Functional Analysis Winter 2018/2019

Transcription

1 Functional Analysis Winter 2018/2019 Peer Christian Kunstmann Karlsruher Institut für Technologie (KIT) Institut für Analysis Englerstr. 2, Karlsruhe These lecture notes contain a summary of the course. They are intended to be used parallel to the lecture. They are not meant to be used as a text book or for self studies. Attending the lectures cannot be substituted by reading these lecture notes. 1

2 1 Metric und normed spaces 1.1. Definition: Let X be a set. A metric on X is a map d : X X [0, ) satisfying (M1) x, y X: d(x, y) = 0 x = y (definiteness); (M2) x, y X: d(x, y) = d(y, x) (symmetry); (M3) x, y, z X: d(x, y) d(x, z) + d(z, y) (triangle inequality). If d is a metric on X then the pair (X, d) is called a metric space. If d only satisfies (N2) and (N3) and d(x, x) = 0 for all x X then d is called a semimetric on X. Remarks: If d is a semimetric on X then x, y, X : d(x, z) d(x, y) d(z, y) (reverse triangle inequality); x, y, u, v X : d(x, y) d(u, v) d(x, u) + d(y, v). The first inequality follows from d(x, z) d(x, y) + d(y, z) and d(x, y) d(x, z) + d(z, y). The proof of the second inequality is similar. Examples: (1) d(x, y) := x y defines a metric on R. (2) d((x 1, x 2 ), (y 1, y 2 )) := x 1 y 1 define a semimetric on R Norms: We always consider vecctor spaces over the field K {R, C}, i.e. real vector spaces for K = R or complex vector spaces for K = C. Definition: Let X be a K-vector space. A norm on X is a map : X R satisfying (N1) x X: x = 0 = x = 0 (definiteness); (N2) x X, α K: αx = α x (homogeneity); (N3) x, y X: x + y x + y (triangle inequality). Then (X, ) is called a normed space. If only satisfies (N2) and (N3) then is called a seminorm on X. Remarks: (1) If is a seminorm on X then x 0 for all x X. [By (N2) we obtain 0 = 0; further 0 = x x x + x = 2 x, hence x 0.] Again, the triangle inequality implies x y x y for all x, y X. (2) If is a norm [resp. seminorm] on X then d(x, y) := x y defines a metric [resp. semimetric] on X. 2

3 Examples: (1) Let n N. Then (x j ) n j=1 1 := n j=1 x j defines a norm on K n. (2) Let M be a set and B(M) := {f : M K : f is bounded }. Then B(M) is a K-vector space (as a linear subspace of the K-vector space K M of all functions M K) and f := sup{ f(ω) : ω M} defines a norm on B(M): (N1) and (N2) are clear. For the proof of (N3) take f, g B(M). For any ω M we then have (f + g)(ω) f(ω) + g(ω) f + g, and taking the sup over ω M yields (N3). (3) Let p (1, ) and n N. Then ( n ) 1/p (x j ) n j=1 p := x j p defines a norm K n. Again, (N1) and (N2) are clear, but (N3) needs some work. j=1 Young s inequality: For all a, b 0 we have ab ap = 1. p q p + bq q where q (1, ) is such that Proof. This is clear for b = 0. So fix b > 0 and consider f : [0, ) R, f(a) := ab a p /p. Then f(0) = 0, f(a) for a, f is differentiable on [0, ) with f (a) = b a p 1. Since f is strictly decreasing, f attains its maximum at the only zero a 0 = b 1/(p 1) of f, and for any a 0: since p p 1 = q. f(a) f(a 0 ) = b 1/(p 1) b bp/(p 1) p = bq q, Hölder s inequality: For all n N and a 1,..., a n, b 1..., b n 0 we have n ( n a j b j j=1 j=1 a p j ) 1/p ( n 1/q. bj) q Proof. This is clear if j ap j = 0 or j bq j = 0. By homogeneity it thus suffices to consider the case j ap j = 1 and j bq j = 1 (otherwise let ã j := a j /( k ap k )1/p and bj := b j /( k bp k )1/p ). We then have by Young j=1 a j b j 1 a p j p + 1 b q j q = 1 p + 1 q = 1. j j j 3

4 Minkowski s inequality: Let n N and x = (x j ) n j=1, y = (y j ) n j=1 K n. Then ( n ) 1/p ( n ) 1/p ( n ) 1/p. x j + y j p x j p + y j p j=1 This finally is (N3) for p on K n. j=1 Proof. By x j + y j x j + y j it suffices to take x j, y j 0. The assertion is clear for j (x j + y j ) p = 0, so we may assume j (x j + y j ) p = 1. Then, by Hölder, 1 = j ( n (x j + y j ) p = j j=1 x p j ) 1/p ( n j=1 z q j x j (x j + y j ) p 1 }{{} ) 1/q + ( n =:z j + j=1 y p j j=1 j y j (x j + y j ) p 1 }{{} =z j ) 1/p ( n ) 1/q, z q j j=1 and z q j = (x j + y j ) qp q = (x j + y j ) p (since pq q = p). Hence j zq j is proved. = 1 and the inequality 1.3. Open and closed sets: Let (X, d) be a metric space. For r > 0, x 0 X let B(x 0, r) := {x X : d(x, x 0 ) < r} open ball with radius r around x 0 B(x 0, r) := {x X : d(x, x 0 ) < r} closed ball with radius r around x 0. Further we define Q X is called open : x 0 Q r > 0 : B(x 0, r) Q. A X is called closed : X \ A is open. Examples: X,, B(x 0, r) are open; X,, B(x 0, r) are closed: The assertions on X and are clear. The triangle inequality shows for x B(x 0, r) that B(x, r d(x, x 0 )) B(x 0, r) and for x B(x 0, r) that B(x, d(x, x 0 ) r) X \ B(x 0, r). Remarks: (1) Arbitrary unions of open sets are open, finite intersections of open sets are open. (2) Arbitrary intersections of closed sets are closed, finite unions of closed sets are closed. (3) If M X then (M, d M M ) is a metric space. A subset Q M is called relatively open [resp. relatively closed] if M is open in the metric space (M, d M M ). Example: [0, 1) is relatively open in [0, 1], but not open in R. Proposition: Let Q M. Then Q is relatively open in M if and only if there exists Q X open in X such that Q = Q M. 4

5 Proof. is clear. For put Q := {B X (x 0, r) : x 0 Q, r > 0, B M (x 0, r) Q}, where B X (x 0, r) = {x X : d(x, x 0 ) < r} and B M (x 0, r) = {x M : d(x, x 0 ) < r}. Observe that B M (x 0, r) = B X (x 0, r) M and thus Q M = {B X (x 0, r) M : x 0 Q, r > 0, B M (x 0, r) Q} = {B M (x 0, r) : x 0 Q, r > 0, B M (x 0, r) Q} = Q, where for the last inclusion we use that Q is relatively open in M. Definition: A subset U X is called a neighborhood of x 0 X if there exists r > 0 such that B(x 0, r) U. Let M X. We define the interior of M end Mon int(m) := {x X : M is a neighborhood of x }, the closure of M and the boundary of M M := {x X : U M for any neighborhood U of x }, M := {x X : U M and U (X \ M) for any neighborhood U of x }. An x X is called interior point of M, if x int(m), and boundary point of M if x M. Remarks: (1) int(m) is the largest open subset of M, M is the smallest closed superset of M, and int(x \ M) = X \ M, X \ M = X \ int(m). (2) M is open int(m) = M; M is closed M = M. (3) M = M X \ M, in particular M is closed. (4) M is the disjoint union of int(m) and M, and X is the disjoint union of int(m), M and int(x \ M) Convergence: Let (X, d) be a metric space and let (x n ) n N be a sequence in X and x X. Then (x n ) is said to be convergent to x (x n x) with respect to d if d(x n, x) 0, and (x n ) is said to be a Cauchy sequence with respect to d if ε > 0 n 0 N n, m n 0 : d(x n, x m ) < ε. By (M1) the limit of a convergent sequence is uniquely determined. Proposition: Let M X, x 0 M. Then x 0 M if and only if there exists a sequence (x n ) in M such that x n x 0. 5

6 Proof. : For any n N we find x n B(x 0, 1/n) M. Then (x n ) is a sequence in M with d(x n, x 0 ) 1/n 0. : Let r > 0. We find n N with d(x n, x 0 ) < r. Then x n B(x 0, r) M. We have shown x 0 M. Examples: (1) If (X, d) is a metric space and x n x, y n y then d(x n, y n ) d(x, y). This follows from d(x n, y n ) d(x, y) d(x n, x) + d(y n, y). (2) If (X, ) is a normed space and x n x, y n y in X and α n α K then x n + y n x + y, α n x n αx and x n x. This follows from x n + y n (x + y) x n x + y n y, α n x n αx α n x n x + α n α x, x n x x n x. In particular, a convergent sequence (x n ) in X is bounded, i.e. satisfies sup n x n <. (X, d) is said to be complete if any Cauchy sequence in X is convergent to some x X. Examples: For the metric induced by, R and C are complete, but Q is not complete. A Banach space is a normed space which is complete for the metric induced by the norm. Examples: (1) Let M be a set. Then (B(M), ) is a Banach space: Let (f n ) be a -Cauchy sequence in B(M). Step 1 Find a candidate: Fix ω M. We have f n (ω) f m (ω) f n f m, so (f n (ω)) is Cauchy in K, hence convergent. Thus we can define f : M K, ω f(ω) := lim n f n (ω). Step 2 Show that f is bounded, i.e. f B(M), and f n f 0: Let ε > 0. By assumption we find n 0 such that f n f m ε for all n, m n 0. Fix n n 0. For any ω M we have f n (ω) f(ω) = lim f n(ω) f m (ω) ε. m }{{} f n f m Taking the sup over ω M shows f n f ε. This implies f = f n0 +(f f n0 ) B(M) and f n f 0. Special case M = N: l := B(N). (2) Let p [1, ). Then l p := { x = (x j ) j N K N : j=1 } x j p < 6

7 is a vector space and x p := ( j=1 x j p ) 1/p defines a norm on l p. Moreover, (l p, p ) is a Banach space. It is clear that 0 l p and that, for x l p and α K, αx l p and αx p = α x p. So let x, y l p. We have to show x + y l p and the triangle inequality. For any N N we have by Minkowski ( N ) 1/p ( N ) 1/p ( N ) 1/p x j + y j p x j p + y j p x p + y p. j=1 Letting N we obtain x + y p x p + y p. ( j=1 j=1 j=1 x j + y j p ) 1/p x p + y p <, hence x + y l p and Completeness: Let (x (n) ) be a p -Cauchy sequence in l p. For any j N, (x (n) j ) n is Cauchy in K and x (0) j := lim n x (n) j exists. Setting x (0) := (x (0) j ) j we have a candidate for the limit. We show x (0) l p and x (n) x (0) p 0: Let ε > 0. By assumption we find n 0 such that x (n) x (m) p ε for all n, m n 0. For n n 0 and any N N we then have ( N j=1 x (n) j x (0) j p ) 1/p = lim m ( N j=1 x (n) j x (m) j p ) 1/p } {{ } x (n) x (m) p This implies for N : x (n) x (0) p < for all n n 0. Hence x (0) = x (n 0) (x (n 0) x (0) ) l p and x (n) x (0) p Lemma: Let (X, d) be a complete metric space and M X. Then (M, d M M ) is complete if and only if M is closed in X. Proof. If M is complete and (x n ) is a sequence in M converging to x X then (x n ) is Cauchy and therefore has a limit x M. But then x = x M, and we have shown that M is closed. If M is closed in X and (x n ) is a Cauchy sequence in M then we find, by completeness of X, an x X such that x n x. By closedness of M we obtain x M and (x n ) has a limit in M. Example: c 0 := {x = (x j ) K N : lim j x j = 0} is a Banach space with respect to, i.e. c 0 is a closed subspace of l : Let (x (n) ) be a sequence in c 0 converging to x (0) l. We have to show x (0) c 0. So let ε > 0. By assumption we find n N with x (n) x (0) ε/2 and by x (n) c 0 we find j 0 N such that x (n) j ε/2 for all j j 0. For any j j 0 we then have x (0) j x (0) j ε. x (n) j + x (n) j x (n) x (0) + ε/2 ε/2 + ε/2 = ε, 7

8 which proves x (0) j 0 as j, i.e. x (0) c 0. Also c := {x = (x j ) K N : (x j ) converges in K } is a closed subspace of l Denseness and separability: Let (X, d) be a metric space and D, B X. Then D is called dense in B if B D, i.e. if an x B can be approximated by a sequence in D. If D is dense in X it is simply called dense. The space (X, d) is called separable if it contains a countable dense subset. Examples: R n is separable since Q n is a countable dense subset. C n is separable since Q n + iq n is a countable dense subset. end Thu Proposition: A normed space (X, ) is separable if and only if it contains a countable subset A such that lin(a) is dense in X. Here lin(a) denotes the linear span of A, i.e. the smallest linear subspace containing A or, in other words, the set of all linear combinations of elements of A. Proof. If D is dense then also lin(d) is dense. So assume that A is countable and lin(a) is dense. Then the set D of all linear combinations of elements of A with rational coefficients is countable and dense in lin(a), and thus also dense in X. Examples: The spaces c 0 and l p, p [1, ), are separable: The space of finite sequences ϕ := {x = (x j ) K N : j 0 N j > j 0 : x j = 0} is dense and we have ϕ = lin{e n : n N} where e n := (δ jn ) j N for any n N denotes the n-th unit vector. The space l is not separable: The set M := {0, 1} N l is uncountable and for any two vectors x, y M with x y we have x y = 1. If D is dense in l then D is dense in M and for every x M we have D x := B(x, 1/4) D, but D x D y = for x y, i.e. D contains uncountably many disjoint non-empty subsets and thus cannot be countable. Lemma: Any subset of a separable space is separable. Proof. left as an exercise Continuity: Let X, Y be metric spaces. A map f : X Y is called continuous if, for any open set Q Y, the pre-image f 1 (Q) is open in X. f is called continuous at x 0 X if for any neighborhood V of f(x 0 ) there exists a neighborhood U of x 0 such that f(u) V. We clearly have: f continuous at x 0 ε > 0 δ > 0 x X : d(x, x 0 ) < δ d(f(x), f(x 0 )) < ε x n x 0 in X f(x n ) f(x 0 ) in X The first condition may be rephrased as ε > 0 δ > 0 : B(x 0, δ) f 1 (B(f(x 0 ), ε)). 8

9 Proposition: A map f : X Y is continuous if and only if it is continuous at every point. Proof. : Let f be continuous, x 0 X, and ε > 0. Then B(f(x 0 ), ε)) is open and thus also M := f 1 (B(f(x 0 ), ε)) is open by assumption. Since x 0 M we find δ > 0 such that B(x 0, δ) M. We have shown that f is continuous at x 0. : Let f be continuous at every point and Q Y be open. Let x 0 f 1 (Q). Then f(x 0 ) Q and we find ε > 0 with B(f(x 0 ), ε) Q and then δ > 0 with B(x 0, δ) f 1 (B(f(x 0 ), ε) f 1 (Q). Hence f 1 (Q) is open. Examples: (1) If (X j, d j ), j = 1, 2, are metric spaces then X 1 X 2 is metric space for the metric d given by d((x 1, x 2 ), (y 1, y 2 )) := d 1 (x 1, y 1 ) + d 2 (x 2, y 2 ). For any metric space (X, d) the metric d : X X [0, ) is continuous. (2) If (X j, j ), j = 1, 2, are normed spaces then X 1 X 2 is a normed space for the norm given by (x 1, x 2 ) = x 1 + x 2. For any normed space (X, ) the maps + : X X X and : K X X are continuous. Moreover, the norm : X R is continuous Uniform continuity: Let X, Y be metric spaces. A map f : X Y is called uniformly continuous if ε > 0 δ > 0 x, y X : d(x, y) < δ d(f(x), f(y)) < ε. Remarks: (1) A uniformly continuous map is clearly continuous. (2) A uniformly continuous map maps Cauchy sequences to Cauchy sequences: If (x n ) is Cauchy in X and ε > 0 we find δ > 0 according to uniform continuity of f and then n 0 such that d(x n, x m ) < δ for all n, m n 0. By choice of δ this implies d(f(x n ), f(x m )) < ε for all n, m n 0. Hence (f(x n )) is Cauchy in Y. (3) A map f : X Y is called Lipschitz continuous if there exists an L > 0 such that d(f(x), f(y)) Ld(x, y) for all x, y X. Any Lipschitz continuous map is unifomrly continuous (one can take δ = ε/l). Examples: (1) If (X, d) is a metric space then d : X X [0, ) is Lipschitz continuous (with L = 1). (2) If (X, ) is a normed space then : X R is Lipschitz continuous (with L = 1) and + : X X X is Lipschitz continuous (with L = 1). If X {0} then : K X X is not uniformly continuous: Choose x 0 X with x 0 = 1. For δ > 0 and k N the pairs (k + δ, (k + δ)x 0 ) and (k, kx 0 ) have distance 2δ but (k + δ) (k + δ)x 0 k kx 0 = (k + δ) 2 k 2 = 2kδ + δ 2 (k ). The following principle has countless applications in praxis. 9

10 1.9. Proposition: Let X, Y be metric spaces, D X, Y complete, and f : D Y uniformly continuous. Then there exists a unique continuous extension f : D Y of f. This extension f is uniformly continuous. Proof. The extension necessarily has to satisfy f(x) = lim n f(x n ) if (x n ) is a sequence in D with x n x. We take this as the definition of f and have to show that f(x) is well-defined, i.e. the limit exists and does not depend on the chosen sequence. If (x n ) is a sequence in D with x n x then (x n ) is Cauchy in D, so (f(x n )) is Cauchy in Y and the limit exists by completeness of Y. If (y n ) is another sequence in D with y n x then also lim n f(y n ) exists. Letting z n := x (n+1)/2 for odd n and z n := y n/2 for even n we have defined a sequence (z n ) in D with z n x and conclude lim n f(x n ) = lim n f(z n ) = lim n f(y n ). Hence f : D Y is well-defined and it rests to show uniform continuity. To this end let ε > 0. We find δ > 0 according to uniform continuity of f. Now let x, y D such that d(x, y) < δ/3. we find sequences (x n ) and (y n ) in D with x n x and y n y and we find n 0 N such that d(x n, x) < δ/3 and d(y n, y) < δ/3 for all n n 0. For n n 0 we then obtain d(x n, y n ) d(x n, x) + d(x, y) + d(y, y n ) < δ 3 + δ 3 + δ 3 = δ, hence d(f(x n ), f(y n )) < ε by choice of δ, and we conclude d( f(x), f(y)) = lim n d(f(x n ), f(y n )) ε, which finishes the proof of uniform continuity Linear operators: Let X, Y be normed spaces (over K) and T : X Y be K-linear. The following are equivalent: end Mon (i) T is continuous, (ii) T is continuous at 0, (iii) there exists C > 0 such that, for all x X, T x C x, (iv) T is Lipschitz-continuous, (v) T is uniformly continuous. Proof. (iv) (v) (i) (ii) is clear. (ii) (iii): If (iii) does not hold we find a sequence (x n ) in X satisfying x n = 1/n and T x n > 1. Then x n 0 but T x n 0, so (ii) does not hold. (iii) (iv): By (iii) and linearity of T we obtain, for all x, y X, Hence T is Lipschitz continuous. T x T y = T (x y) C x y. 10

11 Linear operators satisfying (iii) are called bounded. We let Then L(X, Y ) is a vector space and L(X, Y ) := {T : X Y : X is linear and bounded }. { T x } T := sup x : x X \ {0} defines a norm on L(X, Y ), the operator norm. (For X = {0} we use the convention sup := 0 here). Remark: For T L(X, Y ) we have T = inf{c 0 : (iii) holds } = sup{ T x : x X, x = 1} = sup{ T x : x X, x 1}. Proof. (N1) is clear. (N2) follows from (N2) for the norm of Y. For the proof of (N3) let S, T L(X, Y ) and x X with x 1. Then hence S + T S + T. (S + T )x Sx + T x S + T, Remark: If Z is another normed space and T L(X, Y ), S L(Y, Z) then clearly ST L(X, Z) and ST S T. Definition: In case Y = K the space X := L(X, K) is called the dual space of X. The elements of X are called (linear) functionals on X. Remark: If T L(X, Y ) then the kernel N(T ) = T 1 ({0}) of T is a closed linear subspace of X Proposition: Let X be a normed space and Y a Banach space. Let D X be a dense linear subspace and T L(D, Y ). Then there exists a unique continuous extension T : X Y of T, and for this extension we have T L(X, Y ) and T = T. Proof. Use 1.9, linearity and boundedness of T Proposition: Fix p (1, ). Let q [1, ) such that = 1. We show that p q J : l q (l p ) defined by (Jy)(x) = j=1 x jy j for x = (x j ) l p, y = (y j ) l q, defines an isometry, i.e. J : l q (l p ) is linear and bijective and satisfies Jy (l p ) = y lq for all y l q. 11

12 Proof. If y l q then we have by Hölder (Jy)(x) = x j y j x l p y l q for all x l p. j=1 Hence Jy (l p ) and Jy (l p ) y l q. Conversely, for a given φ (lp ) we let y j := φ(e j ). For any x = (x j ) l p we have x = j x je j where the series converges in l p. Hence φ(x) = j x jφ(e j ) = j x jy j where the series converges in K. In particular, we see that φ is uniquely determined by the sequence y := (y j ). We also have φ(x) x l p y l q which implies φ y l q. We only have to show y l q and y l q φ to finish the proof. To this end we fix N N and define x (N) = N j=1 y j y j q 2 e j. Then we have and We thus obtain N y j q = J(y)(x (N) ) = φ(x (N) ) φ x (N) l p j=1 ( N ) 1/p ( N ) 1 1/q. x (N) l p = y j pq p = y j q j=1 j=1 ( N ) 1/q y j q φ, j=1 which implies for N that y l q and y l q φ. Remark: So for p (1, ), the dual space of l p is l q where the action of y = (y j ) on x = (x j ) is given by J(y)(x) = j=1 x jy j. This is written as duality bracket x, y := x j y j. Further examples: In the same way, using also x, y = j=1 x jy j, one can show (c 0 ) = l 1 and (l 1 ) = l. However, as we shall see later, not every functional on l is given by an l 1 -sequence Proposition: If X is a normed space and Y is a Banach space then L(X, Y ) is a Banach space. In particular, the dual space X of a normed space X is always a Banach space. Proof. Let (T n ) be a Cauchy sequence in L(X, Y ). By T n x T m T n T m x, (T n x) is Cauchy in Y for any x X, hence convergent. Thus we can define T : X Y by x T x := lim n T n x. It is easy to see that T is linear (use 1.4). Boundedness follows from T x = lim T n x lim T n x. n n 12 j=1

13 Let ε > 0. We find n 0 such that T n T m < ε for all n, m n 0. For n n 0 and x X with x 1 we then have T n x T x = lim T nx T m x ε, m }{{} T n T m hence T n T ε. We have shown T n T Proposition (completion): Let X be normed space. There exists a Banach space X and a linear map J : X X such that J(X) is dense in X and Jx = x for all x X. Any such pair ( X, J) is called a completion of X. Proof. We set X := {x = (x j ) K N : (x j ) is Cauchy in X }/{x = (x j ) K N : x j 0} with norm [(x j )] := lim j x j and define J : X X by Jx := [(x) j N ]. Then J is linear and Jx = x for all x X. We show that X is complete. To this end let (z n ) = ([x (n) ]) be Cauchy in X. For any fixed n we find j (n) 0 such that x (n) j x (n) k < 2 n for all j, k j (n) 0. We let x n := x (n) and j (n) 0 z n := J x n. Then we have z n z n = lim x (n) j x n = lim x (n) j j j x (n) 2 n. j (n) 0 In particular, ( z n ) is Cauchy in J(X) X and thus ( x n ) is Cauchy in X. Now we let z 0 := [( x j ) j ] X. Remark: If ( X j, J j ), j = 1, 2, are completions of X then there exists a linear bijection J : X 1 X 2 such that Jx 1 = x 1 for all x 1 X 1. This follows from 1.11, applied to J 2 J1 1, and an approximation argument for the isometric property. Remark: One can also construct completions of metric spaces. end Thu Proposition: Let X be a normed space and Y X be a closed linear subspace. Then the quotient space X/Y = {x + Y : x X} is a vector space and x + Y := inf{ z : z x + Y } defines a norm on X/Y. If X is a Banach space then (X/Y, ) is a Banach space. As we shall see in the proof, is just a seminorm if Y is not closed in X. We will use the following lemma. 13

14 1.16. Lemma: A normed space X is a Banach space if and only if each absolutely convergent series is convergent, i.e. if j=1 x j < implies convergence of j=1 x j in X. Proof. : By m j=n x j m j=n x j the partial sums of an absolutely convergent sequence are a Cauchy sequence, hence convergent. : Let (x j ) be Cauchy sequence in X. We shall find its limit as the sum of an absolutely convergent series. By Exercise Sheet 1 we find a subsequence (x k(j) ) j of (x j ) such that, for y j := x k(j+1) x k(j) we have y j 2 j for all j N. Then j=1 y j is absolutely convergent, hence convergent with sum y X, say. By N j=1 y j = x k(n+1) x k(1) we see that x k(j) y + x k(1) =: x as j. This implies that x j x, and we have shown completeness of X. Proof of As usual we write [x] := x + Y. (N1): If [x] = 0 there exists a sequence (z n ) in [x] with z n 0. By closedness of Y we obtain 0 [x], i.e. [x] = [0]. Clearly [0] = 0. For α 0 we have z x + Y αz αx + Y. This implies (N2). For the proof of (N3) let x j X, j = 1, 2, and ε > 0. We find z j [x j ], j = 1, 2, with z j [x j ] + ε. Then z 1 + z 2 [x 1 ] + [x 2 ] and z 1 + z 2 z 1 + z 2 [x 1 ] + [x 2 ] + 2ε, hence [x 1 ] + [x 2 ] [x 1 ] + [x 2 ] + 2ε, and we obtain (N3) since ε > 0 was arbitrary. Now let X be a Banach space. We use Lemma 1.16 and let j [x j] <. For any j N we find z j [x j ] such that z j [x j ] + 2 j. Hence j z j <, and thus the series converges with sum z X, say. But then j [x j] = j [z j] converges with sum [z], since N N [z] [z j ] z z j 0 j=1 j=1 (N ). Remark: If X is a metric space, x X and = A X then the distance d(x, A) of x to A is given by d(x, A) := inf{d(x, y) : y A}. It is easy to see that d(x, A) = 0 x A and that d(x, A) d(y, A) d(x, y), so d(, A) is Lipschitz-continuous. For the norm we have [x] = inf{ x y : y Y } = d(x, Y ), i.e. the norm of [x] = x + Y in the quotient space X/Y is the distance of x to the closed linear subspace Y. 14

15 Remark: Let X, Z be normed spaces and T L(X, Z). Then N(T ) is a closed linear subspace of X and we can factorize T as T = T q where q : X X/N(T ), x [x] is the quotient map and T L(X/N(T ), Z) is given by T ([x]) := T x. The operator T is injective and satisfies T = T Lebesgue spaces: Let (, Σ, µ) be a measure space, i.e. Σ is a σ-algebra in and µ : Σ [0, ] is σ-additive. Usually we omit Σ in notation as it is implicitly given as the domain of definition of µ. We denote by 1 A the characteristic function of A given by 1 A (ω) = 1 if ω A and 1 A (ω) = 0 otherwise. We denote by the vector space of µ-simple functions. S(µ) := lin{1 A : A Σ, µ(a) < } 1) For ϕ = n j=1 ξ j1 Aj S(µ) let ϕ dµ := n j=1 ξ jµ(a j ). This defines a linear map dµ : S(µ) K, ϕ ϕ dµ. 2) If ϕ S(Σ, µ) and p [1, ) then ϕ p S(µ). We recall that any ϕ S(µ) has representations n ϕ = ξ j 1 Aj where the A j Σ with µ(a j ) < are pairwise disjoint. Then n ϕ p = ξ j p 1 Aj. 3) Seminorms: If p [1, ) then j=1 j=1 ϕ p := ( ϕ p µ defines a seminorm on S(µ) and ϕ p = 0 ϕ(ω) = 0 for µ-a.e. ω. We clearly have, for all ϕ S(µ), ϕ dµ ϕ dµ = ϕ 1. ) 1/p Moreover, if p (1, ) and = 1, then we have, for all ϕ, ψ S(µ), p q ϕψ dµ ϕψ 1 ϕ p ψ q (Hölder s inequality). For the proof we find pairwise disjoint A j and ξ j, η j K such that ϕ = n j=1 ξ j1 Aj and ψ = n j=1 η j1 Aj. Then ϕψ = n j=1 ξ jη j 1 Aj and ϕψ 1 = ( ) 1/p ( ) 1/q ξ j µ(a j ) 1/p η j µ(a j ) 1/q ξ j p µ(a j ) η j q µ(a j ). j j j }{{}}{{} = ϕ p = ψ q 15

16 Remark: As in 1.14 we could construct a completion of the seminormed space (S(µ), p ) by considering equivalence classes of p -Cauchy sequences. But one can find functions that represent this rather abstract concept. 4) Preparation: If (ϕ n ) is a p -Cauchy sequence in S(µ) then it contains a subsequence (ϕ k(n) ) n that converges µ-a.e. to a function f. If (ϕ l(n) ) n is another subsequence converging µ-a.e. to a function g then f = g µ-a.e.. In order to see this we prove a lemma. Lemma: If (ϕ n ) is a sequence in S(µ) such that C := n=1 ϕ n p < then n=1 ϕ n(ω) converges for µ-a.e. ω. Proof. For any N N we have ( N n=1 ) p ϕ n } {{ } =:ρ N dµ = N ( ) p ϕ n p p ϕ n p C p. n=1 Now monotone convergence yields sup N ρ N dµ C p <, hence n=1 ϕ n < µ-a.e., moreover, ( n=1 ϕ n ) p is integrable. 1 Now let (ϕ n ) be p -Cauchy. Then we find a subsequence (ϕ k(n) ) n satisfying n=1 ϕ k(n+1) ϕ k(n) p 2 n for all n N. By the lemma (ϕ k(n+1) ) n := ( ϕ k(1) + n ) (ϕ k(j+1) ϕ k(j) ) converges µ-a.e. to a function f. If (ϕ l(n) ) n is another subsequence that converges µ-a.e. to a function g then we can arrange for ϕ k(n) ϕ l(n) p 2 n (by taking subsequences if necessary). By the lemma n=1 (ϕ k(n) ϕ l(n) ) converges µ-a.e., so ϕ k(n) ϕ l(n) 0 µ-a.e.. On the other hand ϕ k(n) ϕ l(n) f g µ-a.e., so f = g µ-a.e.. 5) Definition: For p [1, ) we define L p (, µ) to be the set of all functions f : K such that there exists a p -Cauchy sequence (ϕ n ) n N such that ϕ n (ω) f(ω) for µ-a.e. ω. In this case we set f p := lim n ϕ n p (recall that the limit exists). j=1 n end Mon One can avoid the use of monotone convergence: We let M := {ω : sup ρ N (ω) = } = {ω : sup ρ N (ω) > k} = N N k N }{{} k N =:B k N N {ω : ρ N (ω) > k} Σ. }{{} =:A N k For any k N we have kµ(a N k ) ρ N dµ C (this is Tchebyshev s inequality), hence µ(a N k ) C/k. By A N k AN+1 k this implies µ(b k ) C/k and we conclude µ(m) inf k (C/k) = 0, i.e. n=1 ϕ n converges µ-a.e.. 16

17 This is well-defined: It suffices to show that ϕ n p 0 holds if (ϕ n ) is p -Cauchy with ϕ n 0 µ-a.e.. Resorting to a subsequence, we may assume that ϕ j+1 ϕ j p 4 j for all j N. This follows from 4) (integrability of ( n=1 ϕ n ) p ) and dominated convergence. 2 We remark that, for fixed k, (ϕ n ϕ k ) n is p -Cauchy and converges µ-a.e. to f ϕ k, hence f ϕ k p = lim n ϕ n ϕ k p. By the Cauchy property of (ϕ n ) this implies f ϕ k p 0 as k, so f represents the equivalence class of the Cauchy sequence (ϕ n ). It is clear that L p (, µ) is a K-vector space. Moreover, p is a seminorm on L p (, µ). 6) Completeness: The seminormed space (L p (, µ), p ) is complete in the sense that, for any p -Cauchy sequence (f n ), there exists f L p (, µ) such that f f n p 0. We find a sequence (ϕ n ) in S(µ) such that f n ϕ n p < 2 n for all n, hence also (ϕ n ) is p -Cauchy. According to 4) and 5) we find f such that a subsequence of (ϕ n ) converges to f w.r.t. p. This implies f f n p 0. 7) Hölder s inequality: If f L p (, µ) and g L q (, µ) then fg L 1 (, µ) and fg 1 f p g q. We find sequences (ϕ n ) for f and (ψ n ) for g in S(µ) as in 5). By Hölder s inequality in 3) we obtain that (ϕ n ψ n ) is 1 -Cauchy. Since ϕ n ψ n fg µ-a.e. we obtain fg L 1 (, µ) and the inequality follows by 3) and a limit argument. 2 One can avoid the use of dominated convergence here. We proceed in two steps. Step 1: By 0 = ϕ 1 + j=1 (ϕ j+1 ϕ j ) (see 4)) outside N Σ with µ(n) = 0, we have ϕ n+1 = j=n+1 (ϕ j ϕ j+1 ) on \ N, and thus, for any α > 0: { ϕ n+1 > α} N Now we use Tchebyshev s inequality again: µ{ ϕ n+1 > α} j=n+1 j=n+1 µ{ ϕ j+1 ϕ j > α2 n j } α p { ϕ j+1 ϕ j > α2 n j }. 2 jp 2 p(j n) ϕ j+1 ϕ j p p α p 2 np j=n+1 j>n }{{} 1 which shows that µ{ ϕ n > α} 0 as n for any α > 0. Step 2: Let ε > 0. We find n such that ϕ m ϕ n p < ε. We let U := {ϕ n 0} and M := max{µ(u) 1/p, ϕ n }. According to Step 1 we find m 0 n such that µ{ ϕ m > ε/m} < (ε/m) p for all m m 0. For m m 0 and P := { ϕ m > ε/m} we then have ϕ m p p = 1 (\P ) U ϕ m p dµ + 1 P U ϕ m p dµ + 1 \U ϕ m p dµ = I 1 + I 2 + I 3. Outside P we have ϕ m ε/m, hence I 1 (ε/m) p µ(u) ε p. Moreover, I 1/p 2 1 P ϕ n p + ϕ n ϕ m p µ(p ) 1/p ϕ n + ε 2ε, and finally I 3 = 1 \U (ϕ m ϕ n ) p p ϕ n ϕ m p p ε p. We have shown ϕ m p 4ε for any m m 0, i.e. ϕ m p 0., 17

18 8) Banach spaces: For p [1, ) we have that N (µ) := {f L p (, µ) : f p = 0} = {f : K : f = 0 µ-a.e. }, is a linear subspace of L p (, µ) and for f, g L p (, µ) with f g N (µ) we have f p = g p. Then the quotient space L p (, µ) := L p (, µ)/n (µ) = {f + N (µ) : f L p (, µ)} is a Banach space with respect to f + N (µ) p := f p. Usually this is phrased as identifying functions that are equal µ-a.e.. Remark: In the same way one can define spaces L p (, µ; X) and L p (, µ; X) for p [1, ), where L p (, µ; X) consists of functions f : X and X is a Banach space, and the Bochner integral f dµ for functions f L 1 (, µ; X). Any f L p (, µ; X) is µ-a.e. limt of a sequence of µ-simple functions (ϕ n ) of the form ϕ n = m n j=1 x(n) j 1 (n) A with x (n) j X and A (n) j Σ, µ(a (n) j ) <. Hence there exist N Σ j with µ(n) = 0 and a separable set A X such that f(ω) A for ω \ N (just take A = {x (n) j : n N, j = 1,..., m n }) Measurable functions: The Borel σ-algebra B = B(K) of K is the smallest σ- algebra in K containing all open subsets of K, i.e. B(K) = σ({q K : Q is open}). For K = R we have B(R) = σ({(a, b] : a, b R, a < b}) = σ({(, b] : b R}) = σ({(a, ) : a R}). 1) Σ-measurable functions: Let (, Σ) be a measurable space, i.e. Σ is a σ-algebra in the set. A function f : K is called Σ-measurable if f 1 (B) Σ for any B B(K). For K = R this is the case if and only if {ω : f(ω) b} Σ for any b R. For K = C a function f is Σ-measurable if and only if its real and imaginary parts are Σ-measurable. Lemma: If (f n ) is a sequence of Σ-measurable functions K that converges pointwise to a function f : K then f is Σ-measurable. If we let S(Σ) := lin{1 A : A Σ} then any Σ-measurable function is the pointwise limit of a sequence in S(Σ). 2) Bounded Σ-measurable functions: We let B(, Σ) denote the set of all bounded Σ-measurable functions f : K. This is a closed linear subspace of (B(), ), hence a Banach space for the sup-norm. Any f B(, Σ) is the uniform limit of a sequence in S(Σ). 3) Remark: Let (, Σ, µ) be a measure space and p [1, ). A function f L p (, µ) is µ-a.e. limit of a sequence in S(µ) S(Σ), but might not be Σ-measurable: For M the function 1 M is Σ-measurable if and only if M Σ, but 1 M = 0 µ-a.e. if and only if there exists N Σ with µ(n) = 0 and M N. This is as bad as it gets: Denoting end Mon Σ := σ ( Σ {M : N Σ : µ(n) = 0, M N} ), 18

19 the measure µ has a unique extension to a measure µ : Σ [0, ] and we have 4) Proposition: Let p [1, ) and f L p (, µ). Then f is Σ-measurable and there exists a Σ-measurable function f 0 such that f = f 0 µ-a.e.. In particular, L p (, µ) can also be represented as {f L p (, µ) : f is Σ-measurable }/{f N (µ) : f is Σ-measurable }. Moreover, L p (, µ) = L p (, µ), N ( µ) = N (µ), and L p (, µ) = L p (, µ). Finally, a function g : K is Σ-measurable if and only if g is µ-measurable, i.e. g is the µ-a.e. limit of a sequence in S(Σ). Remark: For functions f : X where X is a Banach space this last property is used and thus there exists a separable subset A of X and a µ-null set N such that f(ω) A for all ω \ N Essentially bounded functions: A µ-measurable function f : K is called essentially bounded if there exists N Σ with µ(n) = 0 such that 1 \N f is bounded. We denote by L (, µ) the space of all µ-measurable and essentially bounded functions f : K. On this space, f := inf{ sup f(ω) : N Σ, µ(n) = 0} = inf{c > 0 : { f > C} is a µ-null set} ω \N defines a seminorm with f = 0 f N (µ). The quotient space L (, µ) := L (, µ)/n (µ) is a Banach space for the induced norm. 3 The assertion of ) also holds for p = : In every co-class we find essentially bounded Σ-measurable functions, and L (, µ) may also be represented as {f L (, µ) : f is Σ-measurable }/{f N (µ) : f is Σ-measurable }. As for L p with p [1, ) the choice whether to work with Σ-measurable or µ-measurable functions is thus rather a matter of taste or context. Lemma: Let g L (, µ). For all p [1, ], the operator L p (, µ) L p (, µ), f gf, is bounded with norm g Proposition: Let p [1, ) and f : K. Then f L p (, µ) if and only if f is µ-measurable and f p L 1 (, µ). 3 Observe that for each f L (, µ) there exists N Σ with µ(n) = 0 such that f = sup ω \N f(ω). 19

20 Proof. If f L p (, µ) and (ϕ n ) is a p -Cauchy sequence with ϕ n f µ-a.e., then f is clearly µ-measurable. Moreover, ϕ n p f p µ-a.e., and we have, for p = 1, ϕ n ϕ m 1 ϕ n ϕ m 1 by the reverse triangle inequality. For p > 1 we have ϕ n p ϕ m p ϕ n ϕ m p ( ϕ n p 1 + ϕ m p 1) by the mean value theorem. Integration and Hölder yield (with 1 p + 1 q = 1) ϕ n p ϕ m p 1 p ϕ n p ( ϕn p/q p ) + ϕ m p/q p. Since every Cauchy sequence is bounded we obtain that ( ϕ n p ) is 1 -Cauchy, hence f p L 1 (, µ). For the reverse implication let f be µ-measurable and f p L 1 (, µ). Then sgn f 4 is µ-measurable, hence sgn f L (, µ). By the lemma in 1.19 we only have to show f L p (, µ). We find a 1 -Cauchy sequence (ϕ n ) such that ϕ n f p µ-a.e., where we may assume that ϕ n 0 (considering ϕ n otherwise). Then ϕ 1/p n f µ-a.e. and by ϕ 1/p n ϕ 1/p m ϕn ϕ m 1/p and integration we obtain that (ϕ 1/p n ) is p -Cauchy. Hence f L p (, µ). Corollary: Let p, q, r [1, ] with 1 p + 1 q = 1 r. Then (f, g) fg is continuous Lp (, µ) L q (, µ) L r (, µ) and fg r f p g q Proposition: Let (, Σ, µ) be a measure space and p, q (1, ) with 1 p + 1 q = 1. Then J : L q (, µ) (L p (, µ)) given by Jg(f) := fg dµ is well-defined linear and isometric. If is σ-finite then the assertion holds for p = 1 and q =. Here we use the Definition: A subset A is called σ-finite if there exists a sequence (A j ) in Σ such that µ(a j ) < for each j N and A j N A j. Notice that, if p [1, ) and f L p (, µ), then {f 0} is σ-finite. Proof. Assume p [1, ). By Hölder s inequality we have (Jg)(f) fg 1 g q f p, so J is well-defined and linear and Jg (L p ) g q. Now let p (1, ) and g L q (, µ). We use Set f := sgn g g q 1. Then f is µ- measurable and f p = g pq p = g q L 1 (, µ), hence f L p (, µ) and f p p = g q q. Since q = q 1, we obtain p Jg(f) = fg dµ = g q dµ = g q q = g q f p, 20

21 which implies Jg (L p ) g q. Finally let p = 1, let be σ-finite, and let g L (, µ). We may assume g > 0 and fix ε (0, g ). We also assume that g is σ-measurable. Then M 0 := {ω : g(ω) > g ε} Σ satisfies µ(m 0 ) > 0. Since is σ-finite we find a subset M M 0 in Σ such that µ(m) (0, ). Now we let f := (sgn g)1 M. Then f L 1 (, µ) and f 1 = µ(m). Moreover Jg(f) = fg dµ = 1 M g dµ ( g ε)µ(m) = ( g ε) f 1. end Thu We conclude Jg (L 1 ) g. We collect without proof some useful results in Proposition: (a) Let p [1, ). If (f n ) is a sequence in L p (, µ) and f L p (, µ) such that f f n p 0 then there exists a subsequence (f k(n) ) and a function g 0 in L p (, µ) such that f k(n) f µ-a.e. and sup n f k(n) g µ-a.e.. (b) Monotone Convergence Theorem: Suppose (f n ) is a sequence in L 1 (, µ) satisfying f n f n+1 and sup n fn dµ <. Then f := sup n f n < µ-a.e., f1 {f< } L 1 (, µ) and f dµ = sup n fn dµ. (c) Dominated Convergence Theorem: Suppose that (f n ) is a sequence in L 1 (, µ), f is function such that f n f µ-a.e., and there exists a function g L 1 (, µ) such that sup n f n g µ-a.e.. Then f L 1 (, µ) and f dµ = lim n fn dµ Definition: Let (X, d) be a metric space and K X. A family Q of subsets is called an open cover of K if all Q Q are open and K Q = Q Q Q. K is called compact if any open cover Q of K contains a finite subcover, i.e. a finite subset E which is still an open cover of K. K is called relatively compact if its closure K is compact. K is called totally bounded if, for every ε > 0, there exists a finite set F X such that K x F B(x, ε). K is called sequentially compact if every sequence in K contains a convergent subsequence with limit in K. Remark: If K is totally bounded and ε > 0 we can also find a finite set F in the smaller set K such that K x F B(x, ε). By the Proposition in 1.3 it makes no difference whether the Q Q are open in X or relatively open in K, i.e. compactness of K does not depend on the ambient space X. Similarly, total boudedness of K does not depend on he ambient space X. 4 Here sgn f(ω) = sgn (f(ω)) where sgn 0 := 0 and sgn z := z/ z for z C \ {0}. 21

22 1.24. Proposition: Let (X, d) be a metric space and K X. Then the following are equivalent: (i) K is compact. (ii) K is sequentially compact. (iii) K is totally bounded and complete. Proof. We show that (i) implies (ii): If (x n ) is a sequence in K without convergent subsequence then we find, for every x K a radius ε x > 0 such that B(x, ε x ) contains x n only for finitely many n. But then the open cover Q = {B(x, ε x ) : x K} contains a finite subcover x F B(x, ε x) which can only contain x n for finitely many n, a contradiction. Before we proceed we show a lemma. Lemma: K is totally bounded if and only if every sequence in K contains a Cauchy subsequence. Proof. If K is totally bounded and (x n ) is sequence in K then it contains a Cauchy subsequence: We prove this by a diagonal argument. For ε = 1 we find an open ball B 1 of radius 1 that contains a subsequence (x k1 (n)) of (x n ). For ε = 1/2 we find an open ball B 2 of radius 1/2 that contains a subsequence (x k2 (n)) of (x k1 (n)) etc. In other words, subsequently we find, for each j N, an open ball B j of radius 1/j that contains a subsequence (x kj (n)) of (x kj 1 (n)) (where we put k 0 (n) := n). Finally we let l(n) := k n (n). Then (x l(n) ) is a subsequence of (x n ) and for all m n we have x l(n), x l(m) B n, hence d(x l(n), x l(m) ) < 2/n, i.e. (x l(n) ) is Cauchy. Conversely, suppose that K is not totally bounded. then we find a radius ε > 0 such that K cannot be covered by finitely many open balls of radius ε. Choose x 1 K. By assumption we find x 2 K \ B(x 1, ε), then x 3 K \ (B(x 1, ε) B(x 2, ε)) etc. We thus construct a sequence (x n ) in K with x n n 1 j=1 B(x j, ε) for every n. In particular, d(x n, x m ) ε for n m, hence (x n ) does not contain a Cauchy subsequence. By the lemma, (iii) implies (ii). Clearly, (ii) implies completeness of K, so by the lemma, (ii) implies (iii). We finally assume K = X and that (ii) and (iii) hold and show (i). To this end we assume that Q is an open cover of K which contains no finite subcover. We find subsequently finite subsets F k of K such that the open balls B(x, 2 k ), x F k, cover K. By assumption we find x 1 F 1 such that Q contains no finite subcover of U 1 := B(x 1, 2 1 ). Then {U 1 B(x, 2 2 ) : x F 2 } is a finite cover of U 1 and we find x 2 F 2 such that Q contains no finite subcover of U 2 := U 1 B(x 2, 2 2 ) etc. In this way we construct a sequence (x n ) such that Q contains no finite subcover of U n = n j=1 B(x j, 2 j ). In particular, B(x n, 2 n ) B(x n+1, 2 (n+1) ), so d(x n, x n+1 ) < 2 n +2 (n+1) 2 1 n for every n, which implies that (x n ) is Cauchy. By (ii), (x n ) has a convergent subsequence with limit x 0 K, so x n x 0. But then we find Q 0 Q that contains x 0. Since Q 0 is open we find r > 0 such 22

23 that B(x 0, r) Q 0. Then we find n 0 such that d(x n, x 0 ) < r/2 and 2 n < r/2 for n n 0. We conclude that, for n n 0, U n B(x n, 2 n ) B(x, r) Q 0, a contradiction as Q contains no finite subcover of U n. Remark: (a) If K is totally bounded then K is bounded, i.e. sup{d(u, v) : u, v K} <. 5 Proof. We find a finite subset E K such that K x E B(x, 1) and set M := max{d(x, y) : x, y E}. Let u, v K. We find x, y E such that d(x, u) < and d(y, v) < 1. Then d(u, v) d(u, x) + d(x, y) + d(y, v) < 1 + M + 1 = M + 2. end Mon (b) If K X is complete then K is closed in X: If (x n ) is a sequence in K with x n X X then (x n ) is Cauchy in K and hence has a limit in K. Since the limit is unique we conclude x K. (c) By (a) and (b) we have that a compact subset K of a metric space is always closed and bounded. It should be known that the converse holds in (K n, ), and that any non-empty compact subset of R contains a maximal and a minimal element. (d) Any totally bounded subset of a metric space is separable. (e) Any closed subset of a compact set is compact. Any subset of a compact set is relatively compact Proposition: Let X, Y be metric spaces where X is compact and let f : X Y be continuous. Then f is uniformly continuous and f(x) is compact. Proof. Let ε > 0. Then Q := {B(x, δ/2) : x X, δ > 0, f(b(x, δ)) B(f(x), ε/2)} is an open cover of X (since f is continuous). Hence we find a finite subcover E = {B(x j, δ j /2) : j = 1,..., n} Q. We set δ 0 := min j δ j /2, then δ 0 > 0. Let x, y X with d(x, y) < δ 0. Since E covers X we find j {1,..., n} such that d(x, x j ) < δ j /2. Then d(y, x j ) d(x, y) + d(x, x j ) < δ 0 + δ j /2 δ j. Hence d(f(x), f(y)) d(f(x), f(x j )) + d(f(x j ), f(y)) < ε 2 + ε 2 = ε. 5 If (X, ) is a normed space the K X is bounded if and only if sup{ x : x K} <. 23

24 We have proved uniform continuity of f. Now we prove compactness of f(x). Let Q be an open cover of f(x). Then {f 1 (Q) : Q Q} is an open cover of X (since f is continuous). Since X is compact we find a finite subset E Q such that {f 1 (Q) : Q E} covers X. But then E covers f(x). Corollary: If X is a compact metric space then C(X) = C b (X) = BUC(X) and for any f C(X) we find x 0 X such that f = f(x 0 ), i.e. f = max x X f(x) Equivalence of norms: Let X be a K-vector space. Two norms 0 and 1 on X are equivalent if there exist constants c, C > 0 such that c x 1 x 2 C x 1 for all x X, or, in other words, if the identity map I : (X, j (X, 1 j ) is continuous for j = 0, 1. This clearly defines an equivalence relation on the set of norms on X. Moreover, it is clear that open and closed sets, convergent sequences and Cauchy sequences, compact subsets and continuous, uniformly continuous and Lipschitz continuous functions for equivalent norms coincide. Example: For x K n and p (1, ) we have x x p x 1 n x, hence all norms p, p [1, ], are equivalent on K n. Proposition: Let X be a finite-dimensional K-vector space. Then all norms on X are equivalent. Proof. Let be a norm on X and n := dim X. Choosing a basis e 1,..., e n X we define a norm on K n by letting (x 1,..., x n ) := j x je j. Hence we may assume that X = K n, and it suffices to show that is equivalent to the Euclidean norm 2. For x = (x j ) = j x je j we have by Cauchy-Schwarz x = x j e j x j e j ( e j 2) 1/2 x 2. j j j }{{} =:C This implies x y x y C x y 2, which means that is (even Lipschitz) continuous on (X, 2 ). Since S := {y K n : y 2 = 1} is closed and bounded it is a compact subset of (X, 2 ) and the continuous function attains its minimum on S, i.e. we find y 0 S such that c := y 0 y for all y S. If x X \ {0} then x/ x 2 S hence c x/ x 2, i.e. c x 2 x. 24

25 Corollary: If X, Y are normed spaces with dim X < and T : X Y is linear then T is continuous, i.e. T L(X, Y ). Proof. Let n := dim X. We find a basis e 1,..., e n of X. Then j x je j := j x j defines a new norm on X which is equivalent to the original norm. We then have, for any x = j x je j X, T x Y = j which proves T L(X, Y ). x j T e j Y j x j T e j Y max T e j Y x j = C x, j }{{} j =:C Lemma (Riesz): Let X be normed space, Y X a closed subspace, and δ (0, 1). Then there exists x δ X with x δ = 1 and x δ y 1 δ for all y Y. Proof. By 1.15 the quotient space X/Y is a normed space, which is nontrivial by Y X. Hence we find x X such that [x] X/Y = 1 δ and then x [x] with x 1. Letting x δ := x/ x we have x δ = 1 and, for any y Y, x δ y [x δ ] X/Y = [x] X/Y x = 1 δ x 1 δ Proposition: Let X be a normed space. The following are equivalent: (i) dim X <, (ii) B X := {x X : x 1} is compact, (iii) any closed and bounded subset is compact. Proof. We know that (i) implies (iii) and it is clear that (iii) implies (ii). We assume that dim X = and show that B X is not compact. We find x 1 X with x 1 = 1 and let Y 1 := lin{y 1 } and choose δ = 1/2. Then Y 1 is a closed subspace of X (since it is complete) and Y 1 X. By 1.28 we find x 2 X with x 2 = 1 and x 1 x 2 1/2. We set Y 2 := lin{y 1, y 2 } and find, by 1.28, x 3 X with x 3 = 1 and d(x 3, Y 2 ) 1/2, in particular we have x 3 x j 1/2 for j = 1, 2. Continuing like this we find, for any n N with n 2, an x n X with x n = 1 and x n x j 1/2 for all j = 1,..., n 1. Then (x n ) is a sequence in B X that has no Cauchy subsequence. Hence B X is not compact. The following compactness criterion in C(K) is fundamental for various other compactness results. end Thu

26 1.29. Theorem (Arzelà-Ascoli): Let K be a compact metric space and M C(K). Then M is relatively compact if and only if M is bounded and equicontinuous, i.e. for any ε > 0 there exists δ > 0 such that for all x, y K with d(x, y) < δ and all f M we have f(x) f(y) ε. Proof. If M is relatively compact then M is compact, hence bounded. For a given ε > 0 we find a finite subset E M such that M g E B(g, ε/3). Then we find a common δ > 0 for ε/3 according to uniform continuity of g E. Now let x, y K with d(x, y) < δ and f M. Then we find g E such that f g ε/3, and we obtain f(x) f(y) f(x) g(x) + g(x) g(y) + g(y) f(y) 2 f g + ε/3 ε, so M is equicontinuous. Now let M be bounded and equicontinuous and let (f n ) be a sequence in M. Since K is compact we find a countable dense subset D = {x 1, x 2,...} in K. Since (f n (x 1 )) is bounded we find a convergent subsequence (f k1 (n)(x 1 )). Since (f k1 (n)(x 2 )) is bounded we find a convergent subsequence (f k2 (n)(x 2 )) etc. Now we set l(n) := k n (n) for all n N. Then (f l(n) ) is a subsequence of (f n ) and (f l(n) (y)) n converges for each y D. We shall show that (f l(n) ) is -Cauchy. So let ε > 0. For ε/3 we find δ > 0 according to equicontinuity of M, and then a finite E K such that K z E B(z, δ/2). For each z E we find y z D with d(y z, z) < δ/2. We denote the set of all those y z, z E, by D E. Note that D E is a finite subset of D and that K y D E B(y, δ). Now we find n 0 such that, for all n, m n 0 and y D E, we have f l(m) (y) f l(n) (y) ε/3. Now let n, m n 0 and x K. We find y D E with d(x, y) < δ and thus obtain f l(n) (x) f l(m) (x) f l(m) (x) f l(m) (y) + f l(m) (y) f l(n) (y) + f l(n) (y) f l(n) (x) ε. }{{}}{{}}{{} ε/3 ε/3 ε/3 Hence we have shown f l(n) f l(m) ε for all n, m n 0, and (f l(n) ) is -Cauchy. Since (C(K), ) is complete, we have shown that M is relatively compact. Remark: The proof shows that it is sufficient that M is bounded pointwise, i.e. {f(x) : f M} is bounded for every x K. There are versions for functions f : K Y where Y is a metric space. Then one has to require that {f(x) : f M} is a relatively compact subset of Y for every x K Definition: Let X, Y be normed spaces. A linear operator T : X Y is called compact if T (B X ) is relatively compact in Y. The set of all compact operators X Y is denoted by K(X, Y ). An operator T L(X, Y ) is called of finite rank if dim T (X) < 6. The set of all finite rank operators is denoted F(X, Y ). 6 As for matrices, the rank of T is dim T (X). 26

27 Remark: (1) Any compact operator is bounded, i.e. K(X, Y ) L(X, Y ). (2) Any T F(X, Y ) is compact, i.e. F(X, Y ) K(X, Y ). (3) T L(X, Y ) is compact if and only if there exists a compact superset K Y of T (B X ). (4) If Y is a Banach space then T K(X, Y ) if and only if T (B X ) is totally bounded Proposition: Let X, Y, Z be normed spaces. (a) K(X, Y ) and F(X, Y ) are linear subspaces of L(X, Y ). (b) If Y is a Banach space then K(X, Y ) is closed in L(X, Y ), in particular, K(X, Y ) is a Banach space and any limit of a sequence in F(X, Y ) is compact. (c) Let T L(X, Y ) and S L(Y, Z). If T or S is compact then ST is compact (ideal property). Proof. (a) Clearly, T K(X, Y ) [ F(X, Y )] implies λt K(X, Y ) [ F(X, Y )]. If S, T K(X, Y ) then (S + T )(B X ) S(B X ) + T (B X ) which is compact. If S, T F(X, Y ) then (S + T )(X) S(X) + T (X) which has finite dimension. (b) We show that L(X, Y ) \ K(X, Y ) is open. So let T L(X, Y ) be not compact. Then T (B X ) is not totally bounded and we find ε > 0 such that T (B X ) cannot be covered by finitely many balls of radius ε. By the argument we used in the proof of the lemma in 1.24 we find a sequence (x n ) in B X such that T x k T x l ε for all k l. Let S L(X, Y ) with S T ε/3. Then, for all k l we have Sx k Sx l T x k T x l (S T )(x k x l ) ε 2 S T ε 2ε/3 = ε/3, hence (Sx n ) has no Cauchy subsequence and S(B X ) is not totally bounded. (c) If T is compact then K := T (B X ) is compact, hence S(K) is compact. Since (ST )(B X ) S(K), (ST )(B X ) is relatively compact. If S is compact, then S(B Y ) is compact, hence also T S(B Y ) is compact. But by T (B X ) T B Y we obtain (ST )(B X ) T S(B Y ), hence (ST )(B X ) is relatively compact Examples: (1) Let p [1, ] and (a n ) l. Define T by T x = (a n x n ) for x = (x n ) l p. Then T L(l p ) with T = sup n a n. We have{ T K(l p ) if and only if (a n ) c 0 : If (a n ) c 0 define (a (m) n ) n for m N by a (m) an, n m n := 0, n > m and T m by T m x = (a (m) n x n ) n. Then T m F(l p ) and T T m = sup n>m a n, which tends to 0 for m. We conclude that T F(l p ) K(l p ). If (a n ) c 0 we find ε > 0 such that a k(n) ε for a subsequence (k(n)) of (n). Bu then (e k(n) ) n is a sequence in B l p such that (T e k(n) ) = (a k(n) e k(n) ) satisfies T e k(n) T e k(m) p ε for all n m, thus (T e k(n) ) has no convergent subsequence. (2) Let X be a Banach space. Then I X K(X) if and only if dim X < (by 1.28). end Mon