Lecture 1 Real and Complex Numbers

Lecture 1 Real and Complex Numbers Exercise 1.1. Show that a bounded monotonic increasing sequence of real numbers converges (to its least upper bound). Solution. (This was indicated in class) Let (a n ) be a bounded monotonic sequence without loss of generality, say it is monotonic increasing and bounded above by M. Let a = sup{a n }. I claim a n a. Let ɛ > 0 be given. Then there is N such that a N > a ɛ; otherwise, a ɛ would be an upper bound for {a n } smaller than a. Because (a n ) is increasing we have a ɛ < a N a n a for all n N; so a n a by definition. Exercise 1.2. Show that any sequence of real numbers either has an increasing subsequence or else a decreasing subsequence. Deduce the Bolzano-Weierstrass theorem?? (using the previous exercise). Deduce Cauchy s version of completeness from Bolzano-Weierstrass. Solution. Let (a n ) be a sequence of real numbers. Suppose that (a n ) has no maximal element (meaning an element of the sequence that is greater than or equal to all other elements of the sequence). We construct an increasing subsequence of (a n ) by induction; take n 0 = 1 and inductively define n k such that a nk is greater than a 1,..., a nk 1. We assume then that (a n ) has no increasing subsequence, and we shall show that instead it has a (perhaps not strictly) decreasing subsequence. Note that (a n ), and every subsequence, has a maximal element. Choose n 1 such that a n1 a n for all n (that is, a n1 is maximal), and then inductively choose n j such that a nj a n for all n > n j 1. The a nj then form a decreasing subsequence. The previous exercise now shows that any bounded sequence contains a convergent subsequence (Bolzano-Weierstrass). In particular, any Cauchy sequence contains a convergent subsequence. But, if a Cauchy sequence (a n ) contains a subsequence converging to a, the whole Cauchy sequence converges to a. To prove this formally, let ɛ > 0. There is an N such that a n a m < ɛ for n, m > N. In this inequality let n through the subsequence n k. We find that a a m ɛ for m > N, which proves convergence. Exercise 1.3. (Cantor s nested interval theorem) Let [a n, b n ], n = 1, 2,... be a sequence of closed intervals in R that are nested in the sense that [a n+1, b n+1 ] [a n, b n ] and whose lengths b n a n tend to zero. Show that the intersection n [a n, b n ] contains one and only one real number c. Solution. The (a n ) are increasing and bounded above by b 1, so converge (say to a). Similarly the (b n ) converge, say to b. Since a n b n 0, a = b = c, say. We have a n c b n for all n, so c belongs to the intersection of all the intervals. Any real number less than c is also less than some a n, so does not belong to the intersection; any real number greater than c is also greater than some b n, so does not belong to the intersection. Thus n [a n, b n ] = {c}. 1

Exercise 1.4. Use the previous exercise to show that the real numbers are uncountable, i.e. cannot be listed as c 1, c 2,.... Hint: Suppose such a listing is possible. Pick an interval [a 1, b 1 ] of length at most 1 that doesn t contain c 1, then a subinterval [a 2, b 2 ] of length at most 1 2 that doesn t contain c 2, and so on. Apply the nested interval theorem. What can you say about the limit point c? Solution. Follow the construction in the hint. Cantor s nested interval theorem produces a real c n [a n, b n ]. Thus c c n for each n, contradicting the supposed enumeration of the reals. Exercise 1.5. Let z and w be complex numbers of absolute value < 1. Show that the absolute value of z w 1 wz is also less than 1. (Hint: Expand the difference 1 wz 2 z w 2 in terms of complex conjugates, and factorize the resulting mess.) Solution. We have 1 wz 2 = (1 wz)(1 w z) = 1 wz w z + w 2 z 2 and z w 2 = (z w)( z w) = z 2 w z wz + w 2. Taking the difference 1 wz 2 z w 2 = 1 z 2 w 2 + z 2 w 2 = (1 z 2 )(1 w 2 ) > 0 and so z w < 1 wz as required. 2

Lecture 2 Metric spaces Exercise 2.1. Show that a function f as above is continuous at x if and only if, whenever (x n ) is a sequence converging to x, the sequence (f(x n )) converges to f(x). Solution. Suppose that f is continuous at x and let ɛ > 0 be given. By definition of continuity there is δ > 0 such that d(x, x ) < δ implies sd(f(x), f(x )) < ɛ. By definition of convergence there is N such that for all n > N, d(x, x n ) < δ. Thus d(f(x), f(x n )) < ɛ and f(x n ) converges to f(x). Suppose that f is not continuous at x. Then there is ɛ > 0 such that for all δ > 0 there is x with d(x, x ) < δ and d(f(x), f(x ) ɛ. Let x n be a value of x corresponding to δ = 1/n. Then x n x, but d(f(x), f(x n )) ɛ for all n, hence f(x n ) does not converge to f(x). Exercise 2.2. Show that the sequence of functions on [0, 1] defined by f n (t) = nte nt does not converge in the metric space C[0, 1], even though, for all t [0, 1], f n (t) 0 as n. (One says that (f n ) converges pointwise, but not uniformly.) Solution. By elementary calculus the function xe x has a maximum value at x = 1, and this maximum value is 1/e. Thus f n (x) has the same maximum value, 1/e, at x = 1/n. It follows that d(f n, 0) = 1/e for all n, so f n does not converge uniformly to the zero function. However, f n (0) = 0 for all n, and for t > 0, f n (t) = nte nt < nt/ 1 2 n2 t 2 0 as n ; so f n converges pointwise to zero. The figure shows graphs of f 1,..., f 6. 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0.5 1 1.5 2 3

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Lecture 3 Closed sets Exercise 3.1. Show that a A is isolated if and only if there is ɛ > 0 such that B(a; ɛ) A = {a}. Solution. If B(a; ɛ) A = {a} then in any sequence of distinct points of A, all but at most one member must be at distance ɛ from a. Thus no such sequence can converge to A. Conversely, if every ball around a meets A in a point other than a we can construct a sequence a n of distinct points of A tending to a, by the same inductive argument as in the proof of Proposition 3.2 Exercise 3.2. Show that A is the collection of all limits of convergent sequences in A (whether or not the members of the sequence are distinct). Solution. Any sequence of points of A either has a subsequence consisting of distinct points, or else has a constant subsequence. If the original sequence converges any subsequence converges to the same limit. Thus the limit is either a limit point of A (in the first case) or a member of A (in the second case). In either instance it belongs to A. Exercise 3.3. A closed ball in a metric space is a set B(x; r) := {x X : d(x, x ) r}. Show that a closed ball is closed. Must every closed ball be the closure of the open ball of the same center and radius? Solution. If a n is a sequence in X converging to a, and d(a n, x) r for all r, then d(a, x) r also. To see this, let ɛ > 0; there is n such that d(a n, a) < ɛ and therefore d(x, a) < r + ɛ by the triangle inequality; this is true for all ɛ so d(x, a) r. This shows that a closed ball is indeed a closed set. The two point space {0, 1} R has B(0; 1) = {0} which is closed, whereas B(0; 1) = {0, 1}. Thus the closure of an open ball need not be the corresponding closed ball. Exercise 3.4. Show that the intersection of any (finite or infinite) collection of closed sets is closed. Solution. Let F = F α where each F α is closed. A limit point of F is a limit point of each F α, so belongs to each F α, so belongs to their intersection which is F. Thus F contains all its limit points. Exercise 3.5. Verify that if f : A B is onto, and B is uncountable, then A is uncountable. Solution. Suppose not. Let A be enumerated as {a 1, a 2,...}. Then B = {f(a 1 ), f(a 2 ),...}. Enumerate B as follows: b 1 = f(a 1 ), and inductively b n is f(a m ) where m is the least integer such that b 1,..., b n 1 / {f(a 1 ),..., f(a m )} (this construction avoids possible repetitions in our enumeration of B). Thus B is countable, contradiction. 5

Exercise 3.6. Show that any open subset of R is a union of countably many disjoint open intervals (Lindelöf s theorem). (The example of the Cantor set shows that the closed analog of this statement is completely false.) Hint: First show the result without the countability restriction. To get that, note that each interval must contain a rational number. Solution. Let U R be open. Define a relation on U by a b iff (min{a, b}, max{a, b}) U. It is easy to check that this is an equivalence relation, so U is partitioned into disjoint equivalence classes. The equivalence class of a is a union of open intervals all of which contain a, so it is itself an open interval containing a. Thus U is a union of disjoint open intervals. To see that this is a countable union, note that each open interval must contain a rational number, and the rationals are countable. 6

Lecture 4 Compactness Exercise 4.1. Let f : X Y be continuous and surjective. Show that if X is compact then Y is compact also. What has this got to do with the familiar calculus principle that a continuous function on a closed bounded interval is itself bounded and attains its bounds? Solution. Let y n be a sequence in Y. Because f is surjective, each y n = f(x n ) for some x n X. The sequence (x n ) has a convergent subsequence x nk, because X is compact; the sequence y nk = f(x nk ) is then convergent too, because f is continuous. We have shown that an arbitrary sequence y n in Y has a convergent subsequence, so Y is compact. In the case of the calculus example, let f : [a, b] R be continuous. Since [a, b] is compact, so is its image f([a, b]). In particular the image is bounded (which is what we mean by saying that f is bounded), and closed. Since the image is closed it contains sup f([a, b]) and inf f([a, b]), which are accumulation points of f([a, b]); and this is what we mean by saying f attains its bounds. Exercise 4.2. (Connectedness) Show that R cannot be written as the disjoint union of two nonempty open sets. (Let the sets be U and V. Then {U, V } forms an open cover of each closed interval [a, b]. By considering a Lebesgue number for this cover show that [a, b] lies entirely within one of the two sets.) Solution. Let the sets be U and V. Then {U, V } forms an open cover of each compact interval [a, b]. Let δ > 0 be a Lebesgue number for this cover. Then two points of [a, b] which are separated by less than δ are either both in U or both in V. Since any two points of [a, b] can be joined by a finite chain of points whose separations are less than δ, the whole of [a, b] is either in U or in V. Since this is true for any interval [a, b], one of the sets U, V is the whole of R and the other is empty. Exercise 4.3. A map f : X Y between metric spaces is uniformly continuous if for each ɛ > 0 there is δ > 0 such that d(f(x), f(x )) < ɛ whenever d(x, x ) < δ. (The extra information beyond ordinary continuity is that δ does not depend on x.) Show that if X is compact, every continuous f is uniformly continuous. Hint: use Theorem??. Solution. Let ɛ > 0 be given. For each x X there is δ x such that d(f(x), f(x )) < ɛ/2 whenever d(x, x ) < δ x. The open sets U x = B(x; δ x ) form a cover for X. Let δ be a Lebesgue number for this cover. Now, if d(x, x ) < δ, then both x and x belong to the same U x, and therefore d(f(x ), f(x)) < ɛ/2, d(f(x ), f(x)) < ɛ/2. By the triangle inequality, then, d(f(x ), f(x )) < ɛ whenever d(x, x ) < δ. This is uniform continuity. 7

Lecture 5 Complete Spaces Exercise 5.1. Banach s fixed point theorem can be extended as follows: if f : X X is a map and some power f N = f f is a strict contraction, then f has a unique fixed point. Prove this. Solution. Let g = f N. Then g has a unique fixed point x, by Banach s theorem. Notice that g(f(x)) = f N+1 (x) = f(g(x)) = f(x). Thus f(x) is also a fixed point of g. By the uniqueness part of Banach s theorem, f(x) = x; that is, x is a fixed point of f. Exercise 5.2. Consider the complete metric space C R [0, 1] of continuous real-valued functions on [0, 1]. For a, b [0, 1] show that the set of functions f which are nondecreasing on the interval [a, b] is nowhere dense in C R [0, 1]. Using Baire s theorem, deduce that there exist functions f C R [0, 1] that are nowhere monotonic, that is, monotonic on no subinterval of [0, 1]. Solution. Let N[a, b] denote the collection of functions f which are nondecreasing on [a, b], which is to say that f(x) f(y) whenever a x y b. If f n is a sequence in N[a, b] which converges to a function f, then for all x, y as above, f(x) = lim f n (x) lim f n (y) = f(y) so f N[a, b] also. Thus N[a, b] is closed. To show that N[a, b] is nowhere dense it is therefore enough to show that given any f N[a, b] and any ɛ > 0, there is g C[0, 1] \ N[a, b] with d(g, f) < ɛ. Let c = (a+b)/2. Since f is continuous at c there is δ > 0 such that f(x) f(c) < ɛ/2 whenever x c < δ; we may assume without loss of generality that δ < (b a)/4. Let g(x) = f(x) ɛh((x c)/δ), where h(t) = max{0, 1 t }. Then d(f, g) = ɛ, and moreover g(c) = f(c) ɛ, g(c δ) = f(c δ) > f(c) ɛ/2, so g does not belong to N[a, b]. Note that N[a, b] is the space of functions that are nonincreasing on [a, b]. The space of functions that are monotonic on some subinterval of [0, 1] is N[a, b] ( N[a, b]) a,b where the union may be taken over all rational a, b [0, 1] with a < b. This is a countable union of nowhere dense sets, so it is of first category. By Baire s theorem applied to the complete metric space C[0, 1], its complement is dense. In particular the complement is nonempty; that is, there exist nowhere monotonic functions. 8

Lecture 6 Convergent Series Exercise 6.1. Let Ω be an open subset of C. Show that a sequence of functions converges uniformly on compact subsets of Ω if, for every point z Ω, there is a closed disk D(z; r z ) Ω on which the sequence converges uniformly. Solution. Since the closed disks D(z; r z ) are all compact (by the Bolzano-Weierstrass theorem), uniform convergence on compact sets certainly implies uniform convergence on any such disk. Conversely, suppose that K Ω is compact and that the condition of the exercise is satisfied. The open sets D(z; r z ), for z K, form a cover of K. This cover has a finite subcover, say D(z 1 ; r z1,..., D(z k ; r zk ). The sequence converges uniformly on each closed disk D(z j ; r zj ), and hence on their union, which contains K. Exercise 6.2. Cauchy s formula for the radius of convergence is 1/r = lim sup a n 1/n. n Prove this. You need to know the definition of the limit superior of a sequence of real numbers: it is lim sup b n := lim sup b n. n m n m Notice that the limit on the right is the limit of a monotonic sequence, so it always exists (possibly infinite). Solution. Make use of the following property of the limit superior: if b = lim sup b n, then for any ɛ > 0, only finitely many b n are greater than b + ɛ, and infinitely many b n are greater than b ɛ. Applied to Cauchy s formula, this says that for any R > r there are infinitely many n for which a n > 1/R n ; whereas for any ρ < r, the inequality a n < 1/ρ n is satisfied for all but finitely many n. Thus if z > R the series a n z n has infinitely many terms of absolute value greater than 1, so cannot converge. It follows that the radius of convergence is less than or equal to R. If z < ρ then all but finitely many terms of a n z n are dominated by the corresponding terms of the convergent geometric series z n /ρ n. It follows that the radius of convergence is greater than or equal to ρ. Since ρ, R are arbitrary subject to the condition ρ < r < R, we conclude that the radius of convergence is r. Exercise 6.3. Let a C be a constant. The binomial series is the series f a (z) = n=0 a [n] n! zn where a [n] := a(a 1) (a n + 1). Show that the radius of convergence of the binomial series is at least 1. 9

Solution. We use the ratio test. Let b n = a [n] /n! be the coefficients of the series. Then b n+1 /b n = (a n)/(n + 1) 1 as n. It follows that b n ρ n is bounded for any ρ < 1, which in turn tells us that the radius of convergence is at least ρ. Since ρ was arbitrary the radius of convergence is at least 1. Exercise 6.4. Let f a (z) denote the binomial series of the previous exercise. Show that f a+b (z) = f a (z)f b (z) for z < 1. Since f 1 (z) = 1 + z, this justifies the notation (1 + z) a for f a (z). Hint: Start by proving, by induction, the formula (a + b) [n] = n k=0 n! (n k)!k! a[k] b [n k]. Solution. Once we have proved the formula given in the hint, the result follows directly from Proposition 6.7 on multiplying series. To prove the formula by induction, assume it for n. Then (a + b) [n+1] = ( ) n a [k] b [n k] (a + b n). k k We can split the last term on the right to get ( ) n a [k] b [n k] ((b+n k)+(a k)) = k k k ( ) n a [k] b [n+1 k] + k k ( ) n a [k] b [n+1 k] k 1 where we have substituted k 1 for k in the second term. The binomial coefficient identity ( ) ( ) ( ) n + 1 n n = + k k k 1 now completes the induction. Exercise 6.5. Fill in the details of the above argument. (Show that for sufficiently large n, the validity of the inequality for n implies its validity for n + 1.) Solution. Begin with the following observation: if 1 < β < α = 1 + a then ( 1 α ) ( 1 + 1 β 1, for m large. m + 1 m) There are various ways to see this: direct estimation of the binomial series, L Hôpital s rule, use of logarithms.... Apply this to the formula b n+1 /b n = (a n)/(n+1) which is taken from an earlier exercise. Assume that n > a and that b n Kn β. Then b n+1 ( 1 α ) ( ) β n b n Kn β = K(n + 1) β n + 1 n + 1 for n sufficiently large. This gives the desired estimate by induction. 10

Lecture 7 More about power series 11

Lecture 8 The Weierstrass Approximation Theorem Exercise 8.1. In the complex case the Stone-Weierstrass theorem takes the following form: a -subalgebra of C(X) which contains the constants and separates points is dense, where the -condition means that the algebra is closed under (pointwise) complex conjugation. Prove this. Try to give an example to show that the complex theorem is not valid without the extra condition (we shall discuss this in detail later). Solution. Let A be a -subalgebra of C(X). If f = u + iv with u and v real, then u = 1 2 (f + f) and v = 1 2i (f f). Therefore, u and v belong to A. It now follows that the collection B of all real and imaginary parts of members of A is a subset of A, closed under addition, and multiplication and containing the (real-valued) constant functions; in other words it is a subalgebra of C R (X). Moreover, since A separates points, so does B (if f(x) f(y) then either u(x) u(y) or v(x) v(y)). Thus B is dense in C R (X) by the real form of the Stone-Weierstrass theorem, and it easily follows that A is dense in C(X). The desired example may be given by taking X to be the unit circle S 1 = {z C : z = 1}, and A the algebra of (complex) polynomial functions on X. This separates points and contains the constants but it is not a -algebra; and in fact it can be shown that the function z 1 = z is not in the closure of A. 12

Lecture 9 Compact Sets in C(X) Exercise 9.1. Prove the converse of the Arzela-Ascoli theorem: a compact subset of C(X) is closed, bounded and equicontinuous. Solution. All that needs to be proved is equicontinuity, since any compact set is closed and bounded. Let F be a compact subset of C(X) and let ɛ > 0 be given. Then, by compactness, F has a finite cover by ɛ/3-balls, say centered at f 1,..., f n. Moreover, for any x X there is δ > 0 such that, if d(x, x ) < δ, then d(f j (x), f j (x )) < ɛ/3 for each of the (finitely many) functions f j, j = 1,..., n. For any f F there is some f j such that d(f, f j ) < ɛ/3 and it follows that d(f(x), f(x )) d(f(x), f j (x)) + d(f j (x), f j (x )) + d(f j (x )), f(x )) < ɛ as required. 13

Lecture 10 Normed Vector Spaces Exercise 10.1. Let a and b be positive real numbers and let p > 1. Show that inf{t 1 p a p + (1 t) 1 p b p : t (0, 1)} = (a + b) p. Hence (or otherwise) show that the expression is also a norm on R n or C n. x p = ( x 1 p + + x n p ) 1/p Solution. To prove the first statement, one can use standard calculus methods to find the minimum of the function at t = a/(a + b). Alternatively, argue as follows: the function x x p, p > 1, is convex (the graph always lies below its chords) and therefore (a + b) p = [ t a t b + (1 t) 1 t ( a t t ] p ) p + (1 t) ( b 1 t ) p = t 1 p a p + (1 t) 1 p b p with equality at t = a/(a + b). Now to prove that p is a norm, the only nontrivial thing is the triangle inequality. Let x and y be vectors. We have for t (0, 1) x + y p p i ( x i + y i ) p i ( t 1 p x i p + (1 t) 1 p y i p) Taking the infimum over all t we find that x + y p p ( x p + y p ) p, t 1 p x p p + (1 t) 1 p y p p. and taking p th roots gives the result. (See L. Maligranda, Simple proof of the Holder and Minkowski inequalities, American Mathematical Monthly 102(1995), 256 259.) Exercise 10.2. Show that, with the above norm, the collection B(V, W ) of bounded linear maps from V to W is a normed vector space, and that it is a Banach space if W is a Banach space. 14

Solution. We prove the triangle inequality for the norm on B(V, W ) (the other axioms are easier). Let S, T B(V, W ). Then S + T = sup{ Sv + T v : v 1} sup{ Sv + T v : v 1} sup{ Sv : v 1} + sup{ T v : T 1} = S + T. As for completeness, let us re-examine our proof that C(X) is complete for X compact. Notice that this proof really shows that the space of bounded continuous functions from any X to a complete space Y is itself complete. But now, B(V, W ) is a closed subspace of the space of bounded functions from the unit ball of V to W, so it is complete if W is complete. Exercise 10.3. Show that the norm of linear maps is submultiplicative under composition, i.e. S T S T. Solution. We have which gives the result. ST v S T v S T v, 15

Lecture 11 Differentiation Exercise 11.1. Suppose that T : V W is a linear map. Show that T (h) = o( h ) if and only if T = 0. Solution. By definition of the norm, if T > 0 then there is a vector v V such that T v > 1 2 T v. By linearity, any nonzero multiple of v has this same property. Thus it cannot be the case that T (h) / h tends to zero as h 0. We conclude by contraposition that if T (h) = o( h ) then T = 0 and thus T = 0. Exercise 11.2. Show that the derivative of a linear map is always equal to the map itself. Solution. Let T : V W be continuous (thus, bounded) and linear. Then T (x + h) = T (x) + T h which, by definition, shows that the derivative of T at x is T itself. Exercise 11.3. (Open-ended) Formulate precisely the advanced calculus result that the mixed derivatives are symmetric, that is 2 f/ x y = 2 f/ y x, and prove it (under suitable hypotheses) for maps between normed vector spaces. Solution. Let f : V W be a twice-differentiable map from V to W, where V and W are normed vector spaces. The first derivative Df therefore takes values in the vector space B(V, W ), and the second derivative D 2 f takes values in B(V, B(V, W )), which can be regarded as the space of bilinear maps S : V V W which are bounded in the sense that S(v, v ) c v v for some constant c. The statement is that the second derivative, so regarded, is symmetric in the sense that S(v, v ) = S(v, v). Here is a sketch proof. Suppose that f is twice differentiable at 0, that f(0) = 0, and that Df(0) = 0 also. (The general case can easily be reduced to this.) I claim then that f(h + k) f(h) f(k) = D 2 f(0) h k + o(( h + k ) 2 ). The result follows from this, since one the one hand a version of exercise 1 shows that this determines the bilinear map D 2 f(0) uniquely, and on the other hand the left side is obviously symmetric in h and k. To prove the claim let us look at the expression g(k) = f(h + k) f(h) f(k) Df(h) k as a function of k. Clearly it vanishes when k = 0. Its derivative is Dg(k) = Df(h + k) Df(k) Df(h) 16

differentiating the first, third and fourth terms in the definition of g (the second term, considered as a function of k, is constant.) Now we have Df(p) = D 2 f(0) p + o( p ) by definition of the second derivative and the assumption that Df(0) = 0. Substituting these into the expression for Dg above we find that Dg(k) = o(( h + k )). From the mean value theorem, since g(0) = 0, we deduce that On the other hand we have g(k) k o(( h + k )) o(( h + k ) 2 ). Df(h) k D 2 f(0) h k = o(( h + k ) 2 ) from the definition of the second derivative. Putting that together with the previous inequality and the definition of g gives as asserted. f(h + k) f(h) f(k) = D 2 f(0) h k + o(( h + k ) 2 ) 17

Lecture 12 The inverse function theorem Exercise 12.1. Let V be a normed space and let W be the normed space B(V, V ). Show that the mapping i: T T 1 is differentiable (where defined) on W, and that its derivative is Di(T ) H = T 1 HT 1. Solution. This follows from the algebraic identity That is S 1 T 1 = S 1 (S T )T 1. i(t + H) i(t ) = i(t + H)Hi(T ) = i(t )Hi(T ) + o( H ). The o-estimate follows from the continuity of i near T, which we have already proved. 18

Lecture 13 Measure Spaces Exercise 13.1. Show that the σ-algebra of Borel subsets of R is generated by the intervals (a, ), with a Q. Solution. Let B be the σ-algebra generated by the intervals mentioned in the proposition. Clearly B contains all the intervals (a, b) with rational endpoints. But any open set in R is the union of all the intervals with rational endpoints contained in it. Since this union is necessarily countable, any open subset of R belongs to B; the result follows. Exercise 13.2. Let A n be a monotone decreasing sequence of measurable sets, A 1 A 2, and suppose that ν(a 1 ) is finite. Show that then ν ( A n ) = lim ν(a n ). Give an example to show that the finiteness hypothesis is necessary. Solution. Supposing that A 1 has finite measure, let B n = A 1 \ A n. Then the B n are a monotone increasing sequence of sets and B n = A 1 \ A, where A = A n. By previous results ν(a 1 \ A) = lim ν(b n ), that is ν(a 1 ) ν(a) = lim n (ν(a 1 ) ν(a n ) which gives the result. For a counterexample in the infinite measure case take ν to be the counting measure on N and A n = {n, n + 1,...}. Exercise 13.3. Let A n be a sequence of measurable subsets of X. The limit superior lim sup A n is the set of those x X that belong to infinitely many of the A n ; the limit inferior lim inf A n is the set of those x X that belong to all but finitely many of the A n. (a) Show that lim inf A n and lim sup A n are measurable. (b) Show that ν(lim inf A n ) lim inf ν(a n ). (c) Show that ν(lim sup A n ) lim sup ν(a n ) provided that A n has finite measure. (See exercise 6.2 for the definitions of lim inf and lim sup for sequences of real numbers.) Solution. We can write lim inf A n = k n k A n; the unions and intersections are countable, so lim inf A n is measurable. Similarly for lim sup A n = k n k A n. Let B k = n k A n. Then the B k increase monotonically to lim inf A n, and therefore ν(lim inf A n ) = lim ν(b k ). But B k A k so ν(b k ) ν(a k ) and therefore lim ν(b k ) lim inf ν(a k ). Similarly for the supremum. 19

Lecture 14 Integration Exercise 14.1. (Consistency) Verify that a nonnegative simple function is integrable, and that the two senses of integral agree for such functions. Solution. Since the integral is a positive linear functional on the space of simple functions, if g f with g and f simple, then gdν fdν, both integrals being taken in the sense of simple functions. On the other hand, f itself is a simple function f. It follows that sup{ gdν : g simple, g f} = fdν as asserted. Exercise 14.2. Verify that if f, g are nonnegative measurable functions with f g, then fdν gdν. Verify also that cfdν = c fdν for a positive constant c. Solution. If f g, then every simple function f is also g; this gives the first result. As for the second, g is a simple function f if and only if cg is a simple function cf. Exercise 14.3. Verify that the integral of the characteristic function of any measurable A X is equal to the measure of A. (With our definitions, you will need to assume that the measure is σ-finite.) Solution. If A has finite measure, its characteristic function is a simple function and the statement is obvious. If A has infinite measure, let X n be an increasing sequence of finite measure subsets whose union is X. Then A X n increases to A, so ν(a X n ). The characteristic functions of the A X n are nonnegative simple functions dominated by the characteristic function of A, and their integrals are ν(a X n ) ; so the integral of χ A is also. 20

Lecture 15 Integrable functions and the convergence theorems Exercise 15.1. Show that the integrable functions form a vector space and that the integral is a linear functional on this vector space. Solution. If f 1 = g 1 h 1 and f 2 = g 2 h 2 are integrable, so is f 1 + f 2 = (g 1 + g 2 ) (h 1 + h 2 ). Also, if c > 0 is a constant, cf 1 = cg 1 ch 1 is integrable; if c < 0, cf 1 = ( c)h 1 ( c)g 1 is still integrable. In each case it is easy to check that the integral behaves linearly. Exercise 15.2. Give a sensible definition of the integral for complex-valued functions. Solution. The only sensible definition is that if f = u + iv, with u, v real-valued, then f = u + i v. Exercise 15.3. Show that fdν f dν for any integrable function f. Solution. Write f = f + f, as in the proof of lemma 15.2, and notice then that the left side of the inequality is f + f and the right side is f + + f. The ordinary triangle inequality now gives the result. It is worth noticing that this inequality is also true for complex f. Suppose that f = u + iv with u,, v real. Neither side of the inequality is affected if we multiply f by a complex number w of absolute value 1, and by multiplying by a suitable such w we may arrange that fdν is real and positive, which is to say that udν 0 and vdν = 0. Under this hypothesis fdν = udν f dν since u f pointwise. Exercise 15.4. Let f n be a sequence of nonnegative integrable functions on a measure space. Show that if f n dν <, then f n (x) 0 for almost all x. n=1 Solution. Let g m = m n=1 f n be the partial sums. By the monotone convergence theorem g m (x) converges a.e. to g(x) (and g is an integrable function, but that does not matter here). Thus f n (x) = g n+1 (x) g n (x) converges a.e. to zero. Exercise 15.5. If a nonnegative integrable function f has fdν = 0, show that f(x) = 0 for almost all x. (Apply the MCT to the sequence {nf}.) 21

Solution. The hint is the solution, or equivalently apply the preceding exercise to n=1 f(x) (a sum of infinitely many copies of the same function). 22

Lecture 16 Functions and Spaces Exercise 16.1. Give counterexamples to other possible implications among notions of convergence. That is, give examples to show that neither convergence in measure, nor convergence a.e., necessarily imply convergence in L 1, and that convergence a.e. does not imply convergence in measure in case ν(x) =. Solution. Let X be the set of natural numbers equipped with the measure that assigns mass 2 k to the number k. The functions f n defined by { 2 k (k = n) f n (k) = 0 (k n) then tend to 0 in measure, and pointwise (everywhere), but not in L 1. Let Y be the set of natural numbers with the counting measure. Then the functions g n defined by { 1 (k = n) g n (k) = 0 (k n) tend to zero pointwise, but not in measure. Exercise 16.2. Show that if f n f in measure and f n g for some integrable g, then f n f in L 1. Solution. Since f n f in measure, any subsequence converges in measure to f, which means that it has a sub-subsequence converging almost everywhere. By the dominated convergence theorem, this sub-subsequence converges in L 1. Thus, every subsequence of (f n ) has a sub-subsequence converging in L 1. Now use the fact (true in any metric space X) that if x k is a sequence for which any subsequence has a subsubsequence converging to x, then x k itself converges to x. (Proof of fact: Suppose x k does not converge to x. Then there is some δ > 0 for which there exist arbitrarily large k with d(x k, x) > δ. The corresponding x k form a subsequence with no subsubsequence converging to x.) 23

Lecture 17 Constructing Measures I Exercise 17.1. Show that any set of outer measure 0 is measurable. Solution. Let A have outer measure zero. Then, for any B, B A A has outer measure zero by monotonicity. Thus µ (B) µ (B A) + µ (B \ A) = µ (B \ A) µ (B); we must have equality all through, so B is measurable. { Exercise 17.2. Let X be an uncountable set and for A X define µ 0 if A is countable (A) = 1 if A is uncountable. Show that µ is an outer measure. Which subsets of X are µ -measurable and what is their measure? Solution. It is easy to check that µ is an outer measure (use the fact that a countable union of countable sets is countable). By the previous exercise any countable set is measurable (with measure 0), and the complement of a countable set is also measurable (with measure 1). If both A and its complement are uncountable, then A is not measurable, since µ (A) + µ (X \ A) = 2 µ (X). Exercise 17.3. Let (X, B, ν) be a σ-finite measure space. Show that µ (A) = inf{ν(b) : A B, B B} defines an outer measure on X, and that measure produced from µ by Carathéodory s construction agrees with ν on B. Deduce that X can be extended to a complete measure space. Solution. Clearly µ is monotonic. It is also countably subadditive because if A n B n, B n B, then A = A n is a subset of B = B n, and ν(b) ν(b n ). If B B then µ (B) = ν(b) from the definition. We show that B is measurable. Let C be any set and let ɛ > 0. There exists B B such that C B and µ (C)+ɛ ν(b ). Now ν(b ) = ν(b B) + ν(b \ B); the first set on the right contains C B and the second contains C \ B, so we deduce Let ɛ 0 to obtain µ (C) + ɛ ν(b) µ (C B) + µ (C \ B). µ (C) µ (C B) + µ (C \ B). Together with subadditivity, this gives the result. 24

Lecture 18 Lebesgue Measure Exercise 18.1. Show that for a measurable subset A of R (having finite measure) and any ɛ > 0, there is a closed set F A with λ(a) < λ(f ) + ɛ. (The measurability requirement cannot be dropped this is tricky to see.) Solution. Since A is measurable and the sets A n = A [ n, n] form an increasing sequence whose union is A, there is some n such that λ(a n ) > λ(a) ɛ/2. Let B = [ (n + 1), (n + 1)] \ A n. By the previous proposition there is an open set U B such that λ(u) < λ(b) + ɛ/2. Then F = [ (n + 1), (n + 1) \ U is a closed set contained in A n, and λ(f ) (2n + 2) λ(u) > (2n + 2) λ(b) ɛ/2 > λ(a n ) ɛ/2 > λ(a) ɛ as required. Exercise 18.2. Show that a subset A [a, b] is Lebesgue measurable if and only if its outer measure λ (A) is equal to its inner measure λ (A) := (b a) λ ([a, b] \ A). (This connects the definition of Lebesgue measure to the classical exhaustion method of Eudoxus and Archimedes.) Solution. Suppose that A satisfies the given condition. By the previous proposition, there is a Borel set G with A G and λ (A) = λ(g). If we can prove that N = G\A has (outer) Lebesgue measure zero, we shall be finished. Put X = [a, b]. Write the definition of measurability for G, using X \ A as a test set. This gives λ (X \ A) = λ (N) + λ (X \ G). But λ (X \A) = λ(x) λ (A) because outer and inner measures agree for A; and the corresponding fact for G is true because G is measurable. So λ N) = λ(g) λ (A) = 0, as required. 25

Lecture 19 Lebesgue Integration Exercise 19.1. Let f(x) = sin x x. Use the alternating series test from calculus to show R that lim R f(x)dx exists (later, we shall see that the value of this limit is π/2). 0 Nevertheless, show that f is not Lebesgue integrable. Solution. Let a n = (n+1)π nπ sin x x dx. Then the signs of the a n are alternately positive and negative, and we also have a n < frm[o] /n. Thus, by the alternating series test, the series a n converges. Since the partial sums of this series are the integrals nπ (sin x)/x dx, it follows that these 0 integrals converge as n. Now for any R we can find an n (the greatest integer < R/π) such that R 0 sin x nπ x dx 0 sin x x dx < 1 R π and it follows that R (sin x)/x dx converges as R. 0 If the function (sin x)/x were Lebesgue integrable, however, then the series a n would be absolutely convergent. The estimate a n shows that this is not the case. 1 (n + 1)π (n+1)π nπ sin x dx = 2 (n + 1)π Exercise 19.2. ( Differentiating under the integral sign ) Suppose that the function F (x, y) is defined on R R, is integrable (as a function of x) for each fixed y, and is differentiable (as a function of y) for each fixed x; suppose also that its (partial) derivative F y satisfies F y (x, y) g(x) for some integrable function g and all y. Show that then the function G(y) = F (x, y)dx is differentiable and that G (y) = F y (x, y)dx. (Same method as Example??, together with the mean value theorem.) Exercise 19.3. (Normal numbers) Define functions f n on [0, 1] as follows: { 1 if 2 n x is even f n (x) = +1 if 2 n x is odd 26

Show that 1 0 f n (x)f m (x)dx = { 1 (m = n) 0 (m n) and deduce that if g n = (f 1 + + f n )/n, then g 2 n(x)dx = 1/n. Conclude using Exercise 15.4 that g k 2(x) 0 for almost all x as k, and hence that g n (x) 0 for almost all x as n. This conclusion is usually expressed by saying that almost all x [0, 1] are normal in base 2, i.e. have on average as many 0 s as 1 s in their binary expansion. Can you show that almost all x are normal in every base simultaneously? (No explicit example of such an x is known.) Solution. If n = m then f n f m = 1 everywhere and so its integral is equal to 1. If n > m then the integral of f n is zero over every subinterval [k2 m, (k + 1)2 m ] on which f m is constant; so f n f m = 0. Similarly if m > n. Now we can write gn 2 = 1 n n 2 f i f j. i,j=1j The terms with i j vanish and the n terms with i = j yield 1, so g 2 n = 1/n. Thus k (gk 2) 2 <, so g k 2(x) 0 for almost all x by the Borel-Cantelli lemma. To deduce that g n 0 almost everywhere let k be the greatest integer such that k 2 n. Then g n (x) g k 2(x) (1 k 2 n) g k 2(x) + 2k + 1 k 2. This tends to zero as n, and gk 2 (x) tends to zero (a.e.) as n, so we are done. A generalization of the same argument shows that almost all x are normal to any given base m. Since a countable union of null sets is null, it follows that almost all x are perfectly normal, i.e. normal in every base simultaneously. 27

Lecture 20 Set-Theoretic Questions 28

Lecture 21 The Riesz Representation Theorem Exercise 21.1. Show that Lebesgue outer measure is metric. Solution. Let A, B R be separated and let U, V be disjoint open sets containing U, V respectively. By regularity, given ɛ > 0 there is an open set W containing A B with λ(w ) < λ (A B)+ɛ. Then W U and W V are disjoint open sets containing A, B respectively; so λ (A) + λ (B) λ(w U) + λ(w V ) λ(w ) λ (A B) + ɛ. Let ɛ 0 to get λ (A B) λ (A) + λ (B). The opposite inequality follows from subadditivity, so Lebesgue outer measure is metric. Exercise 21.2. The measures ν constructed by the Riesz representation theorem have various regularity properties. Verify that for every ν -measurable set E and every ɛ > 0 there exist an open set U and a closed set K with K E U and ν(u \K) < ɛ. Deduce that every ν -measurable set differs from a Borel set by a set of ν -measure zero. Prove also that the continuous functions form a dense subspace of the normed vector space L 1 (X, ν). Solution. The first part is a general theorem for finite Borel measures, see Extended HW 2. If now E is measurable there exist open sets U n E with ν(u n \ E) 1/n. The intersection G of all the U n is a Borel set containing E and ν(g \ E) = 0. Let E be a Borel set. For ɛ > 0 choose K E U, K closed, U open with ν(u \ K) < ɛ. Choose a continuous, [0, 1]-valued function φ supported in U and equal to 1 on K. Then φ χ E dν = φ χ E dν ν(u \ K) ɛ. X U\K It follows that χ E is an L 1 limit of continuous functions. This extends to measurable E since any such E differs from a Borel set by a set of measure zero. But now it follows that every simple function is an L 1 limit of continuous functions; and simple functions are dense in L 1 by construction. Exercise 21.3. Show that we have uniqueness in the Riesz representation theorem: two Borel measures that determine the same linear functional are the same. (Suggestion: Let A be the class of sets on which the two measures agree. Show that A is a σ-algebra and contains the open sets. 29

Lecture 22 Lebesgue-Stieltjes Measures { 0 (x < 0) Exercise 22.1. Take I = [ 1, 1] and let g(x) =. What is the corresponding Stieltjes measure? Does the value of g at 0 affect your 1 (x > 0) answer? Solution. The Stieltjes measure associated with the distribution function g is the unit mass at the origin (that is, fdg = f(0)). To see this, notice that from the definition, if f 1 = f 2 in a neighborhood of 0, then f 1 dg = f 2 dg. Now given any ɛ > 0 we can find a constant function c = f(0) such that f c < ɛ in a neighborhood of the origin. Then fdg = cdg + O(ɛ) = c + O(ɛ) since integration is a continuous linear functional. Because ɛ is arbitrary we get fdg = c = f(0). The choise of g at 0 does not affect the answer. Exercise 22.2. With dg a Stieltjes measure as above, show that dg(c, d) = g(d ) g(c+) for every open interval (c, d) [a, b]. (Here g(d ) = lim x d g(x) and g(c+) = lim x c g(x); the limits exist because g is monotone.) Find corresponding formulae for closed and half-open intervals. Hence show that if g is continuous from the right, or from the left, then dg determines g uniquely up to an additive constant. Solution. Let f n be the continuous function defined as follows: f n (x) = 0 for x / (c + 1/n, d 1/n), f n (x) = 1 for x [c + 2/n, d 2/n], and f n is linear on [c+1/n, c+2/n] and [d 2/n, d 1/n]. Then f n is a monotone sequence of continuous functions increasing to the characteristic function of (c, d), so f n dg dg(c, d). From the definition of the Stieltjes integral we have g(d 2/n) g(c + 2/n) f n dg g(d 1/n) g(c + 1/n), and as n both sides tend to g(d ) g(c + ). This gives the result. A similar argument for other intervals gives dg[c, d] = g(d + ) g(c ), dg(c, d] = g(d + ) g(c + ), and so on. If g is right continuous then g(x) = g(x + ) for all x. Thus dg(c, d] = g(d) g(c), so dg determines g up to an additive constant. Similarly for left continuity. Exercise 22.3. Conversely, suppose that ν is a finite Borel measure on [a, b], such that ν{a} = 0. Define g(x) = ν(a, x] (with g(a) = 0). Show that ν = dg. Thus every Borel measure on [a, b] arises from the Stieltjes construction. 30

Solution. The function g so defined is right continuous (by the monotone convergence theorem). Thus dg(a) = ν(a) whenever g is a half-open interval (a, x]. The collection of sets A on which the two measures agree contains [a, b] and is closed under countable unions and set-theoretic difference, so it is a σ-algebra. Since the half-open intervals above generate the σ-algebra of all Borel sets, so dg(a) = ν(a) for all Borel sets A. Exercise 22.4. Show that if g is continuous, then dg(e) = λ(g(e)) for any Borel subset E of [a, b]. Solution. By the last exercise but one, dg(e) = λ(g(e)) whenever E is an interval. Now both sides are measures and we can argue as in the previous exercise. Exercise 22.5. Let f : [0, 1] R be a function. One says that f is absolutely continuous if, for any ɛ > 0, there is δ > 0 with the following property: given any (finite or countable) collection [a k, b k ] of disjoint closed subintervals of [0, 1], if k a k b k < δ, then k f(a k ) f(b k ) < ɛ. Show that every absolutely continuous function is continuous and of bounded variation. Show that the Cantor function?? is continuous and of bounded variation, but not absolutely continuous. 31

Lecture 23 Signed measures Exercise 23.1. Let ν 1, ν 2, and µ be finite measures (on the same σ-algebra). Suppose that ν 1 and µ are mutually singular, and that ν 2 and µ are mutually singular. Prove that then ν 1 + ν 2 and µ are mutually singular. Solution. Suppose that we have disjoint decompositions X = A 1 B 1 = A 2 B 2 with µ(a 1 ) = µ(a 2 ) = ν 1 (B 1 ) = ν 2 (B 2 ) = 0. Put A = A 1 A 2, B = B 1 B 2. Then X = A B is a disjoint decomposition, µ(a) µ(a 1 ) + µ(a 2 ) = 0, and (ν 1 + ν 2 )(B) = ν 1 (B) + ν 2 (B) = 0. Thus µ and ν 1 + ν 2 are mutually singular. 32

Lecture 24 Absolute continuity Exercise 24.1. Let A be the σ-algebra of Lebesgue measurable subsets of [0, 1], let µ be the counting measure on A, and let λ be Lebesgue measure. Show that λ is absolutely continuous with respect to µ. Does the Radon-Nikodym theorem apply? If it does, find the R N derivative dλ/dµ; if it does not, say what goes wrong. Solution. The only set of µ-measure zero is the empty set, which certainly has Lebesgue measure zero. However, the Radon-Nikodym theorem does not work here. If it did, there would be a (positive) real-valued function f such that λ(e) = x E f(x). This is impossible: since every one-point set has zero Lebesgue measure, we would have to have f(x) = 0 for all x. The problem is that the measure µ is not finite or even σ-finite. Exercise 24.2. Show that if ν is both absolutely continuous and mutually singular with respect to µ, then it is zero. (This gives the uniqueness in (b) above.) Solution. Take ν to be a positive measure (otherwise, apply the following argument to its absolute value). Let X = A B with ν(a) = µ(b) = 0. If ν µ then ν(b) = 0 also, and so ν(x) = 0. Exercise 24.3. Show that every signed measure is absolutely continuous with respect to its own absolute value. What can you say about the Radon-Nikodym derivative in this case? How does it relate to the Hahn decomposition? Solution. Let ν be a signed measure and let X = P N be a Hahn decomposition, so that ν + (E) = ν(e P ) and ν (E) = ν(e N) are positive measures and ν = ν + + ν. Then ν(e) = ν (P E) ν (N E) = (χ P χ N )d ν which shows that ν is absolutely continuous wrt ν and the Radon-Nikodym derivative dν/d ν equals χ P χ N. E 33

Lecture 25 Inner Products and L 2 (X, µ) Exercise 25.1. Show that the simple functions form a dense subspace of L 2 (X, µ) for any (σ-finite) measure space (X, µ). Solution. Let f L 2 (X, µ). Define f n (x) to equal ±2 n 2 n f(x) (choosing the same sign as that of f(x)) if n f(x) n, and otherwise to equal 0. Then f n is a simple function, f n f, and f n f pointwise. By the dominated convergence theorem, then, f n f 2 = f n f 2 dµ 0 since the integrand is dominated by the integrable function 4 f 2. Exercise 25.2. If µ is a finite Borel measure on the compact space X, show that the continuous functions are dense in L 2 (X, µ). Solution. Let E be a Borel set. By regularity, given any ɛ > 0 there exist a compact K and an open U with K E U and µ(u \ K) < ɛ. Pick a [0, 1]-valued function f equal to 1 on K and supported within U. Then f χ E vanishes except on a set of measure ɛ, where it has absolute value 1, and thus f χ E ɛ 1/2. We conclude that every characteristic function of a Borel set, and therefore every simple function, is a limit of continuous functions. Since the simple functions are dense by the preceding exercise, the result follows. 34

Lecture 26 The Geometry of Hilbert Space Exercise 26.1. In the Hilbert space L 2 [ 1, 1] (with respect to Lebesgue measure) show that the set V of functions f that satisfy f(x) = f( x) a.e. is a closed subspace, and find its orthogonal complement. Solution. The set V of functions f satisfying f(x) = f( x) almost everywhere is certainly a subspace. To see that it is closed, consider the expression Φ(f) = 1 1 f(x) f( x) 2 dx. By Cauchy-Schwarz, Φ(f) depends continuously on f, and the subspace we want is exactly Φ 1 {0}, so it is closed. The orthogonal complement of V is the set W of functions g such that g(x) = g( x) almost everywhere. To prove this note that W is a closed subspace (by the same reasoning) and that V and W are orthogonal (if f V and g W then 0 1 fg = 1 fg, so f, g = 0. Finally, every function is the sum of an even and an odd 0 function, so H = V W and thus V and W are orthogonal complements. Exercise 26.2. Let S be any subset of a Hilbert space. Show that S is the closed linear span of S, that is, the smallest closed subspace that contains S. Solution. For any subset S, S is a subspace, and it is closed because it is the intersection of the closed sets {x : x, s = 0} for all the s S. Moreover S contains S, because anything in S is orthogonal to anything that is orthogonal to everything in S. Thus S is a closed subspace containing S. To see that it is the smallest such subspace, suppose V is another one. Then V contains S, so V is contained in S and thus V contains S. But, because V is a closed subspace by assumption, the projection theorem shows that V = V. The result follows. Exercise 26.3. Show that T T = T 2 for any bounded operator T on a Hilbert space. Solution. Write T x 2 = T x, T x = T T x, x T T x 2 using Cauchy-Schwarz at the last step. Taking the supremum over x in the unit bal;l gives T 2 T T. On the other hand T T T T = T 2 so we have equality all through, which is what is needed. 35

Exercise 26.4. Show that a bounded operator U on a Hilbert space H gives an isometry from H onto itself if and only if U U = I = UU (in this case we say U is unitary). What is the corresponding condition for an operator V to give an isometry of H into (but not necessarily onto) itself? Solution. Let V be an isometry. Then V x 2 = x 2 for all x, which is to say that (V V I)x, x = 0 for all x. Let T = V V I. The polarization identity T (x + y), x + y T (x y), x y = 2 T x, y + 2 x, T y = 4R T x, y (together with the similar identity with y replaced by iy) shows that T x, y = 0 for all x, y and thus that T = 0. Thus V V = I if V is an isometry. If V is in addition onto, then it is bijective and its inverse is also an isometry. But the identity V V = I shows that its inverse (if it exists) must be V ; and then V V = I since V is an isometry. Exercise 26.5. Let (X, µ) be a measure space and let f be a bounded measurable function X C. Show that the multiplication operator M f defined by M f (g) = fg is bounded on L 2 (X, µ). What is the adjoint of M f? Solution. Suppose that f is bounded by the constant C. Then M f (g) 2 = f 2 g 2 C 2 g 2 = C 2 g 2 so M f is bounded, with norm at most C. The adjoint of M f is M f, since M f g, h = fḡh = ḡ fh = g, M f h. 36