1 Functional Analysis

Transcription

1 1 Functional Analysis 1

2 1.1 Banach spaces Remark 1.1. In classical mechanics, the state of some physical system is characterized as a point x in phase space (generalized position and momentum coordinates). An observable (i.e. measurable quantity) is then a function f on phase space and f(x) is the value obtained by measuring f in the state x. For example for the one-dimensional harmonic oscillator the phase space is essentially R 2 and energy is the function E(x, p) := 1 2 x p2 (if units are chosen appropriately). In quantum mechanics, the state space is a Hilbert space whose elements are usually called wave functions. The physical interpretation of such wave functions depends on the model and might not be obvious. The observables are operators on the Hilbert space and possible measurements are their spectral values. Our task in this course will be to understand the mathematical objects mentioned above and how they are interrelated. All the vector spaces mentioned below will be over the field of real (R) or complex (C) numbers. To state results that apply for both cases, we write K for the base field. Definition 1.2. An algebra over K is a K-vector space V with a map V V V called the product or multiplication and denoted by juxtaposition such that for any x, y, z V and λ, ρ K the following hold (i) x(y + z) = xy + xz (ii) (x + y)z = xz + yz (iii) (λx)(ρy) = (λρ)(xy) (iv) (xy)z = x(yz). An algebra is called unital if there is a unit element, i.e. e V such that ev = ve = v for every v V. Remark 1.3. Structures not fulfilling (iv) are also called (non-associative) algebras (e.g. Lie algebras) but all our algebras will be associative. Note that we do not require the product map to be commutative. Definition 1.4. A normed vector space is a pair (V, ) consisting of a vector space V and a map : V [0, ) called the norm fulfilling (i) λx = λ x (positive homogeneity), (ii) x + y x + y (subadditivity/triangle inequality) and (iii) x = 0 if and only if x = 0 (definiteness) 2

3 for any x, y V and any λ K. A map fulfilling all these conditions except for definiteness is called a semi-norm. If V is an algebra, the pair (V, ) is called a normed algebra if for every x, y V. xy x y Remark 1.5. The geometric interpretation of x is the length of the element x V. It also allows us to define the distance of two elements x, y V as x y. Armed with this distance, we can now define the usual topological concepts: (i) a sequence (x n ) V converges against x V if and only if for every ɛ > 0 there is N N such that for every n N the distance x n x < ɛ if and only if x n x converges to 0. (ii) B r (x) = {y V y < r} for r (0, ), (iii) A V is open if and only if for every x A there is r (0, ) such that B r (x) A, (iv) A V is closed if and only if V \A is open if and only if the limit of every convergent sequence (x n ) A is in A, (v) a map f from (V, V ) to (W, W ) is continuous in x V, if and only if for every ɛ > 0 there is δ > 0 such that y x V < δ implies f(y) f(x) W < ɛ, if and only if f 1 (A) is open for every open A W, (vi) the closure A of a set A V is the smallest closed set containing A. Example 1.6. (i) V = K n with any of the following norms for x = (x 1,..., x n ) x 1 = x x n x 2 = x x n 2 x p = ( x 1 p + + x n p ) 1 p for p [0, ) x = sup { x 1,..., x n } In the case n = 1, V is a normed algebra. (ii) with { l 1 := (x n ) n N K } x n < n=1 (x n ) 1 = x n n=1 3

4 (iii) with { l 2 := (x n ) K } x n 2 < n=1 (x n ) 2 = x n n=1 (iv) l := {(x n ) K K [0, ) such that x n K for all n N} with (x n ) = sup x n n N (v) (normed algebra) with (normed algebra) { c 0 := (x n ) K } lim x n = 0 n (x n ) = sup x n n N (vi) A a set, B(A) = {f : A K f bounded} with f := sup f(x) x A (normed algebra) (vii) A R n (or more generally any topological space) with C(A) = {f : A K f bounded and continuous} f := sup f(x) x A 4

5 (normed algebra) For each of these examples, the validity of the norm axioms needs to be verified (exercise). There is a number of other norms we could have considered on the spaces defined above (for example on l 2 ), the particular norms chosen are the usual ones however and we will generally not explicitly mention them. Thus if we are talking of l 2, we will use the norm 2 if nothing else is mentioned. Remark 1.7. Let V be a vector space and a semi-norm on V, then the set V 0 := {v V v = 0} is called the kernel of. It is a vector subspace of V (exercise). The quotien space is the set W = V/V 0 := {v + V 0 v V }. Note that v 1 + V 0 = v 2 + V 0 if and only if v 1 v 2 V 0. The space W inherits a natural vector space structure from V by defining (v 1 + V 0 ) + (v 2 + V 0 ) := v 1 + v 2 + V 0 λ(v 1 + V 0 ) := λv 1 + V 0 for every v 1, v 2 V and λ K (check that these are well defined and fulfill the vector space axioms). The neutral element in this space is then precisely the space V 0 = 0 + V 0. Now we can define v + V 0 := v (we should use a new symbol on the left hand side but we slightly abuse notation here). The equality v 1 + V 0 = v 2 + V 0 implies v 1 v 2 V 0 and hence v 1 v 2 + v 2 v 1 = v 2. Exchanging v 1 and v 2 yields v 1 = v 2 so the map is actually well defined. It inherits the semi-norm properties from the initial semi-norm but it is a norm on W since v + V 0 = v = 0 if and only if v V 0 if and only if v + V 0 = V 0 is the neutral element of W. Example 1.8. Let λ be the Lebesgue measure on R n and A R n (measurable). Define the space of (Lebesgue) integrable functions { } L 1 (A) := f : A K f is measurable and f dλ < A and f 1 := A f dλ. 5

6 For f, g L 1 (A) and κ K we have f + κg 1 = f + κg dλ A A ( f + κ g ) dλ = f 1 + κ g 1 <, which proves the triangle inequality for 1 and that L 1 (A) is a vector space. On the other hand, there are obviously non-zero functions f such that A f dλ = 0. So 1 is only a semi-norm. By quotienting out the kernel of 1 according to the procedure described above, we obtain the space of integrable functions L 1 (A). Note that the elements of this vector space are not functions any more but subspaces or equivalence classes of functions. Often we do not have to care, but there are situations where one needs to be careful. The most important thing to keep in mind is, that the elements of L 1 (A) cannot be evaluated at a single point x A any more. This is due to the fact, that the subspace any f L 1 (A) represents contains functions with arbitrary values at the point x. Remark 1.9. The construction above can be done for any measure space (A, µ). The space l 1 is a special case where A = N and µ is the counting measure. Definition Let (V, ) be a normed space. A sequence (v n ) V is called a Cauchy sequence, if for every ɛ > 0, there is N N such that for every n, m N v n v m < ɛ. The space (V, ) is called complete if every Cauchy sequence converges. A complete normed vector space is called a Banach space. A complete normed algebra is called a Banach algebra. Proposition The normed space (V, ) is complete if and only if every absolutely convergent series k=1 v k in V is convergent. A series is absolutely convergent if and only if k=1 v k <. Proof. Let (V, ) be complete and k=1 v k an absolutely convergent series. Let (s n ) be the sequence of partial sums, i.e. For any ɛ > 0, there is N N such that s n = k=n For every n, m > N ( m > n) we have m s n s m = v k k=n+1 n v k. k=1 < ɛ. m k=n+1 v k < v k < ɛ. k=n 6

7 Thus the sequence (s n ) is a Cauchy sequence and since the space is complete it converges. Suppose now, that any absolutely convergent series in (V, ) converges and let (v n ) V be a Cauchy sequence. Then for each k, there is N k such that for every n, m N k v n v m < 2 k. Without loss of generality, we can choose the sequence (N k ) to be strictly increasing. Define w 1 = v N1 and w k = v Nk v Nk 1 for k > 1. One easily finds k w l = v Nk. (1.1) l=1 Moreover w l = v N1 + v Nl v Nl 1 < v N1 + 2 l 1 < l=1 l=2 l=2 so the series l=1 w l is absolutely convergent and hence convergent. Thus (v Nk ) is a convergent subsequence of (v n ). Any Cauchy sequence with a convergent subsequence is convergent (exercise). Remark All the example spaces from example 1.6 and example 1.8 are complete. We give the proof for one example below and for another in proposition Proposition Let A R n. The space C(A) of bounded continuous functions on A is complete (with respect to the norm f = sup x A f(x) ). Proof. For every function f C(A) and every x A, we have f(x) sup x A f(x) = f. So let (f n ) be a Cauchy sequence in C(A). Fix x A. For every ɛ > 0, there is N N such that for m, n N, f n f m < ɛ so in particular f n (x) f m (x) = (f n f m )(x) f n f m < ɛ. This shows, that (f n (x)) is a Cauchy sequence in K and since K is complete it converges. The point x was arbitrary, so the sequence of functions (f n ) converges pointwise to some limit function f. There is K > 0 such that f n K for every n N since (f n (x)) is a Cauchy sequence (exercise). Thus the sequence (f n (x)) is bounded by K independently of x and so the limit function f is bounded by K as well. We show next, that f n f with respect to the norm. So let ɛ > 0 and choose N N such that f n f m < ɛ for every n, m N. For every n N and x A, this implies f n (x) f m (x) < ɛ for all m N and thus f n (x) f(x) = lim m f n(x) f m (x) < ɛ. 7

8 Taking the supremum with respect to x yields f n f < ɛ for every n N which proves the sought for convergence. Finally, we show that f is continuous. Choose x A and ɛ > 0. There is n N such that f n f < ɛ 3. Since f n is continuous, there is δ > 0 such that f n (x) f n (y) < ɛ 3 for any y A fulfilling x y < δ (note that the last norm is the one on R n ). For any such y we find f(x) f(y) f(x) f n (x) + f n (x) f n (y) + f n (y) f(y) 2 f f n + f n (x) f n (y) < ɛ. This proves the continuity of f in x and since the latter point was arbitrary f is continuous. Definition Let (V, ) be a normed vector space. A set A V is called compact, if for any open cover (O i ) i I of A, i.e. a family of open sets such that A i I O i, there is a finite subcover, i.e. there exist n N and i 1,..., i n I such that A n k=1 O i k. Proposition Let (V, ) be a normed vector space. Any compact set A V is closed and bounded. Proof. For boundedness, consider the cover B 1 (0), B 2 (0),.... For closedness, suppose x is a boundary point of A. For r > 0 let For n N, consider the open sets K r (x) := {y V y x r}. O n := V \K 1 (x). n Since x is a boundary point, no finite set of the O n can cover A, thus by compactness the union O n = V \ {x} n N does not cover A, i.e. x A. Since x was an arbitrary boundary point, A is closed. Remark The Heine-Borel theorem shows, that in K n any closed and bounded set is compact. This is not true in infinite dimensions. A normed vector space (V, ) is finite dimensional if and only if K 1 (0) is compact. Proposition Closed subsets of compact sets are compact. Proof. Exercise. 8

9 Proposition Let (V, ) be a normed vector space and A V. Then A is compact if and only if every sequence in A has a subsequence that converges in A. Proof. We only show the direction =. A sequence (x n ) A has a subsequence converging in A if and only if there is x A such that for every ɛ > 0 the ball B ɛ (x) contains infinitely many sequence elements (exercise). Suppose there is a sequence (x n ) without a convergent subsequence, i.e. for every y A, there is ɛ y > 0, such that O y := B ɛy (y) contains only finitely many of the (x n ). Obviously we have A y A so by compactness, there is a finite subcover O y1,..., O yl. But then A contains only finitely many of the (x n ) which is a contradiction. Definition Let (V, ) be a normed vector space. A set A V is dense in V, if and only if A = V. Proposition Let (V, ) be a normed vector space and A V. The following are equivaltent. (i) A is dense in V, (ii) for every v V and every ɛ > 0, there is w A such that v w < ɛ, (iii) for every v V, there is a sequence (v n ) A that converges to v. Proof. Exercise. Example The set of finite sequences c c := {(x n ) K there is N N such that x n = 0 for n > N} is a dense subspace in c 0, l 1 and l 2 (with respect to the respective norms). We prove the statement for l 1. Let (x n ) l 1 and ɛ > 0. By assumption, (x n ) 1 = O y x n < n=1 so there is N N, such that n=n x n < ɛ. Define (y n ) c c by { x n n < N y n := 0 n N. Then (x n ) (y n ) 1 = x n < ɛ n=n 9

10 holds which by the preceding proposition proves the density. Note that in finite dimensions, subspaces are always closed, so the only dense subspace of R n is R n itself. The space c c is not dense in l since the sequence x n = 1 for n N has -distance at least 1 to any element of c c. Remark Dense subsets are important because they can be substantially easier to deal with than the entire space V. There are many proofs in functional analysis, that show some property for a dense subset first and then show that the property is invariant under norm limits to generalize it to the entire space. Theorem 1.23 (Stone-Weierstraß). Let K R n be compact (more generally any compact topological space). Let A C(K) be a subalgebra fulfilling (i) A contains the constant functions, (ii) for two distinct points x, y K, there is f A such that f(x) f(y) (we say that A separates the points of K), (iii) A is closed under complex conjugation (void if K = R). Then A is dense in C(K). Remark The function t 1 + t is analytic in 0 with the series expansion where a 0 = 1 and ( 1 ) a n = 2 := n a n t n n=0 The series converges to 1 + t for t [ 1, 1). Without proof. ( ) ( 1 2 n + 1). 1 n Lemma Let B be a closed (with respect to ) subalgebra of C(K). Then for non-negative f B, f B. For every g, h B, the functions g, max {g, h} and min {g, h} are in B. Proof. We prove the case K = R, to which the case K = C can be reduced (exercise). Let f B be non-negative. Since B is an algebra, it contains f if and only if it contains λf for any λ > 0. Thus we may assume that f = 1, i.e. 0 f 1. The function g = 1 f also fulfills 0 g 1. Then, using the previous remark, we obtain a n g n n=0 a n g n n=0 a n = 2a 0 + a n ( 1) n+1 = 2, n=0 n=0 10

11 i.e. the series n=0 a ng n converges absolutely and since C(K) is complete it converges. Since B is closed, the limit function f = 1 g = a n g n n=0 is in B. This also proves that h = h 2 B for any h B. Finally max {h, g} = 1 2 (h + g + h g ) and min {h, g} = 1 (h + g h g ) 2 prove the remaining assertion. Iterating the above result, we see that B also contains the max and min over finitely many functions from B. Proof of theorem We prove the case K = R. Let x, y K be distinct points and a, b R. By assuption, there is g such that g(x) g(y). Then the function g := a + b a ( g g(x)) g(y) g(x) is in A and fulfills g(x) = a and g(y) = b. Note that A is also a subalgebra (exercise) to which we can apply the previous lemma. Now choose f C(K) and ɛ > 0. For each x, y K, there is a function g x,y A such that g x,y (x) = f(x) and g x,y (y) = f(y). For fixed x K, define V y := {z K g x,y (z) < f(z) + ɛ} = (g x,y f) 1 (, ɛ) is open (as an inverse image of an open interval under a continuous map) and contains y. So the sets (V y ) y K form an open cover of K and by compactness, there is a finite subcover V y1,..., V yn. Set h x = min {g x,y1,..., g x,yn } A Then h x (x) = f(x) and h x < f + ɛ by the definition of the V y. Now define U x := {z K h x (z) > f(z) ɛ}. By analogous arguments as above, the sets (U x ) x K are an open cover of K and thus there is a finite subcover U x1,..., U xk. Then the function h = max {h x1,..., h xk } A fulfills f ɛ < h < f + ɛ, i.e. f h < ɛ, i.e. f h < ɛ. Since ɛ was arbitrary, f is a limit point of A and since the latter is closed, we find f A. 11

12 Example Let K R n be compact. Then the algebra of polynomials (in n variables) is dense in C(K). Proof. The polynomials are an algebra of continuous functions containing the constants. Let a = (a 1,..., a n ) and b = (b 1,..., b n ) be distinct points of K, i.e. there is i {1,..., n} such that a i b i. Then the polynomial (x 1,..., x n ) x i a i vanishes on a but not on b so the polynomials separate the points of K. Definition A normed space (V, ) is called separable, if it admits a countable, dense subset. Proposition A normed space (V, ) is separable, if and only if there is a countable set A such that is dense. span A = {α 1 a α k a k k N, a 1,..., a k A, α 1,..., α k K} Proof. Let A V be a countable dense subset, then span A A is dense as well. We show the other implication for K = R. For K = C replace Q by Q + iq below. Now let A V be countable such that span A is dense. The set span Q A = {α 1 a α k a k k N, a 1,..., a k A, α 1,..., α k Q} = {α 1 a α k a k } k N a 1 A a k A α 1 Q α k Q is countable (countable union of countable sets). It is moreover dense in span A (exercise) and, since the latter is dense in V, also dense in V. Example By the preceding proposition, example 1.21 and example 1.26, the spaces c 0, l 1, l 2 and C(K) for compact K R n are separable. The space l is not separable. To see this, define the characteristic function (in this case sequence) 1 A of A N by { 1 A 1 n A n = 0 n / A. The set of all such characteristic functions is uncountable (it is in bijection with P (N)) and for A, B N distinct, we have 1 A 1 B = 1. Suppose F l is dense. Then for each A N, there is f A F such that f A 1 A < 1 2. By the triangle inequality, the elements f A and f B cannot coincide for distinct A, B N, hence F must contain at least as many elements as P (N), in particular it cannot be countable. Remark Any normed vector space V can be densely embedded into a larger, complete normed vector space. The latter space is essentially unique and is called the completion of V. Without proof. 12

13 1.2 Hilbert Spaces Definition Let V be a vector space. A map : V V K is called an inner product if (i) v w + λz = v w + λ v z (linearity in the second argument), (ii) v w = w v (conjugate symmetry), (iii) v v 0 (positivity), (iv) v v = 0 if and only if v = 0 (definiteness/non-degeneracy) are fulfilled for any v, w, z V and λ K. The pair (V, ) is called an inner product space. We will also write let V be an inner product space without explicitly mentioning the inner product. We define v := v v (we show below, that this is actually a norm). Remark From the first two conditions, we obtain conjugate linearity in the first argument, i.e. v + λw z = v z + λ w z. The two linearity properties together are called sesquilinearity (for K = C) or bilinearity (for K = R). The sesquilinearity also implies, that v w vanishes, whenever either v or w are zero. Proposition 1.33 (Cauchy-Schwartz-Inequality). Let (V, ) be an inner product space. Then for any v, w V, we have v w v w. Equality holds in the above relation, if and only if v and w are linearly dependent. Proof. Both assertions are trivial, if w = 0, so assume w 0. From the sequilinearity, we have 0 v λw v λw = v 2 λ v w λ w v + λλ w 2 for any λ K. Substituting v w w 2 0 v 2 v w 2 w 2 for λ yields v w w 2 + v w w 2 = v 2 v w 2 w 2 which is the desired inequality. Equality holds if and only if 0 = v λw 2, which by definiteness implies v λw = 0. Remark Note that definiteness is not used in the proof of the inequality. 13

14 Proposition The map v v is a norm on V. It is called the norm induced by the inner product. Proof. We show the triangle inequality: v + w 2 = v + w v + w = v 2 + w v + v w + w 2 The other norm properties are immediate. = v 2 + w 2 + 2R v w v 2 + w v w v 2 + w v w = ( v + w ) 2. Example (i) K n with (x 1,..., x n ) (y 1,..., y n ) = x 1 y x n y n inducing (x 1,..., x n ) = x x n 2 (ii) { l 2 = (x n ) } x n 2 < n N with (x n ) (y n ) = n N x n y n. (1.2) We have to show, that the series actually converges. By the definition of l 2, (x n ) 2 converges for any (x n ). We use the Cauchy-Schartz-inequality on K N to obtain N x n y n n=1 x x n 2 y y n 2 x n 2 y n 2 = (x n ) (y n ). n N n N Since the right hand side is independent of N, this shows that the series in (1.2) is absolutely convergent and since K is complete, this implies convergence. The induced norm coincides with the norm introduced in example 1.6. (iii) Let A R n be a measureable set and define { L 2 (A) := f : A K f measureable with A } f 2 dλ <. 14

15 and f g = A fgdλ. Again we need to show, that this is well defined. In order to do that, apply the arithmetic-geometric mean inequality (pointwise) to the functions f 2 and g 2 : fg = f 2 g 2 f g 2 2 and integrate both sides ( fg 1 dλ f 2 dλ + A 2 A A ) g 2 dλ <. This shows, that fg is integrable and thus the inner product is well defined. The map f g fulfills the properties of an inner product (exercise) except for definiteness. Similarly to example 1.8, define the subspace L 2 0(A) := { f L 2 (A) f = 0 }. Since fulfills the Cauchy-Schwartz inequality f g f g, f g = 0 if at least one of f or g is in L 2 0 (A). On L 2 (A) := L 2 (A)/L 2 0(A) we can define an inner product f + L 2 0 (A) g + L 2 0(A) = f g. Again we should prove, that this is well defined, so let f, g be two other functions representing the sets f +L 2 0 (A) and g+l2 0 (A) respectively, i.e. f f, g g L 2 0 (A). Then f g f g = f g g + f f g = 0 so the definition does not depend on the representing function. The new inner product inherits the inner product properties from the old one, but it is also definite, since implies f L 2 0 (A). 0 = f + L 2 0(A) f + L 2 0(A) = f f = f 2 The space L 2 (A) is called the space of square integrable functions over A and it is the Hilbert space appearing in most models of quantum mechanics. We will from now on write f for elements of L 2 (A) even though the elements are strictly speaking not functions but classes of functions. The comments from example 1.8 about point evaluation of these functions apply. The construction above can be done for any measure space. 15

16 Definition An inner product space H is called a Hilbert space, if it is complete with respect to the induced norm. Proposition The space of square integrable functions L 2 (A) is a Hilbert space. Proof. By proposition 1.11 it suffices to prove, that any absolutely convergent series is convergent. Thus let (f n ) L 2 (A) be such, that n=1 f n = M <. We fix some actual functions representing the f n, that we will also denote by f n. Define n s n := f n. By the triangle inequality we have s n k=1 n f n = k=1 n f n M, i.e. A s2 ndλ M 2. For any fixed x A, the sequence s n (x) is monotonously increasing and hence it converges to s(x) [0, ]. By the monotone convergence theorem we have s 2 dλ = lim n s2 ndλ = lim s 2 n ndλ M 2, A A which shows, that s is square integrable and in particular, that s(x) < for almost all x A. Now define n g n (x) := f k (x). k=1 Since the series k=1 f k(x) is ablsolutely convergent for almost every x A, the functions g n converge almost everywhere to some function g. By the triangle inequality g n k=1 A n f k = s n s k=1 which implies in particular g s and shows, that g n and g are square integrable. Moreover g n g 2 ( g n + g ) 2 4s 2 is an integrable bound and thus by Lebesgue s theorem lim g n g 2 = lim g n g 2 dλ = lim g n g 2 dλ = 0. n n n A This shows, that the series n=1 f n converges with respect to the L 2 -norm. A 16

17 Remark The other inner product spaces we have seen so far, i.e. usual scalar product and l 2 are Hilbert spaces as well. K n with the Remark It is easy to verify (exercise), that in an inner product space the parallelogram identity holds for all v, w V : v + w 2 + v w 2 = 2 v w 2. In fact, any norm is induced by an inner product if and only if it fulfills the above identity, in which case the inner product can be reconstructed using the polarization identity. Applying the parallelogram identity to v u, w u yields the alternative form v w 2 = 2 v u w u (v + w) u 2 2. (1.3) Definition Let V be a vector space, a subset A V is called convex, if for any v, w V, the line segment connecting them is contained in A. {λv + (1 λ)w λ [0, 1]} Definition Let (V, ) be a normed vector space, A V and x V. We define dist(x, A) = inf { x y y A}. Remark In general, there is no element actually minimizing the distance, as is witnessed by the example A = (0, 1), x = 2. In infinite dimensional Banach spaces, this can arise even if A is closed. However, for closed A, dist(x, A) = 0 the existence of a sequence (y n ) A such that y n, x, i.e. y n x. Thus if A is closed, dist(x, A) = 0 implies x A. Theorem 1.44 (Hilbert Projection Theorem). Let H be a Hilbert space A V convex and closed and x H. Then there is a unique element y A that realizes the minimal distance, i.e. dist(x, A) = x y. Proof. By the definition of d := dist(x, A), there is a sequence (y n ) A such that x y n d. Hence for any ɛ > 0, there is N N such that x y n 2 d 2 + ɛ2 2 for n N. Apply (1.3) to get y m y n 2 = 2 x y n x y m (y n + y m ) x 2. 17

18 Since by assumption 1 2 (y n + y m ) A, for n, m N we get the estimate ( ) ( ) y m y n 2 2 d 2 + ɛ2 + 2 d 2 + ɛ2 4d 2 = ɛ. 2 2 Thus (y n ) is a Cauchy sequence and since H is complete, it converges to y. Since A is closed, y is in A. By the continuity of the norm we get x y = lim n x y n = d. Assume z A is another element such that x z = d, then using (1.3) again, we get y z 2 = 2 y x z x (y + z) x 2 2d + 2d 4d = 0, i.e. y = z. Remark In general Banach spaces, the corresponding theorem can fail. Uniqueness can fail even in finite dimensions, e.g. for A = { (α, 0) R 2 α R } and x = (0, 1) for the maximum norm (α, β) = max {α, β}. In infinite dimensions, existence can fail even for the case where A is a closed subspace. Definition Let H be a Hilbert space. Two elements x, y H are called orthogonal if x y = 0. Let A H, then is called the orthogonal complement of A. A := {x H x y = 0 for every y A} Proposition Let H be a Hilbert space and A B H. Then A is a closed subspace and A B. Proof. Exercise. Proposition Let V H be a closed subspace. Then V and V are complementary, i.e. H = V V, i.e. for any x H, there are unique x V and x V such that x = x + x. Proof. Let s show the uniqueness first. So assume we have x, y V and x, z V, such that x = x + x = y + z. Then V x y = z x V and from the definition of V z x 2 = x y z x = 0. By definiteness of the norm we get z x = x y = 0 which is the required uniqueness. The subspace V is in particular convex. So we can apply theorem 1.44 to get the unique x V such that dist(x, V ) = x x. We can define x = x x and it remains to show, that x V. For any y V the function t x x ty 2 = x 2 2tR x y + t 2 y 2 18

19 has a global minimum at t = 0 (since x minimizes the distance from x to V ). But the function is differentiable (even a polynomial) hence it s first derivative in t = 0, i.e. R x y must vanish. If K = R, this already show that x V. For K = C, consider also t x x ity 2 and an analogous argument shows I x y = 0. Thus for any y V, x y = 0 hence x V. Corollary For any subset A of a Hilbert space H, (A ) = span A. For a closed subspace V in particular, we obtain (V ) = V and span A = A. A subspace W H is dense in H, if and only if W = {0}. Proof. Let y be in A. Then y x = 0 for every x A, hence y (A ). Thus A (A ) and, since span A is the smallest closed subspace containing A, also span A (A ). Now let x be in (A ). Applying the previous theorem we get x span A and x span A A such that x = x + x. We have 0 = x x = x x + x x = x x which implies x = 0 and thus x = x span A. Since x was arbitrary, this yields the inclusion (A ) span A. If V is already a closed subspace, i.e. V = span V = span V we get (V ) = V. Now apply this to the closed subspace A to get A = ((A ) ) = span A. Finally, note that {0} = H and H = {0}. So let W be dense, then W = H and thus W = W = H = {0}. If on the other hand W = {0}, then W = (W ) = H, i.e. W is dense. Remark Let H be a Hilbert space, V a closed subspace and x H. Then we can uniquely decompose x a x = x + x with x V and x V. The vector x is called the projection of x to V. Since (V ) = V, x is the projection of x to V. Note that x = x if and only if x V. Remark In the following we will often deal with sequences, that can be finite or countably infinite. In order to speak about both cases at once, I will be either of the index sets N or {1,..., N} for some N N. Some of the assertions are also true for more general index sets (in particular in the case of non-separable Hilbert spaces). Definition A family (e n ) n I is called an orthogonal system if e n and e m are orthogonal whenever n m and an orthonormal system (ONS) if in addition e n = 1 for every n I. An ONS is called an orthonormal basis (ONB) or complete orthonormal system if span {e n n I} is dense in H. 19

20 Example The functions (ϕ n ) n Z, defined by ϕ n (x) = e 2πinx are an orthonormal system in L 2 ([0, 1]) since 1 { 1 ϕ n ϕ m = e 2πinx e 2πimx dx = e 2πi(m n)x 1 m = n dx = 0 m n. 0 These functions are actually an orthonormal basis. The proof of this fact is surprisingly non-trivial and we will not give it here. Remark Let (e n ) n I be an orthonormal system and (α n ) n I, (β n ) n I K sequences of coefficients. Then for any N I N N N α n e n β k e k = α n β n. n=1 k=1 Since the scalar product is continuous, this equation still holds for N = provided the series n=1 α ne n and n=1 β ne n converge. In particular the norm is given by 2 α n e n = α n 2. n I n I 0 n=1 Proposition Let (e n ) n N be an orthonormal system and (α n ) n N K. n N α ne n converges if and only if n N α n 2 <. Then Proof. If n N α ne n converges then there is N N such that 1 > α n e n n=1 N 2 α n e n = n=1 n=n+1 α n e n = n=n+1 α n 2. which implies n N α n 2 <. If n N α n 2 < holds, then for every ɛ > 0, there is N N such that n=n α n 2 ɛ 2. So for M > K N, M K 2 M 2 M α n e n α n e n = α n e n = α n 2 < ɛ 2 n=1 n=1 n=k+1 n=k+1 shows, that the sequence of partial sums is a Cauchy sequence and hence convergent. Proposition 1.56 (Bessel s inequality). Let H be a Hilbert space. Let (e n ) n I be an orthonormal system. Then Bessel s inequality e n x 2 x 2 n I holds for any x H, which implies in particular the convergence of n I e k e k x. 20

21 Proof. The inequality follows from N 2 0 x e n e n x = x 2 = x 2 n=1 N x e n e n x n=1 N e n x 2 n=1 N e k x e k x + n=1 N n=1 k=1 N e n x e k x e n e k which holds for any N I proving Bessels inequality for finite I. For infinite I take the limit for N. The convergence of n I e n e n x follows from Bessel s inequality and the previous proposition. Remark The element y := i I e n e n x is the projection of x to the closed subspace spanned by the orthonormal system. In particular we have x = i I e n e n x if (e n ) is an orthonormal basis. Proof. Define V := span {e n n I}. Obviously x = y + (x y) with y V so by proposition 1.48 it suffices to show, that x y V. For any k I, e k x y = e k x e k x = 0 hence x y {e n n I} = V. Note that this also yields a method to calculate the projection onto an arbitrary subspace V by choosing an orthonormal basis (see below). If the (e n ) are an orthonormal basis, then V = {0} (corollary 1.49) so x = n I e n e n x. Corollary 1.58 (Parseval s Identity). Let (e n ) n I be an orthonormal system in a Hilbert space H. Parseval s identity e n x 2 = x 2 holds for any x H if and only if (e n ) is an orthonormal basis. n I Proof. As we have just seen, x = n I e n e n x if (e n ) is an orthonormal basis. Parseval s identity then follows from remark On the other hand, suppose that Parseval s identity holds for any x H. For x {e n n I}, this implies x 2 = 0 and thus x = 0. Thus {e n n I} = {0} and, taking the complement again H = span {e n n I}. 21

22 Remark Given an orthonormal basis (e n ) of a Hilbert space, for each x we get the unique decomposition with respect to that basis x = n I e n e n x. In example 1.53 this decomposition yields the Fourier series. Note that this series is unconditionally convergent, thus the ordering of the basis elements does not matter. Proposition 1.60 (Gram-Schmidt-Orthogonalization). Let (x n ) n I be a linearly independent sequence in a Hilbert space H, then there exists an orthonormal system (e n ) n I (of the same cardinality) such that span {x 1,..., x n } = span {e 1,..., e n }. for any n I. In the infinite case, we also get span {x n n N} = span {e n n N}. Proof. We show the case I = N. Since the x n are linearly independent, x 1 0. So set e 1 := x 1 x 1.< Then obviously x 1 and e 1 span the same subspace of H. Now we proceed by induction. Suppose we have an orthonormal system e 1,..., e n such that V n := span {x 1,..., x n } = span {e 1,..., e n }. The space V n is closed (all finite dimensional subspaces are). Let z be the projection of x n+1 to V n and e n+1 = x n+1 z x n+1 z. Since the x n are linearly independent, x n+1 is not in V n, so x n+1 z is non-zero and e n+1 hence well defined. Moreover span {x 1,..., x n, x n+1 } = span {e 1,..., e n, e n+1 } holds. Hence the inductively defined sequence (e n ) is an orthonormal system fulfilling the conclusion of the theorem. In the infinite case, we rewrite span {x n n N} = n N span {x 1,..., x n } = n N span {e 1,, e n } = span {e n n N}. Proposition Let H be a separable Hilbert space, then it admits an orthonormal basis. Proof. Since H is separable, there is a sequence (x n ) n N that is dense in H. Choose n 1 = 1 and, given n 1,..., n k inductively defined n k+1 to be the smallest index, such that x nk+1 is not a linear combination of x n1,..., x nk (terminate the process, if there is no such index). The (possibly finite) subsequence y k = x nk so defined is linearly independent and fulfills span {x n n N} = span {y k k N}. 22

23 Now apply the Gram-Schmidt-orthonormalization to this subsequence resulting in an orthonormal system (e k ) k N still spanning the same vector space. Since by assumption span {e k k N} = span {x n n N} is dense in H, the (e k ) form an orthonormal basis. Corollary In a separable Hilbert space, any orthonormal system can be extended to an orthonormal basis. Proof. Let (x n ) n I be an orthonormal system. Then V = {x n n I} is a closed subspace of H and thus also a Hilbert space. The space V is also separable (exercise), so it admits an othonormal basis (y n ) n J. The union of these two bases form a basis for H. Remark In quantum mechanics, the state space of a quantum system is modelled by a separable Hilbert space H (any particular one will do). More precisely the physical states are elements x H fulfilling x = 1. The most common model is L 2 (R 3 ) for the state space of a single quantum mechanical particle (much like R 6 would be the state space of a single classical particle). Usually the function ϕ 2 is the probability density for the location of the particle, however the state ϕ contains more (physically relevant) information than this probability density. If we are interested in other observables than location or dynamics (i.e. time evolution of states), we need to consider operators on the Hilbert space. 23

24 1.3 Linear Operators Definition Let V, W be vector spaces. A linear map from V to W is a map T : V W, such that T (x + λy) = T (x) + λt (y) for all x, y V and λ K. Linear maps from V to V are called (linear) operators or endomorphisms on V. We will often omit the brackets writing T x for T (x). Remark In finite dimensions, all linear maps are automatically continuous (with respect to any choice of the equivalent norms). This is not true for infinite dimensional spaces any more. Proposition Let (V, V ), (W, W ) be normed vector spaces and T : V W a linear map. Then the following are equivalent: (i) T is uniformly continuous (ii) T is continuous, (iii) T is continuous in 0, (iv) there is M R such that T x W M for any x V with x V = 1, (v) there is M R such that T x W M x V for any x V. Proof. The implications (i) = (ii) and (ii) = (iii) are trivial. If T fulfils (iii), then there is δ > 0 such that T x W = T (x) T (0) W < 1 for all x V fulfilling x V = x 0 V < δ. For y V with y V = 1, we have ( ) δ W T 2 y < 1 T y W < 2 δ, which is (iv) If (iv) holds for M R, then for any non-zero x V ( ) T x W = x W x T M V x V shows that T x W M x V (this inequality is trivial if x = 0) and hence (v). Finally suppose (v) holds. For any ɛ > 0 and for any x, y V fulfilling x y V < ɛ M we have which proves the uniform continuity of T. T x T y W M x y V ɛ 24

25 Definition Let (V, V ), (W, W ) be normed vector spaces. An linear map T : V W is called bounded, if it fulfils one of the conditions of the previous proposition. The space of all bounded linear maps between V and W will be denoted by B(V, W ) and we write B(V ) for B(V, V ). On B(V, W ) we define the operator norm T = sup { T x W x V and x V = 1}. (1.4) An linear map with values in the base field K is called a linear functional. We define the dual space V := B(V, K) as the space of bounded linear functionals. Remark i.e. (i) With the pointwise definition of addition and scalar multiplication, (T + S)x := T x + Sx (λt )x := λt x, for any x V and any λ K the space B(V, W ) is a vector space (exercise). (ii) For vector spaces V, W, U and linear maps T : V W and S : W U, the concatenation S T is a linear map as well. We will also write ST for this map. (iii) We will often simply use for the various norms appearing whenever the context is sufficient to disambiguate the symbol. (iv) Note that the notation is slightly ambiguous. A map f : X V is usually called bounded if there is M > 0 such that f(x) M for all x X. If f is linear, it is bounded in this sense if and only if f is constant zero. So bounded linear map or bounded operator always refers to definition (v) The operator norm is indeed a norm. We show the triangle inequality. For any x V with x = 1 we get (T + S)x T x + Sx T + S. Taking the supremum over all such x yields T + S T + S. (vi) In definition (1.4) we can replace x V = 1 by x V 1 without changing it (Why?). (vii) If for some M > 0, T x M x holds for all x V, then T M. On the other hand, we have T x T x for any x V and any T B(V, W ). Thus we have T = inf {M > 0 T x M x for all x V }. 25

26 (viii) For normed vector spaces V, W, U and bounded maps T : V W and S : W U we have (exercise) ST S T. In particular the space B(V ) is a (generally non-commutative) normed algebra. Example (i) Let V be any vector space. Then the identity operator I : V V defined by Ix = x is a linear operator. It is always bounded. (ii) Let A K n m, then is a bounded linear map. T A : K m K n x Ax (iii) Let K R n and x K. The evaluation map is a linear functional on C(K). Since ϕ x : C(K) K f f(x) ϕ x (f) = f(x) f, holds for any f C(K), it is bounded with ϕ x = 1. (iv) Let A R n and f : A K be a bounded, measurable function. Then T f : L 2 (A) L 2 (A) g fg is a linear operator, a so-called multiplication operator. Since T f g 2 = fg 2 dλ f 2 g 2 dλ = f 2 g 2 A holds for any g L 2 (A), T f is bounded and T f f. A (v) On D := { f C 1 ((0, 1)) L 2 ((0, 1)) f L 2 ((0, 1)) } define T : D L 2 ((0, 1)) f f. 26

27 This is a linear operator. If we equip D with the L 2 -norm, the operator T is not bounded since (ϕ n ) D, ϕ n (x) = e inx fulfills ϕ n = 1 but T ϕ n 2 = 1 0 ϕ n (x) 2 dx = 1 0 n 2 ϕ n (x) 2 dx = n 2 diverges to. Unbounded operators in general and the above example in particular do appear in physical applications. They are however more difficult to treat and we will probably not have time to talk about them. (vi) Let A R n be measurable. Then ϕ : L 1 (A) K f fdλ is a linear functional. It is bounded, since for f L 1 (A) we have ϕ(f) = fdλ f dλ = f. A A A Proposition Let V, W be normed vector spaces. space, if W is. Then B(V, W ) is a Banach Proof. Let (T n ) B(V, W ) be such, that n=1 T n <. For any x V T n x n=1 T n x = x T n < n=1 shows, that n=1 T nx is absolutely convergent and hence, since W is complete, also convergent. Define Sx = n=1 T nx. One easily verifies, that S is linear (exercise). Moreover by the generalized triangle inequality ( ) Sx T n x T n x n=1 holds for any x V, hence S is bounded with S n=1 T n. By a similar argument, we get ( ) ( N S T n x = ) T n x T n x n=1 n=n+1 n=1 n=1 n=n+1 for any x V which implies N S T n n=1 T n. n=n+1 27

28 Since the right hand side converges to 0 for N, this proves the convergence of the series n=1 T n with respect to the operator norm. We proved, that every absolutely convergent series in B(V, W ) is convergent which implies completeness of the space by proposition Proposition Let H, G be Hilbert spaces, x H and A : H G a linear map. Then and x = sup { x y y H, y 1} A = sup { x Ay x G, y H, x 1, y 1}. Proof. From the Cauchy Schwarz inequality, we get we get x y x y sup { x y y H, y 1} x. Since moreover x x = x, x we have the first equality. Applying this to Ay, yields Ay = sup { Ay x x G, x 1} A = sup { Ay y H, y 1} = sup { x Ay x G, y H, x 1 y 1}. Definition Let V, W be vector spaces. The kernel of a linear map T : V W is the set The image of T is ker T = {x V T x = 0} V im T = {T x x V }. 28

29 Remark The sets ker T and im T are linear subspaces (exercise). If T is bounded, then ker T is closed, for if (x n ) ker T converges to x, then due to the continuity of T, T x = lim n T x n = 0 and hence x ker T. The image of T need not be closed, even for bounded T. As an example T consider multiplication by ( 1 n (x 1, x 2, x 3...) ) n N on l1, i.e. ). ( x 1, 1 2 x 2, 1 3 x 3,... The operator T is bounded. The image contains the finite sequences c c since T (x 1, 2x 2, 3x 3,..., nx n, 0, 0,...) = (x 1, x 2, x 3,..., x n, 0, 0,...) so it is dense in l 1. However, im T is not the entire space since (1, 12 2, 13 2,... ) / im T since it s alleged pre-image ( 1, 1 2, 1 3,...) is not in l 1. Remark A linear map T : V W is injective if and only if ker T = {0}. It is surjective if and only if im T = W (exercise). Definition Let V, W be normed spaces. An linear map T : V W is called (i) a contraction, if T x x for any x V (equivalently if T 1), (ii) an isometry, if T x = x for any x V, (iii) an isometric isomorphism, if it is a surjective isometry. Remark Any isometry is automatically injective, since T x = x implies, that ker T = {0}. Let T : V W be an isometric isomorphism. Then T is in particular bijective hence it has an inverse map T 1 which is also bijective. For any w 1, w 2 W, there are v 1, v 2 such that T v 1 = w 1 and T v 2 = w 2. Then T 1 (w 1 + λw 2 ) = T 1 (T v 1 + λt v 2 ) = T 1 T (v 1 + λv 2 ) = v 1 + λv 2 = T 1 w 1 + λt 1 w 2 for any λ K which proves the linearity of T 1. Moreover, for w = T v W T 1 w = v = T v = w holds, i.e. T 1 is an isometric isomorphism as well. An isometric isomorphism preserves all the structure that defines a normed vector space. So if two spaces are isometrically isomorphic, then they are essentially the same space, we just chose different names to label their objects. 29

30 If T is an operator between Hilbert spaces fulfilling T x T y = x y for any x, y, then it is clearly an isometry. On the other hand, the scalar product can be expressed in terms of the norm alone (via the polarization identity) so any isometry between Hilbert spaces also preserves the scalar product. Example The spaces (l 1 ) and l are isometrically isomorphic. Proof. We will use the standard unit sequences (e k n) defined by e k n = 1 if and only if k = n and 0 otherwise. Note that these have l 1 -norm 1. For fixed (x n ), define T (x n ) : l 1 K by (T (x n ))(y n ) := n N x n y n. (1.5) Then (T (x n ))(y n ) = x n y n x n y n (x n ) y n = (x n ) (y n ) 1 n N n N n N shows, that the sum in (1.5) converges for (y n ) l 1. The expression (1.5) is linear in (y n ) hence T (x n ) is a bounded linear functional on l 1. We saw already, that T (x n ) (x n ). Since (T (x n ))(e k n) = x k implies T (x n ) x k for any k N, taking the supremum over k yields T (x n ) (x n ), i.e. T (x n ) = (x n ). The expression (1.5) is also linear in (x n ), hence T : l (l 1 ) (x n ) T (x n ) is linear and as we have shown above an isometry. It remains to show, that T is surjective. In order to do so, first note that for (y n ) l 1 N (y n) y k (e k n) = y n 0 n=1 n=n+1 implies that n=1 y k(e k n) converges to (y n ) in the l 1 -norm. So let ϕ (l 1 ) and define (ϕ(e k n)) k N. The sequence is bounded by ϕ and hence in l. For any (y n ) l 1, using the linearity and continuity of ϕ, we get ( ( ( ) ) ) T ϕ(e k n) (y k ) = ϕ(e k n)y k = ϕ y k (e k n) = ϕ(y n ). k N k=1 Since this is true for any (y n ) l 1, we get T ( ϕ(e k n) ) = ϕ and since ϕ was arbitrary, T is surjective. k=1 30

31 Remark Let H be a Hilbert space and x H. Then y x y is a bounded linear functional. The boundedness follows from the Cauchy Schwarz inequality: Hence we can define a map x y x y. T : H H x x. This map is conjugate linear, since the scalar product is conjugate linear in the first argument. We can calculate the norm of T x using proposition 1.71: T x = sup { x y y H} = x. In particular, T is injective, i.e. x y = z y for all y H implies x = z. So T is a conjugate linear isometric map from H into H. Thus we can think of H as a subspace of H. Theorem 1.79 (Riesz Representation Theorem). The map from the previous remark is actually surjective, i.e. for each bounded linear functional ϕ on H, there is a unique x H fulfilling ϕ = x such that ϕ(y) = x y for all y H. Proof. Let ϕ be in H. For ϕ = 0, x = 0 fulfils the requirements. So assume ϕ 0, i.e. ker ϕ H. Then ker ϕ is a closed subspace of H. The space (ker ϕ) cannot be the trivial space {0} for otherwise ((ker ϕ) ) = ker ϕ would be the entire space H. Choose some x 0 from (ker ϕ) and define x = ϕ( x) x (ker ϕ). This element fulfils ϕ(x) = ϕ( x) 2 x 2 = x 2 For y H there are unique y ker ϕ and y (ker ϕ) such that y = y + y. There is κ K, such that ϕ(y ) = κϕ(x) i.e. y κx ker ϕ, but since y and x are in (ker ϕ) this implies y κx = 0. Thus x y = x y + x y = x κx = κ x 2 = κϕ(x) = ϕ(y ) = ϕ(y ) + ϕ(y ) = ϕ(y) holds for any y H. The uniqueness of x is left as an exercise. Proposition Let H be a Hilbert space and A B(H). Then there is a unique A B(H), called the adjoint of A, such that x Ay = A x y holds for any x, y H. The operation : B(H) B(H), A A x 2 31

32 (i) is conjugate linear, (ii) is idempotent, i.e. (A ) = A for any A B(H), (iii) fulfils (AB) = B A, for any A, B B(H), (iv) is an isometry and (v) fulfils A A = AA = A 2 for any A B(H). Proof. Let A be a bounded linear operator on H. For fixed x H, the map y x Ay is a linear functional, so there is a uniquely defined element of H, which we will denote by A x, such that A x y = x Ay for any y H. Varying x, this yields a map x A x which fulfils A (x + λy) z = x + λy Az = x Az + λ y Az = A x z + λ A y z = A x + λa y z for any x, y, z H and λ K. Since the equality holds for any z H, we get A (x + λy) = A x + λa y so A is linear a linear operator. For x, y H, we have x Ay = y A x from which we get A = A taking the supremum over all x, y H with norm 1 and using the norm formula from proposition This shows, that A is a bounded operator. Now suppose B B(H) is another operator fulfilling Bx y = x Ay for any x, y H. Then (A B)x y = A x y Bx y = x Ay x Ay = 0 and taking the supremum over x, y with norm 1 yields A B = 0, i.e. B = A. So the adjoint is unique. The isometry of the map A A was already shown above. For any A, B B(H), λ K and x, y H we have (A + λb) x y = x (A + λb)y = x Ay + λ x By = A x y + λ B x y = (A + λb )x y. Since the adjoint operator is uniquely defined, this implies (A + λb) = A + λb. Similarly we get for any x, y H (AB) x y = x ABy = A x By = B A x y 32

33 and x (A ) y = A x y = x Ay which using the uniqueness of the adjoint yields (AB) = B A and (A ) = A. Finally, we already know, that A A A A = A 2. On the other hand, using the Cauchy Schwarz inequality we get Ax 2 = Ax Ax = x A Ax x A Ax A A x 2. Taking the square root and the supremum over all x with x 1 yields A A A and hence A 2 = A A. Now apply this to A to get the missing equality. Remark A Banach algebra A with a map A A A A fulfilling conditions (i) to (iv) is called a Banach- -algebra. If it also fulfils (v), it is called a C -algebra. Hence we proved, that for a Hilbert space H, B(H) with taking adjoints as the -operation is a C -algebra. Another example of a C -algebra is the space C(K) of continuous functions on some compact set K with complex conjugation as the -operation (verify that as an exercise). It is possible to formulate quantum mechanics purely in terms of (abstract) C -algebras dispensing with Hilbert spaces altogether. This approach is used in particular in axiomatic quantum field theory. Definition Let H be a Hilbert space and A B(H). The operator A is called (i) normal if AA = A A, (ii) self-adjoint if A = A, (iii) unitary if AA = A A = I, (iv) a projection if A 2 = A and (v) an orthoprojection if A is a self-adjoint projection. Remark Self-adjoint operators are the observables of quantum mechanics. Note that for A self-adjoint, x Ax = Ax x = x Ax, so x Ax is real. We will later show, that the spectrum of self-adjoint operators is real as well. For a unitary operator U, we have Ux Uy = U Ux y = x y for any x, y H so U is an isometry. Since x = UU x for any x H, U is also surjective showing that U (and U ) is an isometric isomorphism from H to itself. The importance of unitary operators is thus, that they implement symmetries on the Hilbert space modelling some physical system. 33

34 Example (i) Let T g be multiplication on L 2 (A) by some (measurable) bounded function g : A K. Then for any f 1, f 2 L 2 (A), the equation f 1 T g f 2 = f 1 gf 2 dλ = gf 1 f 2 dλ = T g f 1 f 2 A holds which shows, that (T g ) = T g. Moreover T gh f = ghf = T g T h f for any f L 2 (A) and any bounded g, h, so T g T h = T gh. Thus T g is always normal, it is self-adjoint if and only if g is real-valued (almost everywhere), unitary if and only if g is a function of modulus 1 (almost everywhere) and an orthoprojection if and only if g is the characteristic function of some set B A. (ii) Let x be a real number. For f L 2 (R) we define (U x f)(y) := f(y x). Then for any f, g L 2 (R) we have f U x g = f(y)g(y x)dy = A so (U x ) = U x. This also implies that U x is unitary. f(z + x)g(z)dz = U x f g (iii) Let V be a closed subspace of a Hilbert space and for x H denote the projection of x onto V by P x. We already know, that P is a projection. Also (I P ) 2 = I 2P + P 2 = I P shows, that I P is a projection as well. By definition of the projection P, P x V and (I P )x V for any x H. Hence for any x, y H x P y = P x + (I P )x P y = P x P y = P x P y + (I P )y = P x y shows that P is self-adjoint. Actually any orthoprojection arises in this way (exercise). Proposition Let H be a Hilbert space and A B(H). Then the following equalities hold: (im A) = ker A ker A = im A. Proof. For x ker A and y im A we have y = Az for some z H and hence x y = x Az = A x z = 0. Thus x is in (im A). On the other hand if x (im A) and y H, then 0 = x Ay = A x y. Since y H is arbitrary, the equation implies A x = 0 hence x ker A. Taking the orthogonal complement of the equation just obtained yields im A = ((im A) ) = (ker A ) and we get the second equation by applying this to A. 34