8 Baire Category Theorem and Uniform Boundedness

Transcription

1 8 Bae Categoy Theoem and Unfom Boundedness Pncple 8.1 Bae s Categoy Theoem Valdty of many esults n analyss depends on the completeness popety. Ths popety addesses the nadequacy of the system of atonal numbes. The manne n whch completeness of a metc space X s typcally exploted s that the ntesecton of nested closed balls B(x 1, 1 ) B(x 2, 2 ) B(x n, n ) wth n 0 s non-tval. To see that ths s the case, we obseve that the sequence x n satsfes fo m n, d(x n, x m ) n. (1) Ths mples that ths sequence s Cauchy, and x n x fo some x X. It follows fom (1) that d(x n, x) n, so that the lmt x gves a pont n the ntesecton n 1 B(xn, n ). A vaaton of ths agument also gves: Theoem 8.1 (Bae s Categoy Theoem). Let X be a complete metc space and U n, n 1, a collecton of open dense subsets of X. Then the ntesecton n 1 U n s dense n X. Poof. It s suffcent to show that n 1 U n ntesects non-tvally any ball B(x 0, 0 ) wth adus 0 (0, 1). We constuct a sequence of nested open balls B(x n, n ) nductvely. Gven a ball B(x n 1, n 1 ), we obseve that U n B(x n 1, n 1 ) s non-empty and open. So that we can choose x n, n so that B(x n, n ) U n B(x n 1, n 1 ). We may also choose n s so that n 0. Then by ou pevous dscusson, thee exsts x n 1 B(xn, n ). We obtan that fo evey n, x B(x n, n ) U n B(x 0, 0 ), whch completes the poof. It s natual to thnk about open dense sets (and the countable ntesectons) as lage subsets of X, and the complements as small subsets of X. Ths motvates the followng defnton. Defnton 8.2. Let X be a metc space and M a subset of X. M s called ae (o nowhee dense) f ts closue M has no nteo ponts. M s called meage (o of fst categoy) n X f t s a countable unon of ae sets. Othewse M s called nonmeage (o of second categoy). 1

2 Example. Q s not ae n R because ts closue s all of R. But Q {0} s ae n R 2, ts closue beng R {0}, whch has no nteo ponts as a subset of R 2. Q n s meage n R n, beng a countable unon of ponts. The Bae Categoy Theoem can be also estated as follows: Theoem 8.3 (Bae s Categoy Theoem). A nonempty complete metc space X s nonmeage. Poof. Suppose that n contay X = n 1 M n (2) whee M k s ae closed sets wth empty nteo. Then the sets X\M n ae open and dense. Hence, by Theoem 8.1, U n. n 1 Ths mples that the complement X\ U n = M n X, n 1 n 1 whch gves a contadcton. Hence, X cannot be meage. 8.2 Applcaton: nowhee dffeentable functons The Bae Categoy Theoem s a vey useful tool fo showng exstence of objects wth pecula popetes. We demonstate how ths woks by showng that thee s contnuous functon whch s not dffeentable at even a sngle pont. Theoem 8.4. The subset D = {f C[a, b] : f s dffeentable at some x [a, b]} s a meage subset of C[a, b]. Snce the space C[a, b] s complete, t follows that D C[a, b], and thee exst functons whch ae nowhee dffeentable. In some sense, Theoem 8.4 tells us that most contnuous functons ae nowhee dffeentable. Poof. Fo n, m 1, consde the sets { } A n,m = f C[a, b] : x : f(t) f(x) t x n fo all 0 < x t < 1 m. 2

3 Recall that f s dffeentable at x f the lmt exsts. So that t s clea that f (x) = lm t x f(t) f(x) t x D n,m 1 A n,m. It emans to show that the sets A m,n ae ae. Fst we show that A n,m s closed. Suppose that f f n C[a, b] and all f D. Then thee exsts x [a, b] such that f (t) f (x ) t x n fo all 0 < x t < 1 m. Passng to a subsequence, we may assume that x x fo some x [a, b]. Then f 0 < x t < 1 m, then fo suffcently lage n, we also have 0 < x t < 1 m, and f(t) f(x) t x = lm f (t) f (x ) t x n. Ths poves that the sets A n,m ae closed. Now we show that the sets A n,m have empty nteos. Suppose that A m,n contans an open ball B(f, ɛ) fo some f C[a, b]. Usng unfom contnuty, one shows that the subspace of pecewse lnea contnuous functons s dense n C[a, b]. Hence, wthout loss of genealty we may assume that f s pecewse lnea. Fo pecewse lnea contnuous functons, the one-sded devatves f +(x) and f (x) always exst, and they ae unfomly bounded: fo some M > 0, f ±(x) M fo all x [a, b]. We obseve that evey functon g = f + ɛ 2 φ wth φ C[a, b] such that φ 1 belongs to B(f, ɛ) A n,m. In patcula, t follows that thee exsts x 0 [a, b] such that g ±(x 0 ) n. (3) Fo evey K > 0, one can constuct (execse) a pecewse lnea functon φ such that φ 1 and φ ±(x) > K. Then fo all x [a, b], g ±(x) ɛ 2 φ ±(x) f ±(x) ɛ 2 K M. Takng K = K(ɛ, M) suffcently lage, we obtan g ±(x) > n fo all x, but ths contadcts (3). Hence, the sets A n,m have empty nteos. 3

4 8.3 Unfom boundedness pncple The unfom boundedness pncple answes the queston of whethe a pontwse bounded sequence of bounded lnea opeatos must also be unfomly bounded. Theoem 8.5 (Unfom Boundedness Pncple; Banach Stenhaus). Let X be a Banach space and Y be a nomed space. Suppose that the sequence T n B(X, Y ) of bounded lnea opeatos has the popety that fo evey x X, the sequence T n (x) Y s bounded. Then the sequence of noms T n s bounded. Let us llustate the Unfom Boundedness Pncple by an example. Example 8.6. Let { } X = p(x) = α 0 + α 1 x + α 2 x α d x d α F, d N {0}, be the space of polynomals equpped wth the nom p = max α. We eadly see that ths tuns X nto a nomed space. We gve an example of a sequence of lnea maps T n : X F whch ae pontwse bounded but not unfomly bounded. Let We obseve that T n (p) = α α n 1. T n (p) = α 0 + α α n 1 α α n 1 n p, so that T n n. In fact, ths estmate can be mpoved fo polynomals p of degee d as follows: T n (p) = α 0 + α α n 1 α 0 + α α d α 0 + α α d d p, Ths shows that the sequence (T n (p)) n 1 s bounded fo evey p. Howeve, we clam that T n = n. Indeed, fo p n (x) = 1 + x + x x n 1, we have p n = 1, and T n (p n ) = n, whch mples the clam. In ths example, the sequence (T n (p)) n 1 s pontwse bounded, but not unfomly bounded, contay to the concluson of the Unfom Boundedness Pncple. The Unfom Boundedness Pncple fals hee because the space X s not complete. We gve a concete applcaton of the Unfom Boundedness Pncple. 4

5 Example 8.7. Suppose we have a sequence of complex numbes x = (x n ) n 1 wth the popety that wheneve y = (y n ) n 1 s a sequence satsfng y n 0, we have that the sum n=1 x ny n conveges. Show that n=1 x n conveges. The fst step s of couse to tanslate ths poblem nto a moe convenent functonal analytc settng. The condton that the sequence y conveges to 0 s smply the statement that y s a membe of the space c 0 = {y l : y n 0 as n }. One can show that c 0 s a Banach space when equpped wth the nom. We leave ths as an execse. We ae equed to show that x l 1. Snce we would lke to use the Unfom Boundedness Pncple n some way, let us stat by fndng a sequence T n B(c 0, Y ) whee Y s some nomed space. One natual choce s to let T n (y) = n x y whee y = (y 1, y 2,... ) c 0 be the tuncated sum whch we have assumed conveges. It s clea that T n B(c 0, C) = (c 0 ). Indeed, the calculaton ( n n n ) T n (y) = x y x y x y shows that T n s ae bounded opeatos wth T n n x. In fact, we can do bette: the assumpton that x y conveges fo any y c 0 mples that fo any y c 0, the sequence (T n (y)) n 1 s convegent, hence bounded. So (T n ) n 1 s a pontwse bounded sequence of opeatos. Theefoe, by the Unfom Boundedness Pncple, t s unfomly bounded, meanng that thee s some M > 0 such that T n M fo all n N. We clam that T n = n x. Wthout loss of genealty, we may assume that x 0. We consde y (n) = (y (n) ) 1 c 0 defned by Then y (n) = { x (n) / x (n) f n and x 0, 0 othewse. T n (y (n) ) = n x, and y (n) = 1 unless y (n) = 0. Ths mples that T n n x. The opposte nequalty has aleady been poven above. We conclude that n x M fo all n, and x conveges. 5

6 8.4 Poof of the Unfom Boundedness Pncple The poof of the Unfom Boundedness Pncple s an applcaton of Bae s Categoy Theoem. Let us defne the sets M k = {x X : T n (x) k fo all n}, k 1. Snce T n s ae contnuous, these sets ae closed. Snce fo evey x X, the sequence T n (x) s bounded, we have x M k fo suffcently lage k. Theefoe, X = k 1 M k. Bae s categoy theoem now guaantees that one of these closed sets contans an open ball, say B(x 0, ) M k0. We theefoe have that T n (x) k 0 fo any x B(x 0, ) and n 1. Now let x X, x 0. Then the vecto z = x x x belongs to B(x 0, ) and x = 2 x (z x 0 ). Usng ths, we calculate T n (x) = 2 x T n (z) T n (x 0 ) 2 x 4k 0 x. ( T n (z) + T n (x 0 ) ) Hence, T n 4k 0 bounded. fo all n N, povng that the sequence s unfomly 6