,..., Hn are Hilbert spaces, the set HSF of all Hilbert-Schmidt functionals on H1 × · · · × Hn is itself a Hilbert space when the linear structure, inner product, and norm are defined by


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(aφ + bψ)(x1,...,xn) = aφ(x1,...,xn) + bψ(x1,...,xn) (3.3.17)

φ, ψ = y1∈Y 1 · · · yn∈Y n φ(y1,...,yn)ψ(y1,...,yn) (3.3.18)φ2 =

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2

12

, (3.3.19)

respectively, where Y m is an orthonormal basis in Hm, m = 1,...,n. The sum given by Eq(3.3.18) is absolutely convergent, and the inner product and norm do not depend on the choice of the orthonormal bases Y 1,...,Y n.

For each v(1) in H1, ... v(n) in Hn the equation

φv(1),...,v(n)(x1,...,xn) = x1, v(1) · · · xn, v(n), (x1 ∈ H1,...,xn ∈ Hn) (3.3.20)

defines an element φv(1),...,v(n) of HSF , and

φv(1),...,v(n), φw(1),...,w(n) = w(1), v(1) · · · w(n), v(n) (3.3.21)

φv(1),...,v(n)2 = v(1) · · · v(n). (3.3.22)

The set {φy(1),...,y(n) : y(1) ∈ Y 1,...,y(n) ∈ Y n} is an orthonormal basis of HSF .There is a unitary transformation U from HSF onto l2(Y 1 × · · · × Y n), such that U φis the restriction φ|Y 1 × · · · × Y n when φ ∈ HS F .

Proof. Choose an orthonormal basis Y m in Hm, m = 1,...,n, and then associate witheach bounded multilinear functional φ on H1 × · · · × Hn the complex-valued functionUφ obtained by restricting φ to Y 1 × · · · × Y n. Remember that the condition for aHilbert-Schmidt functional is that if φ is a Hilbert-Schmidt functional if and only if

U φ ∈ l2(Y 1 × · · · × Y n). (3.3.23)

If U φ = 0, then

φ(y1,...,yn) = 0 y1 ∈ Y 1,...,yn ∈ Y n. (3.3.24)

Because Y m is an orthonormal basis, its closed linear span is Hm, then it followsfrom the multi linearity and continuity of φ that φ vanishes throughout H1× · · · × Hn.

Let φ, ψ be Hilbert-Schmidt functionals on H1 × · · · × Hn, then the same istrue of aφ + bψ for all scalars a, b; aφ + bψ is a bounded multilinear functional,Uφ,Uψ ∈ l2(Y 1 × · · · × Y n) and thus

U (aφ + bψ) = aU φ + bUψ ∈ l2(Y 1 × · · · × Y n) (3.3.25)


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The sum, given by Eq(3.3.18), can be rewritten in the form

y∈Y 1×···×Y n

(U φ)(y)(U ψ)(y) (3.3.26)

and it is absolutely convergent with the sum Uφ,Uψ, the inner product inl2(Y 1 × · · · × Y n) of the two restricted functionals U φ and Uψ.

The set HSF of all Hilbert-Schmidt functionals on H1×···×Hnis a complex vectorspace. Eq(3.3.18) then defines an inner product on HS F the restriction U |HSF is aone-to-one linear mapping from HSF into l2(Y 1 × · · · × Y n), and Uφ,Uψ = ∠φ, ψwhen φ, ψ ∈ HSF . The inner product on l2(Y 1 × · · · × Y n) is definite, so is that onHSF ; if φ ∈ HSF and φ, φ = 0, we have Uφ,Uφ = 0, whence U φ = 0 ⇒ φ = 0.From this we see that HSF is a pre-Hilbert space, and it is apparent from Eq(3.3.18)that the norm, denoted 2 in HSF is given by Eq(3.3.19).

From Theorem 3.13, this norm is independent of the choice of the orthonormal

bases Y 1,..,Y n; this is the same of the inner product on HSF .Now we will show that U brings HSF onto the whole of the l2 space. Let f ∈

l2(Y 1 × · · · × Y n) and xm ∈ Hm, m = 1,...,n, the Cauchy-Schwarz inequality andParseval equation gives

y1∈Y 1

· · ·yn∈Y n

|f (y1,...,yn)x1, y1 · · · xn, yn|

≤

y1∈Y 1

· · ·yn∈Y n

|f (y1,...,yn)|2

12

×

y1∈Y 1

· · ·yn∈Y n

|x1, y1|2 · · · | xn, yn|

2 12

= f

y1∈Y 1

|x1, y1|2

12

· · ·

yn∈Y n

|xn, yn|2

12

= f x1 · · · xn (3.3.27a)

From this, the equation

φ(x1,...,xn) = y1∈Y 1 · · · yn∈Y n f (y1,...,yn)x1, y1 · · · xn, yn (3.3.28)defines a bounded multilinear functional φ on H1 × · · · × Hn, with φ ≤ f .

The orthonormality of the sets Y 1,...,Y n, leads to the fact that

(Uφ)(y1,...,yn) = φ(y1,...,yn) = f (y1,...,yn) y1 ∈ Y 1,...,yn ∈ Y n (3.3.29)


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so U φ = f . Furthermore φ ∈ HS F since U φ ∈ l2(Y 1 × · · · × Y n), whence U carriesHSF onto the l2 space.

Since U is a norm preserving linear mapping from HSF onto l2(Y 1 × · · · × Y n)completeness of the l2 space entails completeness of HSF ; so HSF is a HIlbert space,

and U is a unitary operator.When v(1) ∈ H1,...,v(n) ∈ Hn, φv(1),...,v(n) is a multilinear functional on H1 × · ·

· × Hn, and is bounded since

|φv(1),...,v(n)(x1,...,xn)| ≤ v(1) · · · v(n)x1 · · · xn (3.3.30)

by the Cauchy-Schwarz inequality. Furthermore, Parseval’s equation gives

y1∈Y 1

· · ·yn∈Y n

|φv(1),...,v(n)(y1,...,yn)|2

= y1∈Y 1

· · · yn∈Y n

|y1, v(1)|2 · · · | yn, v(n)|2

=

y1∈Y 1

|y1, v(1)|2

· · ·

yn∈Y n

|yn, v(n)|2

= v(1)2 · · · v(n)2. (3.3.31)

Hence φv(1),...,v(n) ∈ HS F and φv(1),...,v(n)2 = v(1) · · · v(n). Using Parseval’sequation again and by absolute convergence,

φv(1),...,v(n), φw(1),...,w(n)

=y1∈Y 1

· · ·yn∈Y n

φv(1),...,v(n)(y1,...,y(n))φw(1),...,w(n)(y1,...,yn)

=y1∈Y 1

· · ·yn∈Y n

y1, v(1) · · · yn, v(n)w(1), y1 · · · w(n), yn

=

y1∈Y 1

w(1), y1y1, v(1)

· · ·

yn∈Y n

w(n), ynyn, v(n)

= w(1), v(1) · · · w(n), v(n) (3.3.32)

When y(1) ∈ Y 1,...,y(n) ∈ Y n, the orthonormality of Y 1,...,Y n implies thatUφy(1),...,y(n) is the function that takes the value 1 at (y(1),...,y(n)) and 0 elsewhereon Y 1 × · · · × Y n. Therefore

{U φy(1),...,y(n) : y(1) ∈ Y 1,...,y(n) ∈ Y n} (3.3.33)

is an orthonormal basis of l2(Y 1 × · · · × Y n), and therefore


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{φy(1),...,y(n) : y(1) ∈ Y 1,...,y(n) ∈ Y n} (3.3.34)

is a basis of HSF .

Let us introduce the notion of the ”conjugate” of a Hilbert space H. For a ”nor-mal” Hilbert space, H has the algebraic structure and inner product defined by themappings

(x, y) → x + y : H × H → H (3.3.35a)

(a, x) → ax : C × H → H (3.3.35b)

(x, y) → x, y : H × H → C. (3.3.35c)

However we define the ”conjugate” Hilbert space with a slight twist.

Definition 23. The conjugate Hilbert space H̄ is the same set H, with the alge-braic structure and inner product defined by the mappings:

(x, y) → x + y : H × H → H (3.3.36a)

(a, x) → ā·x : C × H → H (3.3.36b)

(x, y) → x, y− : H × H → C (3.3.36c)

where

ā·x = āx and x, y− = y, x. (3.3.37)

It is also very easy to see that the conjugate Hilbert space of H̄ is H.A useful aspect that arises is a subset of a Hilbert space is linearly independent,

orthogonal, or orthonormal, or an orthonormal basis of that space, if and only if ithas the same property relative to the conjugate Hilbert space. If H1 and H2 areHilbert spaces and T is a mapping from the set H1 into the set H2, the linearity of T : H1 → H2 is equivalent to linearity of T : H̄1 → H̄2, and corresponds to conjugate-linearity of T : H1 → H̄2 and of T : H̄1 → H2. Of course, continuity of T is the samein all four situations, when T is linear the operators have the same bound, since thenorm on H j is the same as that on H̄ j.

Definition 24. Suppose that H1, ..., Hn and K are Hilbert spaces and L is a mapping from H1 × · · · × Hn into K. L is a bounded multilinear mapping if it is linear in each of its variables and there is a real number c such that

L(x1,...,xn) ≤ cx1 · · · xn. (3.3.38)

The least such constant c is denoted by L.


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By a weak Hilbert-Schmidt mapping from H1 × · · · × Hn into K, we mean abounded multilinear mapping L with the properties:

(i)

Lu(x1,...,xn) = L(x1,...,xn), u ∀u ∈ K (3.3.39)

is a Hilbert-Schmidt functional on H1 × · · · × Hn.(ii) There is a real number d such that Lu2 ≤ du for each u ∈ K.When these conditions are fulfilled, the least possible value of the constant d is

denoted by L2.A bounded multilinear mapping L : H1 × · · · × Hn → K is (jointly) continuous

relative to the norm topologies on the Hilbert spaces. We see that condition (ii)actually flows from (i) by an application of the closed graph theorem to the mapping.

Theorem 3.15. Suppose that H1,..., Hn are Hilbert spaces.(i) There is Hilbert space H and weak Hilbert-Schmidt mapping p : H1 × · · · ×

Hn → H with the following property: given any weak Hilbert-Schmidt mapping L from H1 × · · · × Hn into a Hilbert space K, there is a unique bounded linear mapping T from H into K, such that L = T p; moreover, T = L2.

(ii) If H and p have the properties attributed in (i) to H and p, there is a unitary transformation U from H onto H such that p = U p.

(iii) If vm, wm ∈ Hm and Y m is an orthonormal basis of Hm, m = 1,...,n, then

p(v1,...,vn), p(w1,...,wn) = v1, w1 · · · vn, wn, (3.3.40)

the set { p(y1,...,yn) : y1 ∈ Y 1,...,yn ∈ Y n} is an orthonormal basis of H, and p2 = 1.

Proof. Let H̄m be the conjugate Hilbert space of Hm and let H be the set of allHilbert-Schmidt functionals on H̄1 × · · · × H̄n with the Hilbert space structure givenin Theorem 3.14. When v(1) ∈ H1,...,v(n) ∈ Hn, let p(v(1),...,v(n)) be the Hilbert-Schmidt functional φv(1),...,v(n) defined on the cartesian product of H̄1 × · · · × H̄nby

φv(1),...,v(n)(x1,...,xn) = x1, v(1)− · · · xn, v(n)

−

= v(1), x1 · · · v(n), xn (3.3.41)

Because we let Y j be an orthonormal basis of H j, j = 1,...,n Theorem 3.14 then

says that the set { p(y1,...,yn) : y1 ∈ Y 1,...,yn ∈ Y n} is an orthonormal basis of H,and that

p(v1,...,vn), p(w1,...,wn) = w1, v1− · · · wn, vn

= v1, w1 · · · vn, wn, (3.3.42a)

p(v1,...,vn)2 = v1 · · · vn. (3.3.42b)


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From above we have p : H1 × · · · × Hn → H is a bounded multilinear mapping:we shall prove next that it is a weak Hilbert-Schmidt mapping. Suppose that φ ∈ H,and consider the bounded multilinear functional pφ : H1 × · · · × Hn → C defined by

pφ(x1,...,xn) = p(x1,...,xn), φ. (3.3.43)

With y(1) ∈ Y 1,...,y(n) ∈ Y n, orthonormality of the bases implies that φy(1),...,y(n)takes the value 1 at (y(1),...,y(n)) and 0 elsewhere on Y 1 × · · · × Y n. Thus

pφ(y(1),...,y(n)) = p(y(1),...,y(n)), φ = φy(1),...,y(n), φ

=y1∈Y 1

· · ·yn∈Y n

φy(1),...,y(n)(y1,...,yn)φ(y1,...,yn)

= φ(y(1),...,y(n)), (3.3.44a)

= y(1)∈Y 1

· · · y(n)∈Y n

| pφ(y(1),...,y(n))|2 = φ22. (3.3.44b)

From this we have proven that pφ is a Hilbert-Schmidt functional on H1 × · · · × Hnand that pφ2 = phi2; so p : H1×···×Hn → H is a weak Hilbert-Schmidt mappingwith p2 = 1.

Next, suppose that L is a weak Hilbert-Schmidt mapping from H1 × · · · × Hn intoanother Hilbert space K. Let u ∈ K and Lu is the Hilbert-Schmidt functional givenin Definition 23, while φ ∈ H and let F be a finite subset of Y 1 × · · · × Y n then wehave using Cauchy-Schwarz

|

(y1,...,yn)∈F

φ(y1,...,yn)L(y1,...,yn), u|

≤

(y1,...,yn)∈F

|φ(y1,...,yn)||Lu(y1,...,yn)|

≤

(y1,...,yn)∈F

|φ(y1,...,yn)|2

12 (y1,...,yn)∈F

|Lu(y1,...,yn)|2

12

≤ Lu2

(y1,...,yn)∈F

|φ(y1,...,yn)|2

12

≤ uL2

(y1,...,yn)∈F

|φ(y1,...,yn)|2

12

(3.3.45)

Hence


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(y1,...,yn)∈F

φ(y1,...,yn)L(y1,...,yn)

≤ L2

(y1,...,yn)∈F

|φ(y1,...,yn)|2 12

(3.3.46)

Since

y1∈Y 1

· · ·yn∈Y n

|φ(y1,...,yn)|2 = φ22 < ∞ (3.3.47)

it then follows from Eq(3.3.46) and the Cauchy criterion that the, unordered, sum

y1∈Y 1

· · · yn∈Y n

φ(y1,...,yn)L(y1,...,yn) (3.3.48)

converges to an element T φ ∈ K, and T φ ≤ L2 phi2. Thus T is a boundedlinear operator from H into K, and T ≤ L2. When y(1) ∈ Y 1,...,y(n) ∈ Y n, wehave

T p(y1,...,yn) = T φy(1),...,y(n)

=y1∈Y 1

· · ·yn∈Y n

φy(1),..,y(n)(y1,...,yn)L(y1,...,yn)

= L(y(1),...,y(n)). (3.3.49)

Both L and T p are bounded and multiline and Y m has closed linear span, Hm, m =1,...,n then it follows that L = T p.

The condition that T p = L uniquely determines the bounded linear operator T ,because the range of p contains the orthonormal basis p(Y 1 × · · · × Y n) of H. ∀u ∈ K,Parseval’s equation gives

Lu22 =

y1∈Y 1

· · ·yn∈Y n

|L(y1,...,yn), u|2

= y1∈Y 1

· · · yn∈Y n

|T p(y1,...,yn), u|2

=y1∈Y 1

· · ·yn∈Y n

| p(y1,...,yn), T ∗u|2

= T ∗u2 ≤ T 2u2; (3.3.50)

so we have L2 ≤ T , and thus L2 = T .


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We now prove (ii) of the theorem. Suppose that H and p : H1 × · · · × Hn → H

have the properties given in (i). When we have that K = H and L = p, the equationL = T p is satisfied when T is the identity operator on H, and also when T is theprojection from H onto the closed subspace [ p(H1 × · · · × Hn)] generated by the

range p(H1 × · · · × Hn) of p. From the uniqueness of T ,

[ p(H1 × · · · × Hn)] = H (3.3.51)

furthermore,

p2 = L2 = T = I = 1. (3.3.52)

Using a similar argument and letting K = H and L = p, it follows from theproperties of the Hilbert space H and p given in (i), that theres a hounded linearoperator U : H → H” such that p = U p and

u = L2 = p2 = 1 (3.3.53)

The roles of H, p and H, p can be reversed in this argument, so there is a boundedlinear operator U from H into H such that p = U p and U = 1. Since

U U p(x1,...,xn) = U p(x1,...,xn) = p(x1,...,xn), ∀x1 ∈ H1,...,xn ∈ Hn, (3.3.54)

while

[ p(H1 × · · · × Hn)] = H (3.3.55)

it follows that U U is the identity operator on H; and similarly U U is the identityoperator on H. Finally,

x = U U x ≤ U x ≤ x, ∀x ∈ H (3.3.56)

so U x = x, and U is an isomorphism from H onto H.

4 Stinespring’s Dilation Theorem

Theorem 4.1. Let A be a unital C ∗- algebra, and let φ : A → B(H) be a completely positive map. Then there exists a Hilbert space K, a unital ∗-homomorphism π : A →B(K), and a bounded operator V : H → K with φ(1) = V 2 such that

φ(a) = V ∗π(a)V (4.0.57)


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Proof. Consider the algebraic tensor product A ⊗ H, of the unital C ∗-algebra andthe Hilbert Space H, and define a symmetric bilinear function , on this space bysetting

a ⊗ x, b ⊗ y = φ(b∗a)x, yH, (4.0.58)

and extending linearly, where , H is the inner product on H. From the definition,φ is completely positive thereby ensuring that , is positive semidefinite since

n j=1

a j ⊗ x j,n

i=1

ai ⊗ xi

=

φn((a

∗a))

x1...

xn

,

x1...

xn

H(n)

≥ 0, (4.0.59)

where , H(n) denotes the inner product not the direct sum H(n) of n copies of H,

given by

x1...

xn

,

y1...

yn

H(n)

= x1, y1H + · · · + xn, ynH (4.0.60)

A result of positive semidefinite bilinear forms is that they satisfy the Cauchy-Schwarz inequality,

|u, v|2 ≤ u, u · v, v (4.0.61)

Therefore we have that

{u ∈ A ⊗ H|u, u = 0} = {u ∈ A ⊗ H|u, v = 0∀v ∈ A ⊗ H} (4.0.62)

is a subspace, N , of A ⊗ H. The induced bilinear form on the quotient spaceA ⊗ H/ N defined by

u + N , v + N = u, v (4.0.63)

will be an inner product. We let K denote the Hilbert space that is the completionof the inner product space A ⊗ H/ N .

If a ∈ A, define a linear map π(a) : A ⊗ H → A ⊗ H by

π(a)

ai ⊗ xi

=

(aai) ⊗ xi (4.0.64)

A matrix factorization shows that the following inequality in M n(A)+ is satisfied:

(a∗i a∗aa j) ≤ a

∗a · (a∗i a j), (4.0.65)


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and consequently,

π(a) a j ⊗ x j, π(a) ai ⊗ xi=i,j

φ(a∗i a∗aa j)x j, xiH ≤ a

∗a ·i,j

φ(a∗i a j)x j, xiH

= a2 ·

a j ⊗ x j ,

ai ⊗ xi

. (0.0.12)

Therefore, π(a) leave N invariant and consequently induces a quotient linear trans-formation on A ⊗ H/ N , which we will still denote by π(a). The above inequality alsoshows that π(a) is bounded with π(a) ≤ a. Thus, π(a) extends to a boundedlinear operator on K, which we will still denote by π(a). It is straightforward to verify

that the map π : A → B(K) is a unital ∗ - homomorphism.Now define V : H → K via

V (x) = 1 ⊗ x + N (4.0.67)

Then V is bounded, since

V x2 = 1 ⊗ x, 1 ⊗ x = φ(1)x, xH ≤ phi(1) · x2. (4.0.68)

Indeed, it is clear that V 2 = sup{φ(1)x, xH : x ≤ 1} = φ(1).To complete the proof, we only need to observe that

V ∗π(a)V x , yH = π(a)1 ⊗ x, 1 ⊗ yK = φ(a)x, yH, ∀x, y ∈ H, (4.0.69)

and so V ∗π(a)V = φ(a).

Quantizing Analysis

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