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s invertible (if we restrict the codomain appropriately), and the inverse is also strictly increasing. It is also clear that these conditions are necessary for an inverse to exist. However, if we relax the conditions a bit, we can get some sort of “pseudoinverse” (some categorists may call them “left adjoints” (and wil... |
ce. Suppose that (fn), f are measurable functions. We say fn → f in measure if for each ε > 0, we have µ({x ∈ E : |fn(x) − f (x)| ≥ ε}) → 0 as n → ∞, then we say that fn → f in measure. If (E, E, µ) is a probability space, then this is called convergence in probability. In the case of a probability space, this says P(|... |
ral on simple functions, and then extend the definition to more general measurable functions by taking the limit. When we do the definition for simple functions, it will be obvious that the definition satisfies the nice properties, and we will have to check that they are preserved when we take the limit. Definition (Simple ... |
le functions. Then µ(lim inf fn) ≤ lim inf µ(fn). Note that a special case was proven in the first example sheet, where we did it for the case where fn are indicator functions. Proof. We start with the trivial observation that if k ≥ n, then we always have that By the monotonicity of the integral, we know that inf m≥n f... |
Actually, everything still works for σ-finite measure spaces, as we can just reduce to the finite case. However, things start to go wrong if we don’t have σ-finite measure spaces. Proof. One might be tempted to just apply the Caratheodory extension theorem, but we have a more direct way of doing it here, by using integra... |
= c(m) = c(E[X]). So done. 56 4 Inequalities and Lp spaces II Probability and Measure We are now going to use Jensen’s inequality to prove H¨older’s inequality. Before that, we take note of the following definition: Definition (Conjugate). Let p, q ∈ [1, ∞]. We say that they are conjugate if 1 p + 1 q = 1, where we take... |
nal expectation). Suppose we have a probability space (Ω, F, P), and (Gn) is a collection of pairwise disjoint events with n Gn = Ω. We let G = σ(Gn : n ∈ N). The conditional expectation of X given G is the random variable Y = ∞ n=1 E[X | Gn]1Gn, where E[X | Gn] = E[X1Gn ] P[Gn] In other words, given any x ∈ Ω, say x ∈... |
for complex valued Borel functions on Rd with f p = 1/p |f |p Rd < ∞. The integrals of complex-valued function are defined on the real and imaginary parts separately, and satisfy the properties we would expect them to. The details are on the first example sheet. Definition (Fourier transform). The Fourier transform ˆf : ... |
s t → 0 by the bounded convergence theorem, since we know that e is bounded. Finally, we have f ∗ gt − f p ≤ f ∗ gt − h ∗ gtp + h ∗ gt − hp + h − f p ≤ = ε 3 2ε 3 + ε 3 + h ∗ gt − hp + h ∗ gt − hp. Since we know that h ∗ gt − hp → 0 as t → 0, we know that for all sufficiently small t, the function is bounded above by ε. ... |
n determines the law of X. (iv) We start off with a boring Gaussian vector Y = (Y1, · · · , Yn), where the Yi ∼ N (0, 1) are independent. Then the density of Y is fY (y) = (2π)−n/2e−|y|2/2. We are now going to construct X from Y . We define ˜X = V 1/2Y + µ. This makes sense because V is always non-negative definite. Then ... |
Birkhoff’s theorem, we know a.e. Also, we know f − gp < ε 3 . Sn(g) n → ¯g Sn(g) n ≤ M for all n. So by bounded convergence theorem, we know Sn(g) n − ¯g p → 0 as n → ∞. So we can find N such that n ≥ N implies Sn(g) n − ¯g p < ε 3 . Then we have ¯f − ¯g p p = lim inf n Sn(f − g) n Sn(f − g) n p p dµ dµ ≤ lim inf + ¯g −... |
= [N ] and each |Ai| < δN . Our next theorem will mix arithmetic and graph-theoretic properties. Consider a colouring c : N(2) → [2]. As before, we say a set X is monochromatic if c|X (2) is constant. Now we want to try to find a monochromatic set with some arithmetic properties. The first obvious question to ask is — ca... |
ct. It turns out this is essentially the only way we know for proving this theorem. One possible variation is to use the “first non-zero digit” to do the colouring, but this is harder. Let’s now try and do the other direction. Before we do that, we start by doing a warm up. Last time, we proved that if we had 1 λ, then ... |
· · · ∪ Br c(i) = i∈Bs qisc(i) i∈B1∪...∪Bs−1 for some qis ∈ Q. These qis only depend on the matrix. In other words, we have where disc(i) = 0, dis = −qis 1 0 i ∈ B1 ∪ · · · Bs−1 i ∈ Bs otherwise For a fixed s, if we scan these coefficients starting from i = n and then keep decreasing i, then the first non-zero coeffi... |
monochromatic arithmetic progression of length r. Proof. Let c : Z → [k]. Consider (¯c, L). By topological van der Waerden, we can find x ∈ ¯c and n ∈ N such that ρ(x, Linx) < 1 for all i = 1, · · · , r. In particular, we know that x and Linx agree at 0. So we know that x(0) = Linx(0) = x(in) for all i = 0, · · · , r. W... |
→ [k + 1] with x(i) = x(−i) = c(i) if x = 0, and x(0) = k + 1. Then it suffices to find an infinite an infinite A ⊆ Z such that F S(A) is monochromatic with respect to x. We apply topological Hindman to (¯x, L) to find a minimal colouring y such that x and y are proximal. Then either inf n∈N ρ(Lnx, Lny) = 0 or inf n∈N ρ(L−nx... |
inal proof (arxiv:1605.01469) is a bit less magical, and used concepts from topological dynamics. Interested people can find the original paper to read. However, we will sacrifice “deep understanding” for having a more elementary proof that just uses van der Waerden. Proof of theorem. Let N = C1 ∪ · · · ∪ Cr. We build th... |
ove type and that all bounded intervals are of the form listed in (1) together with intervals of the form (a, b). 1. For a # 0 and b t 0, show that (ab)-1 = a-1b-1. 2. Verify the following: PROBLEMS (i) For each real number a * 0, a2 > 0. In particular, 1 > 0 since 1 * 0 and 1 = 12. (ii) For each positive number a, its... |
e correspondence between {1.... , N} and A. Thus A is finite. Now consider the case that B is countably infinite. Let f be a one-to-one correspondence between N and B. Define g(1) to be the first natural number j for which f (j) belongs to A. Arguing as in the first case, we see that if this selection process terminate... |
E. If E contains all of its points of closure, that is, E = E, then the set E is said to be dosed. Proposition 10 For a set of real numbers E, its closure t is closed. Moreover, E is the smallest closed set that contains E in the sense that if F is closed and E C F, then E C F. Proof The set E is closed provided it co... |
pen set is an FQ set. 1.5 SEQUENCES OF REAL NUMBERS A sequence of real numbers is a real-valued function whose domain is the set of natural numbers. Rather than denoting a sequence with standard functional notation such as f : N--> R, it is customary to use subscripts, replace f (n) with an, and denote a sequence Secti... |
E/c responds to the c challenge regarding the criterion for the continuity of f at x. Not all continuous functions are Lipschitz. For example, if f (x) = lx- for 0 < x < 1, then f is continuous on [0, 1] but is not Lipschitz. We leave as an exercise the proof of the following characterization of continuity at a point ... |
...... 2.6 Nonmeasurable Sets ................................. 2.7 The Cantor Set and the Cantor-Lebesgue Function ............... 49 29 31 34 43 47 2.1 INTRODUCTION The Riemann integral of a bounded function over a closed, bounded interval is defined using approximations of the function that are associated with parti... |
n of sets, disjoint or not, then m*IUEkEm*(Ek). 00 k=1 k=1 Proof If one of the Ek's has infinite outer measure, the inequality holds trivially. We therefore suppose each of the Ek's has finite outer measure. Let c > 0. For each natural number k, there is a countable collection {Ik,i}°O1 of open, bounded intervals for w... |
tion of all the a-algebras of subsets of R that contain the open sets is a a-algebra called the Borel a-algebra; members of this collection are called Borel sets. The Borel a-algebra is contained in every a-algebra that contains all open sets. Therefore, since the measurable sets are a a-algebra containing all open set... |
possesses the following continuity properties: (i) If {Ak}k 1 is an ascending collection of measurable sets, then m O Ak1) = klim m(Ak). k=1 1 +00 (ii) If {Bk}k11, is a descending collection of measurable sets and m( B1) < oo, then \1 M I I Bk I I k=1 m(Bk) k OO (12) Section 2.5 Countable Additivity, Continuity, and th... |
n the construction of the Cantor set is to subdivide I into three intervals of equal length 1/3 and remove the interior of the middle interval, that is, we remove the interval (1/3, 2/3) from the interval [0, 1] to obtain the closed set C1, which is the union of two disjoint closed intervals, each of length 1/3 : C1 = ... |
s empty.) 43. Use the preceding two problems to provide another proof of the uncountability of the Cantor set. 44. A subset A of R is said to be nowhere dense in R provided that for every open set 0 has an open subset that is disjoint from A. Show that the Cantor set is nowhere dense in R. 45. Show that a strictly incr... |
nverse image of each open set is measurable. Let 0 be open. Then (fog)-1(0) =g 1(f-1(d)) Since f is continuous and defined on an open set, the set U = f-1(0) is open.3 We infer from the measurability of the function g that g-1 (U) is measurable. Thus the inverse image (f o g)-1(0) is measurable and so the composite fun... |
tion Lemma, applied to the restriction of f to En and with the choice of e = 1/n, we may select simple functions (pin and on defined on En which have the following approximation properties: 0 < (pin < f < af/n on En and 0 < On - (pin < 1/n on En. Observe that 0 <(pn < f and0< f -(pin <a/1n-(pin <1/n on En. (1) Section ... |
n=1 The set F is closed since it is the intersection of closed sets. Each fn is continuous on F since F C Fn and f, = gn on F. Finally, (fn} converges to f uniformly on F since F C F0. However, the uniform limit of continuous functions is continuous, so the restriction of f to F is continuous on F. Finally, there is a ... |
sjoint and the ai's being distinct. Definition For a simple function 0 defined on a set of finite measure E, we define the integral of 0 over E by f=ai-m(Ei), i=1 E 72 Chapter 4 Lebesgue Integration where 0 has the canonical representation given by (1). Lemma 1 Let {Ei}" 1 be a finite disjoint collection of measurable ... |
fA Therefore limn , fE fn fE f- 5 L if - fn1 <- [e/m(E)l . m(E) = E. This proposition is rather weak since frequently a sequence will be presented that converges pointwise but not uniformly. It is important to understand when it is possible to infer from that (fn} -+ f pointwise a.e. on E lim n-+oo L r f 1 f f. We refe... |
omains and the observation that, by Proposition 9, the integral of a nonnegative function over a set of measure zero is zero. The following lemma will enable us to establish several criteria to justify passage of the limit under the integral sign. Fatou's Lemma Let [f,) be a sequence of nonnegative measurable functions... |
y assume that f and g are finite on E. To verify linearity is to show that f[f+gft -fE[f+g] =Lf f+ - E E f J+[f 8+- E fE 9 1. fE (19) But (f+g)+- (f+g) =f+g=(f+- f-)+ (g+ -g-) onE, and therefore, since each of these six functions takes real values on E, (f+g)++f +g =(f+g) +f++g+onE. We infer from linearity of integrati... |
se m (E) < oo, if for each e > 0, there is a S > 0 for which (26) holds, then f is integrable over E. Proof The theorem follows by establishing it separately for the positive and negative parts of f. We therefore suppose f > 0 on E. First assume f is integrable over E. Let e > 0. By the definition of the integral of a ... |
mann integrable over a closed, bounded interval if and only if it is continuous at almost all points in its domain. 5.1 UNIFORM INTEGRABILITY AND TIGHTNESS: A GENERAL VITALI CONVERGENCE THEOREM The Vitali Convergence Theorem of the preceding chapter tells us that if m (E) < oo, If,,) is uniformly integrable over E and ... |
rl > 0 and E > 0 there is an index N such that for all m, n > N, m {x E E l I.fn(x) - .fm(x)I > rl} < E. Show that if {f,} is Cauchy in measure, then there is a measurable function f on E to which the sequence { f,} converges in measure. (Hint: Choose a strictly increasing sequence of natural numbers {nj} such that fo... |
(Diffhf)o<h<1, is uniformly integrable. Therefore, by the Vitali Convergence Theorem, (i) follows for f absolutely continuous by taking the limit as h -+ 0+ in its discrete formulation. If f is monotone and (i) holds, we prove that f must be absolutely continuous. From the integral form of the fundamental theorem, (i),... |
sure zero. Fix rationale a, /3 with a > /3 and set E = Ea,p. Let e > 0. Choose an open set 0 for which ECOC(a, b) andm(O) <m*(E)+E. (10) Let .E be the collection of closed, bounded intervals [c, d] contained in 0 for which f (d) - f (c) < /3 (d - c). Since D f < R on E, F is a Vitali covering of E. The Vitali Covering ... |
s an upper bound of the set of all variations off with respect to a partition of [a, b] and hence TV(f) < c [b - a]. Example Define the function f on [0, 1] by x cos(?r/2x) 0 if 0 <x < 1 ifx 0 Then f is continuous on [0, 1]. But f is not of bounded variation on [0, 1]. Indeed, for a natural number n, consider the parti... |
pen subintervals of (a, b) for which 2k=1[dk - ck] < S. For 0 < h < 1 and 1 < k < n, by (14), E f(dk) AVh - AVh f (ck) = dk J ck Diffh f. Therefore n Y, I Avh f(dk) - Avh f (ck) I< k=1 n dk k=1 Ck I Diffh f l= f J E I Diffh fl, where E = Uk=1(ck, dk) has measure less than S. Thus, by the choice of S, n I IAvh f(dk)-Avh... |
uppose (35) holds. We claim that f = 0 for all measurable sets E C [a, b]. JE (34) (35) (36) Indeed, (36) holds for all open sets contained in (a, b) since integration is countably additive and every open set is the union of countable disjoint collection of open intervals. The continuity of integration then tells us th... |
(P (u ) :5 P (u+) :!s (v) - u(u) < -P (V-):5 V(v+) (40) 132 Chapter 6 Differentiation and Integration Corollary 17 Let p be a convex function on (a, b). Then (P is Lipschitz, and therefore absolutely continuous, on each closed, bounded subinterval [c, d] of (a, b). Proof According to the preceding lemma, for c < u < v ... |
union of a countable collection of sets of measure zero also is of measure zero, the sequence {gn} also converges pointwise a.e. on E to g. To state that a function f in LP[a, b] is continuous means that there is a continuous function that agrees with f a.e. on [a, b]. Since complements of sets of measure zero are den... |
LP (E) in the sense that there is a constant M for which 11f 11p < M for all f in.F. Then the family .F is uniformly integrable over E. Proof Let E > 0. We must show there is a S > 0 such that for any fin F, fA If I < E if ACE is measurable and m(A) < S. Let A be a measurable subset of E of finite measure. Consider LP(... |
vergent series of positive numbers such that II fk+i - fkIIp s Ek for all k, (9) and therefore Ifk+1 - fk I p < Ekp for all k. Fix a natural number k. Since, for x e E, I fk+l (x) - fk (x) I > Ek if and only if Ifk+l (x) fk (x) I P > EP , we infer from (10) and Chebychev's Inequality that JE (10) m{xEE I Ifk+1(x)-fk(x)... |
he LP Spaces: Completeness and Approximation which, if we define U Uk-1 Ik, then the symmetric difference A AU = [A ^- U] U [U n A] has the property that m(AAU) <e'. Since U is the union of a finite disjoint collection of open intervals, Xu is a step function. Moreover, IIXA - Xull p = [m(Ai U)]1"p. (14) Therefore IIXA... |
the Dual of LP,1 -<p < oo 157 We leave it as an exercise to show that IITII*=sup{T(ffEX, 11f11<_1 (8) and use this characterization of II II* to prove the following proposition. Proposition 1 Let X be a normed.linear space. Then the collection of bounded linear functionals on X is a linear space on which II II: is a n... |
near space X. Show that T is bounded if and only if the continuity property (7) holds. 4. A functional T on a normed linear space X is said to be Lipschitz provided there is a c > 0 such that IT(g)-T(h)I <clig-hllforallg,hEX. The infimum of such c's is called the Lipschitz constant for T. Show that a linear functional ... |
n-+00 x J n sin nt dt = 0 for all x c: I. Explicit calculation of these integrals shows that this is true. On the other hand, observe that for each n, ir flsinnr2dt = f sin2ntdt=7T. 7r Thus no subsequence of (fn } converges strongly in L2(I) to f = 0. A similar estimate shows no subsequence converges strongly in any L... |
elley in 1912 for the special case X = C[a, b], normed by the maximum norm. In his 1932 book, Stefan Banach observed, providing a one-sentence proof, that the result holds for any separable normed linear space. 172 Chapter 8 The LP Spaces: Duality and Weak Convergence We inductively continue this selection process to o... |
al, continuity is equivalent to boundedness. In general, these concepts are unrelated. Definition A real-valued functional T defined on a convex subset C of a normed linear space X is said to be convex provided whenever f and g belong to C and A E [0, 1], T(Af + (1- A)g) 5 AT(f) + (1- A)T(g). In any normed linear space... |
of a function or mapping. We first consider these general concepts. We then study metric spaces which possess finer structure: those that are complete, compact, or separable. 9.1 EXAMPLES OF METRIC SPACES Definition Let X be a nonempty set. A function p : X X X -+ R is called a metric provided for all x, y, and z in X... |
n ball B(x, S) is contained Ot n 02. Therefore 01 n 02 is open. 188 Chapter 9 Metric Spaces: General Properties The following proposition, whose proof we leave as an exercise, provides a description, in the case the metric space X is a subspace of the metric space Y, of the open subsets of X in terms of the open subset... |
B(f (x), c) for n > N. Thus the sequence If (xn) } converges to f (x) and therefore f : X--> Y is continuous at the point x. Proposition 8 A mapping f from a metric space X to a metric space Y is continuous if and only if for each open subset 0 of Y, the inverse image under f of 0, f -1(0), is an open subset of X. Pro... |
n} is Cauchy, the sequence {Fn} is contracting. Thus, by assumption, there is a point x in X for which {x} _ fl 1 Fn. For each index n, x is a point of closure of (Xk I k > n) and therefore any ball centered at x has nonempty intersection with {xk I k > n}. Hence we may inductively select a strictly increasing sequence... |
et Pk be a partition of [-a, a] for which each partition interval has length less than 1/k. Then Pk X Pk induces a partition of [-a, a] X [-a, a] into closed rectangles of diameter at most 12-1k. Choose k such that 12-1k < E. Consider the finite collection of balls of radius E with centers (x, y) where x and y are part... |
(x), f (x')) < E/2. Therefore, setting Ox = B(x, Sx ), by the triangle inequality for a-, 0-(.f(u), f(v)) <o(.f(u), f(x))+o(f(x), f(v)) <Eifu,vEOx. (5) Let S be a Lebesgue number for the open cover {Ox}x E x. Then for u, v E X, if p(u, v) < S there is some x for which u E B (v, S) C Ox and therefore, by (5), o-(f (u),... |
r a function fin C(X ), define Illllmax = maxIf(x)I. XEX This defines a norm, as it did in the special case X = [a, b] we first considered in Chapter 7. This maximum norm induces a metric by Pmax(g, h) = IIg-hllmaxforallg,hEC(X). We call this metric the uniform metric because a sequence in C(X) converges with respect t... |
s equicontinuous. 5. A real-valued function f on [0, 1] is said to be Holder continuous of order a provided there is a constant C for which Define the Holder norm If(x)-f(Y)I CIx - ylaforall x,yE[0, 1]. Illlla=max{If(x)I+I1(x)-f(Y)I/IX -Yla I x,yE[0, 1],x#y}. Show that for 0 < a < 1, the set of functions for which Ilfl... |
pace X is said to be of the first category (or meager) if E is the union of a countable collection of nowhere dense subsets of X. A set that is not of the first category is said to be of the second category (or nonmeager), and the complement of a set of first category is called residual (or co-meager). The Baire Catego... |
two variables g, if a continuous function f : I - R has the property that (x, f (X)) E O for each x E I, then f is a solution of (14) if and only if f(x)=yo+Jxg(t, f(t))dtforallxEl. (15) 218 Chapter 10 Metric Spaces: Three Fundamental Theorems As we will see in the proof of the next theorem, this equivalence between so... |
for a given set of points, and in such cases we sometimes use the symbol X to denote both the set of points and the topological space (X, T). When greater precision is needed, we make explicit the topology. Proposition l A subset E of a topological space X is open if and only if for each point x in X there is a neighbo... |
of a topological space X. For a subset E of X, show that 0 is disjoint from E if and only if it is disjoint from E. 8. For a collection S of subsets of a nonempty set X, show that there is a topology Ton X that contains the collection S and has the property that any other topology that contains S also contains T: it i... |
and every subspace of a second countable space is second countable. 20. Show that the Moore Plane is separable (see Problem 10). Show that the subspace R X {0} of the Moore Plane is not separable. Conclude that the Moore Plane is not metrizable and not second countable. 21. Show that the Sorgenfrey Line is first counta... |
he case of measurable functions and the inheritance of measurability from the measurability of restrictions. 38. Show that for any two numbers a and b, Isgn (a) 11.5 COMPACT TOPOLOGICAL SPACES We have studied compactness for metric spaces. We provided several characterizations of compactness and established properties ... |
formly. 11.6 CONNECTED TOPOLOGICAL SPACES Two nonempty open subsets of a topological space X are said to separate X if they are disjoint and their union is X. A topological space which cannot be separated by such a pair is said to be connected. Since the complement of an open set is closed, each of the open sets in a s... |
the normally ascending collection of open sets {OA}A E A,, Observe that the union of this countable collection is a normally ascending collection of open sets parametrized by A, each of which is a neighborhood of F that has compact closure contained in U. Proof of Urysohn's Lemma By Lemma 2, applied with F = A and U =... |
rding collections of sets that possess the finite intersection property. Lemma 5 Let A be a collection of subsets of a set X that possesses the finite intersection property. Then there is a collection 8 of subsets of X which contains A, has the finite intersection property, and is maximal with respect to this property;... |
ich f (xo) :f- f (y). The function gyL L 2 f-f(xo) 11f - f(x0)Jimax J J 2The proof we present is due to B. Brasowski and F. Deutsch, Proceedings of the American Mathematical Society, 81 (1981). Many very different-looking proofs of the Stone-Weierstrass Theorem have been given since the first proof in 1937 by Marshal S... |
ndedness Principle ....................... 268 We have already examined important specific classes of normed linear spaces. The most prominent of these are: (i) for a natural number n, Euclidean space R"; (ii) for a Lebesgue measurable subset E of real numbers and 1 < p < oo, the LP(E) space of Lebesgue measurable func... |
ore, by the e - S criterion for continuity at u Proof Let X and Y be normed linear spaces and T : X -+ Y be linear. If T is bounded, (3) tells us that T is continuous. Now suppose T : X -* Y is continuous. Since T is linear, 1, we may choose T (O) S > 0 such that IIT(u) - T(0)II < 1 if IIu - 011 < S, that is, IIT(u)II ... |
existence of these two constants follows where II from the observation that x - II T(x) II defines a norm on Rn, which, since all norms on Rn are equivalent, is equivalent to the Euclidean norm. Corollary6 Any finite dimensional normed linear space is complete and therefore any finite dimensional subspace of a normed l... |
ed whenever {x,,) is a sequence in X if yo, then T(xo)=yo The graph of a mapping of T : X --). Y is the set ((x, T (x)) E X X Y I X E X). Therefore an operator is closed if and only if its graph is a closed subspace of the product space X X Y. The Closed Graph Theorem Let T : X -+ Y be a linear operator between the Ban... |
{Tn : X - Y} is uniformly bounded. Furthermore, the operator T: X -> Y defined by T(x) = lim T,, (x) for all forallx E X n aoo is linear, continuous, and IITII <_ liminf IITII Proof The pointwise limit of a sequence of linear operators is linear. Thus T is linear. We infer from the Uniform Boundedness Principle that th... |
to a point in the set. It is easy to see that a sequence {xn } in X converges to x E X with respect to the F-weak topology if and only if lim f(xn) = f(x)forall f E.T. n- 00 (4) A function on X that is continuous with respect to the F-weak topology is called .T-weakly continuous. Similarly, we have .F-weakly open sets,... |
(Y1 -z)+(y2+z)) <- P(Y1 -z)+P(Y2+z), and therefore I'01) -P(Y1 -z) <-402)+P(Y2+z). As we vary y, and Y2 among all vectors in Y, any number on the left-hand side of this inequality is no greater than any number on the right. By the completeness of R, if we define 4,(z) to be the supremum of the numbers on the left-hand ... |
to be positive (with respect to the cone C) provided f > 0 on C. Let Y be any subspace of X with the property that for each x E X there is a y c Y with x < y. Show that each positive linear functional on Y may be extended to a positive linear functional on X. (Hint: Adapt the Hahn-Banach Lemma and use Zorn's Lemma wit... |
weakly, then it converges pointwise. 38. For X and Y normed linear spaces and an operator S E C(X, Y), define the adjoint of S, S* E ,C(Y*, X*) by [S*(4r)](x) = ir(S(x)) for all 41 E Y*, X E X. (i) Show that II S* II = II S II and that S* is an isomorphism if S is an isomorphism. (ii) For 1 < p < oo and X = LP(E), whe... |
which f (0) # R. 44. Let X be a normed'linear space and W a subspace of X* that separates points. For any X is continuous, where X has the W-weak topological space Z, show that a mapping f : Z topology, if and only if qr o f : Z -+ R is continuous for all 41 E W. 45. Show that the topology on a finite dimensional local... |
z E KO. Define K to be the strong closure of KO. Then x does not belong to K. The strong closure of a convex set is convex. Moreover, K is convex since KO is convex. Therefore, by Mazur's Theorem, K is weakly closed. Since {xn} converges to x with respect to the weak topology, x is a point of closure of K with respect ... |
of X* that is weak-* closed is weak-* compact. From Alaoglu's Theorem and Mazur's Theorem we then infer that any strongly closed bounded convex subset of a reflexive Banach space is weakly compact. PROBLEMS 66. Find the extreme points of each of the following subsets of the plane R2: (i) {(x, y) I x2 + y2 =1}; (ii) {(x... |
from B, with the weak topology, onto B**, with the weak-* topology. But by Alaoglu's Theorem, applied with X replaced by X*, B** is weak-* compact, so any topological space homeomorphic to it also is compact. In particular, B is weakly compact. Now assume B is weakly compact. The continuous image of compact topological... |
gent subsequence (see Problem 11), but every bounded sequence of continuous linear functionals on C(K) has a subsequence that converges pointwise to a continuous linear functional on C(K). 15.4 METRIZABILITY OF WEAK TOPOLOGIES If the weak topology on the closed unit ball of a Banach space is metrizable, then the Eberle... |
oralltER. The quadratic polynomial in t defined by the right-hand side fails to have distinct real roots and therefore its discriminant is not positive, that is, the Cauchy-Schwarz Inequality holds. Proposition 1 For a vector h in an inner product space H, define Then II II is a norm on H called the norm induced by the... |
(hnk, (Id-P)[h]) = (h0, (Id-P)[h]) = 0 for all h E H. Therefore lim (hnk, h) = (ho, h) for all h E H. k- oo Thus {hnk } converges weakly to h0 in H. We gather in the following proposition some properties regarding weakly convergent sequences which we established earlier for general Banach spaces but which, because of ... |
for all u E H, where c = II T11 I I v II . According to the Riesz-Frechet Representation Theorem, there is a unique vector h E H such that (T(u), v) = (h, u) = (u, h) for all u E H. We denote this vector h by T* (v). This defines a mapping T*: H -* H that is determined by the relation c (T(u), v)=(u, T*(v))forall u,vE... |
ts of Nonnegative Symmetric Operators) Let T E ,C(H) be a nonnegative symmetric operator. A nonnegative symmetric operator A E C(H) is called a square root of T provided A2 = T. Use the inductive construction in the preceding problem to show that T has a square root A which commutes with each operator in ,C(H) that com... |
of Let S be the unit sphere in H. According to the Hilbert-Schmidt Lemma, we may choose a vector 411 E S and µi c- R for which K(1) = µt 4,1 and Iµ11=sup I (K(h ), h) I hES Since K :f- 0, we infer from Proposition 16 that µ1 t- 0. Define H1 = [span t4,1)1-1. Since K(span 14,11) C span (411 }, it follows from Propositio... |
t Spaces Corollary 23 (the Fredholm Alternative) Let H be a Hilbert space, K E L(H) compact, and µ a nonzero real number. Then exactly one of the following holds: (i) There is a nonzero solution of the following equation p.h-K(h)=0,heH. (ii) For every h0 E H, there is a unique solution of the equation µh-K(h)=ho,heH. D... |
set in M that contains xo and 0 to a set that does not contain xo: this defines the Dirac measure space (X, M, Sxo). A slightly bizarre example is the following: let X be any uncountable set and C the collection of those subsets of X that are either countable or the complement of a countable set. Then C is a if-algebr... |
ng that may occur is that v is not always nonnegative. Moreover, v(E) is not even defined for E E M such that µ1(E) =,u2(E) = no. With these considerations in mind we make the following definition. Definition By a signed measure v on the measurable space (X, M) we mean an extended real-valued set function v: M [-oo, oo... |
ts in S, then n µ(E) 1 tl(Ek) k=1 To see this, set Ek = 0 for k > n. In particular, such a set function µ is monotone in the sense that if A and B belong to S and A C B, then µ(A) < µ(B). Definition A set function µ*: 2X and p* is countably monotone. [0, oo] is called an outer measure provided µ*(0) = 0 Section 17.3 Th... |
induced by the set function µ: S -> [0, oo) and the o--algebra of measurable sets. 22. On the collection S = {0, [1, 2]} of subsets of R, define the set functionµ: S -* [0, oo) as follows: µ(0) = 0, µ([1, 2]) = 1. Determine the outer measure µ* induced by µ and the o-algebra of measurable sets. 23. On the collection S ... |
prove uniqueness it suffices to show that µ and µ1 agree on the measurable sets contained in each Xk. Let E be measurable with E C E0, where Eo E S and µ(Eo) < oo. We will show that µ(E) = µ1(E) (12) According to Proposition 10, there is a set A E Ss for which E C A and µ(A - E) = 0. We may assume that A C E0. However,... |
a real-valued function on X. Then f is measurable if and only if for each open set 0 of real numbers, f-1(0) is measurable. For a measurable space (X, M) and measurable subset E of X, we call an extended real-valued function f that is defined on E measurable provided it is measurable with respect to the measurable spa... |
0 on X and g = XE. Show that f = g a.e. on X while f is measurable and g is not. 3. Suppose (X, M, µ) is not complete. Show that there is a sequence { f } of measurable functions on X that converges pointwise a.e. on X to a function f that is not measurable. 4. Let E be a measurable subset of X and f an extended real-... |
simple function for which 0 < f on X, then jPdP$liminfffnd/L. (12) Let cp be such a function. This inequality clearly holds if fx dµ = 0. Assume fx p dµ > 0. Case 1: fx (P dA = no. Then there is a measurable set X,,, C X and a > 0 for whichA (X,,,,) = no and cp = a on X,,,,. For each natural number n, define An={xEX I... |
ose points in X at which h and g take infinite values of opposite sign. Theorem 12 Let (X, M, µ) be a measure space and f and g be integrable over X. (Linearity) For real numbers a and 0, a f + lag is integrable over X and f[af+/3gJd/L= x a fx f dµ + l3 fx gdµ. (Monotonicity) If f < g a.e. on X, then ffd,Lfgd/L. x 374 ... |
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