Math 61CM  Solutions to homework 9


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1 Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ > 0 so tht for ny x (c δ, c) we hve f(x) l <. The right limit lim x c + f(x) is defined very similrly. Let f(x) be monotonic function on n intervl [, b]. (i) Let c (, b). Show tht the left nd right limits l(c) = lim x c f(x), r(c) = lim f(x) x c + exist. (ii) Show tht for ny n N there exist finitely mny points c in [, b] such tht l(c) r(c) > 1/n. (iii) Show tht f is continuous on [, b] except possibly t finite or countble set of points. Recll tht n infinite set S is countble if there is onetoone mp (tht is, bijection) between S nd N. Solution: We will ssume tht f is nondecresing; the other cse is similr. (i) We will only prove tht the left limit exist; the proof for the right limit is similr. Define the number ξ = sup x<c f(x); we will show tht ξ = lim x c f(x). Let > 0. Since ξ is supremum, it is lest upper bound, hence there exists d < c such tht f(d) > ξ. Tke δ = c d. Then for ny y (c δ, c) = (d, c), we hve f(d) f(y) ξ by monotonicity nd definition of supremum. It follows tht f(y) ξ < f(d) ξ <, which proves tht ξ = lim x c f(x). (ii) Assume by contrdiction tht for some n N, there exists infinitely mny points c such tht l(c) r(c) > 1/n. Notice tht we must hve l(c) r(c). Fix N N, nd pick N such points; 1
2 lbel them so tht they re in incresing order. Then, we cn write f(b) f() = f(b) f() + N ( r(c k ) + r(c k ) l(c k ) + l(c k )) N = f(b) r(c N ) + (r(c k ) l(c k )) 0 >1/n N/n. (l(c k+1 ) r(c k )) 0 N l(c 1 ) f() 0 Here ll the inequlities 0 follow from the definition of l(c) nd r(c) s infimum nd supremum s in prt (i), nd from the monotonicity of f. But f is defined on the closed intervl [, b], so f(b) f() must be finite number. Therefore, it cn not be lrger thn N/n for every N. (iii) Notice tht function g is continuous t x if nd onlf l(x) nd r(x) exist, nd l(x) = g(x) = r(x). Since f is monotone, we therefore know tht f is continuous t x if nd onlf l(x) = r(x), since we know by (i) tht these limits exist, nd lso we hve l(x) f(x) r(x). Let y be point of discontinuity of f. Then, we must hve r(y) l(y) > 0; let n y be the smllest nturl number such tht r(y) l(y) > 1/n y. Now, define sets D n = {y [, b] f is discontinuous t y nd n y = n}. Notice tht ech D n is finite, by the point (ii). Furthermore, the set of points of discontinuity of f is included in the union of ll the sets D n. Since countble union of finite sets is finite or countble, the set of points of discontinuity of f is finite or countble. This lst fct is esy to prove: number ll the elements of the first set, then continue nd number ll the elements of the second set, etc. Problem 2: A set S of rel numbers hs mesure zero if for ny > 0 there exists cover of S by t most countbly mnntervls I k such tht Note tht the intervls depend on. I k <. (i) Show tht ny countble subset of R hs mesure zero. (ii) Show tht union of countbly mny sets S k tht ech hve mesure zero, hs mesure zero. (iii) Give n exmple of set of mesure zero tht is not countble. Solution: (i) Let S be countble subset of R: cll its elements k for k N. Let > 0. Consider the intervls ( I k = k 2 k+2, k + ) 2 k+2. 2
3 Then the sets I k cover S since k I k for every k, nd furthermore I k = proving tht S is countble. I k = 2 k+1 = 2 <, 2 k+1. Therefore, (ii) Let > 0. Since S k hs mesure 0, there exists t most countbly mnntervls I n S k, nd such tht I n < 2 k+1. n=1 Then the collection of ll the intervls I n I = n,k n n=1 tht cover for ll n nd k covers S, nd the sum of their lengths is I n < 2 k+1 = 2 <. It remins to prove tht countble union of finite or countble sets is finite or countble. Since countble union of finite or countble sets is the union of finite or countble union of finite sets nd of finite or countble union of countble sets, it suffices to prove tht finite or countble union of finite sets is finite or countble nd tht finite or countble union of countble sets is finite or countble 1. A finite or countble union of finite sets is finite or countble: see prt (iii) of question 2. A finite or countble union of countble sets is finite or countble: if the union is finite, we cn lbel it by {1,, N} N, which we cn lbel by N by counting the first row, then the second row, etc. If the union is countble, we cn lbel it by N 2, which we cn lbel by N by doing zigzg strting from the origin. (iii) The Cntor set ( hs mesure zero but is not countble. I invite you to red the Wikipedi pge, but here is sketch of wht is going on. The Cntor set C is defined recursively, strting from the intervl C 0 = [0, 1]. We obtin C 1 by removing the open centrl third of C 0, i.e. C 1 = [0, 1/3] [2/3, 1]. Then, C 2 is obtined by removing the open centrl third of ech of the two intervls of C 1 ; therefore, C 2 hs 4 intervls. Keep going this wy, nd define C s the limit: C = n C n. Since the totl length of C n is (2/3) n, nd since C is contined in C n for every n, the 2 n intervls tht compose C n form cover for C, tht hs length equl to (2/3) n. By tking n lrge enough, this cn be mde smller thn ny. This proves tht C hs mesure 0. To see tht C is not countble, notice tht to ech point x in C we cn ssocite sequence of 0 s nd 1 s: the first term is 0 if x is in the left piece of C 1, nd 1 otherwise. Then, in C 2, the intervl of C 1 in which x is gets cut in two; define the second term of the sequence s 0 if x is in the left piece nd 1 if x is in the right piece. Keep going this wy. This defines mp from C to ll sequences of 0 s nd 1 s; furthermore, it is surjective mp (by the nested intervls lemms or something). Now, usul digonl rgument shows tht the set of ll such sequences is not countble: if it is countble, lbel them by N, nd crete sequence whose n th term is not the n th term of the n th sequence. This new sequence cn not be in the list, by construction. A contrdiction follows. Therefore, since C mps surjectively to noncountble set, C is noncountble. 1 I ve lwys wnted to write this sentence. Del with it. 3
4 Problem 3: Let f nd g be continuous functions defined on n intervl [, b]. (i) Show tht there exists c [, b] so tht f(x)dx = f(c)(b ). (ii) Show tht if g(x) 0 for ll x [, b] then there exists c [, b] so tht f(x)g(x)dx = f(c) g(x)dx. Solution: (i) This follows from prt (ii) with g(x) = 1, but we will prove it independently (lthough in the sme wy). Notice tht m f(x) M for every x [, b], where m = min x [,b] f(x) nd M = mx x [,b] f(x) (both exist nd re chieved by some point in [, b] becuse f is continuous on compct set). It follows tht the functions f(x) m nd M f(x) re both nonnegtive nd continuous; it follows from point (ii) (positivity) of Proposition 10.9 in the notes tht (f(x) m)dx 0 nd n (M f(x))dx 0. It follows then from point (i) of the sme proposition tht m(b ) = mdx f(x)dx Mdx = M(b ). Now, since f is continous on compct set, there exists x [, b] nd x + [, b] such tht f(x ) = m nd f(x + ) = M. Since f(x ) = m f(x)dx b M = f(x + ), it follows from the intermedite vlue theorem (recll tht f is continuous) tht there exists c [, b] between x nd x + such tht f(c) = f(x)dx b. (ii) Let m, M, x nd x + be defined s before. We hve mg(x)dx f(x)g(x)dx Mg(x)dx If g(x) = 0 for ll x, then the equlits clerly true. So we cn ssume tht g(x) 0 for some x, which implies tht g(x)dx > 0 since g is continuous nd nonnegtive. So we cn divide bt: f(x ) = m f(x)dx b g(x)dx M = f(x +), nd the conclusion follows s before. 4
5 Problem 4: Let function f be continuously differentible on n intervl [, b]. Show tht f cn be represented s difference of two nondecresing functions on [, b]. Solution: Let g(x) = mx(f (x), 0) nd h(x) = min(f (x), 0); these re two nonnegtive continuous functions (since f is continuously differentible, nd minimum or mximum of continuous functions is continuous). Notice lso tht g(x) h(x) = f (x). Now, using the fundmentl theorem, we cn write f(x) = f() + f (y)dy = f() + This proves tht f is the difference of the functions nd G(x) = f() + H(x) = g(y)dy h(y)dy. g(y)dy h(y)dy. Furthermore, since g nd h re nonnegtive, the functions G nd H re nondecresing (this is esy to see using the properties (i) nd (ii) of the Proposition 10.9). Problem 5: (i) Show, without using the contrction mpping principle, tht if ll eigenvlues λ k, k = 1,... n, of symmetric n n mtrix A stisfy λ k < 1, then the mp F (x) = y + Ax : R n R n with some y R n fixed, hs unique fixed point. (ii) Give n exmple of mpping A tht mps complete metric spce X to itself, so tht d(a(x 1 ), A(x 2 )) < d(x 1, x 2 ) for ll x 1, x 2 but A does not hve fixed point. Hint: let l 1 be the spce of ll sequences {x n } with x n R such tht x k <, with the norm x = x k. Fix y l 1 nd consider mp A : l 1 l 1, with [A(x)] = y k + λ k x k. Find y l 1 nd numbers λ k so tht A(x) A(u) l1 < x u l1 for ll x, u l 1 but there is no x l 1 so tht A(x) = x. Solution: (i) By the spectrl theorem, there exists bsis {v 1,, v n } of R n consisting of eigenvectors of A, where v i is n eigenvectors corresponding to λ i. Denote by (y 1,, y n ) nd (x 1,, x n ) the coordintes of y nd x in tht bsis, i.e. x = x i v i nd y = v i. Then the eqution x = y+ax rewrites s x i = + λ i x i for ll i, which we cn rewrite s x i = 1 λ i. This mkes sense, since λ i < 1, hence we cn divide by 1 λ i. It follows tht is the unique fixed point of F. x = n i=1 5 1 λ i v i
6 (ii) Let us first see wht exctly we re looking for. For ll sequences x, denote x i its i th component. The eqution A(x) = x then rewrites s x i = 1 λ i for ll i, nd the inequlity A(x) A(u) l1 < x u l1 rewrites s λ i x i u i < i=1 x i u i. i=1 The inequlits certinly stisfied if λ i < 1 for ll i. Concerning the first condition, we just need to find nd λ i such tht y l 1, but ( yi 1 λ i ) in l 1 since the series i 1/i2 is summble, but x is not in l 1 becuse x i = summble. / l 1. Tke = 1 nd λ i 2 i = 1 1 i. Then s 1 λ i = 1/i2 1/i = 1 i is not Proving tht l 1 is complete is not very difficult, nd quite similr to the proof tht R n is complete. Here is sketch of how it works: if (y n ) n is Cuchy sequence of elements in l 1 (i.e. ech y n is itself sequence), then ech component must lso form Cuchy sequence (see my solutions to problem 6 of homework 4). So we know wht the limit of (y n ) n should be: it is the sequence formed by the limit of ech component. Then it is esy to show tht the limit is ctulln l 1 itself. Similrly, it is not difficult to show tht y n must ctully converge in l 1 to this limit, using the fct tht the components of ech y n must go to 0 by definition of l 1. 6
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