Normed and Hilbert Spaces

Chapter 2

2.1 Topics to be covered

Normed spaces • `p spaces, Holder inequality, Minkowski inequality, Riesz-Fischer theorem The space C(X) Quotients and conditions for completeness, the 2/3’s theorem Finite dimensional normed spaces, equivalence of norms Convexity, absolute convexity, the bipolar theorem Consequences of Baire’s theorem: Principle of Uniform Boundedness, Resonance Principle Open mapping, closed graph and bounded inverse theorems Hahn-Banach theorem Krein-Milman theorem Dual spaces and adjoints The double dual Weak topologies, weak convergences

Hilbert spaces • Cauchy-Schwarz inequality Polarization identity, Parallelogram Law Jordan-von Neumann theorem

17 18 CHAPTER 2. NORMED AND HILBERT SPACES

Orthonormal bases and Parseval identities Direct sums Bilinear maps and tensor products of Banach and Hilbert spaces Inﬁnite tensor products and quantum spin chains

2.2 Banach Spaces

All vector spaces will be over the ﬁeld R or C, when we wish to consider a notion that applies to both ﬁelds we shall write F.

Definition 2.1. Let X be a vector space over F.Thenanorm on X is a function, : X R+ satisfying: k·k ! 1. x =0 x =0, k k () 2. x = x , F, x X, k k | |·k k 8 2 8 2 3. (Triangle Inequality) x + y x + y , x, y X. k kk k k k 8 2 We call the pair (X, )anormed linear space or n.l.s., for short. k·k Proposition 2.2. Let (X, ) be a n.l.s.. Then d(x, y)= x y is a metric k·k k k on X (called the “induced metric”) that satisfies d(x + z,y + z)=d(x, y) (“translation invariance”) and d(x, y)= d(x, y) (“scaled”). | | Conversely, if X is a vector space and d is a metric on X that is scaled and translation invariant, then x = d(0,x) is a norm on X. k k Definition 2.3. An.l.s.(X, )isaBanach space if and only if it is k·k complete in the induced metric, d(x, y)= x y . k k Some examples are in order.

Example 2.4. Let X = CR([0, 1]) denote the vector space of continuous real-valued functions on the unit interval. Given f X we set 2 f =sup f(t) :0 t 1 . k k1 {| |   } We leave it to the reader to verify that this is a norm. We claim that (X, ) is a Banach space. Note that a sequence of functions fn X || · ||1 { }✓ is Cauchy if and only if for each ✏>0thereisN so that for n, m > N,

d (fn,fm)=sup fn(t) fm(t) :0 t 1 <✏. 1 {| |   } 2.2. BANACH SPACES 19

Now we recall some arguments from undergraduate analysis. For each 0  t0 1, fn(t0) fm(t0) fn fm and so fn(t0) is a Cauchy sequence  | |k k1 { } of real numbers. Hence, lim fn(t0) exists and we deﬁned f(t0)tobethis value. Next one shows that

lim sup f(t) fn(t) :0 t 1 =0, n {| |   } so that the sequence f converges uniformly to f and recall from under- { n} graduate analysis that this implies that f is continuous. Thus, f X and 2 the last equation shows that

f fn 0. k k1 ! Thus, (X, ) is a Banach space. k·k1 The next example is a vector space that is not a Banach space.

Example 2.5. Let X = CR([0, 1]) as before, but now set

1 f = f(t) dt. k k1 | | Z0 Note that if f X and f =0then,sincef is continuous, there is a >0 2 6 and a small interval [a, b] on which f(t) . Hence, f (b a). This | | k k1 proves that f = 0 if and only if f = 0, and we have veriﬁed one property k k1 of a norm. We leave it to the reader to verify that the remaining properties of a norm are met. We claim that (X, ) is not complete. To see this consider the fol- k·k1 lowing sequence of functions. For each n 3set 1 1 0, 0 t 2 n , 2nt n+2 1  1  1 1 fn(t)= , t + , 8 4 2 n   2 n <>1, 1 + 1 t 1. 2 n   > For n, m > N the functions: fn and fm are equal except on a subset of the interval [ 1 1 , 1 + 1 ], and on this subinterval they can di↵er by at most 2 N 2 N 1. Hence, 2 f f , k n mk1  N and so the sequence is Cauchy. Now check that no continuous function can be the limit of these functions in . k·k1 20 CHAPTER 2. NORMED AND HILBERT SPACES

n Example 2.6. For 1 p<+ and n N, and x =(x1,...,xn) R ,set  1 2 2 n (x ,...,x ) = x p 1/p, k 1 n kp | j| j=1 X p n then this is a norm called the p-norm and we write `n to denote R endowed with this norm. It is not so easy to see that this satisﬁes the triangle inequality, this depends on some results below. For p =+ we set 1 (x ,...,x ) = max x ,..., x . k 1 n k {| 1| | n|} Example 2.7. More generally, for 1 p<+ ,welet  1 + 1 `p = (x ,x ,...): x p < + , { 1 2 | j| 1} Xj=1

p + p 1/p and for x ` we set x = 1 x . 2 k kp j=1 | j| For p =+ ,weset 1 P

`1 = (x1,x2,...):sup xj < + , { j | | 1} and deﬁne x =supj xj . k k1 | | Then for 1 p + ,(`p, ) are all Banach spaces. This fact relies   1 k·kp on a number of theorems that we will state below.

1 Problem 2.8. Prove that (` , 1) and (`1, ) are Banach spaces. k·k k·k1 Problem 2.9. Let c = (x ,x ,...):lim x =0. Prove that c is a 0 { 1 2 n n } 0 vector subspace of `1 and that it is closed in the metric induced by the norm. k·k1 1 1 Lemma 2.10 (Young’s Inequality). Let 1

Proposition 2.12 (Holder’s Inequality). Let 1

Problem 2.13. Prove Holder’s inequality. Theorem 2.14 (Minkowski’s Inequality). Let 1 p + , and let x, y   1 2 `p, then x + y `p and x + y x + y . 2 k kp k kp Proof. We shall only prove the case that 1

p p 1 x + y x + y ( x + y ). | i i| | i i| | i| | i| Also, p + q = pq so that (p 1)q = p. From this it follows that the sequence x + y p 1 is q-summable. By Holder’s inequality, | i i| p p 1 q 1/q p 1/p p 1/p x + y ( x + y ) ( x ) +( y ) | i i|  | i i| | i| | i| i i i i X X X X = x + y p 1/q ( x p)1/p +( y p)1/p . | i i| | i| | i| i i i X X X Cancelling the common term from each side and using 1 =1 1 yields the q p result.

Theorem 2.15 (Riesz-Fischer). The spaces (`p, ) are Banach spaces. k·kp Given a normed space (X, ), a sequence of vectors x and a vector k·k { n} x,wewrite 1 x = xn, n=1 X to mean that lim x N x =0, i.e., that the partial sums converge N k n=1 nk in norm. Similarly, we write x = n N xn to mean that the net of finite P 2 sums sF = n F xn converges in the metric induced by the norm, i.e., that 2 P given ✏>0, there is a finite set F0 such that whenever F is a finite set with P F F ,then x s <✏. 0 ✓ k F k We shall let e , also denoted i , be the vector that is 1 in the i-th entry i | i and 0 in every other entry. Given x =(x ,x ,...) `p, we can write 1 2 2 1 x = xiei. Xi=1 22 CHAPTER 2. NORMED AND HILBERT SPACES

For 1 p<+ this notation makes sense as the series does converge.  1 In fact, one can show the even stronger statement that x = i N xiei. However, this notation is somewhat misleading for p = . For example,2 if 1 P we set xi = 1 for all i,thenx +(1, 1,...) `1 and x = 1. But 2 k k1 N x xiei =1 k k1 Xi=1 so the series does not converge to x. In fact, any two partial sums of this series are distance 1 apart, so the partial sums are not even Cauchy. Later we will see that there is a di↵erent topology on `1 for which this series does converge to x in that topology. Problem 2.16. Prove that for x =(x ,...) `p, 1 < + ,wehavex = 1 2 1 1 i=1 xiei = i N xiei. Thus verifying the above claim. 2 ProblemP 2.17.P Let x =(x ,...) ` . Prove that the series 1 x e 1 2 1 i=1 i i converges to x (in norm) if and only if x c0. What about i N xiei ? 2 2 P Another useful test for determining if a normed space is aP Banach space is given in terms of convergent series. Theorem 2.18. Anormedspace(X, ) is a Banach space if and only if k·k whenever xn X is a sequence such that n1=1 xn < + then there is { }✓ + k k 1 an element x X such that 1 xn = x. 2 n=1 P This theorem is sometimesP stated as a normed space is a Banach space if and only if every absolutely convergent series is convergent. Here a series 1 x is called absolutely convergent provided that 1 x < + . n=1 n n=1 k nk 1 One ﬁnal example of a family of Banach spaces. P P Example 2.19. Let (K, ) be a compact Hausdor↵space and let C(K) T denote the space of continuous real or complex valued functions on K.This is clearly a vector space. If we set f =sup f(x) ; x K ,thenthis k k1 {| | 2 } deﬁnes a norm and (C(K), ) is a Banach space. Completeness follows k·k1 from the fact that convergence in this norm is uniform convergence and the fact that uniformly convergent sequences of continuous functions converge to a continuous function.

2.2.1 Bounded and Continuous Proposition 2.20. Let (X, ) and (Y, ) be normed spaces and let k·k1 k·k2 T : X Y be a linear map. Then the following are equivalent: ! 2.2. BANACH SPACES 23

1. T is continuous,

2. T is continuous at 0,

3. there is a constant M such that Tx M x , x X k k2  k k1 8 2 Deﬁnition 2.21. Let X and Y be normed spaces and let T : X Y be ! a linear map. Then T is bounded if there exists a constant M such that Tx M x , x X.Weset k k k k 8 2 T := sup Tx : x 1 < k k {k || k k } 1 and this is called the operator norm of T . Some other formulas for this value include

T =sup Tx : x =1 =inf M : Tx M x , x X . k k {k k k k } { k k k k 8 2 } 2.2.2 Equivalence of Norms Suppose that ,i =1, 2 are norms on X and that d ,i =1, 2 are the k·ki i metrics that they induce. Recall that we say the metrics are equivalent if and only if the maps id :(X, d ) (X, d ) and id :(X, d ) (X, d ) are 1 ! 2 2 ! 1 both continuous. But since these are normed spaces this is the same as requiring that both maps be bounded, which is the same as both metrics being uniformly equivalent. Thus, the metrics that come from norms are equivalent if and only if they are uniformly equivalent. This leads to the following deﬁnition. Deﬁnition 2.22. Let (X, ),i =1, 2 be two norms on X.Thenwe k·ki say that these norms are equivalent provided that there exists constants, A, B > 0 such that

A x x B x , x X. k k1 k k2  k k1 8 2 Note that these last inequalities are the same as requiring that there are constants C, D > 0 such that x C x and x D x .Ifwethink k k2  k k1 k k1  k k2 geometrically then this is the same as requiring that A B , ·B1 ✓B2 ✓ ·B1 where denotes the unit balls in each norm. Bi Finally, note that “equivalence of norms” really is an equivalence relation on the set of all norms on a space X, that is, it is a symmetric ( equivalent k·k1 to if and only if is equivalent to ) and transitive ( equivalent k·k2 k·k2 k·k1 k·k1 to and equivalent to implies that is equivalent to ) k·k2 k·k2 k·k3 k·k1 k·k3 relation. 24 CHAPTER 2. NORMED AND HILBERT SPACES

Proposition 2.23 (Reverse Triangle Inequality). Let (X, ) be a normed k·k space. Then x y x y x, y X. |k kk k| k k8 2 Theorem 2.24. All norms on Rn are equivalent.

Proof. It will be enough to show that an arbitrary norm on Rn is k·k equivalent to the Euclidean norm 2. Let ei = i denote the standard n k·k | i basis for R , so that x =(x1,...,xn)=x1e1 + + xnen.Then, ··· x x e + + x e x 2 1/2 e 2 1/2 C x , k k| 1|k 1k ··· | n|k nk | i| k ik  k k2 i i X X where C = e 2 1/2. i k ik Let f : Rn R be deﬁned by f(x)= x . By the reverse triangle P ! k k inequality we have that,

f(x) f(y) x y C x y , | |k k k k2 which shows that f is continuous from Rn to R when Rn is given the usual Euclidean topology. Now K = x : x =1 is a compact set and so for some x X, { k k2 } 0 2 A =inf f(x):x K = f(x )= x =0. { 2 } 0 k 0k6 Now for any non-zero x n, x K and hence, R x 2 2 k k 2 x x A = k k , k x k x k k2 k k2 so that A x x and we are done. k k2 k k 2.3 Consequences of Baire’s Theorem

Many results are true for Banach spaces that are not true for general normed spaces. In particular, this is because Banach spaces are complete metric spaces and so we have Baire’s theorem as a tool. In this section we present several facts about Banach spaces that rely on Baire’s theorem. The version of Baire’s theorem that we shall use most often is the version that says that in a complete metric space, a countable union of nowhere dense sets is still nowhere dense. 2.3. CONSEQUENCES OF BAIRE’S THEOREM 25

2.3.1 Principle of Uniform Boundedness We begin with an example to show why the following result is perhaps surprising. For n 2, let fn :[0, 1] R be deﬁned by ! n2t, 0 t 1/n 2   fn(t)= 2n n t, 1/n t 2/n , 8   <>0, 2/n t 1   > so that each fn is continuous.: We have fn = n and so each function is bounded. Also for each x k k1 we have that for n large enough, fn(x) = 0. Hence, we also have that for each x there is a constant M so that f (x) M , n. However, clearly x | n | x 8 sup fn =+ . {k k1} 1 Note that in this case the domain is a complete metric space. Theorem 2.25 (Principle of Uniform Boundedness). Let X be a Banach space, Y a normed space, and let be a set of linear mappings from X to F Y . If for each A there is a constant C such that Ax C x (i.e., 2F A k ||  Ak k each A is bounded) and for each x X there is a constant M so that 2 x Ax M (i.e., the collection is bounded at each point), then k k x F sup A : A < . {k k 2F} 1 Proof. Let E = x : Ax n, A . Since each A is continuous, it n { k k 8 2F} is not hard to see that each En is closed. By the hypothesis of pointwise boundedness, X = E . Hence, one of these sets is not nowhere dense, [n n say for n0.SinceEn0 is already closed, there must exist x0 and a constant r>0 so that y : y x

The following is just the contrapositive statement of the above theorem, but it is often useful. 26 CHAPTER 2. NORMED AND HILBERT SPACES

Theorem 2.26 (Resonance Principle). Let X be a Banach space and let Y be a normed space. If is a set of bounded linear maps from X to F Y with sup A ; A =+ , then there is a vector x X such that {k k 2F} 1 2 sup Ax : A =+ . {k k 2F} 1 The vector given by the above result is often called a resonant vector for the set . F Here is an example to show what can go wrong if the domain X is not a Banach space.

Deﬁnition 2.27. Let C00 = span en : n N `1 { 2 }✓ Thus, x =(x ,x ,...) C if and only if there is an N so that x =0 1 2 2 00 x n for all n>Nx.ThenumberNx varies from vector to vector.

Example 2.28. Deﬁne A : C C by n 00 ! 00

An(x)=(x1, 2x2, 3x3,....,nxn, 0, 0,...).

Clearly, each A is a bounded linear map with A = n. Also for each n k nk x C we have an N as above and so sup A x N < + . 2 00 x n k n k x 1 Problem 2.29. Let X and Y be normed spaces with X Banach, and let A : X Y be a sequence of bounded linear maps such that for each x X n ! 2 the sequence An(x) converges in norm to an element of Y . Prove that if we set A(x)=limn An(x)thenA deﬁnes a bounded linear map. (The sequence A is said to converge strongly to A and we write A S A. { n} n !

2.3.2 Open Mapping, Closed Graph, and More! The following results are also all consequences of Baire’s theorem applied to Banach spaces. In fact, given any one as the starting point, it is possible to deduce the others as consequences. These require that the domain and range both be Banach spaces.

Lemma 2.30. Let X and Y be Banach spaces, and let A : X Y be a ! bounded linear map that is onto. Then there exists >0 such that

(0; ):= y Y : y < A (0; 1) , BY { 2 k k }✓ BX where (0; 1) = x X : x < 1 . BX { 2 k k } 2.3. CONSEQUENCES OF BAIRE’S THEOREM 27

Theorem 2.31 (Bounded Solutions). Let X and Y be Banach spaces and let A : X Y be a bounded linear map. If for each y Y there is x X ! 2 2 solving Ax = y, then there is a constant C>0 such that for each y Y 2 there is a solution to Ax = y with x C y . k k k k Theorem 2.32 (Bounded Inverse Theorem). Let X and Y be Banach spaces and let A : X Y be a bounded linear map. If A is one-to-one and onto ! then the inverse map A 1 : Y X is bounded. ! Theorem 2.33 (Equivalence of Norms). Let X be a space with two norms and , such that X is a Banach space in both norms. If there is a k·k1 k·k2 constant such that x C x , x X, then the norms are equivalent. k ||1  k k2 8 2 As an application of the equivalence of norms theorem and Baire’s theorem, one can prove the following.

Theorem 2.34 (Open Mapping Theorem). Let X and Y be Banach spaces, and let A : X Y be a bounded linear map. If A is onto, then for every ! open set U X the set A(U) is open in Y . ✓ If X and Y are vector spaces, then their Cartesian product is also a vector space with operations, (x1,y1)+(x2,y2)=(x1 + x2,y1 + y2) and (x, y)=(x, y). This space is sometimes denoted X Y . There are many equivalent norms that can be put on X Y ,wewilluse (x, y) := x + y . k k1 k k k k It is fairly easy to see that a subset C X Y is closed in the norm ✓ k·k1 if and only if x x 0 and y y 0 implies that (x, y) C. k n k! k n k! 2 Given a linear map, A : X Y its graph is the subset G := (x, Ax): ! A { x X . 2 } Theorem 2.35 (Closed Graph Theorem). Let X and Y be Banach spaces, and let A : X Y be a linear map. If x x 0 and Ax y 0 ! k n k! k n k! implies that Ax = y (this is equivalent to the statement that GA is a closed subset), then A is bounded.

In general, a linear map A : X Y is called closed provided that G ! A is closed. When X and Y are Banach spaces, then closed implies bounded by the above. But if X is not Banach then the map need not be bounded. Here is an example.

Example 2.36. Consider the following function space,

1 C ([0, 1]) = f :[0, 1] R : f 0 exists and is continuous . { ! } 28 CHAPTER 2. NORMED AND HILBERT SPACES

Set f = f and let Y = C([0, 1]), also with . Let D : C1([0, 1]) k k k k 1 k·k1 ! C([0, 1]) by D(f)=f 0. By Rudin, Theorem 7.1.7, if fn f 0 and k k1 ! fn0 g 0, then f is di↵erentiable and g = f 0. This result is equiva- k k1 ! lent to saying that the graph of D is closed. But D is not bounded since D(tn) = n tn . What goes wrong is that C1([0, 1]) is not complete k k1 k k1 in . This last fact can be seen directly, or it can be deduced as an k·k1 application of the closed graph theorem.

Problem 2.37. Let X be a Banach space and let Y and Z be two closed subspaces of X such that Y Z = (0) and Y + Z := y + z : y Y,z \ { 2 2 Z = X. Thus, each x X has a unique representation as x = y +z.Deﬁne } 2 x = y + z . Prove that (X, ) is a Banach space and that and k k1 k k k k k·k1 k·k are equivalent norms on X. k·k1

2.4 Hahn-Banach Theory

The Hahn-Banach theorem is usually stated for norm extensions, but it is really more general with the proofs being no harder, so we state it in full generality.

Deﬁnition 2.38. Let X be a vector space. Then a function p : X R is ! sublinear if:

1. p(x + y) p(x)+p(y), x, y X,  8 2 2. for all t>0, p(tx)=tp(x).

Note that, unlike a norm, it is not required that p( x)=p(x). Lemma 2.39 (One-Step Extension). Let X be a real vector space, Y X ✓ a subspace, p : X R a sublinear functional and f : Y R alinear ! ! functional satisfying f(y) p(y), y Y .Ifx X Y and Z = span Y,x ,  8 2 2 \ { } then there exists a linear functional g : Z R such that g(y)=f(y), y ! 8 2 Y (i.e., g is an “extension” of f)andg(z) p(z), z Z.  8 2 Proof. Since every vector in Z is of the form z = rx+y for some r R every 2 possible extension of f is of the form g↵(rx + y)=r↵ + f(y). So we need to show that for some choice of ↵ we will have that r↵+f(y) p(rx+y), r, y.  8 First note that if y ,y Y ,then 1 2 2 f(y + y ) p(y + y ) p(y + x)+p(y x). 1 2  1 2  1 2 2.4. HAHN-BANACH THEORY 29

Hence, f(y ) p(y x) p(y + x) f(y ), y ,y Y. 2 2  1 1 8 1 2 2 Thus,

↵ := sup f(y) p(y x):y Y inf p(y + x) f(y):y Y =: ↵ . 0 { 2 } { 2 } 1 The proof is completed by showing that for any ↵ with ↵ ↵ ↵ we 0   1 will have that g is an extension with g (rx + y) p(rx + y). ↵ ↵  Theorem 2.40 (Hahn-Banach Extension Theorem for Sublinear Functions). Let X be a real vector space, p : X R a sublinear functional, Y X a sub- ! ✓ space and f : Y R a real linear functional satisfying f(y) p(y), y Y . !  8 2 Then there exist a linear functional g : X R that extends f and satisﬁes, ! g(x) p(x) < x X.  8 2 The idea of the proof is to use the one-step extension lemma and then apply Zorn’s lemma to show that an extension exists whose domain is maximal among all extensions and argue that necessarily this maximal extension has domain equal to all of X. Zorn’s lemma is equivalent to the Axiom of Choice, so this is one theorem that requires this additional axiom, which almost every functional analyst is comfortable with assuming. Next we look at complex versions. The following result is often useful for passing between real linear maps and complex linear maps.

Proposition 2.41. Let X be a complex vector space.

1. If f : X R is a real linear functional, then setting ! g(x)=f(x) if(ix) deﬁnes a complex linear functional.

2. If f : X C is a complex linear functional, then Ref(x)=Re(f(x)), ! is a real linear functional and

f(x)=Ref(x) iRef(ix).

Theorem 2.42. Let X be a complex vector space, p : X R be a function ! that satisﬁes, p(x + y) p(x)+p(y) and p(x)= p(x), C. If Y X  | | 8 2 ✓ is a complex subspace and f : Y C is a complex liner functional satisfying ! f(y) p(y), y Y , then there exists a complex linear functional g : X | | 8 2 ! C that extends f and satisﬁes g(x) p(x), x X. | | 8 2 30 CHAPTER 2. NORMED AND HILBERT SPACES

Proof. Consider the real linear functional Ref, observe that it satisfies Ref(y) p(y). Apply the previous theorem to extend it to a real linear  functional h : X R satisfying h(x) p(x). !  Now set g(x)=h(x) ih(ix) and check that g works. Given a bounded linear functional f : Z F we define f =sup f(z) : ! k k {| | z 1 . k k } Theorem 2.43 (Hahn-Banach Extension Theorem). Let X be a normed linear space over F, let Y X be a subspace, and let f : Y F be a ✓ ! bounded linear functional. Then there exists an extension g : X F that is ! also a bounded linear functional and satisfies, g = f . k k k k Proof. Apply the above result with p(x)= f x . k k·k k Corollary 2.44. Let X be a normed space and let x X. Then there exists 0 2 a bounded linear functional g : X F with g =1such that g(x0)= x0 . ! k k k k

Proof. Let Y be the one dimensional space spanned by x0 and let f(x0)= x . Check that f = 1 and apply the last result. k 0k k k 2.4.1 Convex Sets Convex sets play an important role in many settings. We gather a few facts about these sets here. Many of these facts are ultimately consequences of the Hahn-Banach theorem, especially the version for sublinear functions. Recall that a subset K of a real vector space X is called convex if whenever x, y K,thentx+(1 t)y K, 0 t 1. Writing tx+(1 t)y = 2 2 8   y + t(x y), we see that K is convex if and only if whenever x, y K,then 2 the line segment from y to x is in K. Note that if K is convex and x ,x ,x K then 1 2 3 2 s tx +(1 t)x +(1 s)x 0 t 1, 0 s 1 = { 1 2 3|     } 3 t x + t x + t x t 0, t =1 , { 1 1 2 2 3 3| i i } Xi=1 is a subset of K. Inductively, one can show that if K is convex, x ,...,x K 1 n 2 and t 0, 1 i n with n t = 1, then n t x K.Suchasum i   i=1 i i=1 i i 2 is called a convex combination of x1,...,xn. It readily follows that K is P P convex if and only if every convex combination of elements of K is again in K. 2.4. HAHN-BANACH THEORY 31

Given a non-empty subset S of a real vector space X, the smallest convex set that contains S is called the convex hull of S and is denoted by convh(S). Using this last fact it is not hard to see that

n n convh(S)= t x t 0,x S, 1 i n, t =1, with n arbitrary . { i i| i i 2   i } Xi=1 Xi=1 For an inﬁnite dimensional vector space the number of terms needed in these sums is, generally, not bounded, but in ﬁnite dimensions there is a bound.

Theorem 2.45 (Caratheodory). Let S be a non-empty subset of Rn, then

n+1 n+1 convh(S)= t x t 0,x S, 1 i n +1, t =1 . { i i| i i 2   i } Xi=1 Xi=1 This has a nice geometric interpretation. Given a non-empty set S in a vector space, we let `(1)(S)= tx +(1 t)y x, y S, 0 t 1 ,which { | 2   } is the union of all line segments joining points is S, and let `(k+1)(S)= `(1)(`(k)(S)). Caratheodory’s result says that for S Rn, this process ter- ✓ minates with convh(S)=`(n)(S)=`(m)(S), m n. 8 One corollary of Caratheodory’s result which is very useful is the following.

Corollary 2.46. Let S be a non-empty compact subset of Rn, then convh(S) is also compact. Proof. Set

n+1 K = (t ,...,t ,x ,...,x ) t bbR, t 0, t =1,x S . { 1 n+1 1 n+1 | i 2 i i i 2 } Xi=1 2 Since (n + 1) + (n + 1)n =(n + 1)2, K is a subset of R(n+1) , and is closed and bounded. Deﬁne L : K Rn by ! n+1 L((t1,...,tn+1,x1,...,xn+1)) = tixi, Xi=1 which is continuous. By Caratheodory, conv(S)=L(K) and since continuous images of compact sets are compact, the result follows.

In particular, in ﬁnite dimensions, the convex hull of a compact set is automatically closed. This is false in inﬁnite dimensions, the convex hull of 1 the compact set S = 0 en/n : n N ` is not closed. { }[{ 2 }✓ 32 CHAPTER 2. NORMED AND HILBERT SPACES

Deﬁnition 2.47. Given a real normed space X, a bounded linear functional f : X R and ↵ R we set ! 2 H := y X : f(y) ↵ , f,↵ { 2  } and we call Hf,↵ a closed 1/2-space. The following result is a consequence of the more general sublinear functionals version of the Hahn-Banach theorem. Theorem 2.48 (Hahn-Banach Separation Theorem). Let X be a real normed space and let K X be a convex subset. ✓ 1. If inf x y : y K =0, then there exists a bounded linear func- {k k 2 }6 tional f : X R and ↵ R such that f(y) ↵, y K,while ! 2  8 2 ↵convex set.  } The Minkowski functional of the set K✏ is deﬁned by

1 p(z)=inf t>0:t z K . { 2 ✏} One proves that this is a sublinear functional with p(K ) 1

For example in R2, the extreme points of the unit square are its four corners, while the extreme points of the closed unit disk is the unit circle. Note that the open unit disk is convex but has no extreme points. The following theorem tells us that for compact convex sets, there are always plenty of extreme points. 2.4. HAHN-BANACH THEORY 33

Theorem 2.50 (Krein-Milman, normed version). Let K be a compact, convex subset of a normed linear space, then K is the closure of the convex hull of its extreme points.

Even when K is a convex and compact subset of R3,thesetExt(K) need not be a closed set, but it is always bounded. However, there is a better theorem in fnite dimensions.

Theorem 2.51 (Caratheodory-Minkowski). Let K Rn be compact and ✓ convex. Then every point in K is a convex combination of at most n +1 extreme points.

2.4.2 Separable and Entangled States and Entanglement Wit- nesses In this section we give some applications of the above ideas to quantum information theory. For these applications we assume that the reader has some familiarity with matrix theory. n Given vectors x =(a1,...,an) and y =(b1,...,bn)inC ,wesettheir inner product equal to n x y = a b . h | i i i Xi=1 Note that this is linear in the ﬁrst variable and conjugate in the second variable. The (Euclidean) length of a vector is

x = x x . k k h | i An n n matrix P =(p ) is called positivep semidefinite(denoted P 0), ⇥ i,j provided n n x Px = pi,jajai 0, x C . h | i 8 2 i,jX=1 We shall write M to denote the set of n n matrices and M + to denote n ⇥ n the set of positive semidefinite matrices. Note that if P, Q M +,then 2 n P + Q M +. Also if P M +,thenP is self-adjoint, i.e., is equal to its 2 n 2 n own conjugate transpose, and every eigenvalue of P must be non-negative. For each unit vector x =(a1,...,an) the matrix Px =(aiaj) satisfies P (y)= y x x, x h | i and thus is the orthogonal projection onto the 1-dimensional space spanned by x. 34 CHAPTER 2. NORMED AND HILBERT SPACES

Finally, the trace of a matrix is

n Tr(P )= pi,i. Xi=1 Recall that Tr(P ) is equal to the sum of the eigenvalues of P . A matrix P is called a density matrix if P M + and Tr(P ) = 1. The 2 n states of a quantum system are usually identiﬁed with density matrices. It is readily seen that the set of density matrices is a closed convex set. Hence by Krein-Milman, the set of density matrices will be the convex hull of its extreme points.

Problem 2.52. Use the fact that self-adjoint matrices are diagonalizable to show that the extreme points of the density matrices in Mn are the projections onto 1-dimensional subspaces and that every density matrix is the convex combination of n extreme points.

2 Since Mn is a 2n -dimensional real vector space, by Caratheodory’s theorem we know that we should need at most 2n2 +1 extreme points, but the problem shows that for density matrices we can use considerably fewer. We now look at the case of bipartite quantum systems. States of such systems are typically represented by density matrices in a tensor product. So we need to discuss the tensor product of the space of matrices. One of the simplest ways to view tensor products of matrices is through the concept of block matrices.Givenn2 matrices, A , 1 i, j n with each A a i,j   i,j k k matrix, we form an nk nk matrix A as follows: ⇥ ⇥ A A 1,1 ··· 1,n . .. . A := (Ai,j)=2 . . . 3 . A A 6 n,1 ··· n,n7 4 5 We write M (M ) to indicate that we are considering nk nk matrices as n k ⇥ written in such blocks. Clearly every element of Mnk can be considered as an element of M (M ) by just introducing parentheses around every k k n k ⇥ block. Block matrices are also another way to think about the tensor product M M .IfweletE , 1 i, j n denote the n n matrix that is 1 in the n ⌦ k i,j   ⇥ (i, j)-entry and 0 elsewhere, then these form a basis for the vector space Mn. Consequently, every element of the tensor product M M has a unique n ⌦ k representation as n E A for some set of matrices A M . i,j=1 i,j ⌦ i,j i,j 2 k In this manner we have identified, M = Mn(M )=Mn M . Note P nk k ⌦ k that with this identification, if we start with B =(b ) M and C M i,j 2 n 2 k 2.5. DUAL SPACES 35 then we have that n n B C =( b E ) C = E (b C)=(b C), ⌦ i,j i,j ⌦ i,j ⌦ i,j i,j i,jX=1 i,jX=1 where the latter matrix is the block matrix with (i, j)-th block equal to bi,jC. This block matrix representation of B C is often called the Kronecker ⌦ tensor. Given two densities matrices P =(p ) M and Q M it is not hard i,j 2 n 2 k to see that P Q =(p Q) is a density matrix in M M = M (M )= ⌦ i,j n ⌦ k n k Mnk, and hence, the convex hull of all such density matrices is again a set of density matrices. Definition 2.53. The set of matrices in M M = M (M )=M that n ⌦ k n k nk are in the convex hull of the matrices of the form P Q,withP M ,Q ⌦ 2 n 2 Mk density matrices, is called the set of separable density matrices. Any density matrix in Mn(Mk) that is not separable is called an entangled density matrix. Since the set of density matrices is a compact set, it follows that the set of all matrices of the form P Q with P and Q density matrices is also a ⌦ compact set. Hence, by the corollary to Caratheodory’s theorem, the set of separable density matrices is a compact subset of the 2n2k2 real dimensional vector space, Mn(Mk). Problem 2.54. Prove that the extreme points of the separable density n k matrices are the matrices of the form Px Py where x C and y C ⌦ 2 2 where x and y are unit vectors. Conclude that every separable density matrix is a convex combination of at most 2n2k2 + 1 such matrices. By the Hahn-Banach separation theorem, the set of separable density matrices is an intersection of closed 1/2-spaces. In particular, if R is an entangled density matrix, then there must be a real linear functional f and ↵ R, such that f(S) ↵ for every separable density matrix and ↵

2.5 Dual Spaces

Let X be a normed linear space. We let X⇤ denote the set of bounded linear functionals on X. It is easy to see that this is a vector space and that 36 CHAPTER 2. NORMED AND HILBERT SPACES

f =sup f(x) : x 1 is a norm on this space. The normed space k k {| | k k } (X⇤, ) is called the dual space of X. k·k d Other notations used for the dual space are X0,X and sometimes X†.

Proposition 2.55. Let (X, ) be a normed space, then (X , ) is a k·k ⇤ k·k Banach space.

Here are some examples of dual spaces.

1 1 Example 2.56 (The dual of ` ). Given y `1 deﬁne fY : ` R by 2 ! fy(x)= n xnyn, i.e., the dot product! It is not hard to see that fy is 1 bounded and that fy = y . This gives a map from `1 into (` )⇤. P k k k k1 It is easy to see that this is a linear map, i.e., that fy1+y2 = fy1 +fy2 and fy = fy. Also we have seen that it satisﬁes fy = y , i.e., that it is k k k k1 isometric and hence must be one-to-one. Finally, we show that it is onto. To 1 see this start with f : ` R and let yn = f(en)whereen is the vector that ! is 1 in the n-th entry and 0 elsewhere. We have that y f e = f , | n|k kk nk1 k k so that y =(y ,...) ` . Now check that f (x)=f(x) for every x so that 1 2 1 y fy = f and we have onto. 1 The above results are often summarized as saying that (` )⇤ = `1,when what they really mean is that there is the above linear, onto isometry between these two spaces.

Example 2.57 (The dual of `p, 1

However, this fails when p =+ , as we shall show. 1 Example 2.58 (The space C ). We set C = x =(x ,x ,...) ` : 0 0 { 1 2 2 1 lim x =0 . It is easy to show that this is a closed subspace. In fact C n n } 0 is the closed linear span of en : n N. Given a bounded linear functional 2 f : C0 R if we set yn = f(en), then the same calculations as above show ! that y `1 and that f = f . This leads to C = `1. 2 y 0⇤ 2.5. DUAL SPACES 37

1 Example 2.59 ((`1)⇤ = ` ). Now let f : `1 R be a bounded linear 6 ! functional. As before let y = f(e ) and show that y `1. The functional n n 2 f f has the property that (f f )(e ) = 0 and hence, (f f )(C ) = 0. y y n y 0 But this does not guarantee that f = fy. We now construct an element of `1 that is not of the form fy for any y `1. Let e ` be the vector of all 1’s. It is easy to check that 2 2 1

inf e x : x C0 =1. {k k1 2 }

Now let W = re + x : r R,x C0 , which is easily seen to be a { 2 2 } vector subspace of `1. We deﬁne a linear functional g : W R by setting ! g(re+ x)=r.Ifr = 0, then for any x C , re+ x = r e + r 1x r . 6 2 0 k k | |k k| | Hence, g(re + x) = r re + x , so that g is a bounded linear functional | | | |k k and g 1. Since 1 = g(e)= e , we have that g = 1. Now by the k k k k k k Hahn-Banach theorem, we can extend g to a linear functional f : `1 R ! with f = 1. However, f(e )=g(e ) = 0, so if f = f then y would need k k n n y to be the 0 vector. Since f = 1 we see that f = f for any y `1. k k 6 y 2 Thus, not every bounded linear functional on ` is of the form f ,y `1. 1 y 2 What is true is that every f ` ⇤ decomposes uniquely as f = f +h where 2 1 y y `1 and h ` ⇤ with h(x) = 0 for every x C . 2 2 1 2 0

2.5.1 Banach Generalized Limits It is often useful to be able to take limits even when one doesn’t exist! This is one reason that we use lim inf and lim sup but these two operations are not linear. For example if we let xn be 1 for odd n and 0 for even n and let yn be one for even n and 0 for odd n,thenlimsupn xn =limsupn yn =1= lim sup(xn + yn). Banach generalized limits are one way that we can assign ”limits” in a linear fashion.

Proposition 2.60 (Banach Generalized Limits). There exists a bounded linear functional, G : `1 R such that: ! G =1, •k k when x =(x ,x ,...) and lim x exists, then G(x)=lim x , • 1 2 n n n n x ` , lim inf x G(X) lim sup x , •8 2 1 n n   n n given K if we let y =(x ,x ,...), then G(y )=G(x), i.e., the • K K+1 K+2 K generalized limit doesn’t care where we start the sequence. 38 CHAPTER 2. NORMED AND HILBERT SPACES

The existence of such functional follows from the Hahn-Banach theorem and a Banach generalized limit has the property that G(C0)=0sothese are examples of functionals that do not come from `1 sequences. Example 2.61. For those of you familiar with measure theory. Given any (X, ,µ) measure space and 1 p<+ ,thesetLp(X, ,µ) of equivalence B  1 B classes of p-integrable functions (essentially bounded functions in the case p =+ ) is a Banach space in the p-norm. 1 The Lp-spaces and their duals behave like the `p spaces and their duals. In fact, `p = Lp(N, 2N,µ)whereµ is counting measure, so some authors treat the `p spaces as just a special case of the Lp-theory. A measure space (X, ,µ) is called -ﬁnite, provided that there is a B countable collection of sets B with µ(B ) < + and X = B . n 2B n 1 [n n Theorem 2.62 (Riesz Representation Theorem). Let 1

Moreover, it is not hard to see that every linear functional on Mm,n is of this form for some B.Thenon-commutative Holder inequality says that sup Tr(BA) : A 1 = B , {| | k kp  } k kq where p and q are any Holder conjugates. Thus, the map from (M , n,m k· ) (M , ) given by B f deﬁnes an onto linear isometry and kq ! m,n k·kp ⇤ ! B so as in the case of the `p-spaces we can identify,

(M , )⇤ =(M , ). m,n k·kp n,m k·kq We should note that some authors prefer to deﬁne fB : Mm,n C by ! f (A)=Tr(BtA), where Bt is the transpose of B, so that B M B 2 m,n instead of in Mn,m. Other authors set fB(A)=Tr(B⇤A), imitating the inner product of vectors. But in this case the map B f is conjugate ! B linear.

2.5.2 The Double Dual and the Canonical Embedding

Given a normed linear space, X,weletX⇤⇤ =(X⇤)⇤, i.e., the bounded linear functionals on the space of bounded linear functionals. Given x X we 2 deﬁne a bounded linear functionalx ˆ : X⇤ R by settingx ˆ(f)=f(x), i.e., ! reversing the roles of dependent and independent variables. Thus,x ˆ X . 2 ⇤⇤ The map J : X X deﬁned by J(x)=ˆx is easily seen to be linear, ! ⇤⇤ one-to-one and an isometry. This map is called the canonical injection or the canonical embedding of X into X⇤⇤. 40 CHAPTER 2. NORMED AND HILBERT SPACES

Deﬁnition 2.65. A normed space X is called reﬂexive provided that J(X)=X⇤⇤, which is often written simply as X = X⇤⇤.

Note that if X is reﬂexive, then so is X . The spaces `p and Lp(X, ,µ) ⇤ B are reﬂexive for 1

2.5.3 The Weak Topology Deﬁnition 2.66. Give a normed space X, we say that a subset U X is ✓ weakly open provided that whenever x U then there exists a ﬁnite set 0 2 f ,...,f X and ✏ ,...,✏ > 0 such that 1 n 2 ⇤ 1 n := x X : f (x x ) <✏, 1 i n U. Bf,✏i { 2 | i 0 | i 8   }✓ The set of all weakly open subsets of X is a topology on X called the Tw weak topology.

We say that a net converges weakly to x0 if it converges to x0 in the weak topology.

Theorem 2.67. A net x D X converges in the weak topology to { } 2 ✓ x X if and only if 0 2

lim f(x)=f(x0), f X⇤. 8 2

We write x w x to indicate that a net converges weakly to x . ! 0 0 Example 2.68. Let 1

2.5.4 The Weak* Topology Definition 2.69. Given a normed space X,wedefineasubsetU X ✓ ⇤ to be weak* open provided that for every f U there exists a finite set 0 2 x ,...,x X and ✏ ,...,✏ > 0 such that 1 n 2 1 n

f X⇤ : f(x ) f (x ) <✏, 1 i n U. { 2 | i 0 i | i 8   }✓

The set w of all weak* open subsets of X⇤ is a topology on X⇤ called the T ⇤ weak* topology.

A net of functionals converges weak* to f0 provided that it converges in the weak* topology to f0.

Theorem 2.70. A net f D converges in the weak* topology to f0 if and { } 2 only if lim f(x)=f0(x), x X. 8 2 w We write f ⇤ f to indicate that a net converges in the weak* topology ! 0 to f0. Example 2.71. In parallel with the last example, consider e `p for n 2 1

Theorem 2.72 (Banach-Alaoglu). Let X be a normed space and let K = f X : f 1 . Then K is compact in the weak* topology. { 2 ⇤ k k } 42 CHAPTER 2. NORMED AND HILBERT SPACES

Consequently, any time that we have a bounded net in X⇤,wemayselect a weak* convergent subnet. Note that when X is reﬂexive, then regarding X =(X⇤)⇤ we have a weak* topology on X, but this is also the same as the weak topology on X. So weak convergence and weak* convergence mean the same thing. In particular, in `p, 1

2.5.5 Completion of Normed Linear Spaces We discussed earlier that every metric space (X,⇢) has a canonical complete metric space (X,ˆ ⇢ˆ) that it is isometrically contained in as a dense subset. So if (X, ) is a normed space then it is contained in a complete metric k·k space, Xˆ. But there is nothing in the earlier theorem that says that Xˆ can be taken to be a vector space and that the metric⇢ ˆ can be taken to come from a norm. Both of these things are true and can be shown with some care. However, the following gives us an easier way to see all of this. Given a normed space (X, ), we have that X =(X ) and so it is a Banach k·k ⇤⇤ ⇤ ⇤ space. The map J : X X is a linear, isometric embedding. So if we just ! ⇤⇤ take the closure of J(X) inside the Banach space X⇤⇤ we will get a closed subspace and that subspace will be a Banach space. Because the completion of a metric space is unique, this shows that the completion of X actually has the structure of a vector space.

2.6 Banach space leftovers (Placeholder Section)

These are currently in no particular order. It is not hard to see that the induced metric on a n.l.s. satsﬁes: n ! = x x and x x, y y = x + y x + y.If(X is a ) n ! n ! n ! ) n n ! vector space and ⇢ is a complete metric on X satsfying these two properties, then (X,⇢) is called a Frechet space. The vector space C(R)withthe metric ⇢ corresponding to uniform convergence on compact subsets is an example of a Frechet space that is not a Banach space.

2.6.1 Quotient Spaces A concept that plays a big role in functional analysis are quotient spaces. Recall from algebra, that if we have a vector space X and a subspace Y then 2.6. BANACH SPACE LEFTOVERS (PLACEHOLDER SECTION) 43 there is a quotinet space, denoted X/Y consisting of cosets, X/Y := x + Y : x Y , { 2 } and defining, (x1 + Y )+(x2 + Y )=(x1 + x2)+Y,(x + Y )=(x)+Y gives well-defined operations that makes this into a vector space. One often used fact is that if T : X Z is a linear map and Y = ker(T )thenthereis ! a well-defined linear map Tˆ : X/Y Z given by Tˆ(x + Y )=T (x). ! To form quotients of normed spaces, we want one more condition. Let (X, ) be a normed space and let Y be a closed subspace. On the quotient k·k space X/Y ,wedefine x + Y := inf x + y : y Y =inf x y : y Y . k k {k k 2 } {k k 2 } The second formula shows that x + Y is the distance from x to the set k k Y . This is the reason that we want Y closed, otherwise there would be a point in the closure that is not in Y and any such point would be distance 0 from Y . Thus, such a vector would be one for which x + Y =0+Y but 6 x + Y = 0. This quantity is called the quotient norm. k k Proposition 2.73. Let (X, ) be a normed space and let Y be a closed k·k subspace, then the quotient norm is a norm on X/Y . The vector space together with the quotient norm is called the quotient space or if we want to be very clear the normed quotient space.

2.6.2 Conditions for Completeness Next is one of many 2outof3theorems. Theorem 2.74 (The 2/3’s Theorem). Let (X, ) be a normed space, let k·k Y X be a closed subspace and let X/Y be the normed quotient space. If ✓ any 2 of these normed spaces is a Banach space, then the 3rd space is also a Banach space. Proof. We only prove the case, that if Y and X/Y are Banach spaces, then so is X.Tothisendlet x be a Cauchy sequence in X and let [x ]=x +Y { n} n n be the image of this sequence in X/Y .Since [x ] [x ] x x , k n m kk n mk this is a Cauchy sequence in X/Y , and hence, converges to some [x] X/Y. 2 Since [x x ] 0 we can pick y Y such that x x y 0. Note k n k! n 2 k n nk! that

y y = (x + y x) (x + y x)+(x x ) k n mk k n n m m m n k x + y x + x + y x + x x , k n n k k m m k k n mk 44 CHAPTER 2. NORMED AND HILBERT SPACES since the ﬁrst two terms tend to 0 and the 3rd is Cauchy, it is easy to see that y is a Cauchy sequence in Y and hence y y. { n} n ! This implies that

x =(x + y x)+x y x y, n n n n ! and so X is complete.

To see how this last result can be used we will show that every ﬁnite dimensional normed space is automatically complete(although this is not the shortest way to prove this fact).

Theorem 2.75. Every ﬁnite dimensional normed space is complete.

Proof. We do the case that the ﬁeld is R, the case for C is identical. This is easy to see when the dimension is one. To see pick any x X with 2 x = 1 Then every vector is of the form x and x x = . k k k 1 2 k | 1 2| Thus, a sequence nx is Cauchy in X i↵ n is Cauchy in R. { } { } Now, inductively assume that it is true for spaces of dimension n and let X be (n + 1)-dimensional take a n-dimensional subspace Y .Itwillbe complete by the inductive hypothesis and hence closed. Also the quotient X/Y is one-dimensional so it is complete. Hence by the 2/3 theorem X is complete.

2.6.3 Norms and Unit Balls Let (X, ) be a normed space, we call x : x < 1 the open unit ball k·k { k k } and x : x 1 the closed unit ball. It is not hard to see that these { k k } sets are in fact open and closed, respectively and that the closed unit ball is the closure of the open unit ball. We wish to characterize norms in terms of properties of these balls.

Deﬁnition 2.76. AsubsetC of a vector space V is called:

convex if whenever x, y C and 0 t 1thentx +(1 t)y C, • 2   2 i.e., provided that the line segment from x to y is in C,

absolutely convex if whenever x ,...,x C and + + = 1, • 1 N 2 | 1| ··· | n| then x + + x C, 1 1 ··· n n 2 absorbing if whenever y V then there is a t>0 such that ty C. • 2 2 A ray is any set of the form ty : t>0 for some y V . { } 2 2.6. BANACH SPACE LEFTOVERS (PLACEHOLDER SECTION) 45

Proposition 2.77. Let (X, ) be a normed space and let be the open(or k·k B closed) unit ball, then is absolutely convex, absorbing and contains no rays. B Moreover, 1 1 x =inf t>0:t x = sup r>0:rx . k k { 2B} { 2B} Conversely, if C X is any absolutely convex, absorbing set that contains ✓ no rays and we define 1 x =inf t>0:t x C , k k { 2 } then is a norm on X and k·k x : x < 1 C x : x 1 . { k k }✓ ✓{ k k } This latter fact gives us a way to generate many norms, by just defining sets C that satisfy the above conditions. Also notice that x 1 x 2 i↵ k k kn k and only if the unit balls satisfy 2 1. Since the p-norms on R satisfy B ✓B p p = x x we see that .  0 )kkp k kp0 Bp ✓Bp0 Suppose that and are the unit balls of two norms and B1 B2 k·k1 k·k2 on X. It is easy to see that satisfies the conditions needed to be a B1 \B2 unit ball. What is the norm? Consider K = (x, y):xy 0 and x2 + y2 1 . Find the smallest set {  } containing K that is absolutely convex, absorbing find the norm that this set defines.

2.6.4 Polars and the Absolutely Convex Hull Given any set K the intersection of all absolutely convex sets containing K is the smallest absolutely convex set containing K, this set is called the absolutely convex hull of K. Its closure is called the closed absolutely convex hull of K. Given any set K Rn its polar is the set ✓ K0 := y : x y 1 . { | · | } The set K00 := (K0)0 is called the bipolar Theorem 2.78 (The Bipolar Theorem). Let K Rn. Then K0 is a closed, ✓ absolutely convex set. The bipolar K00 is the closed absolutely convex hull of K. n 00 If i R are closed unit balls for norms i,then = 1 2) B ✓ k·k B B [B is the closed unit ball for a norm called the decomposition norm and it satisﬁes x =inf y + z : x = y + z . k k {k k1 k k2 } 46 CHAPTER 2. NORMED AND HILBERT SPACES

2.7 Hilbert Spaces

2.7.1 Hilbert Space Completions

Deﬁnition 2.79. Given a vector space V over F a map, < >: V V F | ⇥ ! is called an inner product provided that it for all vectors v ,v ,w ,w V 1 2 1 2 2 and F, it satisﬁes: 2 1. v + v ,w = v ,w + v ,w , h 1 2 1i h 1 1i h 2 1i 2. v ,w + w = v ,w + v ,w , h 1 1 2i h 1 1i h 1 2i 3. v ,w = v ,w , h 1 1i h 1 1i 4. v ,w = v ,w , h 1 1i h 1 1i 5. v ,w = w ,v , h 1 1i h 1 1i 6. v ,v 0, h 1 1i 7. v ,v =0 v = 0. h 1 1i () 1 The pair (V, ) is called an inner product space. If only satisifes 1–6, h|i h|i then it is called a semi-inner product. Some examples of inner products:

n Example 2.80. The space F with (x1,...,xn), (y1,...,yn) = xiyi h i i Example 2.81. The vector space C([a, b]) of continuous F-valuedP functions on [a, b]with b f,g = f(t)g(t)dt. h i Za 2 Example 2.82. The vector space ` with x, y = 1 x y . h i n=1 n n Proposition 2.83 (Cauchy-Schwarz Inequality). LetPV be an inner product space then v, w v, v w, w . |h i| h i h i Corollary 2.84. Let V be an innerp productp space, then x = x, x k || h i deﬁnes a norm on V . p This is called the norm induced by the inner product. Deﬁnition 2.85. An inner product space is called a Hilbert space if it is complete in the norm induced by the inner product. 2.8. BASES IN HILBERT SPACE 47

Examples 8.2 and 8.4 are Hilbert spaces. To see that 8.3 is not, consider the case of [0,1]. It is enough to consider the sequence

0, 0 t 1/2,   fn(t)= n(t 1/2), 1/2 t 1/2+1/n, , 8   <>1, 1/2+1/n t 1   which can be shown to:> be Cauchy but not converge to any continuous function.

Problem 2.86. Let V be a semi-inner product space. Prove that = v : N { v, v =0 is a subspace, that setting v + ,w+ = v, w is well-deﬁned h i } h N Ni h i and gives an inner product on the quotient V/ . (Hint: First show that N the Cauchy-Schwarz inequality is still true.)

Proposition 2.87. Let V be an inner product space and let be the norm k·k induced by the inner product. Then:

Polarization Identity x, y =1/4 3 ik x + iky 2, h i k=0 k k Parallelogram Law x + y 2 + Px y 2 =2 x 2 +2 y 2. k k k k k k k k Theorem 2.88 (Jordan-von Neumann). Let (X, ) be a Banach space k·k whose norm satisfies the parallelogram law. Then defining via the polar- h|i ization identity, defines an inner product on X and the norm on X is the norm induced by the inner product. Thus, any Banach space satisfying the parallelogram law is a Hilbert space.

Corollary 2.89. Let (V, ) be an inner product space, then there exists a h|i Hilbert space and a one-to-one inner product preserving map J : V H !H such that J(V ) is a dens subspace of . H The space is called the Hilbert space completion of V . For ex- H ample, the Hilbert space completion of 8.3 is the space L2([a, b], ,) of M equivalence classes of measurable square integrable functions with respect to Lebesgue measure.

2.8 Bases in Hilbert Space

Deﬁnition 2.90. Given an inner product space a set of vectors S is called orthonormal if v S implies v = 1 and whenever v, w S and v = w 2 k k 2 6 then v, w = 0. We will often write v w v, w = 0. h i ? () h i 48 CHAPTER 2. NORMED AND HILBERT SPACES

Proposition 2.91 (Pythagoras). Let V be an inner product space and let x :1 n N be an otrhonormal set. Then { n   } N N x 2 = x ,x 2 + x x ,x x 2. k k |h n i| k h n i nk n=1 n=1 X X Proposition 2.92 (Bessel’s Inequality). Let xa : a A be an orthonormal 2 { 2 }2 set. Then a A xa,x converges and is less than x . 2 |h i| k k DeﬁnitionP 2.93. We say that an orthonormal set e : a A in a Hilbert { a 2 } space is maximal if there does not exist an orthonormal set that contains H it as a proper subset. A maximal orthonormal set is called an orthonormal basis(o.n.b.).

Maximal orthonormal sets exist by Zorn’s Lemma(which is equivalent to the Axiom of Choice). The following explains the reason for calling these bases.

Theorem 2.94 (Parseval’s Identities). Let be a Hilbert space and let H e : a A be an orthonormal basis. Then for any x, y , { a 2 } 2H 2 2 1. x = a A ea,x , k k 2 |h i|

2. x = aPA ea,x ea, 2 h i

3. x, yP= a A x, ea ea,y . h i 2 h ih i Problem 2.95.P Prove the 3rd Parseval identity.

Deﬁnition 2.96. Two sets S and T are said to have the same cardinality if there is a one-to-one onto function between them and we write card(S) = card(T). When there exists a one-to-one function from S into T we write that card(S) card(T).  It is a fact that card(S)=card(T ) Card(S) card(T ) and card(T ) ()   card(S). For each natural number 0, 1,... we have a cardinal number. We { } set 0 := card(N) and we set 1 := card(R). The continuum hypothesis @ @ asks if there can be a set with card(N)

Cardinal arithmetic is deﬁned as follows: Given disjoint sets A and B we set card(A)+card(B):=card(A B) and we set card(A) card(B):= [ · card(A B), where A B = (a, b):a A, b B is the Cartesian product. ⇥ ⇥ { 2 2 } Theorem 2.97 (Basis Theorem). Let be a Hilbert space and let e : a H { a 2 A and f : b B each be an orthonormal basis for . Then card(A)= } { b 2 } H card(B).

Deﬁnition 2.98. We call the cardinality of any orthonormal basis for H the Hilbert space dimension of . We denote it by dim ( ). H HS H Example 2.99. For every cardinal number, there is a Hilbert space of that dimension. Here is a way to see this fact. Given any non-empty set A set

2 2 ` := f : A F : f(a) < + . A { ! | | 1} a A X2 On this vector space we deﬁne

f,g = f(a)g(a). h i a A X2 The fact that this converges and defines an inner product goes much like the proof for `2. Now define e `2 by a 2 A 1,t= a ea(t)= . (0,t= a 6 2 Then the set ea : a A can be shown to be an o.n.b. for À. Hence, 2 { 2 } dimHS(À)=card(A). Proposition 2.100 (Gram-Schmidt). Let x :1 n N be a linearly { n   } independent set in a Hilbert space. Then there exists an orthonormal set e :1 n N such that { n   } span x ,...,x = span e ,...,e , k, 1 k N. { 1 k} { 1 k} 8  

Proposition 2.101 (Gram-Schmidt). Let xn : n N be a linearly in- { 2 } dependent set in a Hilbert space. Then there exists an orthonormal set en : n N such that { 2 }

span x1,...,xk = span e1,...,ek , k N. { } { } 8 2 50 CHAPTER 2. NORMED AND HILBERT SPACES

One produces this set by the Gram-Schmidt orthogonalization process. In the finite case. For the general case one uses the finite case together with the Principle of Recursive Definition. Recall that a metric space is called separable if it contains a countable dense subset. Theorem 2.102. Let be a Hilbert space. Then is separable(as a metric H H space) if and only if dim ( ) , i.e., any o.n.b. for is countable. HS H @0 H 2.8.1 Direct Sums Given two vector spaces V and W we can make a new vector space by taking the Cartesian product

(v, w):v V,w W , { 2 2 } and defining (v1,w1)+(v2,w2)=(v1 + v2,w1 + w2) and (v, w)=(v,w). Even though it uses the Cartesian product, this space is generally denoted V W . One reason is that in the finite dimensional case if v ,...,v is { 1 n} a basis for V and w ,...,w is a basis for W ,then (v , 0),...,(v , 0) { 1 m} { 1 n }[ (0,w ),...,(0,w ) is a basis for V W so that dim(V W )=dim(V )+ { 1 m } dim(W ). Given Hilbert spaces ( , , ),i =1, 2 on their Cartesian product we Hi h ii define (h ,h ), (h0 ,h0 ) = h ,h0 + h ,h0 , h 1 2 1 2 h h 1 1i1 h 2 2i2 and it is easy to check that this is an inner product and that the Cartesian product is complete in this inner product. To see this one checks that (h ,h ) is Cauchy if and only if h is Cauchy in and h is Cauchy 1,n 2,n 1,n H1 2,n in . Hence, there are vectors such that h =lim h and h =limh . H2 1 n 1,n 2 2,n Now one checks that (h ,h ) (h ,h ) 0. k 1 2 1,n 2,n k! This Hilbert space is denoted by . Note that in this space, H1 H2 (h , 0) (0,h ) and that (h , 0) = h while (0,h ) = h so that 1 ? 2 k 1 k k 1k1 k 2 k k 2k2 these subspaces can be identified with the original HIlbert spaces and they are arranged as orthogonal subspaces of . H1 H2 It is not hard to verify that if e : a A is an o.n.b. for and { a 2 } H1 f : b B is an o.n.b. for ,then { b 2 } H2 (e , 0) : a A (0,f ):b B , { a 2 }[{ b 2 } is an o.n.b. for .Thus,dim ( )=dim ( )+dim ( ), H1H2 HS H1H2 HS H1 HS H2 where the addition is either ordinary in the finite dimensional case or cardinal arithmetic in general. 2.8. BASES IN HILBERT SPACE 51

Note that when we form the direct sum, we can identify and H1 H2 canonically as subspaces of the direct sum in the following sense. Namely, if we look at = (h , 0) : h and = (0,h ):h ,thenwe H1 { 1 1 2H1} H2 { 2 2 2H2} have that, as vector spaces these are isomorphic to the original spaces and for any h ,he , e 1 10 2H1 h h = (h , 0) (h , 0) , 1 10 1 1 10 1 2 h | iH h | iH H and similarly for any pair of vectors in . We also have that H2 ? = . H1 H2 We now generalize this constructionf f to arbitrary collections. Given a family of Hilbert spaces ( , , ),a A. We deﬁne a new Hilbert space Ha h ia 2 which is a subset of the Cartesian product via

2 a A a = (ha)a A : ha a < + , 2 H { 2 k k 1} a A X2 and inner product (h ), (k ) = h ,k . h a a i h a aia a A X2 2 Proofs similar to the case of `A show that this is indeed a Hilbert space.

2.8.2 Bilinear Maps and Tensor Products Given vector spaces V,W,Z a map B : V W Z is called bilinear ⇥ ! provided that it is linear in each variable, i.e.,

B(v + v ,w)=B(v ,w)+B(v ,w), • 1 2 1 2 B(v, w + w )=B(v, w )+B(v, w ) • 1 2 1 2 B(v, w)=B(v, w)=B(v, w). • Tensor products were created to linearize bilinear maps. This concept takes some development but, in summary, given two vector spaces their tensor product,

V W = span v w : v V,w W , ⌦ { ⌦ 2 2 } where v w is called an elementary tensor and the rules for adding these ⌦ mimic the bilinear rules: 52 CHAPTER 2. NORMED AND HILBERT SPACES

v w + v w =(v + v ) w, • 1 ⌦ 2 ⌦ 1 2 ⌦ v w + v w = v (w + w ), • ⌦ 1 ⌦ 2 ⌦ 1 2 (v w)=(v) w = v (w). • ⌦ ⌦ ⌦ If V and W have bases v : a A and w : b B ,respectively, { a 2 } { b 2 } then v w : a A, b B is a basis for V W . Thus, the vector space { a ⌦ b 2 2 } ⌦ dimension satisﬁes

dim(V W )=dim(V ) dim(W ), ⌦ · where this is just the product of integers in the ﬁnite dimensional case and is true in the sense of cardinal arithmetic in the general case.

Proposition 2.103. There is a one-to-one correspondence between bilinear maps, B : V W Z and linear maps T : V W Z given by T (v w)= ⇥ ! ⌦ ! ⌦ B(v, w).

When V and W are both normed spaces, we would like the tensor product to be a normed space too. There are several “natural” norms to put on the tensor product and on bilinear maps.

Deﬁnition 2.104. Let X, Y and Z be normed spaces and let B : X Y ⇥ be a bilinear map. Then we say that B is bounded provided that

B =sup B(x, y) : x 1, y 1 < + , k k {|| || k k k k } 1 and we call this quantity the norm of B. For u X Y the projective 2 ⌦ norm of u is given by

u =inf xi yi : u = xi yi , k k_ { k k·k k ⌦ } X Xi where the inﬁmum is taken over all ways to express u as a ﬁnite sum of elementary tensors. The completion of X Y in this norm is denoted ⌦ X Y and is called the projective tensor product of X and Y. ⌦_ Here is the relationship these two concepts.

Proposition 2.105. Let X, Y and Z be Banach spaces. Then B : X Y ⇥ ! Z is bounded if and only if the linear map T : X Y Z is bounded when B ⌦ ! X Y is given the projective norm. In this case B = T and T can ⌦ k k k Bk B be extended to a bounded linear map of the same norm from X Y into Z. ⌦_ 2.8. BASES IN HILBERT SPACE 53

The extension of TB to the completion is, generally, still denoted by TB. There is one other tensor norm that is very important. Note that if f : X F and g : Y F are linear, then there is a linear map f g : ! ! ⌦ X Y F given by f g(x y)=f(x) g(y). To see this just check that ⌦ ! ⌦ ⌦ · B(x, y):=f(x)g(y) is bilinear.

Deﬁnition 2.106. Let X and Y be normed spaces. Given u X Y ,the 2 ⌦ injective norm of u is given by the formula

u := sup f g(u) : f X⇤,g Y ⇤, f 1, g 1 . k k^ {| ⌦ | 2 2 k k k k } This is indeed a norm on X Y and the completion in this norm is denoted ⌦ by X Y and is called the injective tensor product of X and Y. ⌦^ Grothendieck was the first to systematically study and classify tensor norms. He defined a reasonableness condition for tensor norms and proved that if X and Y are normed spaces, then any reasonable tensor norm on X Y satisfies, u u u . ⌦ k k^ k kk k_ .

2.8.3 Tensor Products of Hilbert Spaces All of the above discussion is preliminary to defining a norm on the tensor product of Hilbert spaces that yields a new Hilbert space. Let and be Hilbert spaces. We wish to define an inner product on H K .Givenu = h k and v = v w two finite sums representing H⌦K i i ⌦ i j j ⌦ j vectors in ,weset H⌦K P P

u, v = hi vj ki wj . h i h | iHh | iK Xi,j It is not hard to see that if this formula is well-deﬁned, i.e., does not depend on the particular way that we write u and v as a sum of elementary tensors, then it is sesquilinear.

Proposition 2.107. The above formula is well-defined and gives an inner product on . H⌦K Proof. We omit the details of showing that it is well-defined. As noted above, given that it is well-defined it is elementary that it is sesquilinear. So we only prove that it satisfies u =0 = u, u > 0. 6 )h i 54 CHAPTER 2. NORMED AND HILBERT SPACES

To this end let u = n h k . Let = span k :1 i n and i=1 i ⌦ i F { i   } let ej :1 j m be an o.n.b. for so that m n. Write each {   } P F  ki = j ai,jej. Then we have that P u = h (a e )= (a h ) e = v e , i ⌦ i,j j i,j i ⌦ j j ⌦ j Xi Xj Xj Xi Xj where v = a h . Now since u = 0 we must have that some v = 0. j i i,j i 6 j 6 Using the fact that the inner product is independent of the way that we P express u as a sum of elementary tensors, we have that

2 u u = vi vj ei ej = vi =0, h | i h | iHh | iK k k 6 Xi,j Xi and we are done.

Deﬁnition 2.108. Given Hilbert spaces and as above, we let H K H⌦K denote the completion of in the above inner product and call it the H⌦K Hilbert space tensor product.

Many authors use to denote both the vector space tensor product H⌦K AND its completion, or they sometimes invent a non-standard notation to denote the vector space tensor product, like ,reserving for the HK H⌦K completion.

Proposition 2.109. Let and be Hilbert spaces, let e : a A be an H K { a 2 } o.n.b. for and f : b B be an o.n.b. for . Then H { b 2 } K e f : a A, b B . { a ⌦ b 2 2 } is an o.n.b. for . Hence, H⌦K dim ( )=dim ( ) dim ( ). HS H⌦K HS H · HS K In a similar way one can form a Hilbert space tensor product of any ﬁnite collection of Hilbert spaces ,..., , denoted .Theinner H1 Hn H1⌦···⌦Hn product on any two elementary tensors is given by

v1 vn w1 wn = v1 w1 1 vn wn n , h ⌦···⌦ | ⌦···⌦ i h | iH ···h | iH and then is extended linearly to ﬁnite sums of elementary tensor products. Unlike for the direct sum construction, there is no natural way that we can regard and as subspaces of . In particular, h 0:h = H K H⌦K { ⌦ 2H} 2.8. BASES IN HILBERT SPACE 55

(0 0). One way that we can regard as a subspace of the tensor product ⌦ H is to pick a unit vector k . Then we have that := h k : h is 2K H { ⌦ 2H} isomorphic to as a vector space, since (h + h ) k = h k + h k and H 1 2 ⌦ 1 ⌦ 2 ⌦ (h) k = (h k) and has the property that for anye h ,h , ⌦ ⌦ 1 2 2H

h1 h2 = h1 k h2 k . h | iH h ⌦ | ⌦ iH⌦K Similarly, fixing a unit vector in h allows us to include into the tensor 2H K product as := h k : k . K { ⌦ 2K} However, unlike the direct sum these two subspaces are not orthogonal, in fact, theire intersection is the one-dimensional subspace spanned by h k. ⌦ 2.8.4 Infinite Tensor Products Many physical models call for tensor products of infinitely many Hilbert spaces. This arises for example where one has an infinite lattice of electrons. Each electron has a state space that is a Hilbert space and the models Ha say that the state space of the whole ensemble should be the tensor product over all electrons. The first problem that one encounters is that if one has two elementary tensor products of infinitely many vectors, v v and w w , 1 ⌦ 2 ⌦··· 1 ⌦ 2 ⌦··· then one would like to set

1 v v w w = v w , h 1 ⌦ 2 ⌦···| 1 ⌦ 2 ⌦···i h i| ii Yi=1 but such an infinite product need not converge. Second, if we had a set of unit vectors v ,... and we sat u = v (v /2) (v /3) , then we would 1 1 ⌦ 2 ⌦ 3 ⌦··· have that 1 1 u 2 = u u = =0. k k h | i i2 Yi=1 However, in physics one usually has a lot more. Each electron has a distinguished unit vector in its state space called the ground state and it turns out that given Hilbert spaces and distinguished unit vectors Ha a 2 , then this is enough data to form an infinite tensor product. Ha The key is, remember that when we have a unit vector then we can “include” into by sending h h k,wherek is some fixed unit H H⌦K ! ⌦ vector. We form the infinite tensor product of a collection ( , ),a A,where Ha a 2 is a unit vector as described below. a 2Ha 56 CHAPTER 2. NORMED AND HILBERT SPACES

For each finite subset F A there is a well-defined tensor product ✓ Hilbert space of these finitely many spaces, which we denote by := HF a F a(we do not need to complete at this time). ⌦ 2 H Now given two finite sets F F A,sayF = a ,...,a and 1 ✓ 2 ✓ 1 { 1 n} F = a ,...,a ,...a with m>n, we can regard as a subspace of 2 { 1 n m} HF1 HF2 by identfying u with u . Formally, this 2HF1 ⌦ an+1 ⌦···⌦ am 2HF2 means that we have a linear map, V : . F1,F2 HF1 !HF2 Now take the union over all finite subsets, F A F , and define a relation [ ✓ H on this union by declaring u equivalent to u provided 1 2HF1 2 2HF2 that there exists a finite set F with F F and F F such that 3 1 ✓ 3 2 ✓ 3 V (u )=V (u )( we will write u u for this relation). It is not F1,F3 1 F2,F3 2 1 ⇠ 2 hard to show that this is an equivalence relation and we will write [u] for the equivalence class of a vector.

It is not hard to show that = F A F / = [u]: F, u F K [ ✓ H ⇠ { 9 2H } is an inner product space with operations deﬁned as follows. Given F 2 and [u]set[u]=[u]. Given [u ], [u ]withu and u pick 1 2 1 2HF1 2 2HF2 F with F ,F F and set 3 1 2 ✓ 3

[u1]+[u2]=[VF1,F3 (u1)+VF2,F3 (u2)].

Finally, the inner product is deﬁned by

[u1] [u2] = VF1,F3 (u1) VF2,F3 (u2) . h | iK h | iHF3

Showing that all of these operations are well-defined, that is, only depend on the equivalence class, not on the particular choices, and that the last formula is an inner product is tedious, but does all work. The completion of the space is denoted K a A( a, a). ⌦ 2 H In the end this tensor product does not depend on the choice of vectors, in the following sense. If we also choose unit vectors , then for each a 2Ha Hilbert space there is a unitary map U : such that U ( )= a Ha !Ha a a a(we will discuss unitaries in a later section) and these define unitaries U : via U (h h )=U (h ) U (h ). Together F HF !HF F a1 ⌦···⌦ an a1 a1 ⌦···⌦ an an these define a unitary U on by for u setting U([u]) = [U (u)]. Then K 2HF F this unitary satisfies U([ ]) = [ ], and so induces a a1 ⌦··· an a1 ⌦···⌦ an unitary on the completions, U : a A( a, a) a A( a,a). ⌦ 2 H ! ⌦ 2 H