Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.


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1 Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points nd ϕ be nondecresing function on [, b]. If ρ is convex function on [0, ), show tht ρ(f) R ϕ [, b]. Moreover, if ϕ(b) ϕ() = 1, show tht (3) Q4 of Tutoril on integrls. (4) Let ρ( f(x)dϕ) ρ(f(x))dϕ. 1 f(x) = if x = m, m, n N nd m nd n hve no common divisor n n 0 otherwise Determine if f is Riemnn integrble on [0, 1]. (5) Let f : [, b] IR be continuous function nd n = ( tht sup{ f(x) : x [, b]} = lim n n. f(x) n dx) 1/n ). Show (6) Let f be bounded function on [, b] such tht f R[c, b] for ll < c < b nd lim c + c f(x)dx exists. Show tht f R[, b] nd f(x)dx = lim c + c. f(x)dx. 51
2 52 RiemnnStieltjes integrls ( quick introduction) We will follow Apostol Mthemticl Anlysis (but we will not cover ll). You could lso red the first chpter of Mesure nd integrl by Wheeden nd Zymund for more detil (if you re seriously interested nd hve time to spre). Let ϕ, f : [, b] IR be bounded functions. We would like to extend the Riemnn integrl to f(x)dϕ so tht it is just the usul Riemnn integrl when ϕ(x) = x. Definition: One simple wy is to define this extension s fdϕ =: lim f(ξi )(ϕ(x i ) ϕ(x i 1 )) (if it exists) P 0 where the limit cn be defined s before for the limit of Riemnn sums. However, for esier tretment, we will tke the limit in the following weker sense : for ll ε > 0, there exists prtition P ε of [, b] such tht if P = (P, ξ), P =: = x 0 < x 1 < < x n = b, ξ = {ξ i : i = 1,, n} with ξ i [x i 1, x i ] is tgged prtition of [, b] such tht P is refinement of P ε, then We will write S( P, f, ϕ) γ = f(ξ i )(ϕ(x i ) ϕ(x i 1 )) γ < ε. lim S( P, f, ϕ) = lim f(ξi )(ϕ(x i ) ϕ(x i 1 )) = γ P 0 P 0 nd f R ϕ [, b]. We will sy the Riemnn Stieltjes integrl fdϕ is equl to γ. Unfortuntely, this definition is not equivlent to the one we use before (or the one used in Mesure nd integrl by Wheeden nd Zymund). But they re equivlent in mny cses, for exmple, when ϕ(x) = x (nd do you think they re equivlent when ϕ is continuous?) Uniqueness of limits
3 53 Obviously, there is no reson to expect the limit to exist in generl nd we should expect the limit ro exist if both f nd ϕ re resonbly nice. For exmple, ϕ is usully ssumed to be of bounded vrition. Tht is, n sup{ ϕ(x i ) ϕ(x i 1 ) : = x 0 < x 1 < < x n = b, n N} <. i=1 Note tht piecewise differentible function on [, b] will be of bounded vrition on [, b]. On the other hnd, there is continuous function on [, b] tht is not of bounded vrition. A function is sid to be piecewise differentible on [, b] if there exists prtition P =: = x 0 < x 1 < x n = b such tht f is differentible on (x i 1, x i ) nd f is uniformly continuous on (x i 1, x i ) for ll i. Proof: It suffices to show the cse f is uniformly continuous on (, b). Clssifiction of functions of bounded vrition A function on [, b] is of bounded vrition if nd only if it is difference of two monotone functions on [, b]. Remrk We shll skip the proof s it is stndrd subject treted in Honours yer Anlysis. Exercise: Let ϕ be step function on [, b], tht is there exists prtition P s bove such tht ϕ is constnt on (x i 1, x i ) for ll i. Then show tht for ny continuous function f on [, b], n 1 n fdϕ = f(x i )(ϕ(x + i ) ϕ(x i )) + f(x i )(ϕ(x i ) ϕ(x i )). i=0 i=1 Give n exmple of f nd ϕ such tht f is not Riemnn Stieltjes integrble. Cn you give generl criteri for tht?
4 54 The following theorems re esy consequences of the definition of the Riemnn Stieltjes integrl. Linerity of RiemnnStieltjes integrls (9.2,9.3 Apostol) (i) If fdϕ exists, then so do cfdϕ nd fd(cϕ) for ny constnt c; moreover, cfdϕ = fd(cϕ) = c fdϕ. (ii) If both f 1dϕ nd f 2dϕ exist, then so does (f 1 + f 2 )dϕ, moreover, (f 1 + f 2 )dϕ = f 1 dϕ + f 2 dϕ. (iii) If both fdϕ 1 nd fdϕ 2 exist, then so does fd(ϕ 1 + ϕ 2 ), moreover, Proof. fd(ϕ 1 + ϕ 2 ) = fdϕ 1 + fdϕ 2.
5 Theorem (9.4 Apostol) : Let c (, b). If two of the following three integrls in ( ) exist, then the third lso exists nd we hve 55 fdϕ = c fdϕ + c fdϕ ( ). Remrk Unfortuntely, this is not true under the other definition of Riemnn Stieltjes integrls (see Wheeden nd Zygmund). Tht is, there re functions f nd ϕ such tht both 1 0 fdϕ nd 0 1 fdϕ exist but 1 1 fdϕ does not exist (under the other definition, the exmple cn be found in Wheeden nd Zygmund ). In generl, under our definition, if f R ϕ [, b], then f R ϕ [, c] nd lso in R ϕ [c, b]. Theorem (Integrtion by prts, Apostol 9.6) ϕdf nd b ϕdf = f(b)ϕ(b) f()ϕ() If fdϕ exists, then so does fdϕ. Proof. Here is key observtion: n n+1 ϕ(ξ i )(f(x i ) f(x i 1 )) = f(b)ϕ(b) f()ϕ() f(x i 1 )(ϕ(ξ i ) ϕ(ξ i 1 )) i=1 i=1 = f(b)ϕ(b) f()ϕ() n+1 i=1 with ξ 0 = nd ξ n+1 = b. Note tht f(x i 1 )[(ϕ(x i 1 ) ϕ(ξ i 1 )) + (ϕ(ξ i ) ϕ(x i 1 )] ( 1) P o =: ξ 0 = x 0 ξ 1 x 1 ξ n x n = ξ n+1 = b is prtition of [, b] nd refinement of P =: x 0 < < x n = b. Thus if P ε is prtition of [, b] such tht S(f, P, ϕ) γ < ε for ny tgged prtition P tht is refinement of P ε ( 2)
6 56 We my ssume P =: x 0 < < x n = b, ξ i [x i 1, x i ], i = 1,, n. Then P o s bove is refinement of P (nd hence refinement of P ε ). By ( 2), we hve n+1 i=1 f(x i 1 )[(ϕ(x i 1 ) ϕ(ξ i 1 )) + (ϕ(ξ i ) ϕ(x i 1 )] γ < ε. Theorem (9.26 Apostol) [, b]), then fdϕ exists. If f is continuous nd ϕ is of bounded vrition (on Chnge of vribles (9.7 Apostol) Let fdϕ exist nd let g be strictly monotonic continuous function on [c, d] (or [d, c] with = g(c), b = g(d). Let h = f(g(x)) nd β(x) = ϕ(g(x)). Then d c hdβ exists nd = integrl will pply. fdϕ. Note tht in cse c > d, stndrd interprettion of the Theorem (Apostol 9.8) If fdϕ exists nd ϕ is continuously differentible (on [, b]), then fϕ dx exists nd = fdϕ.
7 Riemnn integrls nd sequences of functions Let (f n ) R ϕ [, b] nd (f n ) converges uniformly to f on [, b]. If ϕ is of bounded vrition on [, b], then f 57 R ϕ [, b] nd Proof. fdϕ = lim f n (x)dϕ. n Exmple nd ppliction ( 1) k. Derivtive of 1 0 k sin(xt2 )dt. k=1
8 58 Finl Remrks (1) Uniform convergence is too stringent, indeed, the following is true (but difficult to prove using Riemnn theory). Bounded convergence theorem Let (f n ) R[, b] nd (f n ) converges (pointwise) to f on [, b]. If f R[, b] nd there exists M > 0 such tht f n (x) M for ll x [, b] nd n N. Then fdx = lim f n (x)dx. n Its proof is very simple using Lebesgue s theory but complicted using only elementry tool. One could lso chnge uniformly bounded to monotone. Note tht it is necessry to ssume its limit function to be Riemnn integrble (s limit function in these two cses need not be Riemnn integrble nd this is the min problem of Riemnn theory). I m sorry tht you re not llowed to use these two fcts (2) Necessry nd sufficient conditions for Riemnn integrbility A bounded function f on [, b] is Riemnn integrble if nd only if it is continuous t lmost every point in [, b]. Tht is, f is continuous on [, b] \ E such tht E hs mesure 0 (i.e., for ny ε > 0, there exists {( i, b i )} i=1, E i=1( i, b i ) nd i=1 b i i < ε. Exmple: the following function is Riemnn integrble on [0, 1]: 1, if x = 1/n, n N f(x) = 0, otherwise. As the proof using Lebesgue theory of integrtion is much simpler, we shll not prove it in this module. However, if you re interested (nd hve time to spre), you could tke look of proof of Theorem 6.18 nd 6.19 (which is elementry but complicted). I m sorry tht you re not llowed to use this theorem in this module.
9 59 Improper integrls It seems too stringent to require function to be bounded before we could define integrls. Menwhile, we would lso like to define integrl on unbounded intervls. For these two purposes, We will need to extend Riemnn integrls to unbounded functions nd/or unbounded domins. First, we will define integrls for (possibly) unbounded functions. Let us recll n obvious fct: If f R[, b], then f(x)dx = lim f(x)dx. We will now define improper inte c + c grls. Let f : (, b] IR be such tht f R[c, b] for ll < c < b. We could define the integrl by tking limit: f(x)dx = lim f(x)dx r + r nd we will sy the improper integrl converges if the limit exists. Otherwise, we shll sy the improper integrl diverges (even if the limit equls to or ). Exmple f(x) = sin x x, g(x) = 1, h(x) = 1 x x.
10 60 Of course, Similr definition cn be defined if f R[, c] for ll < c < b. Exmple Next, if there exists c (, b) such tht f R[, c ε] nd f R[c + ε, b] for ll ε > 0 such tht < c ε nd c + ε < b, we will define r fdx = lim f(x)dx + lim f(x)dx if both limits exist. r c r c + c We could then extend the bove definition to more complicted cses. Next, we define integrls on unbounded intervls. Let us ssume f : [, ) IR such tht f R[, b] for ll b >. Then we define K f(x)dx = lim f(x)dx if the limit exists K nd we shll sy the improper integrl converges. Similr definition cn be defined for functions on (, b].
11 However, on (, ), if f R[, b] for ll [, b] (, ) nd tht both the limits we define K lim K 0 f(x)dx, lim K 0 K f(x)dx exist K 0 f(x)dx = lim f(x)dx + lim f(x)dx K 0 K K nd we will then sy the improper integrl exists. Note tht the choice of 0 s intermedite point is just for convenient, one could certinly just use ny point other thn Exmple Finl Remrk: Essentil questions (1) Why the choice of intermedite point cn be rbitrry? (2) Why integrl test for infinite series work? (3) Is there n lternting integrl test insted of lternting series test? Hint: use integrtion by prts. (4) Why does Cuchy criterion work for improper integrl? It is needed when we do not know the limit nd for comprison test to work.
12 62 Tutoril (1) Let lim f(x)/g(x) = L 0 nd f(x), g(x) re nonnegtive continuous func x b tions on [, b). If g(x)dx converges, show tht (2) Test of convergence of ech of the following improper integrls: 1 xdx 1 (i) 0 1 (ii) (log x) 2 dx x 0 (iii) e t2 t 2 sin x dt (iv) dx x (3) (v) (vii) 0 1 e cos x x dx (vi) sin x cos 2x dx 0 x e t sin tdt (viii) e t2 t 2 dt f(x)dx lso converges. (ix) log xdx (x) x log xdx 0 0 ( ) sin x 2 (xi) dx 0 x f(x, t)dx is sid to converge to F (t) uniformly on I if given ny ε > 0, there exists N > 0 such tht r f(x, t)dx F (t) < ε for ll r N, t I. Of course we re ssuming f t (x) = f(x, t) R[, c] for ll c > nd t I. Similr definition cn then be defined for improper integrls on [, b]. Test of uniform convergence of ech of the following improper integrls on the indicted intervls: (i) (ii) (iii) (iv) 1 0 (log t) 3 dt, x 0; 1 + xt x sin xy dy, x for ny > 0; x + y te xt dt, x for ny > 0; sin xt dt, t x IR.
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