0.1 Properties of regulated functions and their Integrals.


 Percival Welch
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1 MA244 Anlysis III Solutions. Sheet 2. NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!. Properties of regulted functions nd their Integrls.. (Q.) Pick ny ɛ >. As f, g re regulted, there exist φ, ψ S[, b]: f φ < ɛ/2, g ψ < ɛ/2. Notice tht ψ + φ S[, b] (Assignment ). But Therefore f + g R[, b]. f + g φ ψ f φ + g ψ < ɛ. To show tht fg R[, b], notice tht regulted functions re bounded. (Assignment.) Therefore there exist M, N > : f < M, g < N. Let φ, ψ S[, b]: f φ < δ/m, g ψ < δ/n, Notice tht φψ S[, b]. (Assignment.) Then fg φψ = f(g ψ) + g(f φ) + (f φ)(ψ g) f(g ψ) + (f φ)g + (f φ)(ψ g) 2δ + δ 2 /(MN). For ny ɛ > the eqution 2δ+δ 2 /(MN) = ɛ hs positive solution nd we re done. The lst inequlity bove uses the following property of the supnorm: b = sup x (x) b(x) sup x (x) sup b(y) = b y 2. (Q.2) sin 2 (t ) is continuous hence regulted on [, b]. By Question, sin 2 (t )φ R[, b], so the question is well posed. Recll tht sin 2 (tx) = cos(2tx). By the 2 RL lemm, cos(2tx)φ(x)dx s t. Therefore, lim t sin 2 (tx)φ(x)dx = 2 φ(x)dx. 3. (Q.3) Let h := g f. Then h() = nd h(x) = sign(x sin(/x)) = sign(sin(/x)) for x >. So, h(x) = for x (, ) nd h(x) = for x 2π(n+/2) 2πn (, ). Such function is not regulted s for ny step function φ in 2πn 2π(n+/2) I, h φ > min c R ( c, + c ). 4. (Q. 4) Let f be defined s follows: f( + (b )/n) = /n, n N; f(x) = if x / +(b )/N. Note tht f R[, b]: consider φ k S[, b] : φ k (x) = f(x) for x +(b )/k nd φ k (x) = for x < +(b )/k. Then f φ k /k s k. Therefore, the sequence of step functions {φ k } converges to f uniformly nd f is regulted by definition. Of course, f itself is not step function s the crdinlity of its rnge is infinite. The integrl: f = lim k φ k =.
2 5. (Q.5) Let f(x ±) = lim y x ± f(y). According to the Theorem cited in the hint, if f R[, b], then f(x±) exists for ny x (, b) nd the corresponding onesided limits exist t the boundry points. The first step is to show tht the set of points S ɛ = {x [, b] : f(x+) f(x ) > ɛ} is finite. Notice tht S ɛ is the set of ll points of discontinuity of f with the jump size greter thn ɛ. As f is regulted, there exists φ S[, b] such tht for ny x [, b], f(x) φ(x) < ɛ/4. Tking onesided limits of the bove double inequlity, we find fter simple mnipultion: f(x+) f(x ) ɛ/2 φ(x+) φ(x ) f(x+) f(x ) + ɛ/2. Tke x S ɛ. As f(x+) f(x ) > ɛ, the bove inequlity implies tht φ(x+) φ(x ). But the set of discontinuities of step function is finite. As we hve just showed, tht S ɛ is contined in the set of discontinuities of φ, we cn conclude tht S ɛ <. Finlly, let us consider the sequence of sets S /n, n N. As it is esy to check, ny point of discontinuity of f is in S /n for some n. Note tht S S /2 S /3.... The enumertion of the set of the set of ll discontinuities cn be done inductively: the points of S cn be enumerted s S is finite. Assume tht ll points in S /n hve been enumerted by, 2,... N n. As the set S /(n+) \ S /n is finite, we cn enumerte its points strting with N n +. The result is the enumertion of ll points in S /(n+). As ny discontinuity of f is contined in S /n for some n, we hve estblished onetoone correspondence between the set of ll discontinuities of f nd N, which mens tht this set is countble by definition..2 Integrtion. 6. (Q.6) Let P N = 5 N {,, 2,... N} be sequence of prtitions of [, 5]. Let φ N : φ N () =, φ N (5/N)(k,k] = ((5/N) k) 2, k =, 2,..., N be sequence of step functions. Then f φ N mx k N (((5/N) k) 2 ((2/N) (k ) 2 )) 5/N s N. Therefore, φ N s converge to f uniformly. By definition, 5 f = lim N 5 5 φ N = lim N N But N k= k2 = x 2 N k= ekx N k= ( ) 3 5 N (5/N k) 2 == lim k 2 N N k= x= = 2 x e(n+)x e x x= = (/3)N 3 + (/2)N 2 + (/6)N. Substituting this into the previous formul nd computing the limit, we get 5 f = 53 3.
3 dx. Ech of the terms is differ log(x) 7. (Q.7) By dditivity, I = 2 entible by FTC2. As F (b) =, I = / log(), b I = / log(b). dx + log(x) 2 8. (Q.8) I(x) = J((x)) + J(b(x)), where J() = / log(t)dt. As J is differentible on the domin of, b nd, b re differentible on R, the function I 2 is differentible on R. So, we cn pply the chin rule: 9. (Q.9) I (x) = J ((x)) (x) + J (b(x))b (x) = (x) log (x) + b (x) log b(x). () F (x) = log (e x )e x = x e x (b) G (x) = 2x + ( + x 2 ) 4 2x + x 8 (c) H (x) = 2x(e x + e x ) = 4xe x (d) I (x) = 2 x cos(x2 ) + x 2 cos(/x 4 ). (Q.) For ll the integrls below the integrnds re continuous on the respective intervls of integrtion nd we cn pply the fundmentl theorem of clculus for their evlution. () Substitution: t = log(x). Then e sin(log(x))/xdx = sin(t)dt = cos(). (b) log(3) / log(2) cosh2 (x)dx = ( log(3) sinh(x) log(2) cosh(x)) dx = tnh(log(3)) tnh(log(2)) = /5. (c) Substitution: x = sin(t). Then / 2 x 2 dx = π/4 sign(cos(t))dt = π/4. (d) Substitution: t = log(x). Then e 2 e dx = 2 x log(x) dt = log(2). t (e) x e x dx = s the integrnd is odd nd the limits of integrtion re symmetric..3 Improper integrls.. (Q.) Let b n = n F, n n = n f. Note tht n n n f b n n. The fct tht F converges implies tht the series b n < n= As n < b n, the series n= n converges bsolutely by the comprison test N for series (Anlysis I). Therefore, lim N f exists nd is finite. To finish the proof, we need to show tht R n f converges for ny choice of sequence R n : lim n R n =. Let x be the smllest integer greter or equl thn x. Then R R R R R f f f f b R, R R
4 s n= b n converges. Therefore, R lim R f = lim R R where the right hnd side hs lredy been shown to converge. 2. (Q.2) The integrls e x3 nd e x3 dx diverge nd converge simultneously. But e x3 e x for ny x. Therefore, e x3 f, e x dx =. We conclude tht e x3 dx converges by the comprison principle of Question. 3. (Q.3) ( x 4 ) = ( x 2 )( + x 2 ) = ( x)( + x)( + x 2 ) ( x) for x [, ). Therefore, dx <. x 4 x Therefore the elliptic integrl converges by the comprison principle. 4. (Q.4) R R x3 dx = x4 4 R R =. However, xdx is divergent, s does not exist. lim R R x 3 dx + lim R 2 R2 x 3 dx 5. (Q.5) () dx 3 dx = lim x /3 +2x /4 +x 3 x /3 ɛ dx = lim ɛ x /3 ɛ. So the integrl converges by the comprison principle. 2 x2/3 ɛ < (b) The integrl sin(x) converges. (Use the comprison principle with x 2 F (x) = /x 2.) Therefore, sin(x) nd sin(x) dx diverge nd converge x 2 x 2 simultneously. But ( sin(x) sin(x) ) lim = lim cos(x) ɛ ɛ x 2 ɛ dx ɛ x ɛ x ( sin(x) ) = lim ɛ cos(x) log(x) ɛ sin(x) log(x)dx, ɛ x which diverges logrithmiclly (verify tht log(x) sin(x)dx converges!). Therefore, sin(x) dx diverges. x 2 ɛ.4 Uniform convergence. 6. (Q.6) For ny x (, ), lim n n k= xk = x f n (x) = n k= xk = xn x. Choose x n = /n (, ). Then lim f n(x n ) n x = lim n x n n := f(x) pointwise. Let x = lim n( n /n)n = +.
5 So choosing ɛ = in the negtion of the definition of the uniform convergence, we cn lwys find n nd x (, ) such tht f(x) f n (x) >. Therefore, k xk does not converge uniformly in (, ). On the other hnd, for ny x [ R, R], where R <, f n (x) f(x) Rn R for n, which immeditely implies the uniform convergence on [ R, R]. 7. (Q.7) Without restricting generlity, we cn consider symmetric intervls [ R, R]. For ny x [ R, R], n k= x k ex k=n Rk R k k= = Rk R k k=n R k n (n + )(n + 2)... (n + (k n)) = Rk er s k, where the lst step cn be justified using Stirling s formul. Therefore, the exponentil sequence converges uniformly on [ R, R], hence on ny finite subintervl of R. 8. (Q.8). If f n f uniformly on A, then for ny ɛ > there is N N: for ny n > N nd ny x A, f(x) f n (x) < ɛ/2. Therefore, sup x A f n (x) f(x) < ɛ nd we get the definition of lim n sup x A f n (x) f(x) =.. If sup x A f n (x) f(x) < ɛ, then for ny x A f n (x) f(x) < ɛ nd the the uniform convergence follows immeditely from lim n sup x A f n (x) f(x) =. 9. (Q.9) Let S n = x 2 x 4 + x 4 x x 2n x 2n+2 = x 2 x 2n+2. Clerly, lim n S n (x) = x 2 for x < nd lim n S n (x) = for x =. Consider x n = /n (, ). Then for ny δ >, there is n δ N such tht for ny n > n δ x 2 n S n = x 2n+2 n = ( /n) 2n+2 ( /n) n = e n log( /n) e δ.. Therefore, lim n sup x (,) S n (x) x 2 nd the convergence cnnot be uniform by Question (Q.2) Suppose the convergence F n F were not uniform. Negting the definition of the uniform convergence, we conclude tht there must exist c > nd strictly incresing sequence {p n } n N such tht sup (F pn (x) F (x)) > c, n =, 2,... x [,b] As F pn F is continuous on the closed intervl [, b], it reches its mximl vlue t some x pn [, b]. The sequence {x pn } n is bounded. By the Boltzno Weierstrss theorem it contins subsequence {x pnk } k, which converges to point x [, b]. By construction, F pnk (x pnk ) F (x pnk ) > c, k =, 2,...
6 As F n (x) > F m (x) for ny x nd m > n, we conclude tht F pn (x pnk ) F (x pnk ) > c. for ech n < n k. Tking the limit k in the bove inequlity nd using the continuity, we find tht F pn (x ) F p (x ) c >, n =, 2, 3,..., which contrdicts the pointwise convergence F n F t x. Therefore, F n must converge to F uniformly. November the 9th, 25 Sergey Nzrenko nd Oleg Zboronski.
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