Homework 11. Andrew Ma November 30, sin x (1+x) (1+x)


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1 Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin b +b converges to bsolutely (one cn compre the bsolute vlue to which converges to ), the epression converges to. Therefore sin (+) d. (+b) + d = Clim: (stting thm.) Suppose F, G re differentible functions on [, b] for ny rel b >, F = f R nd G = g R, F ()g() d, f()g() d eist, nd lim b F (b)g(b) F ()G() =. Then F ()g() d = f()g() d Pf: For fied b, by thm. 6. (integrtion by prts) F ()g() d = F (b)g(b) F ()G() f()g() d By pplying limits to both sides you get, b lim b F ()g() d = lim b F (b)g(b) F ()G() f()g() d = F ()g() d = lim b F (b)g(b) F ()G() lim b f()g() d = f()g() d which shows the clim. Clim: sin (+) d converges bsolutely sin Pf: (+) (+) by thm. 6. sin (+) d (+) d for ll b. Since b (+) d = +b + by pplying the limit s b this epression converges to, so tht (+) d converges. Since the this lrger integrl con verges the smller integrl, Clim: + sin (+) d converges. d converges bsolutely. Pf: Consider lower step function f() = { / + π+nπ+ 5 6 π [π/ + nπ + π/6, π/ + nπ π], n elsewhere (I chosen this step function in the hopes tht it will be ectly under ech of
2 the pes creted by the function + ). I m struggling to show tht this step function relly is bdd. bove by the function +. I d lie to rgue tht the function is concve down on the intervls [π/+nπ+π/6, π/+nπ+ 5 6π], n N (this could be shown by ting two dedrivtives of the function nd seeing tht the second derivtives re positive over the intervls). Then I could sy, let = π/ + nπ + π/6, b = π/ + nπ π nd then = ( λ) + λb for some λ nd then by concvity ( λ) cos + λ cos b = / (by using trig identities to bre up cos nd cos b ppropitely). Then since bounds / on these intervls, the function f() is bdd. by + on ech of these intervls. Finlly, if you consider the integrl of the step function, tht is f() d = n= / + π+nπ+ 5 6 π 3 π (becuse 3π is the width of ech intervl). Net note tht this sum is greter or equl to some constnt, I ll cll times the series n= n. Since the series n= n diverges the integrl f() d diverges, nd therefore the integrl + diverges. Problem () Clim: If u, v then uv up p + vq q with equlity iff up = v q Pf: First I ll show the inequlity. Since the inequlity if u or v were, ssume both re positive. uv = e ln uv = e ln u+ln v = e p ln up + q ln vq. Becuse e is conve function, e p ln up + q ln vq p eln up + q eln vq = up p + vq q Now I wts the iff. I believe this comes from the fct tht e is not simply montoniclly incresing, but it is lso strictly incresing, so tht < y = e < e y (to see this one could te derivtive of e nd see tht the derivtive is lwys positive). In this cse, e p ln up + q ln vq = p eln up + q eln vq iff ln u p = ln v q, which hppens iff u p = v q however I relize tht this is ll very nonrigorous. I hve been hving trouble finding wy to do this proof rigorously. (b) Clim: If f, g R(α), {, } nd f p dα = gq dα = then fg dα Pf: First since the f, g R(α) the product is lso integrble. Now consider fg dα by prt (), fg dα f p p + gq q dα = f p p dα + g q q dα = b p f p dα + b q gq dα = p + q =. (c) Clim: fg dα ( f p dα) /p ( g q dα) /q Pf: Let A = f p dα nd B = g q dα. If A, B = then the inequlity would be trivil, so ssume tht both A, B >. Net note tht ( f ) p dα = A /p b ( g ) q dα =. Then by prt (b), f g = ( B /q A /p B /q ( f ) p dα) /p ( A /p ( g ) q dα) /q = B /q
3 A /p B /q b f g A/p ( f p dα) /p B /q ( g q dα) /q = f g ( f p dα) /p ( g q dα) /q. Now since fg dα f g dα ( f p dα) /p ( g q dα) /q this shows the clim. (d) This inequlity cn be shown by pplying limits to both sides of the inequlity. On the LHS, becuse fg dα is the composition of cts. functions, lim b fg dα = lim b fg dα = fg dα. On the RHS, this is the product nd compositions of cts. functions, therefore lim b ( f p dα) /p ( (lim b f p dα) /p b (lim b g q dα) /q = ( f p dα) /p ( Putting this together, one gets which is wht I ws sed to show. 3 Problem fg dα ( g q dα) /q = g q dα) /q. f p dα) /p ( g q dα) /q, Clim: f h f g + g h Pf: This sttement is equivlent to showing ( f h ) / dα ( f g dα) / + ( g h dα) / which is equivlent to showing f h dα f g dα + ( f g dα) / ( g h dα) / + g h dα which is the epression one gets by squring both sides. To see tht this lst inequlity holds, relize tht f h dα f g + g h dα by using the tringle inequlity (which comes from the Cuchy Schwrz inequlity). Net see tht f g + g h dα = f g dα + g h dα f g dα + ( f g dα) / ( g h dα) / + g h dα (becuse ( f g dα) / ( g h dα) / ). Then by trnsitivity, we get tht f h dα f g dα+( f g dα) / ( g h dα) / + g h dα which is the sttement I ws trying to show. 4 Problem Clim: {f n + g n } converges uniformly. Pf: Sy f n f nd g n g. Then given ɛ >. Te N s.t. n N = f n f < ɛ/ nd te N s.t. n N = g n g < ɛ/. Let N = m{n, N }. Then n N = f n + g n (f + g) f n f + g n g < ɛ. Therefore {f n + g n } converges uniformly to f + g. Clim: {f n g n } converges uniformly. Pf: Becuse {f n }, {g n } converge nd re sequences of bdd. functions, by problem in chpter 7 (it ws n unssigned problem), {f n }, {g n } re both unifmorly bdd. So let M be the uniform bdd. for {f n } nd let M be the uniform bdd. for {g n }. Now given ɛ >. Te N s.t. n N = f n f < ɛ M 3
4 nd te N s.t. n N = g n g < ɛ M. Let N = m{n, N }. Then n N = f n g n fg = f n g n f n g+f n g fg f n g n f n g + f n g fg M g n g + M f n f < ɛ. 5 Problem 3 Let f n () = n + nd note tht f n converges uniformly to f() =, becuse given n ɛ > te N s.t. N < ɛ. Then f n f = n < ɛ. Also let g n() = nd note tht g n () clerly converges to. Consider the sequence of functions {h n } which is the product of the two sequences, tht is h n = ( n + )( ) = n +. It is cler tht the h n () converge pointwise to the function however the h n s don t converge uniformly to this function becuse given n ɛ > nd fied N, for n N h n () = n + = n nd this epression cn be mde to be lrger thn ɛ by ting n s.t. < 6 Problem 4 Clim: The series converges bsolutely for ny vlue of, Z. Pf: It is cler tht the series doesn t converge bsolutely for = nd if = Z, then the denomintor in one of the terms of the series will be indefined. So now I wts tht the series will converge bsolutuely for ny other vlue. For fied, Z the term +n =. Note tht + n > for n lrge enough s.t. this inequlity, I cn rewrite the sum n= nɛ. +n < +n = n where K is some constnt. Since the series i nfty n= n n. Becuse of +n < K + i nfty n= n converges, the series i nfty n= n converges bsolutlely. converges nd therefore the originl series n= +n Clim: The series will converge uniformly only on closed intervls tht don t contin, Z Pf: Clerly the series won t even converge for closed intervl contining n =, Z s eplined previously. So I wts tht the series will converge on ny other closed intervl. Consider closed intervl [, b] which does not contin ny of the bd points described erlier. let α = if < b or let α = b if b < nd then note tht [, b] + n α + n > n for n lrge enough. Apply well ordering, nd sy tht is the first integer for which this inequlity holds. Then pply Thm. 7. nd let M n = +n for n < nd let M n = b n Mn = n= + n= b n thm. 7. the oringl series converges uniformly. for n. Then series M n converges becuse which converges. Becuse M n converges, by 4
5 Clim: The series is cts. everywhere it converges Pf: Since the prtil sums re cts. over every intervl where the series converges nd it converges uniformly on these intervls toom by thm. 7. the series will be cts. Clim: The series in generl is not bdd. Pf: if you use = then the series becomes n infintie series of s, which clerly diverges. 7 Problem 7 Clim: f n () = +n converges uniformly to f() =. Pf: I believe tht it is cler tht the f n () converges pointwise to (to see this one could pply L Hopitl s rule), so I ll try to show tht the function converges uniformly. So consider M n = sup f n () f() = sup f n (). Clim: M n = n Note tht f n () = +n = +n. Now note tht + n n >, becuse ( n ) = n n + = n + n = + n n. Since wlog, I cn ssume in the epression +n is positive (I could fctor out minus otherwise) I cn sy tht +n n with equlity when = n. Therefore M n = n. Since M n = n nd the lim n M n =, by thm. 7.9 the functions f n () converge uniformly. Clim: f () = lim n f n() is correct if = nd flse if. Pf: Since f() = then f () =. By pplying the quotient rule, f n() = n (+n ). If = then f n() = nd in this cse lim n f n() = f (). If then I cn pply L Hopitl s rule to this limit to sy tht n = lim n lim n (+n ) = lim n (+n ) (+n ) =. Therefore lim n f n() = = f (). 8 Prt of Problem 3 () Following the hint, by thm. 7.3 there is subseq. {f ni } which converges uniformly on ll rtionl numbers, r to sy f(r). Define f() = sup r f(r). Clim: f() = lim i f ni () t every where f() is cts. Pf: Note tht f ni (r) f ni () f ni (q) for ll r, q Q s.t. r < < q (by montonicity). Net pply limits to this inequlity to get lim i f ni (r) lim i f ni () lim i f ni (q) = f(r) lim i f ni () f(q). Since 5
6 f is cts. t s rtionl numbers r, q f(r), f(q) f(). Thus the inequlity becomes f() lim i f ni () f(). This shows the clim. Now since the number of pts. of discontinuity of f is countble, by thm. 7.3 gin, there is subseq. of {f ni }, I ll cll this seq. {f n } which converges to f. This completes the proof to prt (). 9 Other Problems () Clim: If f is cts on [, ] nd differentible t some pt. in (, ) then there is some N s.t. f A N. Pf: Sy is pt. where f is differentible. Let f () = c then te N > c. Then there eists ɛ > s.t. c + ɛ < N. δ > s.t. t < δ = f(t) f() < c + ɛ. Then te N s.t. N < δ. Let N = m{n, N }. Then for ll t < f(t) f() N = < c + ɛ < N. Therefore f A N. (b)since f is cts. over closed intervl, f is lso uniformly cts. So given ɛ >, te δ > s.t. y < δ = f() f(y) < ɛ/. Using δ prtition the intervl [, ] into intervls of size less thn δ. Cll this prtition P = { = < < n = }. Then define g() on ech intervl [ i, i+ ] s g() = ( λ)f( i ) + λf( i+ ) where λ = i i+ i for [ i, i+ ]. Clim: For ech intervl [ i, i+ ], f g < ɛ. Pf: Te ny [ i, i+ ]. By the choosing of the intervl nd construction of g(), g() g( i ) < ɛ/. Additionlly, by the choosing of the intervl, g( i ) f() < ɛ/. Then by the tringle inequlity, f() g() g() g( i ) + g( i ) f() < ɛ. Since ws rbitrry, this shows the clim. Since there re only finite number of intervls nd on ech intervl f g < ɛ, it is sfe to sy tht f g < ɛ on the entire intervl [, ]. (c) I will me n wful ttempt t constructing h, disclimer: it is.m. s I m writing this. Becuse g is piecewise liner g will be differentible on intervls. Let i = g () for [ i, i+ ] nd let = m i n i. Then given N nd ɛ >, te rel number n ɛ so smll s.t. n N +.Then let n = min{n, N }. Now consider the function s() = M sin π n, where M is ten to be so lrge so tht h(t) h() M = ɛ, nd let h() = g() + s(). Now for ny in [, ], s(t) s() g(t) g() (using reverse tringle inequlity) s(t) s(). By the wy s() ws constructed, s() will trvel t lest one full period in nbhd. of rdius /N round the pt.. Therefore there will be some t in the nbhd. round s.t. s(t) s() ɛ. Then note tht ɛ ɛ n N +. Using this fct, I cn sy tht s(t) s() N + = N. So overll I hve shown 6
7 tht for ny in [, ], h(t) h() N. Therefore, h() / A N. (d) I m not sure how to do this one!!! I ws thining of doing proof to show tht for ny open set G C I would show tht there is n open subset G G s.t G A N =. To do this I would te cts. function from G nd then by prts (b) nd (c), construct new function, h(), in G s.t. h() / A N. Then I would te nbhd. of size ɛ round h() so smll s.t. ny function f s.t. f h < ɛ = f / A N. However I could not figure out how to precisely show this impliction with delt epsilon rgument. (e) Without (d) I did not thin much bout this problem, however, if I hd to te guess, it would be becuse piecewise liner function would me it esier to tl bout the steepness or even the slope of the function over n intervl. This would be crucil to doing prt (c), s dding something zigzggy to the function needs to te into ccount the contribution of the originl function to the slope. 7
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