Kerala School of Mathematics, Kozhikode Instructional Workshop on Ergodic Theory October 17 - November 4, 2011

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1 Kerala School of Mathematics, Kozhikode Instructional Workshop on Ergodic Theory October 7 - November 4, 20 Measure Theoretic Ergodic Theory M. G. Nadkarni Indian Institute of Technology, Indore

2 Chapter Preliminaries and the Poincaré Lemma Borel Spaces.. Let X be a non-empty set. A σ-algebra B on X is a non-empty collection of subsets of X which is closed under countable unions and complements. A set together with a σ-algebra B is called a Borel space or a Borel structure (X, B)..2. The intersection of any family of σ-algebras on X is again a σ- algebra. If A is any collection of subsets of X, then the intersection of all the σ-algebras on X which contain the collection is again a σ-algebra. It is the smallest σ-algebra containing A. It is called the σ-algebra generated by A..3. If (X, B) is a Borel space then a subcollection N B is called a σ-ideal if. N is closed under countable unions 2. B B and N N implies that B N N. For example if m is a countably additive measure on B, then the collection of sets in B of m measure zero forms a σ-ideal. We will come across other σ-ideals later. If E B is any collection then there is a smallest σ-ideal containing E, namely, the intersection of all the σ-ideas containing E. We call it the σ-ideal generated by E. It is formed by taking all sets of the form B E, B B, E E and taking countable unions of such sets. If N B is a σ-ideal and A, B belong to B, then we write A = B(mod N ) if A B = (A B) (B A) N..4. An interesting σ-algebra on a complete separable metric space X is the σ-algebra of sets with the property of Baire. A set A is said to have the property of Baire if A can be expressed in the form A = G P where G is open and P is of first category.

3 2 Basic Ergodic Theory.5. Theorem. A set A in a complete separable metric space X has the property of Baire if and only if it can be expressed in the form A = F Q where F is closed and Q is of the first category. Proof. If A = G P, G open, P of first category, then N = G G is a nowhere dense closed set, and Q = N P is of first category. Let F = G. Then A = G P = (G N) P = G (N P ) = F Q. Conversely if A = F Q, where F is closed and Q is of first category, let G be the interior of F. Then N = F G is nowhere dense, P = N Q is of first category, and A = F Q = (G N) Q = G (N Q) = G P. This proves the theorem..6. Corollary. If A has the property of Baire then so does its complement. Proof. For any two sets A and B, (A B) c = A c B. Hence if A = G P, G open, P of first category, then A c = G c P which again has the property of Baire by the above theorem..7. Theorem. The class of sets having the property of Baire is a σ-algebra. It is the σ-algebra generated by open sets together with the sets of first category. Proof. Let A i = G i P i, (i =, 2, 3,...) be any sequence of sets having the property of Baire. Put G = G i, P = P i, A = A i. Then G is open, P is of first category, and G P A G P. Hence G A P is of first category, and A = G (G A) has the property of Baire. This result and the corollary above show that the class in question is a σ-algebra. It is evidently the smallest σ-algebra that includes all open sets and all sets of first category. This proves the theorem. Note that the first category sets form a σ-ideal in the σ-algebra of sets with the property of Baire..8. Two Borel spaces (X, B ), (X 2, B 2 ) are said to be isomorphic if there a one-one map φ of X onto X 2 such that φ(b ) = B 2. The map φ is called a Borel isomorphism between the two Borel spaces..9. If (X, B) is Borel structure and A is a non-empty subset of X,

4 Preliminaries and the Poincaré Lemma 3 then the collection of sets of the form A B with B B is a σ-algebra on A called the induced σ-algebra on A and denoted by A B or B A. Two sets A, B X are said to be Borel isomorphic if there is a one-one map φ of A onto B such that φ(b A ) = B B, i.e., the Borel structures (A, B A ), (B, B B ) are isomorphic. Standard Borel Spaces.0. Let X be a complete separable metric space and B X the σ- algebra generated by the collection of open sets in X. B X is called the Borel σ-algebra of X. The following results are known from descriptive set theory. (see K.R.Parthasarathy [4]). A set in B X is either countable or has the cardinality c of the continuum. 2. If A and B in B X are of the same cardinality, then A and B are Borel isomorphic. 3. If Y is another complete separable metric space of the same cardinality as X and B Y its Borel σ-algebra, then (X, B X ) and (Y, B Y ) are Borel isomorphic. 4. It follows from (), (2) and (3) that if A B X and B B Y have the same cardinality, then the Borel spaces (A, A B X ) and (B, B B Y ) are Borel isomorphic... A Borel space isomorphic to the Borel space of an uncountable complete separable metric space is called standard. Such a space is, in view of the results above, isomorphic to the Borel space of the unit interval equipped with the σ-algebra generated by its usual topology. If a standard Borel space is equipped with a finite or a σ-finite measure m then the resulting measure space is called a standard measure space. In particular if m(x) = then such a measure space is called a standard probability space..2. We know that the forward image of a measurable set under a measurable map need not be measurable, in general. However a theorem of Lusin in classical descriptive set theory states that if f is a measurable

5 4 Basic Ergodic Theory function on a standard Borel space into another such space and if f is countable to one in the sense that the inverse image of every singleton is at most countable, then the forward image under f of any Borel set is Borel. In particular if such an f is one-one and onto then it is a Borel isomorphism..3. In a complete separable metric space every Borel set has the property of Baire since the σ-algebra of sets with the property of Baire includes the Borel σ-algebra. It should be noted that if X is a complete separable metric space and if A is a Borel set in X, then A can be expressed as G P where G is open and P is not only of first category but also a Borel set in X. This is because the class of sets of the form G P, G open, P a Borel set of first category forms a σ-algebra which coincides with the Borel σ-algebra of X. Borel Automorphisms.4. A one-one measurable map τ of a Borel space (X, B) onto itself such that τ is also measurable is called a Borel automorphism of X. If (X, B) is a standard Borel space then a measurable one-one map of X onto X is a Borel automorphism in view of.2. In what follows, we will carry out an elementary analysis of a Borel automorphism on a standard Borel space..5. Let τ be a Borel automorphism of the unit interval X = [0, ] equipped with its Borel σ-algebra. For any x X, the set {τ n x n Z} is called the orbit of x under τ and denoted by orb (x, τ). A point x X is said to be periodic if there is an integer n such that τ n x = x, and the smallest such positive integer is called the period of x under τ. If n is the period of x under τ, then the set x, τx, τ 2 x,..., τ n x consists of distinct points of [0, ]. Let E = {x τx = x} E 2 = {x τx x, τ 2 x = x}. E = {x τ n x x for all integers n} The set E n for n < is made up of precisely those points in X which have period n. Each E n is in B, E m E n = if m n, and the union

6 Preliminaries and the Poincaré Lemma 5 of all the E i, i =,, 2, 3,... is X. A set in X is said to be invariant under τ or τ-invariant if τa = A. It is clear that all the sets E n are τ-invariant..6. Let us consider E n for n <. If x is in E n, then x, τx,... τ n x are all distinct and if y {x, τx,..., τ n x} then {x, τx,..., τ n x} = {y, τy,..., τ n y}. If further y = min{x, τx,..., τ n x}, then y < τy, y < τ 2 y,..., y < τ n y, τ n y = y. We put B n = {y E n y < τy, y < τ 2 y,,..., y < τ n y}. B n is a measurable subset of E n which contains exactly one point of the orbit of each x in E n. We may view the restriction of τ to E n pictorially as in Figure.. τ n x τ n 2 x τ n B n τ n 2 B n. τx x τb n B n Figure. E n is viewed as the union of n horizontal lines B n, τb n,..., τ n B n. A point x B n moves one step up with each application of τ until it reaches τ n x τ n B n. One more application of τ brings it back to x. Now, X E = E n where the union is over n < = n= n τ k B n

7 6 Basic Ergodic Theory and we may view τ on X E pictorially as in Figure.2. τ n x τ 2 x τx B x B 2 x B 3 x B n Figure.2 As before, a point x B n moves one step up with each application of τ with τ n (x) = x. The set B = k= B k is a Borel set and has the property that orbit of any point in X E intersects B in exactly one point..7. Let us now consider τ on E. In this case there is, in general, no neat way in which we can find a measurable set B which intersects the orbit of each x in exactly one point. The set {τ n x n Z} being infinite in this case, we can no longer conclude that inf{τ n x n Z} is in the set {τ n x n Z}. Indeed it can happen that inf{τ n x n Z} = 0 for all x E. (For example, the orbit of every point in E may be dense in [0, ].) We can use the axiom of choice to select one point from each orbit and thus form a set B which intersects the orbit of each x E in exactly one point. But such a B may not be measurable, and we are not interested in sets which are not measurable. We give below two examples. In the first example X = E and there is a B which is measurable. In the second example, the so called irrational rotation of the circle, E = X and there is no B which is measurable.

8 Preliminaries and the Poincaré Lemma 7.8. Example. X = R and τx = x +. In this case the Borel set B = [0, ) has the property that the orbit of every point in R intersects [0, ) in exactly one point. Moreover, τ n x x for any n and the union of τ n [0, ) over n in Z is R..9. Example 2. X = the unit circle = {e iϑ 0 ϑ < 2π }. Let α be an irrational number and β = e 2πiα. Define τ by setting τe iϑ = βe iϑ. Now τ n e iϑ = β n e iϑ cannot be equal to e iϑ for any n 0, for if τ n e iϑ = e iϑ for some integer n 0, then β n = e 2πiαn =, i.e., nα is an integer which contradicts the irrationality of α. Thus τ has no periodic points. Next we show that τ admits no measurable B. Suppose B X is a set which intersects every orbit in exactly one point. Then τ n B, n Z, are pairwise disjoint with union X. Let l denote the Lebesgue measure on X. If B were measurable then l(τ n B ) = l(b ) for all n, in view of the invariance of l under rotation. Since τ n B, n Z, are pairwise disjoint with union X we have l(x) = 2π = l(τ n B ), which is a contradiction because l(τ n B ) are all the same = l(b ). Thus this τ admits no measurable B which intersects each orbit in exactly one point..20. Let us return to the consideration of a general Borel automorphism τ on a standard Borel space X which we may assume to be [0, ] without loss of generality. Let c n (τ) denote the cardinality of the set of orbits of points in E n = cardinality of B n, n <. Let c (τ) denote the cardinality of the set of orbits of points in E. The sequence of integers {c (τ), c (τ), c 2 (τ),...} is called the cardinality sequence associated to τ..2. Definition. A Borel automorphism τ is said to be an elementary Borel automorphism if there exists a measurable set B which intersects the orbit of each point in E in exactly one point, equivalently, τ is elementary if and only if there exists a set B which is measurable and intersects each orbit in exactly one point. The equivalence of the two formulations above is obvious because if a measurable B exists, then we can take B = B k, where the union is taken over k. On the other hand if B as postulated in the definition above exists, then we can take B = B E k, where the union is taken over k <.

9 8 Basic Ergodic Theory Orbit Equivalence and Isomorphism.22. Definition. Two Borel automorphisms τ and τ 2 on Borel spaces (X, B ), (X 2, B 2 ) respectively are said to be isomorphic if there is a Borel isomorphism φ : X X 2 such that φτ φ = τ 2. Definition. We say that τ and τ 2 are weakly equivalent or orbit equivalent if there is a Borel isomorphism φ : X X 2 such that for all x X, φ(orb (x, τ )) = orb (φ(x), τ 2 ). It is clear that if two Borel automorphisms are isomorphic then they are also orbit equivalent. However the converse is not true in general as will emerge in the sequel..23. Exercise. If τ and τ 2 are orbit equivalent then the associated cardinality sequences (see.20.) are the same. Exercise 2. If τ and τ 2 are elementary and the associated cardinality sequences are the same, then τ and τ 2 are isomorphic, hence also orbit equivalent..24. Given a Borel automorphism τ on a standard Borel space (X, B) we say that the orbit space of τ admits a Borel cross-section (or simply that τ admits a Borel cross-section ) if there is a Borel set B which intersects each orbit in exactly one point. Clearly, the statements τ is elementary and τ admits a Borel cross-section are equivalent. In contrast to exercise 2 above, the question when are two non-elementary Borel automorphisms isomorphic? has so far not found a simple answer although there are some deep theorems in ergodic theory dealing with this question in a measure theoretic setting. The question when are two Borel automorphisms orbit equivalent? has now a complete solution and will be discussed in chapter. Poincaré Recurrence Lemma.25. The study of Borel automorphisms which do not admit Borel cross-sections is intimately connected with both topological and measure theoretic ergodic theory. However some of the basic concepts such

10 Preliminaries and the Poincaré Lemma 9 as recurrence, induced automorphisms, dissipative and conservative automorphisms etc., are in essence set theoretic in nature and can be explained without any reference to measure or topology. One of the very first and very basic results is the Poincaré recurrence lemma which we give below in a rather distilled form (see J.C.Oxtoby [5])..26. Definition. A measurable set W is said to be wandering with respect to a Borel automorphism τ if τ n W, n Z, are pairwise disjoint. The σ-ideal generated by all wandering sets in B will be denoted by W τ and called the Shelah-Weiss ideal of τ. It is clear that if W is a wandering set then W intersects the orbit of any point in at most one point. Moreover W never intersects the orbit of a periodic point. If a measurable B (see.7) exists then it is a wandering set. A subset A of orb (x, τ) is called bounded below (bounded above) if the set of integers n such that τ n x A is bounded below (bounded above). Bounded and unbounded subsets of orb (x, τ) are defined similarly. If A orb (x, τ) is bounded below then there is a smallest integer n such that τ n x A; we call such τ n x the smallest element of A. Similarly we define the largest element of A in case A is bounded above. A very useful sufficient condition for a set N B to belong to W τ is that for all x X, orb (x, τ) N is either bounded below or above. For then, firstly, we can prescribe a well defined procedure for choosing a point from each orbit which has a non-empty intersection with N : choose the least element of orb (x, τ) N if there is one, otherwise choose the largest element; secondly, the set W of points thus chosen from each orbit having non-empty intersection with N is a wandering Borel set whose powers under τ cover N so that N belongs to the Shelah- Weiss ideal of τ. For any set A in B the set M of points x in A such that orb (x, τ) A is bounded below or above belongs to the Shelah- Weiss ideal of τ and for any x A M, the points τ n x return to A M for infinitely many positive and infinitely many negative n. This is, in short, the Poincaré recurrence lemma which we formally prove below.(the Borel nature of M needs to be proved)..27. Poincaré Recurrence Lemma. Let τ be a Borel automorphism of a standard Borel space (X, B). Then given A B there exists N W τ such that for each x A N the points τ n x return to A for

11 0 Basic Ergodic Theory infinitely many positive n and also for infinitely many negative n. Proof. Consider W = {x A τ k x / A for all k } = A τ k A k= W = {x A τ k x / A for all k } = A τ k A k= Now W is a wandering set for if τ k W τ l W, for some k < l, then W τ l k W, so that there is an x W A such that τ l k x W A, which is a contradiction since l k is positive. Similarly we can prove that W is a wandering set. The set τ k (W W ) = N, (say) thus belongs to W τ. We show that if x A N, then τ n x A for infinitely many positive n and also for infinitely many negative n. Indeed if τ n x does not return to A for infinitely many positive n, then there is a largest positive integer m = m(x) such that τ m x A. Clearly then τ m x W, so that x τ m W N which is a contradiction. So τ n x returns to A for infinitely many positive n. Similarly τ n x returns to A for infinitely many negative n. This proves the lemma. Remark. It is to be noted that if x A N then τ k x in fact returns to A N for infinitely many positive and infinitely many negative n because N is invariant under τ and x / N. Remark 2. If A 0 is a set such that for each x A 0 the points τ k x return to A 0 for infinitely many positive k and also for infinitely many negative n, then it is easy to see that τ k A 0 = τ k A 0 k= If A B, and if we set A 0 = A N, where N is as in the Poincaré recurrence lemma then τ k A = ( τ k A 0 ) N = ( τ k A 0 ) N ( τ k A) N k= k=

12 Preliminaries and the Poincaré Lemma whence for any A B A = τ k A (mod W) k=.28. Suppose N B is a σ-ideal such that τn = τ N = N and W N. Then it is clear that given A B, there is an N N such that for each x A N, τ n x returns to A N for infinitely many positive n and for infinitely many negative n. Let us consider some concrete cases of such σ-ideals..29. Let m be a probability measure on B invariant under τ which means that m τ = m. In this case every wandering set W has measure zero for if m(w ) > 0 then the set τ n W ( union taken over all n Z) has measure m(τ n W ) =, since m(τ n W ) = m(w ) > 0. Since m is finite this gives a contradiction. If we write N = the σ-ideal of m-null sets in B, then we see that W N, τn = τ N = N. We have thus proved: Poincaré Recurrence Lemma (Measure theoretic version): If τ is a Borel automorphism on (X, B) which preserves a probability measure on B,and if A B is given, then for almost every x A the points τ n x return to A for infinitely many positive n and for infinitely many negative n. Example 2. We say that τ is dissipative with respect to a finite or σ-finite measure m on B if there exists a wandering set W in B such that m is supported on n= τ n W. On the other hand if m(w ) = 0 for every wandering set W, then τ is said to be conservative with respect to m. If τ is conservative with respect to m, then clearly W N = σ-ideal of m-null sets in B..32. We have thus proved the Poincaré Recurrence Lemma For Conservative Automorphisms: If τ is conservative with respect to m and if A B is given, then for almost every x A the points τ n x return to A for infinitely many positive n and also for infinitely many negative n.

13 2 Basic Ergodic Theory.33. Let X be a complete separable metric space and C the σ-algebra of subsets of X with the property of Baire. The class N of subsets of X of first category forms a σ-ideal in C. Let τ be a homeomorphism of X which does not admit a wandering non-empty open set. For such a τ every wandering set with the property of Baire is necessarily of first category. For if W in C is wandering and if W = U F, where U is open and F is of first category, then it is easy to see that U is wandering under τ, which is not possible unless U is empty. Thus every wandering set in C belongs to N, hence the σ-ideal generated by wandering sets in C is contained in N. We have proved:.34. Poincaré Recurrence Lemma (Category Version): If τ is a homeomorphism of a complete separable metric space which does not admit a wandering non-empty open set, then for every A X with the property of Baire (in particular for any Borel set A) there exists a first category set N (which is Borel if A is Borel) such that for each x A N, the points τ n x return to A N for infinitely many positive n and for infinitely many negative n. Exercise. Show that if τ is a homeomorphism of a metric space X which is conservative with respect to a σ-finite measure m on Borel subsets of X and if m assigns positive measure to every non-empty open set, then almost every x X is recurrent in the sense that τ n x returns to every neighbourhood of x for infinitely many positive and infinitely many negative n. Exercise 2. Show that for the irrational rotation τ (discussed in.9) both the measure theoretic and the category version of the Poincaré recurrence lemma apply. Show that the orbit of every point is dense in S. Further for any non-empty open arc C, every point of C is recurrent. Exercise 3. Call a homeomorphism τ of a Polish space X topologically transitive if there is point in X whose orbit is dense in X. Show that if X has no isolated points and τ is topologically transitive then there is no non-empty open set wandering under τ..35. For deeper investigations into the phenomenon of recurrence and its connection with combinatorial number theory, we refer the reader to H. Furstenberg[3].

14 Preliminaries and the Poincaré Lemma 3 Asides.36. A Borel automorphism is periodic if every point is periodic although the period may differ from point to point. We have discussed periodic automorphisms in some detail because in a very deep sense the periodic motions bear the same kind of relation to the totality of motions that repeating doubly infinite sequence of integers to 9 such as 2323 do to the totality of all such sequences. ( Birkhoff []). Approximation by periodic automorphisms is a powerful tool in ergodic theory and the pictures we have drawn above will reappear with an important change when we discuss induced automorphisms. Concerning the Poincaré recurrence lemma we express our appreciation of it by quoting Birkhoff and Koopman, and Oxtoby, who were among the originators and early contributors to mathematical ergodic theory: Introduction of the Modern Theory of Real Variables into Dynamics:- In his discussion of recurrent motion, H. Poincaré introduces the fundamental notion of a dynamical property which, without being true for all possible motions, has a probability of one of being realized. Poincaré wrote before Lebesgue s great work, but the very steps of his proof, as well as the formulation of his theorem, are all in almost an exact form for interpretation in terms of the theory of measure. Such an interpretation was accomplished by C. Carathéodory, who renders exceptions of probability zero as exceptions forming in the phase space a set of measure zero; and such is the first entrance into the realm of dynamics of the modern theory of real variables.( Birkhoff and Koopman [2].) In the course of his studies in celestial mechanics, Poincaré discovered a theorem which is remarkable both for its simplicity and for its far reaching consequences. It is noteworthy also for having initiated the modern study of measure-preserving transformations, known as ergodic theory. From our point of view, this recurrence theorem has a special interest, because in proving it Poincaré anticipated the notions of both measure and category. Publication of his treatise, Les méthodes nouvelles de la mécanique céleste antedated slightly the introduction of either notion... The category assertion has to be read between the lines of Poincaré s discussion. He began by showing that recurrent points of an open set are dense in the open set. His proof involved the construction of a nested sequence of regions; it may be interpreted as amounting to a

15 4 Basic Ergodic Theory proof of Baire s theorem for the case in hand. Since it is a trivial matter to show that the set of points recurrent with respect to an open set is a G δ set, the category assertion may properly be ascribed to Poincaré even though he makes no such explicit statement... The measure assertion of the recurrence theorem was formulated by Poincaré in terms of Probability. In this part of his proof he tacitly assumed the countable additivity of Probability, although this had not been properly justified at the time he was writing. However, when read against an adequate background of measure theory, his argument is perfectly sound. It was formulated in modern terms by Carathéodory(Über den Wiederkehrsatz von Poincaré,S.B.Preüss.Acad.Wiss.(99), )... Poisson had attempted to establish this kind of stability in the restricted problem of three bodies by an inconclusive argument based on the kind of terms that can appear in certain series expansions. Poincaré established the conclusion rigorously and by a revolutionary new kind of reasoning. (J. C. Oxtoby [5].) Finally it is worth mentioning that in his proof of the pointwise ergodic theorem Birkhoff appeals to the Poincaré recurrence lemma to define the function t(p ), the first return time of a point P into a set, and proves the ergodic theorem for this particular function. The same proof carries over for any integrable function as we shall see in the next chapter. REFERENCES [ ] G. D. Birkhoff. Recent Advances in Dynamics, Science, January 6, 920, vol. 5, no. 307, Birkhoff: Collected Mathematical Papers, vol. 2, 06 0, Dover, New York, 968. [2 ] G. D. Birkhoff and B. O. Koopman. Recent Contributions to Ergodic Theory. Proc. Nat. Acad. Sci. U. S. A., 8 (932), Birkhoff: Collected Mathematical Papers, vol. 2, , Dover, New York, 968. [3 ] H. Furstenberg. Recurrence in Ergodic Theory and Combinatorial Number Theory. Princeton University Press, 98. [4 ] K. R. Parthasarathy. Probability Measures on Metric Spaces. Academic Press, 967.

16 Preliminaries and the Poincaré Lemma 5 [5 ] J. C. Oxtoby. Measure and Category. Springer-Verlag, New York, 980. æ

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18 Chapter 2 Ergodic Theorems of Birkhoff and von Neumann 2.. The ergodic theorem of G. D. Birkhoff [] is an early and very basic result of ergodic theory. Simpler versions of this theorem will be discussed before giving two well known proofs of the measure theoretic case. A third purely set theoretic proof will be given in Chapter 0. Ergodic Theorem for Permutations 2.2. If X is a finite set then by permutation of X we mean a map σ : X X which is one-one and onto. We write σ σ to denote the composition of σ with itself. If σ is composed with itself n times then we denote the resulting permutation by σ n. If x X, then the elements x, σ x, σ 2 x,..., σ n x,... cannot all be distinct because then the collection (x, σ x, σ 2 x,..., σ n x,...) would be an infinite collection, and since X is a finite set, this cannot happen. Thus if x X, then for some l, k, σ k x = σ l x, i.e., σ k l x = σ l k x = x. So there is a first positive integer p such that σ p x = x. We call this p the period of x under σ. It depends on x and we may denote this dependence by writing p = p x Consider the following permutations on X = {, 2, 3, 4, 5}.. σ () = 2, σ (2) = 3, σ (3) = 4, σ (4) = 5, σ (5) =. 2. σ 2 () = 3, σ 2 (3) = 5, σ 2 (5) = 4, σ 2 (4) = 2, σ 2 (2) =. 3. σ 3 () = 2, σ 3 (2) = 3, σ 3 (3) =, σ 3 (4) = 5, σ 3 (5) = 4. In examples and 2 we see that σ 5 (x) = x for x {, 2, 3, 4, 5}, so all elements have period 5. In example 3, σ 3 () =, σ 3 (2) = 2, σ 3 (3) = 3, σ 2 (4) = 4, σ 2 (5) = 5. Thus in example 3, periods p = p 2 = p 3 = 3 whereas p 4 = p 5 = 2.

19 8 Basic Ergodic Theory 2.4. Let us return to the general case of a finite set X, which we assume has N elements. If σ is a permutation on X and x X, then the set {x, σx, σ 2 x,...} is the orbit of x under σ. Clearly the orbit of x will have p x elements {x, σ x,..., σ p x}, p = p x. Also the orbits of two elements in X are either identical or disjoint, so that the collection of orbits of points in X forms a partition of X A permutation σ on X is called irreducible if there is an x X such that p x = N = number of elements in X. This means that as k runs over integers, σ k x runs over all the elements of the set X. In our examples above, the permutations σ and σ 2 are irreducible but σ 3 is not an irreducible permutation. If σ is an irreducible permutation on X, then indeed p x = N for all x X. An irreducible permutation is also called a cycle or a cyclic permutation If σ is a permutation on X, not necessarily irreducible, and if we restrict σ to orb (x, σ) = orbit of x under σ, then this restriction of σ is an irreducible permutation since orb (x, σ) has p x elements and σ px x = x. We thus see that for a permutation σ of X, the collection of orbits of points in X forms a partition of X and the restriction of σ to any of these orbits is an irreducible permutation The ergodic theorem for an irreducible permutation σ on a finite set X = {x, x 2,..., x N } with N elements may be stated as follows: Theorem. If f is a real valued function on X, then lim n n n f (σ k x) = N (f(x ) + + f(x N )) i.e., the limit on the left hand side exists for each x X and can be identified with the value on the right hand side which is the average of the function on the set X. Proof. By the division algorithm we can write n = N l + r, 0 r < N. Since N is fixed and 0 r < N we see that l as n. Now σ is irreducible. Hence for any x X, σ N x = x, so that

20 Ergodic Theorems of Birkhoff and von Neumann 9 σ 2N x = x,..., σ ln x = x. Hence n f(σk x) = f(x) + + f(σ N x) +f(x) + + f(σ N x). +. +f(x) + + f(σ N x) +f(x) + f(σx) + + f(σ r x) l-times Therefore, N n f(σk x) = Nl+r l (f(x) + + f(σ N x) ) ( f(x) + + f(σ r x) ). + Nl+r Since r is bounded by N and l as n, we see that n lim f(σ k x) = n n N ( f(x) + + f(σ N x) ). As σ is irreducible, x, σx,..., σ N x form the set {x,..., x N }. Thus the limit is equal to N (f(x ) + + f(x N )). This proves the ergodic theorem for irreducible permutations. We can borrow the language of classical statistical mechanics and agree to call the sum N N k= f(x k) the space average and the limit of the averages n n f(σn x) the time average. The above theorem states that the time average of a function over iterates of an irreducible permutation equals the space average If σ is not an irreducible permutation, then we know that the restriction of σ to any orbit is an irreducible permutation. If x X has period p = p x under σ, then σx x,..., σ p x x, σ p x = x and we see as before that n lim f(σ k x) = p x f(σ k x) n n p x If we write S(x) = constant on orbits. p x px f(σk x), then S(σx) = S(x), i.e., S is

21 20 Basic Ergodic Theory 2.9. We have thus proved: Ergodic Theorem (for permutations): Let σ be a permutation on a finite set X with N points and let f be a real valued function on X, then lim n n (f(x) + + f(σn x)) exists and is equal to p x (f(x) + + f(σ px x)) where p x is the period of x under σ. The limit function is constant on orbits of σ. Easy Generalisations 2.0. It is easy to remove the condition of finiteness on X if we impose some condition on σ or f. Indeed the above proof shows that n f(σk x) exists whenever f(σ k x) as a function of k is lim n n periodic, and if p be its period, i.e., if f(σ p x) = f(x) and p is the smallest positive such integer, then the limit in question is equal to p (f(x) + + f(σp x)). If σ is a one-one onto self map of a set X (not necessarily finite) such that for each x X, σ p x = x for some smallest positive integer p = p x (depending on x) then f(σ k x) as a function of k is periodic with period p x so that exists and is equal to n lim f(σ k x) n n p x p x f(σ k x) 2.. Let us now consider the situation where X is an infinite set and σ : X X is one-one onto map such that σ k x x for all k 0 and for all x. Such maps are called free. Here are some examples of such maps :. X = Z = all integers, σx = x X = R = all real numbers, σx = x X = [0, ), σx = (x + 2)(mod ).

22 Ergodic Theorems of Birkhoff and von Neumann 2 Examples and 2 are easily seen to be free. A proof is needed for example 3 and it depends on the irrationality of 2. In example 3 if σ k x = x(mod ) for some x and some k 0, then it means that x + k 2 = x + n for some integer n, i.e., 2 = n k a rational number, which is a contradiction. Thus in example 3 also σ k x x for any k 0 and any x Consider the following example. Take X = Z and σx = x +. Write f(x) = ( ) x x. We see that f(0) + f() + + f(n ) = ( ) n (n ). That is, f(0) + f() + + f(n ) = n { m if n = 2m m + 2m if n = 2m + so that lim n n f(σk 0) does not exist. (It is 2 if n over even integers and + 2 if n over odd integers). This shows that n lim n n f(σk x) need not exist without some additional restriction on f when σ is a free map of X. Almost Periodic Functions 2.3. Consider a real valued function f on Z. There is a notion of almost periodicity of f which is a kind of generalization the notion of periodic function. Almost periodicity of f, roughly speaking, means that if we fix l and look at f(k + l) for various values of k, then f(k + l) comes arbitrarily close to f(l) with a certain degree of regularity. One such well known definition of almost periodicity is due to H. Bohr [6] which we give below: Let f be a real valued function on Z and ε > 0. A positive integer τ = τ ε is called a translation number for f pertaining to ε if for all k Z, f(k + τ) f(k) < ε. Such a number τ is also called ε-period of f. A function on Z may or may not admit an ε-period Definition. A real valued function f on Z is said to be uniformly almost periodic if for every ε > 0 there exists a positive

23 22 Basic Ergodic Theory integer L(ε) such that every set of L(ε) consecutive integers contains a translation number for f pertaining to ε. We will show below ( ) that if f is uniformly almost periodic n then lim n n f(k) exists and is finite. The limit is called the Bohr mean of f A uniformly almost periodic function f is bounded. We see this as follows: Take ε = and let L be a positive integer such that every set of L consecutive integers contains an ε-period. Let M = max { f( L),..., f(0),..., f(l) }. Let x Z. For some n, nl x < (n+)l. The L consecutive integers between nl and (n + )L contains an ε-period τ ε since f is uniformly almost periodic. This implies that f(x τ ε ) f(x) ε =. Now L x τ ε L so that f(x τ ε ) M whence f(x) M +. Since x is any point in Z, f is bounded Lemma. Let f be uniformly almost periodic on Z and ε > 0 be given. Let u be an ε-period larger than L(ε), where L(ε) is as in the definition of uniformly almost periodic function. Then for all n, u (n+)u k=nu f(k) u f(k) < 2ε. u Proof. Since u > L(ε) we can choose an ε-period v of f between nu and (n + )u. Now, (n+)u k=nu = u f(k) v k=nu f(k) (n+)u (n+)u v f(k) + f(k) f(k) k=v u k=(n+)u v f(k). The difference between the middle two terms can be estimated as follows: (n+)u k=v (n+)u v f(k) (n+)u v f(k) = (f(k + v) f(k))

24 Ergodic Theorems of Birkhoff and von Neumann 23 (n+)u v f(k + v) f(k) ε ((n + )u v) (2.) where we have made use of the fact that v is an ε-period. The difference between the first and the last term can be estimated as follows: v u f(k) f(k) k=nu = = = v k=nu v k=nu v k=nu k=(n+)u v f(k) f(k) u k=(n+)u v u+v k=(n+)u (f(k) f(k + v)) f(k) (f(k) f(k + u)) u k=(n+)u v u k=(n+)u v Now both u and v are ε-periods so that v k=nu f(k) u k=(n+)u v u k=(n+)u v (f(k) f(k + v)) (f(k) f(k + v)) f(k + v) f(k) 2ε(v nu). (2.2) From ( ) and ( 2) and the fact that v nu u, we see that (n+)u k=nu so that u f(k) u This proves the lemma. f(k) ε ((n + )u v + v nu) + ε(v nu) (n+)u k=nu 2εu. f(k) u f(k) 2ε. u Now, 2ε u (n+)u k=nu f(k) u f(k) 2ε. u

25 24 Basic Ergodic Theory Summing over n from 0 to (m ) and dividing by m we get Hence, Thus we have, 2ε mu m n=0 (n+)u k=nu 2ε mu f(k) mu u 2ε lim sup m mu mu f(k) u f(k) 2ε. u u f(k) 2ε. f(k) u f(k) 2ε. u When this is combined with a similar inequality for lim inf, we have 4ε lim sup m mu mu f(k) lim inf m mu mu f(k) 4ε. (2.3) Now if M = mu + r, 0 r < u, then m as M. Further, f(k) is made up of at most u bounded terms. Hence, M k=mu lim sup M Similarly, M M f(k) = lim sup m lim inf M Thus, from ( 3) we get 4ε lim sup M = lim sup m = lim sup m M M M M mu+r mu + r f(k) mu m(u + r/m) ( f(k) + mu mu f(k) = lim inf m f(k) lim inf M f(k). mu mu M M M k=mu f(k)) f(k). (2.4) f(k) 4ε.

26 Ergodic Theorems of Birkhoff and von Neumann 25 Since ε is arbitrary we see that the lim sup and lim inf agree We have thus proved the: Theorem If f is uniformly almost periodic on Z then f is bounded and the averages M M f(k) converge as M. More generally, the limit as M of the averages M M f(x + k) exists and is independent of x Corollary. If σ is a one-one onto self map of a set X and f is a real valued function on X such that for all x X the function k f(σ k n x) is uniformly almost periodic, then lim n n f(σk x) exists for all x and if S(x) be this limit then S(σ k x) = S(x), i.e., S is constant on orbits of σ If X = [0, ), σ(x) = x + 2(mod ) and if f be a continuous function on X with f(0) = f() then it can be shown that the function k f(σ k x), k Z, is uniformly almost periodic so that the limit of the averages n n f(σk x) exists for each x X Exercise. For the map σ : x x + 2(mod ) on [0, ), show that given ε > 0 and x [0, ] there is an integer L such that any set of L consecutive points in the orbit {σ k x} k Z forms an ε-net in [0, ]. Use this together with the uniform continuity of f to prove the uniform almost periodicity of the map k f(σ k x), k Z. (A subset of [0, ] is an ε-net in [0, ] if every interval of length ε contains a point from the set). For more on the elementary theory of almost periodic functions the reader is referred to N. Wiener s book [2] from which the account given above is adapted. Birkhoff s Ergodic Theorem 2.2. The averages n n f(σk x) are called ergodic averages. We have seen above some simple cases where such averages converge as n. Let us now prove Birkhoff s ergodic theorem in its usual setting

27 26 Basic Ergodic Theory of measure preserving transformations. To this end let (X, B, m) be a measure space and σ : X X a measurable transformation. Such a transformation is said to be measure preserving if m(σ (A)) = m(a) for all A B. We do not require σ to be one-one. However if σ is one-one and onto and if σ is measurable then we say that σ is invertible Example. Take X = S = unit circle = {e iϑ : 0 ϑ < 2π} equipped with its Borel σ-algebra and Lebesgue measure. Define σ(z) = z 2, z X. Then σ is measure preserving but not invertible. Example 2. Take X = R, the real line equipped with its Borel σ- algebra and Lebesgue measure. Then the map σ(x) = x + is invertible and measure preserving. Example 3. Let X be a finite or infinite set equipped with the power set of X as the σ-algebra. The measure of a set A X defined to be the number of points in A. Then any σ : X X, one-one and onto, is an invertible measure preserving transformation The ergodic theorem of G. D. Birkhoff is as follows : Theorem. Let σ be a measure preserving transformation of the measure space (X, B, m) and let f L (X, B, m). Then the ergodic averages n f(σk x) converge for almost every x to a limit function f (x) n which is again in L (X, B, m). The function f is constant on orbits, i.e., f (σ k x) = f (x) a.e. In case m(x) < we also have X fdm = X f dm To prove this theorem we need a lemma. Put n s n (x) = f(σ k x), s 0 (x) = 0 for all x. Birkhoff s ergodic theorem is equivalent to proving that n s n(x) converges a.e. to a limit function f (x) The lemma we need, known as the Maximal Ergodic Theorem, is as follows:

28 Ergodic Theorems of Birkhoff and von Neumann 27 Lemma. Let A = {x : sup n 0 s n (x) > 0}. Then f(x)dm 0. A Proof. Therefore, where We have max s k(σx) = max{0, f(σx),..., f(σx) + + f(σ n x)} 0 k n f(x) = = max{f(x), f(x) + f(σx),..., f(x) + f(σx) + + f(σ n x)} f(x). max s k(x) max s k(σx) = φ n+(x) φ n (σx) k n+ 0 k n φ n(x) = max{s (x),..., s n (x)}, φ n (x) = max{0, s (x),..., s n (x)}. Suppose A n = {x X : φ n (x) > 0}. Then, since φ n+(x) φ n(x), f(x)dm A n φ n(x)dm A n φ n (σx)dm. A n Now for x A n, φ n(x) = φ n (x) and for x / A n, φ n (x) = 0. Hence φ n(x)dm = φ n (x)dm = φ n (x)dm = φ n (σx)dm A n A n X X in view of the measure preserving nature of σ. Thus f(x)dm φ n (σx)dm φ n (σx)dm A n X A n = φ n (σx)dm X A n 0 since φ n 0. Now A n A n+ and A = n= A n. Hence f(x)dm = lim f(x)dm 0, A n A n and the lemma is proved. This proof of the maximal ergodic theorem is due to A.M. Garsia: A simple proof of Eberhard Hopf s maximal ergodic theorem, J. Math. Mech 4, No. 3, (965). (See Ergodic Theory, I. P. Cornfeld,

29 28 Basic Ergodic Theory S. V. Fomin, Y. I. Sinai from which this presentation (2.25) and what follows (2.26) is borrowed.) Proof of Birkhoff s Ergodic Theorem: For any two rational numbers a, b, a < b, write E a,b = {x X : lim inf n n s n(x) < a < b < lim sup n n s n(x)}. Obviously E a,b B and is σ-invariant. To prove the existence of lim n n s n(x) a.e. it suffices to show that m(e a,b ) = 0 for all a, b. Fix a, b and put E = E a,b. Consider the function { f(x) b for x E g(x) = 0 for x / E. Applying the maximal ergodic theorem to this function we get g(x)dm 0 (2.5) A(g) where A(g) = {x X : sup n n s n(x, g) > 0} = {x X : sup n n s n(x, f) > b}. Clearly A(g) E. Since E is invariant and g vanishes outside E we see that s n (x, g) = 0 for x X E, i.e., A(g) E. Therefore A(g) = E, and we can write ( 5) in the form f(x)dm bm(e) (2.6) In a similar way consider the function { g a f(x) for x E (x) = 0 if x / E. Then E A(g ) = {x X : sup n n s n(x, g ) > 0} = {x X : inf n n s n(x, f) < a}

30 Ergodic Theorems of Birkhoff and von Neumann 29 Again A(g ) = E for reasons similar to above and f(x)dm am(e). (2.7) E From ( 6) and ( 7) we see that m(e) = 0 since a < b. Thus the ergodic averages converge a.e. to a limit function which we denote by f. Now, n f(σ k x) dm n f(σ k x) dm = f(x) dm n n X By Fatou s lemma we see that f dm lim inf n f(σ k x) dm n n X X X X X f(x) dm so that f is in L (X, B, m). Now assume that m(x) <. If f is a bounded function then the functions n s n(x, f) are also bounded by the same constant which bounds f. By the bounded convergence theorem we have X f dm = lim n X n s n(x, f)dm = lim n = X X fdm. n f(σ k x)dm n If f is not bounded but merely in L (X, B, m), then we can find a sequence (f k ) k= of bounded functions converging to f in L norm. Now f n s n(x, f) f f k + f k n s n(x, f k ) + n s n(x, f k ) n s n(x, f) (f f k ) + f k n s n(x, f k ) + n f k f n. It follows from this that n s n(x, f) f in L -norm so that n s n(x, f)dm f dm. X X

31 30 Basic Ergodic Theory But X n s n(x, f)dm = X fdm. Hence X f dm = X fdm Remark. Suppose there is no measurable invariant subset of X with positive finite measure, then the function f (x) = 0 a.e. To see this note that for any ε > 0, the set {x : f (x) > ε} is invariant since f is an invariant function. If it were of infinite measure, then X f (x) dm would be infinite which contradicts the fact that f L (X, B, m). Thus for every ε > 0, the set {x : f (x) > ε} is of measure zero, since there are no invariant measurable sets of positive finite measure. Thus f (x) = 0 a.e Our second proof of the ergodic theorem is for the case of an invertible measure preserving transformation on a finite measure space. Except for some minor modifications it is the same as the original 93 proof due to Birkhoff [] for so-called strongly transitive systems. Apart from the assumption of strong transitivity (which is same as ergodicity in current terminology) Birkhoff proves the theorem for a particular function, namely, the return time of points to a set. But his proof is general and one can dispense with both these assumptions Theorem. Let σ be a measure preserving automorphism on a finite measure space (X, B, m). Let f be a bounded non-negative measurable function on X. Then the ergodic averages of f converge to a function f which is constant on orbits and has the same integral as f. Proof. Write s n (x) = n f(σk x), s 0 (x) = 0. Let f(x) = lim inf n n s n(x), and f(x) = lim sup n n s n(x). Then f f and both functions are constant on orbits. If we can show that X f X f f, then it will follow that f = f a.e. so that X the ergodic averages converge to a function f which is same as f or f. Also, it will follow that f is constant on orbits and has the same integral as f. We show that f f. X Fix ε > 0. From the definition of lim sup we know that for infinitely many k, k s k(x) > f(x) ε, i.e., for infinitely many k, s k (x) > k(f(x) ε). We X

32 Ergodic Theorems of Birkhoff and von Neumann 3 define, as Birkhoff does, the sets U = {x s (x) > f(x) ε} U 2 = {x s 2 (x) > 2(f(x) ε)} U.. U k = {x : s k (x) > k(f(x) ε)} U U 2 U k.. The sets U k are pair-wise disjoint and k= U k = X. If x U k, then k l f(σ j x) > k(f(x) ε), and f(σ j x) l(f(x) ε), l k j=0 whence by subtraction k f(σ j x) = j=l k l j=0 j=0 f(σ j σ l x) > (k l)(f(x) ε). We infer that, if x is a point of U k then σ l x, for l < k, is a point of one of the sets U k l, U k l,..., U, hence not in U k. Thus, the sets U k, σu k, σ 2 U k,..., σ k U k are pairwise disjoint (for if not for some i and j, i < j < k, σ i U k σ j U k, i.e., U k σ j i U k, which is a contradiction as 0 < j i < k). Write V k = U k, V k = U k (V k σv k... σ k V k ) then V k, σv k,..., σ k 2 V k are pairwise disjoint since V k is a subset of U k. Moreover each of σ i V k, i k 2 is disjoint from σ j V k, j k. Generally, for 0 l k we define l V k l = U k l (V k j σv k j σ k j V k j ) j=0 for 0 l k. We note as before that σ i V k l σ j V k p = for 0 i k l, 0 j k p unless i = j and k l = k p. We

33 32 Basic Ergodic Theory have k U j = j= = k (V j σv j σ j V j ) j= k j= ( j i=0 σi V j ) where each union is a pairwise disjoint union. What we have done above is the usual disjointification of sets (in the given order) U k, σu k,..., σ k U k, U k,..., σ k 2 U k,..., U k l,..., σ k l U k,..., U The identity k U j = j= k j= ( j i=0 σi V j ) may be written in the form W k = k j= X j, where W k = k j= U j and X j = j i=0 σi V j. Now, k f = f W k X j = j= k j= k j= V j s j V j j(f ε) where the last step is valid because V j U j on which s j is bigger than j(f ε). Now f(σx) = f(x) and the measure m is invariant under σ, whence j j(f ε) = (f ε) V j i=0 σ i V j = (f ε). X j

34 Ergodic Theorems of Birkhoff and von Neumann 33 Thus W k f = k j= X j (f ε) W k f εm(w k ) Let k. We have, (since W k increases to X) f f εm(x). X X Since ε is arbitrary, we see that X f f. We can carry out similar X analysis by defining sets Ũ i = {x : s i (x) < i(f(x) + ε)} Ũ Ũi and show that X f f. Thus we see that X f f f and the proof is complete. X X Remark. If f is a non-negative measurable function, not necessarily bounded, then we can define f n (x) = f(x) if f(x) n, f n (x) = 0 otherwise. Then f n (x) f(x) and f n (x) f(x) in monotonically increasing fashion, whence by the monotone convergence theorem f = lim f n lim f n = f lim f X n X n X X n n = f X X Thus f = f and the ergodic averages of f converge when f is nonnegative and measurable, not necessarily bounded. Finally decomposing any integrable function into its positive and negative parts we can apply the above consideration to each part. Thus we have proved: 2.3. Theorem. If f is an integrable function on a finite measure space (X, B, m) acted upon by a measure preserving automorphism σ, then the ergodic averages of f with respect to σ converge a.e. to an integrable function f which is constant on orbits and has the same integral as f. X

35 34 Basic Ergodic Theory von Neumann [9] has described Birkhoff s proof of the ergodic theorem as depending on an extremely astute method in the domain of set theory, which is indeed the case since our proof crucially depends on the analysis leading up to the identity k k U j = ( j i= σi V j ). j= j= It should be remarked that Birkhoff s original set analysis is slightly more intricate, but then it carries over to the case when σ is not invertible although Birkhoff deals in a situation where σ is invertible. von Neumann Ergodic Theorem Let us return briefly to the case when X is a finite set with N elements and σ : X X is a permutation. If f is a real valued function on X, we know that lim n n (f(x) + + f(σn x)) = (f(x) + + f(σ px x)) p x where p x is the period of x under σ. Now let X,..., X m be the members of the partition of X into orbits of σ. Suppose we wish to minimise the quantity x X (f(x) ξ(x))2 over all functions ξ on X which are constant on orbits of σ. If ξ takes values c on X, c 2 on X 2..., c m on X m, then we wish to minimise m φ(c,..., c m ) = ( (f(x) c i ) 2 ). i= x X i If we solve φ c i = 0 for c i we immediately get c i = p x x X i f(x) where p i = number of elements in X i. Thus the minimum over ξ of x X (f(x) ξ(x))2 is attained for the function ξ which has the constant value p i x X i f(x) on X i, i.e., ξ is same as the limit of the ergodic averages of f. That this situation prevails in the more general setting of measure spaces and square integrable functions on them is one of the conclusions of the mean ergodic theorem of von Neumann [9] which we prove next von Neumann proved his ergodic theorem in the context of unitary groups arising from Hamiltonian dynamical systems, but the

36 Ergodic Theorems of Birkhoff and von Neumann 35 result is entirely a Hilbert space proposition. Let therefore H be a Hilbert space on the field of complex numbers and let (f, g) denote the inner product of two elements f and g in H. Let U be an isometry on H, i.e., U is a bounded linear operator on H such that U U = I, U = adjoint of U We will need the following Lemma. {f : Uf = f} = {f : U f = f}. Proof. If Uf = f, then applying U on both sides we get U Uf = U f, i.e., f = U f since U U = I. Thus, {f : Uf = f} {f : U f = f}. Now suppose f is such that U f = f. Then Uf f 2 = (Uf f, Uf f) = (Uf, Uf) (Uf, f) (f, Uf) + (f, f) = (U Uf, f) (f, U f) (U f, f) + (f, f) = (f, f) (f, f) (f, f) + (f, f) = 0. Thus Uf = f so that {f : U f = f} {f : Uf = f}. The lemma is proved. The subspace {f : Uf = f} is the space of elements invariant under U. Let us denote it by K von Neumann Ergodic Theorem. n n U n f P K f as n where P K denotes the orthogonal projection on the subspace K of elements of H invariant under the isometry U. Proof. The elements of the form g Ug form a linear subspace which need not be closed. Let L denote the closure of this subspace. We first observe that H = K L. For if k K and g Ug L, then (k, g Ug) = (k, g) (k, Ug) = (k, g) (U k, g) = (k, g) (k, g) = 0. The subspace L being closure of the set of elements of the form g Ug, we see that K and L are mutually orthogonal. Now suppose f is orthogonal to L. Then (f, g Ug) = 0 for all g H. This gives us (f, g) = (U f, g) for all g H, whence U f = f, i.e., f K. Thus K is the orthogonal complement of L in H and H = K L. Now let f H and suppose