Math 09B Homework Edward Burkard 3.5. Functions of Bounded Variation. 3. Signed Measures and Differentiation Exercise 30 Construct an increasing function on R whose set of discontinuities is Q. Let Q {q n } be an enumeration of the rationals. Define f by: fx) n χ [q n, )x) Let x, y R such that y < x and let q k be a rational between y and x, then: fx) fy) n χ [q n, )x) n χ [q n, )y) n n k > 0 n Thus f is an increasing function. Let q n be any rational number. Then given any δ > 0 we have: fq n + δ) fq n δ) n > 0 and thus f is discontinuous on Q since the limit of f as x q n is always bigger from the right. Now let j I. For any n we can find a η > 0 such that q i / j η, j + η) for i,..., n. Thus we have: fj + η) fj η) in+ i n+ i0 i n+ n+ n+ Letting n we have fj + η) fj η) 0; thus f is continuous at j. Therefore, f is continuous on I and discontinuous on Q. { x Exercise 3 Let fx) sin x ), x 0 0, x 0 and gx) { x sin x ), x 0 0, x 0. a. F and G are differentiable everywhere including x 0). b. F BV [, ]), but G / BV [, ]). a. Clearly both F and G are differentiable away from zero, so it remains to check that they are differentiable at zero. Check F : F 0 + h) F 0) h sin h lim lim 0 h sin h lim 0 h 0 this is proven using the squeeze theorem on the inequality: x x sin x ) x.) Check G: G0 + h) G0) h sin h 0 h sin lim lim h lim 0 h 0 this is proven using the squeeze theorem on the inequality: x x sin x ) x.) b. First I prove Example 3.5.c: If F is differenitable on R and F is bounded, then F BV [a, b]) for < a < b < ). Suppose that F K. Then by the mean value theorem on [a, b] we get that c [a, b] such that: F b) F a) F c)b a)
and thus F b) F a) Kb a). Let P {a x 0 < x < < x n b} be a partition of [a, b]. Then on each subinterval [x j, x j ] we can apply the MVT to get F x j ) F x j ) Kx j x j ). Thus we have: F x j ) F x j ) Kx j x j ) Kb a). j j T F b) T F a) sup{ n j F x j) F x j ) n N, a x 0 < x < < x n b} Kb a) <. Thus F is of bounded variation. { x sin Observe that f x) x cos x, x 0 0, x 0. Then we can see that f x) 4 on [, ]. Thus by the previous statement f BV [, ]). Now choose the partition of [, ] given by the points x i i+)π for i,..., n and including the points x 0 and x n let g) gx n ) + gx ) g0) k n ). Then: gx j+ ) gx j ) g) gx n ) + gx ) g0) + j0 n k n + gx j+ ) gx j ) j gx j+ ) gx j ) j n ) k n + )j j + )π + j )π j n ) k n + j + )π + j )π j Letting n we get a series that does not converge by comparison to the series n ). Thus g / BV [0, ]). Exercise 37 Suppose F : R C. There is a constant M such that F x) F y) M x y for all x, y R that is, F is Lipschitz continuous) iff F is absolutely continuous and F M a.e. ) Suppose that F is Lipschitz continuous. Then if a, b ),..., a n, b n ) are disjoint intervals such that b i a i < δ we get i F b i ) F a i ) i Thus given any ɛ > 0, if we choose δ < ɛ M Mb i a i ) Mδ. i we see that F is absolutely continuous. Observe: F F x + h) F x) M x + h x M h x) lim lim lim M h 0 h and thus F x) M. ) Assume that F is absolutely continuous and that F M. Also, WLOG assume x y. Then by theorem 3.35c we get: x x x F x) F y) F t) dt F t) dt M dt Mx y) and thus F is Lipschitz continuous. y y y
3 Exercise 40 Let F denote the Cantor function on [0, ], and set F x) 0 for x < 0 and F x) for x >. Let {[a n, b n ]} be an enumeration of the closed subintervals of [0, ] with rational endpoints, and let F n x) F x an b n a n ). Then G n F n is continuous and strictly increasing on [0, ], and G 0 a.e. Use Exercise 39. - If {F j } is a sequence of nonnegative increasing functions on [a, b] such that F x) F jx) < for all x [a, b], then F x) F j x) for a.e. x [a, b].) Let M n. Notice that Fnx) n n for all x R and n N. Since the series n n n converges, by Theorem 7.0 in Baby Rudin we have that n converges uniformly to G. Since the Cantor function is continuous, notice that each Fnx) is continuous, and thus the partials sums s n j x) : j F nx) n n unif. are continuous for all j. By the previous statement we ahve that s j G, and thus, by theorem 7. in Baby Rudin G is continuous. Since each Fnx) is a non-negative increasing function on R, and Gx) < for n all x R by exercise 39 of chapter 3 in Folland we have that G x) ) Fnx) F n n n x) n. But n F nx) 0 a.e. for all n. Thus we have that G x) 0 a.e. Exercise 4 A function F : a, b) R a < b ) is called convex if F λs + λ)t) λf s) + λ)f t) for all s, t a, b) and t 0, ). Geometrically, this says that the graph of F over the interval from s to t lies underneath the line segment joining x, F s)) to t, F t)).) a. F is convex iff for all s, t, s, t a, b) such that s s < t and s < t t, F t) F s) F t ) F s ) t s. b. F is convex iff F is absolutely continuous on every compact subinterval of a, b) and F is increasing on the set where it is defined). c. If F is convex and t 0 a, b), there exists β R such that F t) F t 0 ) βt t 0 ) for all t a, b). d. Jensen s Inequality) If X, M, µ) is a measure space with µx), g : X a, b) is in L µ), and F is convex on a, b), then ) F g dµ F g dµ. Let t 0 g dµ and t gx) in c), and integrate). [] a. ) Suppose that F is convex. Suppose we have s, t, s, t a, b) such that s s < t and s < t t. Let lx) t s [t x)f s) s x)f t )] be the line segment joining the points s, F s)) and t, F t )). Then we have that ls) F s) and lt ) F t ). Also, since F is convex, we have F x) lx) x [a, b]. Thus: F t) F s) the slope of a line is constant lt) F s) lt) ls) lt ) ls ) t s lt ) F s ) t s F t ) F s ) t s ) Assume for all s, t, s, t a, b) such that s s < t and s < t t F t) F s), t s F t ) F s ) t s. Let lt) be defined as above. Let s s and suppose there is a t [s, t ] such that F t) > lt). Then:
4 F t) F s) which contradicts our original assumption. F is convex. lt) F s) > lt) ls) lt ) ls) t s F t ) F s) t s F t ) F s ) t s b. ) Suppose F is convex on a, b). Let [s, t] a, b), and let M lim x s + F x) F t) lim x t x t. Then for δ sufficiently small we get F s + δ) F s) δ < M + and F t) F t δ) < M + δ Since F is convex, if a j, b j ) [s, t], with δ j b j a j, by part a), we have F s + δ j ) F s) δ j F b j) F a j ) δ j F t) F t δ j) δ j Then, if we let δ j < δ and M maxm +, M + ), we have F b j ) F a j ) maxf s + δ j ) F s), F t) F t δ j )) < Mδ j F x) F s) x s and M So if we choose intervals a j, b j ) such that b j a j δ, we have that each δ j < δ and so F bj ) F a j ) < Mδ j Mδ So F is absolutely continuous on [s, t]. Now, if a < s < s + h < t < t + h < b, by part a), we have and taking limits gives So F is increasing. F s + h) F s) h F t + h) F t) h F F s + h) F s) F t + h) F t) s) lim lim ) For the converse, let lx) be the line connecting a, F a )) and b, F b )), for a < a < b < b. Thus to show that F is convex, we only need to show that F x) lx) for x [a, b ]. If F is absolutely continuous, and F is increasing, then if there existed a t [a, b ] with F t) > lt), applying the mean value theorem twice would yield a c [a, t] and a d [t, b ] such that and F c) F t) F a ) t a > lt) la ) t a lb ) la ) b a F d) F b ) F t) b < lb ) lt) t b lb ) la ) t b a So c t d but F c) > F d) which contradicts the fact that F is increasing. So we must have that F is convex.
5 c. Suppose F is convex on a, b) and a < t 0 < b. Then we fix s < t 0 < t. Then by part a), if t 0 < t, we have F t 0 ) F s ) t 0 s F t) F t 0) F t0 ) F s ) ) F t) F t 0 ) t t 0 t 0 s t t 0 ) and if t < t 0, by part a) we have F t) F t 0 ) F t ) F t 0 ) t t 0 t F t) F t 0 ) t 0 ) So if we choose β min F t ) F t 0) t t 0, F t0) F s ) t 0 s then F t) F t 0 ) βt t 0 ) F t ) F t 0 ) t t 0 ) t t 0 ) for all t a, b). d. Let X, M, µ) be a measure space, with µx), g : X a, b) is in L µ), and F convex on a, b). If we let t 0 gdµ and t gx), and apply part c) we have ) F gx) F gdµ β gx) ) gdµ Since µx), we have that dµ, so integrating leaves constants unchanged. With this in mind, we integrate both sides and we get ) F gdµ F gdµ β gdµ β gdµ 0 So F ) gdµ F gdµ 4.. Topological Spaces. 4. Point Set Topology Exercise If cardx), there is a topology on X that is T 0 but not T. Let X be a set with more than elements. Let x X. Let T be the topology on X generated by the basis consisting of the empty set, the whole set, and all the one point sets except for the set x. Then T is a topology that is T 0 since given any point y X, y x, there is always an open set containing y but not x. But there is never an open set containing x but not y. Exercise 3 Every metric space is normal. If A, B are closed sets in the metric space X, ρ), consider the sets of points x where ρx, A) < ρx, B) or ρx, A) > ρx, B).) Let X, d) be a metric space. Let A and B be closed, disjoint subsets of X. Define à {x X ρx, A) < ρx, B)} and B {x X ρx, B) < ρx, A)}. Since A and B are closed, their complements are open, and thus if x A B C we have dx, B) > 0 which gives A à since if x A, dx, A) 0; similarly if x B we have dx, A) > 0, thus B B. Suppose that x B. Let ɛ dx, A) dx, B) > 0. Thus we can place the open ball B ɛ x) in B. We can 4 do this because given any y B ɛ x), by the triangle inequality: dy, B) dy, x) + dx, b), where b B. 4 Taking the infimum over all b B we arrive at: dy, B) inf dy, B) inf dy, x) + dx, b) inf dy, x) + inf dx, b) b B b B b B b B dy, x) + dx, B) < ɛ 4 + dx, B) < ɛ 4 + dx, A) ɛ dx, A). 4 Thus y B, therefore B is open. Similarly we have that à is open, and thus à and B are open sets surrounding A and B respectively. Thus X, d) is a metric space.
6 4.. Continuous Maps. Exercise 4 If X and Y are topological spaces, f : X Y is continuous iff fā) fa) for all A X iff f B) f B) for all B Y. Identify the properties as follows: a) f : X Y is continuous b) fā) fa) for all A X c) f B) f B) for all B Y a) b) Assume that f is continuous. Let A X. Let V be an open neighborhood of fx). Then f V ) is an open set in X containing x. Thus y A, y x such that y f V ). But then fy) V fa), and thus fx) fā). b) c) Assume b). Let B Y. Then f B) X. Thus by our assumption: ff B)) ff B)) B. Apply f to both sides and we get f B) f B) as desired. c) a) Assume c). Let C Y be a closed set. Then by c) we have f C) f C) f C). Thus f C) f C), which implies that f C) is closed in X. f is continuous. Exercise 5 If X is a topological space, A X is closed, and g CA) satisfies g 0 on A, then the extention of g to X defined by gx) 0 for x A C is continuous. { gx) if x A Denote the extention by gx) 0 if x A C. Let U gx) ga) be open. If 0 / U then g U) g U) which is clearly open. But if 0 U, g U) g U) A C. However, since A C is the complement of a closed set, it is open and thus g U) g U) A C is open. g is continuous. Exercise 6 Let X be a topological space, Y a Hausdorff space, and f, g continuous maps from X to Y. a. {x fx) gx)} is closed. b. If f g on a dense subset of X, then f g on all of X. a. Let C {x fx) gx)}. It suffices to show that C C : A is open. Let x A, then fx) gx). Since Y is T there are disjoint open neighborhoods U and V around fx) and gx) respectively. By the continuity of f and g, f U) and g V ) are open neighborhoods of x. Thus O : f U) g V ) is an open neighborhood of x on which fy) gy) y O. Thus y A and A is open. C is closed. b. Let the dense set be D. Then D C from part a). We know that D X so D C D X C C since C is closed. But then we have X C, and thus f g on all of X. 4.4. Compact Spaces. Exercise 43 For x [0, ), let a nx) n a n x) 0 or ) be the base- decimal expansion of x. If x is a dyadic rational, choose the expannsion such that a n x) 0 for n large.) Then the sequence a n in {0, } [0,) has no pointwise convergent subsequence. Hence {0, } [0,), with the product topology arising form the discrete topology on {0, }, is not sequentially compact. It is, however, compact, as we shall show in 4.6.) a n x) is the n th binary digit of x. Since binary decimal expansions are not always unique, choose them in a way so that they are i.e. for 0.a a...a n 0... instead choose the equivalent
7 0.a a...a n 000...). Now suppose {a n } has a convergent subsequence, say {a nj }. Choose y [0, ) so that { 0, j odd a nj y), j even Then {a nj } {0,, 0,, 0,,...}, which does not converge. 4.5. Locally Compact Hausdorff Spaces. Exercise 5 The one-point compactifcation of R n is homeomorphic to the n-sphere {x R n+ x }. First I will prove the following useful lemma: Lemma. A homeomorphism between locally compact Hausdorff spaces extends to a homeomorphism between their one-point compactifications. proof: Let X and Y be locally compact T spaces. Let f : X Y be a homeomorphism. Define X X {p} and Y Y {q} be the one-point compactifications of X and Y respectively. Define: fx) { fx), x X q, x p. It is clear that f is a bijection by construction. Recall the topology on the one-point compactification of a space T : τ {U U T open} {T \ C C T compact}. Now to show that f is continuous. Suppose that U Y is open. Then f U) f U) is open in X and thus open in X. Now suppose that C Y is compact. Then f Y \ C) f Y ) \ f C) X \ f C) which is open in X since f C) X is compact. Thus f is continuous. Moreover, f is a continuous bijection from a compact space to a T space, and thus is a homeomorphism. So by this lemma it suffices to show that R n is homeomorphic to S n \ {p}. Note that the one-point compactification of S n \ {p} is S n. Consider the map ϕ : S n \ {p} R n given by: ϕx) ϕx,..., x n+ ). x,..., x n ). This map is commonly known as Stereographic Projection. It has a well known inverse ψ : R n S n \ {p} given by ψy) ψy,..., y n ) + y y,..., y n, y ). Clearly these are both continuous as they are just quotients of polynomials. To see that they are bijective I show that their composition each way is the identity: ) y ϕψy)) ϕψy,...y n )) ϕ + y,..., y n + y, y ) y y + + y,..., y n + y ψϕx)) ψϕx,..., x n+ )) y + x + +x n x n+) x,..., x n++x n+ +x + +x n x n+) ) x x,..., x n, x + +x n x n+) x x n,...,, x ) n+ ) ) x n, x n+ )
8 This gives a homeomorphism between R n and S n \ {p}. Thus by the lemma above, we have that the one-point compactification of R n is homeomorphic to S n. Exercise 54 Let Q have the relative topology induced from R. a. Q is not locally compact. b. Q is σ-compact it is a countable union of singleton sets), but uniform convergence on singletons i.e., pointwise convergence) does not imply uniform convergence on compact subsets of Q. a. Suppose Q were locally compact. Let q Q. Suppose that K is a compact neighborhood of q in Q. But we have an open interval I Q such that q I I K. Now let j I. There is a sequence of rational numbers {r n } n entirely contained in I converging to j in R. Since this is a convergent sequence, all of its convergent subsequences converge to the same thing, namely j. Now notice that none of these subsequences can converge in Q because j / Q. Thus I is not sequentially compact, as it should be being a compact metric space. Thus all compact subsets of Q have empty interior which guarantees that Q cannot be locally compact. b. Consider the subset H { n n N} {0} of Q. This is compact since if you take any open cover of H, the element of the cover that contains 0 must contain all but finitely many elements of H since Q has the subspace topology, the rest follows easily. Now consider the sequence {f n } of functions on H given by f n x) x n. This converges pointwise i.e. uniformly on singletons) to fx) {, x 0, otherwise. If this were uniform convergence on H, the function it converged to would have to be continuous by Theorem 7. of Baby Rudin) since each of the f n s are continuous, however, clearly f is not continuous, thus f n unif. f. Zhang Problems ) A function F : [a, b] R is an indefinite integral i.e. F x) F a) x a F t)dt) iff F is absolutely continuous. ) Extend F to R by setting F x) 0 for x < a and F x) F b) for x > b. Then F x) 0 x / [a, b]. Thus F L m). Then by corollary 3.33 we have that F is absolutely continuous. ) Suppose that F is absolutely continuous on [a, b]. Extend F to R by setting F x) 0 for x < a and F x) F b) for x > b. Then by lemma 3.34, F BV ; and actually, moreover, F NBV since it is right continuous and F ) 0. Now by corollary 3.33, since F NBV and is absolutely continuous, F L m) and F x) x F t) dt. But F x) 0 on, a) so: F x) F a) + x where F is now viewed as a function in L [a, b], m). a F t) dt ) Show that B {B r x) r Q, x R n } forms a basis for the standard topology on R n. Why are the rationals sufficient? Let B r x) be any element of B where x x,..., x n ). Then it determines the open ball {y R n x y < r}. Consider the element x r 3, x + r 3 ) x n r 3, x n + r 3 ). This is an element of the standard basis for the standard topology on R n. Now choose any open set a, b ) a n, b n ) in the standard basis for R n. Let ɛ min i n {b i a i }. Let r be any rational number smaller than ɛ a+b 3, and let x,..., an+bn ). Then the open ball B r x) a, b ) a n, b n ). Thus this basis generates the standard topology on R. The rationals are sufficient because they are dense in R.
9 References [] Hemenway, Brett. http://psych.dnsalias.org/ brett/solutions/45b/hw.pdf