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Normed and Hilbert Spaces Chapter 2 Normed and Hilbert Spaces 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. λx = λ x , λ F, x X, k k | |·k k 8 2 8 2 3. (Triangle Inequality) x + y x + y , x, y X. k kk 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(λx, λy)= λ 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 δ>0 2 6 and a small interval [a, b] on which f(t) δ. Hence, f δ(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 <p,q<+ with p + q =1. p q 1 Then for any a, b 0 we have ab a + b . ≥ p q Problem 2.11. Prove this inequality. Proposition 2.12 (Holder’s Inequality). Let 1 <p,q<+ with 1 + 1 =1, 1 p q and let x =(x ,...) `p,y=(y ,...) `q, then (x y ,...) `1 and 1 2 1 2 1 1 2 x y x y . | n n|k kpk kq n X 2.2. BANACH SPACES 21 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<+ for succinctness. First 1 note that 2 xi when xi yi xi + yi | | | |≥| | . | |(2 yi when xi yi | | | || | p p p p p Hence, xi +yi max 2 xi , 2 yi , from which it follows that i xi + p | | { | | | | } | yi < + . | 1 P Next notice that 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.
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