f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that

Transcription

1 Lecture 15 We have see that a sequece of cotiuous fuctios which is uiformly coverget produces a limit fuctio which is also cotiuous. We shall stregthe this result ow. Theorem 1 Let f : X R or (C) be a sequece of cotiuous fuctios. Let A X o which {f } coverges uiformly. The {f } coverges o the closure Ā of A to a fuctio f which is cotiuous. Proof: Let us fix a poit x 0 Ā. We must first of all show that the sequece {f (x 0 )} is coverget. Eough to show it is Cauchy. Give ɛ > 0 there exist 0 such that, m > 0 implies f (x) f m (x) < ɛ/3 for all x A. By cotiuity of f ad f m we ca fid δ > 0 such that d(x, x 0 ) < δ implies that f m (x) f m (x 0 ) + f (x) f (x 0 ) < 2ɛ/3. Now sice x 0 Ā, there exists x B δ(x 0 ) A. With the help of this x, we have f (x 0 ) f m (x 0 ) f m (x) f m (x 0 ) + f (x) f (x 0 ) + f (x) f m (x) < ɛ. Therefore, we have got a fuctio f : Ā R which is the limit of {f } ad the covergece is uiform o A. We ow wat to show that f is cotiuous at x 0. f(x) f(x 0 ) f(x) f (x) + f (x) f (x 0 ) + f (x 0 ) f(x 0 ) Give ɛ > 0 we ca choose N 1 such that > N 1 implies f(x) f (x) + f (x 0 ) f(x 0 ) < 2ɛ/3, for all x A.

2 Fix oe such. The by cotiuity of f we ca fid δ > 0 such that d(x, x 0 ) < δ implies f (x) = f((x 0 ) < ɛ/3. Oce agai sice B δ (x 0 ) A, has to be used to coclude the cotiuity of f at x 0. Remark 1 What about differetiabilty uder uiform covergece? We should be careful here as illustrated by the example: f (x) = x 1+x 2 o [0, 1]. This sequece coverges uiformly to the fuctio which is idetically 0. However the derived sequece f (x) = to a fuctio which is ot eve cotiuous. 1 x2 (1+x 2 ) 2 coverges It is also true that a uiform limit of a sequece of smooth fuctios ca be cotiuous but ot differetiable, or differetiable but ot cotiuously differetiable or... ad so o. O the positive side, we shall ow see that by cotrolig the limitig process of the derived sequece itself we get better results: Theorem 2 Let f : [a, b] R be a sequece of differetiable fuctios such that f coverges uiformly i [a, b] to a fuctio g. Also suppose for some x 0 [a, b], the sequece {f (x 0 )} is coverget. The the sequece f coverges uiformly to a fuctio f ad f = g = lim f. Proof: First we wat to show that f is uiformly coverget ad for this it is eough to show that it is uiformly Cauchy, i.e., give ɛ > 0 we must fid 0 such that, m > 0 implies f (x) f m (x) < ɛ, x [a, b] (1) Usig the hypothesis we get 1 such that, m > 1 implies f (x) f m(x) < ɛ, x [a, b]. (2) 2(b a) 1

3 Put φ m = f f m. Therefore by Mea Value theorem applied to φ m, we have φ m (x 1 ) φ m (x 2 ) x 1 x 2 < This is the same as ɛ 2(b a), x 1, x 2 [a, b], m, > 1. (3) f (x 1 ) f m (x 1 ) f m (x 2 ) + f (x 2 ) < x 1 x 2 2(b a) ɛ/2. (4) We ow use the fact that f (x 0 ) is coverget ad hece fid 2 such that, m > 2 implies f (x 0 ) f m (x 0 ) < ɛ/2. (5) Combiig the above two iequalities we coclude that f is uiformly Cauchy, f (x) f m (x) < ɛ, m, > max{ 1, 2 } (6) as required. Let ow f(x) = lim f (x). To show that f = g : Now fix a x 2 [a, b] ad put h (x 1 ) = f(x 1) f (x 2 ) x 1 x 2. The (11) implies that h is uiformly Cauchy i [a, b] \ {x 2 } ad hece coverges to a cotiuous fuctio h(x 1 ) which is othig but f (x) f (x 2 ) lim = f(x 1) f(x 2 ). x 1 x 2 x 1 x 2 Therefore the limit fuctio is cotiuous o the closure of [a, b] \ {x 2 } which is [a, b]. We ca ow iterchage the takiog limit with respect to with limit with respect to x, i.e., = lim x 2 x 1 g(x 1 ) = lim f f (x 2 ) f (x 1 ) (x 1 ) = lim lim x 2 x 1 x 2 x 1 f (x 2 ) f (x 1 ) lim x 2 x 1 = lim x 2 x 1 f(x 2 ) f(x 1 ) x 2 x 1 = f (x 1 ). 2

4 Lecture 16 Cesaro Summability Give a sequece {a } of complex umbers, a method T first associates aother sequece {t } to it ad the takes the limit of {t }. If this limit exists ad is equal to L the we say {a } is T -coverget to the T-limit L ad write T lim a = L; OR lim a = L(T ). Example 1 Series summatio is such a method i which t is just the th partial sum of the give sequece. Aother method is called (C, 1)-summable (Cesero-1) i which t = a = a 1+ +a. Note that if the sequece a L, the it is (C, 1)-summable to the sum L. Proof: a L, which is the same as sayig lim (a L) = 0. Give ɛ > 0 there is a N 0 such that a L < ɛ/2 for N 0. Also, the sequece {a L} is bouded ad so there is M > 0 such that a L < M for all. Therefore t L = a 1+ +a L N 0M+( N 0 )ɛ/2 ad so o. N 0M + ɛ/2 Aother example is a = ( 1). Of course the sequece is ot coverget. But it is (C, 1)-summable to 0. The (C, 1)-limit is a good represetatio of the average. Example 2 More geerally, give k 1, we defie a sequece {a } to be (C, k)-summable to L if the sequece 1 t = ) ( + k 1 j j j=1 ( +k 1 1 ) a j L It is ot hard to check that if {a } is (C, k)-summable to L the it is (C, k + 1)-summable to L. Also, there are sequeces which are (C, k + 1)-summable but ot (C, k)-summable. For istace the sequece 1, 1, 2, 2, 3, 3,..., is ot (C, 1) summable but is (C, 2). Similarly the sequece 1, 2, 3, 4, 5, 6,... is ot (C, 2) summable but (C, 3). 3

5 Example 3 (Geeral Weighted Averages) Eve more geerally, give a sequece of positive real umbers P = {p 1, p 2,..., p,...}, we put P = j=1 p j ad we defie P-summability of a sequece {a } if the sequece j=1 t = a jp j P coverges to a limit L ad say P lim a = L. Check that each (C, k) i ideed a P method for some sequece P. Thus each Cesaro sum ca be thought of as a combiatorial (biomial) average. Defiitio 1 We say a summability method T is regular if wheever lim a = L the T lim a = L. What we have see above is that each (C, k) is regular. O the other had the series method is ot regular. Theorem 3 P is regular iff for each k, lim p k P = 0. (7) Proof: Suppose P is regular. Take a = 0, k ad = 1 for = k to see (7). Coversely, suppose (7) holds ad let a L. WLOG we may assume that L = 0. Give ɛ > 0 fid N 0 such that a < ɛ for N 0. The for each k N 0 fid N k such that p k P < ɛ/n 0 for N k. Take N = max{n 0,..., N N0 }. The for N we have t < ɛ(m + 1), where M is a boud { a }. Remark 2 I this sese series is ot a regular summability, whereas, all Cesaro summabilities are. Defiitio 2 Give a series a with partial sums {s }, we say that a is (C, 1)-summable to S if lim s = S(C, 1). 4

6 Ad the we write a = S. (C, 1). =0 Example 4 Cosider the series ( 1). Here the sequece of partial sums is 1, 0, 1, 0, 1, 0,... which is (C,1)-coverges to 1/2. Therefore we write ( 1) = 1/2 (C, 1). Notice that the (C, 1) limit of the sequece {( 1) } is equal to 0. So, you must pay attetio to this defiitio properly. Defiitio 3 A sequece {a } is called square summable OR is said to be of class l 2 if a2. We ca add two square summable sequeces to get aother such. Ideed square summable sequeces form a vector space. {1/} is i l 2 whereas { 1/} is ot i l 2. Thus there are several importat summatio methods oe ca use depedig upo oes requiremet. We shall meet (C, 1) summability agai while studyig Fourier series. You may cosult Goldberg s book for a elemetary expositio of this subject beyod what we have see so far. Coectedess Defiitio 4 Let X be ay topolgical space. We say X is coected if the oly subsets A X which are both ope ad closed i X are X ad. We say a subset A X is coected if the subspace A of X is coected. Theorem 4 Let X be a topological space. The the followig are equivalet: 5

7 (a) X is coected. (b) A B = X, both A ad B are ope, A B the A B. (c) A B = X, both A ad B are closed, A = B the A B. (d) A X is both ope ad closed the A = X. Theorem 5 A subset of R is coected iff it is a iterval. Proof: Suppose A R which is ot a iterval. This meas there exist x < z < y such that x, y A but z A. Put F = A (, z); G = A (z, ). The both F, G are ope i A oempty ad the uio is A. This is a cotradictio. Coversely, let A be a iterval i R, A = F G, x F, y G x < y. Assume that both F, G are closed i A. We shall show that F G. Put w = sup F [x, y]. The w A ad sice F is closed w F. Clearly, w y. Now for ay z such that w < z y, the z F ad hece z G. This meas w is a limit poit of G. Sice G is closed w G. Theorem 6 Let f : X Y be a cotiuous fuctio, A X is coected. The f(a) is coected. Proof: If ot we ca write f(a) = U V where U, V are both o empty ope i f(a) ad U V =. But the f 1 (U) ad f 1 (V ) are o empty ope i A ad A = f 1 (U) f 1 (V ) ad f 1 (U) f 1 (V ) =. THis meas A is ot coected. Theorem 7 Itermediate Value Property Let f : [a, b] R be a cotiuous fuctio. Let f(a) < z < f(b). The there exists a < c < b such that f(c) = z. Proof: Sice [a, b] is coected this implies f[a, b] is coected ad hece is a iterval. Therefore all real umbers betwee the two values f(a) ad f(b) are also i this iterval. 6

8 Remark 3 IVP is equivalet to itervals beig coected. Example 5 (i) Every path is coected. (ii) Every path coected space is coected. But coverse is ot true. (iii) R is coected. (iv) Every cell i R is coected. (v) Complemet of a coutable set i R, 2 is coected. (vi) Complemet of a vector subspace of codimesio 2 i R is coected. (vii) Every covex subset is coected. (viii) Spheres ellipsoids etc are coected. Not ec. hyperboloids. Lecture 18 Uiform limits of fuctios We have see that a sequece of cotiuous fuctios which is uiformly coverget produces a limit fuctio which is also cotiuous. We shall stregthe this result ow. Theorem 8 Let X be ay metric space. Let f : X R or (C) be a sequece of cotiuous fuctios. Let A X o which {f } coverges uiformly. The {f } coverges o the closure Ā of A to a fuctio f which is cotiuous. Proof: Let us fix a poit x 0 Ā. We must first of all show that the sequece {f (x 0 )} is coverget. Eough to show that it is Cauchy. To begi with we have the sequece is uiformly coverget o A. Therefore, give ɛ > 0 there exist 0 such that, m > 0 implies f (x) f m (x) < ɛ/3 7

9 for all x A. By cotiuity of f ad f m we ca fid δ > 0 such that d(x, x 0 ) < δ implies that f m (x) f m (x 0 ) + f (x) f (x 0 ) < 2ɛ/3. Now sice x 0 Ā, there exists x B δ(x 0 ) A. With the help of this x, we have f (x 0 ) f m (x 0 ) f m (x) f m (x 0 ) + f (x) f (x 0 ) + f (x) f m (x) < ɛ,, m > 0. (8 Let us carefully examie what we have doe ow. We have got 0 satisfyig (8) without depedig o what x 0 we have chose i Ā. This just meas that the sequece {f } is uiformly Cauchy o Ā. Therefore, we have got a fuctio f : Ā R which is the limit of {f } ad the covergece is uiform o A. Therefore the coclusio of the theorem follows. Remark 4 What about differetiability uder uiform covergece? We should be careful here as illustrated by the example: f (x) = x 1+x 2 o [0, 1]. This sequece coverges uiformly to the fuctio which is idetically 0. (To see this fid the maxima of f i [0, 1].) However the derived sequece f (x) = 1 x2 (1+x 2 ) 2 coverges to a fuctio which is ot eve cotiuous. It is also true that a uiform limit of a sequece of smooth fuctios ca be cotiuous but ot differetiable, or differetiable but ot cotiuously differetiable or... ad so o. O the positive side, we shall ow see that by cotrollig the limitig process of the derived sequece itself we get better results: Theorem 9 Let f : [a, b] R be a sequece of differetiable fuctios such that f coverges uiformly i [a, b] to a fuctio g. Also suppose for some x 0 [a, b], the sequece {f (x 0 )} is coverget. The the sequece f coverges uiformly to a fuctio f ad f = g = lim f. 8

10 Proof: First we wat to show that f is uiformly coverget ad for this it is eough to show that it is uiformly Cauchy, i.e., give ɛ > 0 we must fid 0 such that, m > 0 implies f (x) f m (x) < ɛ, x [a, b] (9) Usig the hypothesis we get 1 such that, m > 1 implies f (x) f m(x) < ɛ, x [a, b]. (10) 2(b a) Put φ m = f f m. Therefore by Lagrage Mea Value Theorem applied to φ m, we have φ m (x 1 ) φ m (x 2 ) x 1 x 2 This is the same as φ m(c) = f (x) f m(x) < f (x 1 ) f m (x 1 ) f m (x 2 ) + f (x 2 ) < x 1 x 2 2(b a) ɛ 2(b a), x 1, x 2 [a, b], m, > 1.(11) ɛ/2. (12) We ow use the fact that f (x 0 ) is coverget ad hece fid 2 such that, m > 2 implies f (x 0 ) f m (x 0 ) < ɛ/2. (13) Put x 1 = x 0 ad x 2 = x ad combiig the above two iequalities (12),(??), we coclude that f is uiformly Cauchy: f (x) f m (x) < ɛ, m, > max{ 1, 2 } (14) as required. Let ow f(x) = lim f (x). To show that f = g : Now fix x 2 [a, b] ad put h (x 1 ) = f(x 1) f (x 2 ) x 1 x 2. The (11) implies that h is uiformly Cauchy i [a, b]\{x 2 } ad hece coverges to a cotiuous fuctio h(x 1 ) which is othig but f (x) f (x 2 ) lim = f(x 1) f(x 2 ). x 1 x 2 x 1 x 2 9

11 Therefore the limit fuctio is cotiuous o the closure of [a, b] \ {x 2 } which is [a, b]. We ca ow iterchage the order of takig limit with respect to with the limit with respect to x, i.e., = lim x 2 x 1 g(x 1 ) = lim f f (x 2 ) f (x 1 ) (x 1 ) = lim lim x 2 x 1 x 2 x 1 f (x 2 ) f (x 1 ) lim x 2 x 1 = lim x 2 x 1 f(x 2 ) f(x 1 ) x 2 x 1 = f (x 1 ). Remark 5 There are certai properties of real valued fuctios defied o itervals, which is peculiar to the 1-variable fuctios oly. For istace, let f : [a, b] [c, d] be a bijectio. The it is ot hard to see that f is cotiuous iff it is strictly mootoe. (For istace, assume that f is order preservig. The f 1 is also order preservig ad show that image of ay closed iterval is a closed iterval uder f 1. It would follow that image of a ope iterval is a ope iterval ad hece f is cotiuous.) I particular it follow that if f : [a, b] [c, d] is a cotiuous bijectio the its iverse is also cotiuous. (That meas f is homeomorphism.) There is othig sacrosact about takig closed itervals. The statemet holds for ope itervals ad for the whole of R as well. Remark 6 O the other ad, there are some geeral results about topological spaces which have special properties such as compactess ad coectedess. For example cosider a cotiuous bijectio f : X Y where X is a compact space ad Y is a metric space. We ca prove that f 1 is cotiguous very easily as follows: It is eough to show that f is a ope mappig this is equivalet to show that f is a closed mappig (because f is a bijectio). If F is a closed subset of X, F will be also compact. Therefore f(f ) is compact. Beig a compact subset of metric space Y, f(f ) is closed. 10

12 Thus we get a alterative easy proof of the fact that a cotiuous bijectio [a, b] [c, d] is a homeomorphism. 11