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2 Banach Spaces and Hilbert Spaces

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2 Banach Spaces and Hilbert Spaces 2 Banach spaces and Hilbert spaces Trying to do analysis in the rational numbers is difficult for example consider 2 the set fx 2 Q : x ≤ 2g: This set is non-empty and bounded above but does not have a least upper bound in the rationals. In the real numbers things are much better since we have completeness (one of the axioms for the real numbers) which says that every set of reals bounded above has a least upper bound. An alternative way to state this axiom is that every Cauchy sequence in the real numbers is convergent to a real number. Similarly with normed spaces it will be easier to work with spaces where every Cauchy sequence is convergent. Such spaces are called Banach spaces and if the norm comes from an inner product then they are called Hilbert spaces. Definition 2.1. Let V be a normed space with norm k·k and fxngn2N a sequence in V . We say that fxngn2N is a Cauchy sequence if for all > 0 there exists N 2 N such that for all n; m ≥ N we have that kxn − xmk ≤ . Definition 2.2. Let V be a normed space with norm k·k. We say that V is a Banach space if every Cauchy sequence fxngn2N in V is convergent (i.e. there exists x 2 V such that limn!1kxn − xk = 0). In other words the metric defined by the norm is complete. Definition 2.3. Let V be an inner product space. We say that V is a Hilbert space if V is a Banach space with the norm defined by the inner product (i.e. kvk = phv; vi). In this rest of the section we will look at some examples of normed spaces which are Banach spaces and some which are not. We'll then go on to look at the geometry of Hilbert spaces in more detail. To show that a normed space V is a Banach space we simply need to show that every Cauchy sequence xn is convergent. To do this there is a standard stragedy. step 1 Find a candidate y for the limit of xn. step 2 Show that y 2 V . step 3 Show that limn!1 xn = y. Example. Let V = C([0; 1]) be the space of all complex valued continuous functions with the norm kfk1 = supx2[0;1]fjf(x)j. We show that V is a Banach space using the above stragedy. Let fn be a Cauchy sequence in V . step 1. We need to find a candidate f for the limit of fn. Fix x 2 [0; 1] and note that for all n; m 2 N we have that jfn(x) − fm(x)j ≤ kfn − fmk1: Thus fn(x) is a Cauchy sequence in C and since C is complete we know that limn!1 fn(x) exists. So we let f(x) = limn!1 fn(x). 1 step 2. We need to show that f is continuous. Fix x 2 [0; 1] and let > 0 we can choose N such that for all n; m ≥ N we have that kfn − fmk1 ≤ /3. We fix δ > 0 such that if jx − yj ≤ δ then jfN (x) − fN (y)j ≤ /3. For any n; m ≥ N and y such that jx − yj ≤ δ we have that jf(x) − f(y)j = jf(x) − fn(x) + fn(x) − fN (x) + fN (x) − fN (y) + fN (y) − fm(y) + fm(y) − f(y)j ≤ jf(x) − fn(x)j + kfn − fN k1 + jfN (x) − fN (y)j + jfN (y) − fm(y)j + jfm(y) − f(y)j ≤ 3 + jf(x) − fn(x)j + jfm(y) − f(y)j: If we let n; m ! 1 and use that 0 = limn!1 jf(x) − fn(x)j = limm!1 jfm(y) − f(y)j then we have that jf(x) − f(y)j ≤ and we can conclude that f is continuous. step 3 Finally we need to show that limn!1kfn − fk1 = 0. Again let > 0 and choose N such that for n; m ≥ N we have that kfn − fmk ≤ . For n; m ≥ N and any x 2 [0; 1] we can write jfn(x) − f(x)j ≤ jfn(x) − fm(x)j + jfm(x) − f(x)j ≤ + jfm(x) − f(x)j: So if we let m ! 1 we get that jfn(x) − f(x)j ≤ . Since this holds for all x 2 [0; 1] we have that kfn − fk ≤ and we can conclude tha tin V , limn!1 fn = f. Thus V is a Banach space. Theorem 2.4. For all 1 ≤ p ≤ 1 we have that `p with respect to the norm k·k is a Banach space. Proof. The case p = 1 is left a san exercise so we'll fix 1 ≤ p < 1 and p follow the same three steps. We let (xn)n2N be a Cauchy sequence in ` and write (1) (1) xn = (xn ; xn ;:::): (k) step 1. Fix k 2 N and consider the sequence (xn )n2N in F . For any n; m 2 N we have that 1 !1=p (k) (k) (k) (k) p 1=p X (k) (k) p jxn −xm j = (jxn −xm j ) ≤ jxn − xm j = kxn−xmkp: k=1 (k) (k) Thus xn is a Cauchy sequence in F and so has a limit y . So let y = (y(1); y(2);:::). 2 p P1 (k) p step 2. We now need to show that y 2 ` . So we need to show that k=1 jy j < 1. We let N 2 N satisfy that for any n; m ≥ N we have kxn−xmk ≤ . Thus for j; n 2 N with n ≥ N we have that by Minkowski's inequality j !1=p j !1=p X (k) p X (k) (k) (k) (k) (k) p jy j = jy − xn + xn − xN + xN j k=1 k=1 j !1=p X (k) (k) p ≤ jy − xn j + kxn − xN kp + kxN kp: k=1 We have that for any n ≥ N, kxn − xN k ≤ 1 and j !1=p X lim jy(k) − x(k)jp = 0: n!1 n k=1 So we can conclude that for any j 2 N, j !1=p X (k) (p) jy j ≤ 1 + kxN kp k=1 and thus 1 X jy(k)jp < 1: k=1 step 3 Finally we have to show that limn!1kxn − ykp = 0 and we proceed similarly to step 2. Let > 0 and choose N such that for all n; m ≥ N we have that kxn − xmkp ≤ . For n; m ≥ N and j 2 N we have that using Minkowski's inequality j !1=p j !1=p X (k) (k) p X (k) (k) p jy − xn j ≤ jy − xm j + kxm − xnkp: k=1 k=1 For any n; m ≥ N we have that kxn − xmk ≤ and j !1=p X lim jy(k) − x(k)jp = 0: m!1 m k=1 Thus j !1=p X k (k) p jy − xn j ≤ k=1 and since this bound is independent of j we have that ky − xnkp ≤ . Thus we have shown that limn!1kxn − ykp = 0. 3 We now turn to an example which is not a Banach space. Example. Let C([0; 1]) be the space of continuous functions f : [0; 1] ! C. We define Z 1 1=2 2 kfk2 = jf(x)j dx : 0 This is a normed space where the norm comes from an inner product but it not a Banach space. To see this we define for n ≥ 2 8 1 1 0 if 0 ≤ x ≤ 2 − n < 1 1 1 1 1 fn(x) = n x − 2 + n if 2 − n < x ≤ 2 : 1 1 if 2 < x ≤ 1: p Let > 0 choose N such that 1= N ≤ . We then have that for n; m ≥ N Z 1 Z 1=2 2 jfn(x) − fm(x)j dx ≤ 1dx = 1=N: 0 1=2−1=N p Thus kfn − fmk2 ≤ 1= N ≤ . So fn is a Cauchy sequence. Now suppose there exists f 2 C([0; 1]) such that limn!1kfn − fk2 = 0. In this case we have that Z 1=2 Z 1 2 2 lim jfn(x) − f(x)j dx = lim jfn(x) − f(x)j dx = 0: n!1 0 n!1 1=2 Since fn(x) is constantly equal to 1 on [1=2; 1] this means that f(x) = 1 for all x on [1=2; 1]. Now since f is continuous this means there exists δ > 0 such that f(x) > 1=2 for all x 2 [1=2 − δ; 1=2]. We now choose N 2 N such that N > 2/δ. Thus for all n ≥ N we have that fn(x) = 0 for any x 2 [1=2 − δ; 1=2 − δ=2]. Thus Z 1=2 2 jfn(x) − f(x)j dx ≥ δ=4: 0 So fn cannot converge to f and we can conclude that the norm k·k2 is not complete on C([0; 1]) and we do not have a Hilbert space. Remark. It is in fact possible to extend the space C([0; 1]) so we do get a Hilbert space. We could first extend to Riemann integrable functions, however this is still not complete and k·k2 is no longer a norm. To get round this we need to extend to the space of functions where jfj2 is Lebesgue integrable. To get round the problem that k·k2 is not a norm we need to identify functions which only disagree on measure zero sets (if you're not familiar with measure don't worry).
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