Fall 2013 MTH431/531 Real analysis Section Notes

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1 Fall 013 MTH431/531 Real aalysis Sectio Notes Yi Su Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters i this sequece: ad x. So whe we take the limit as we should discuss the behavior of this sequece f (x) for differet values of x i the domai. Defiitio of poitwise covergece. (8.1.1) f coverges to f poitwise o A if f (x) f (x) as for each x A. That is, for every x A, f (x) is a sequece of real umbers ad this sequece coverges to f (x) i the regular sese. Notice that we oly require f (x) f (x) for ay choice of x. We do t have ay requiremet for the pace of the covergece. That meas, possibly for some x A, f (x) coverges to f (x) very fast, but for some other x, f (x) approaches f (x) very slowly. Example 1. The sequece f (x) = x for x [0, 1] coverges poitwise to the fuctio f (x) =. Because if 0 x < 1, x will 0, x 1 1, x = 1 approach 0 but if x = 1, ay power of 1 is still 1 so the sequece f (1) is a costat sequece of 1 with limit 1. This is a example of poitwise covergece. That meas for ay choice of x [0, 1], the sequece f (x) will approach f (x). But for differet x [0, 1], the pace of covergece of f (x) is quite differet. Sice whe x 1, the limit fuctio is all 0, we look at whe the sequece f (x) will reach the iterval [0, 0.01]. That is, we look for the smallest umber s.t. f (x) = x Take x = 0.5. The f (1/) = (1/). The idex such that f (x) lies i [0, 0.01] is log So f 7 (1/) is the first term which stays i [0, 0.01]. Figure 1: From the left: x, x, x 3, x 5, x 10, x 0, x 50 Take x = 0.9. Doig similar calculatio, we obtai taht the idex s.t. f (0.9) lies i [0, 0.01] is log So f 44 (0.9) is the first term which stays i [0, 0.01]. Take x = We have that the idex s.t. f (0.99) lies i [0, 0.01] is log So f 459 (0.99) is the first term which stays i [0, 0.01]. We ca expect that whe x gets closer to 1 but is ot equal to 1, f (x) will coverge to f (x) = 0 more slowly because it takes loger to compress the sequece ito a small iterval. From here you might get some feelig that f (x) approaches f (x) i differet pace, although soo or later they will all approach the limit fuctio f (x). The umber 0.01 we choose is like the ɛ i the defiitio of sequetial limit ad the first umber we look for is like the N i the defiitio such that for all N, all future terms will stay i the strip of legth ɛ. 1

2 Figure : From the left: x 4, x 5, x 6, x 7, x 8, x 9, x 10 Figure 3: From the left: x 40, x 43, x 45, x 48, x 50 Figure 4: From the left: x 455, x 458, x 459, x 460, x 463, x 465 So ow it s atural to ask for a better covergece i which for all x A, f (x) coverges to f (x) i the same pace. This is called uiform covergece. Defiitio of uiform covergece. (8.1.4) Say that f coverges to f uiformly if for ay ɛ > 0, there exists N (idepedet of x) s.t. for each N, we have f (x) f (x) < ɛ for all x A. Example 1 (cot.) For fixed ɛ = 0.01, we ca ot fid a commo N s.t. for all N, f (x) for all x will stay i [0, 0.01]. Ideed, for x = 0.1, this N is 7, for x = 0.9, this N is 44, ad for x = 0.99 this N is 459. So this N will get larger if we pick larger x value. Thus the N we ca choose really depeds o the x value we use. That meas a commo N workig for all x is ot possible to fid. Therefore x o [0, 1] is ot uiformly coverget to f. To be precise, we have

3 Prop 1. f is ot uiformly coverget to f if there exists ɛ 0 > 0 s.t. for every we ca fid a x A s.t. f (x ) f (x ) ɛ 0. Example 1 (cot.) We still take ɛ = For ay, we wat to fid x s.t. f (x ) f (x ) We assume x 1 so f (x ) = 0. The what we eed is f (x ) = (x ) Thus we ca set x = The this x will guaratee f (x ) = Therefore this sequece x o [0, 1] is ot uiformly coverget. Questio. We have show f does t coverge uiformly to f i the above example. But ca f coverge uiformly to aother fuctio? The aswer is o, because Prop. If f f o A, the f f o A. This property simply says that uiform covergece is stroger tha poitwise covergece. The if f coverges uiformly, the uiform limit fuctio must be the same as the poitwise limit fuctio. We ca apply this to aswer the above questio. Suppose f g for some fuctio g other tha f. The g must be also the poitwise limit fuctio. But we already showed that f is the limit fuctio. So such fuctio g ca t exist.. Examples. Trivial examples: (1) If f are idepedet of. That is, all f are the same fuctio of x. The {f } is uiformly coverget. Examples: f (x) = x o, f (x) = x o, f (x) = 1 o (0, ). x () If f are idepedet of x ad f 0 as. The { f } is uiformly coverget. This is because for ay ɛ, the differece f (x) f (x) is idepedet of x thus a uiform N is possible to fid. Examples: f (x) = 1 o, f (x) = 1 o. To prove the uiform covergece, sometimes we have to use the defiitio. That meas we eed to cotrol f (x) f (x) for all x A. Or i other words, we ca try to fid the least upper bouded sup f (x) f (x). The lemma (8.1.8) i textbook says we ca use this to rewrite the defiitio of uiform covergece: Defiitio of uiform covergece (ii) f coverges to f uiformly if for ay ɛ > 0, there exists N (idepedet of x) s.t. for each N, we have sup f (x) f (x) < ɛ. Importat ote. The two equivalet defiitios of uiform covergece give us two ways to prove the uiform covergece: (1). For each, try to fid a upper boud for f (x) f (x) which does t deped o x (oly depeds o ) but approaches 0 as. (). For each, if all f are differetiable, use calculus to fid the maximal value of f f o A (depedig o ) ad show the maximum approaches 0 as. (The maximum value is exactly the least upper boud sup f (x) f (x).) Example. The sequece f (x) = x e x is uiformly coverget o [0, + ). Easy to see that o [0, ), the poitwise limit fuctio is the zero fuctio f = 0. The poitwise covergece ca be obtaied by usig L Hospital rule. We may assume that x 0 because if x = 0, the sequece f (0) is all 0 which does t affect the uiform covergece. 3

4 Proof followig (1). We wat to cotrol f (x) f (x) = x e x = x. So we eed to fid a upper boud for e x this fuctio idepedet of x. To do this we may use the Taylor expasio of e x which is e x = 1+x + (x) +. Note that all terms i the expasio are positive so we have e x (x), the third term i the expasio. Thus the fractio will get larger: f (x) f (x) = x e x = x e x x x / = Now the upper bouds 0 obviously. The for ɛ > 0, we ca choose N > ɛ, the < ɛ for all N ad hece by our aalysis f x f (x) < ɛ, x [0, ) Therefore we have f f = 0. Proof followig (). For each fixed, we try to fid the maximum of f o [0, ) usig Calculus. By the derivative tests, the maximum occurs at critical poits or boudary poits. Sice f (x) = xe x x e = ( x)xe x, we kow that the critical poits are x = 0 ad x =. Easy to verify x = is the absolute maximum with value f (x) = ( ) e = 4 e. Agai we see that the maximum 4 e 0. So choose suitable N for arbitrary ɛ. We ca guaratee that f (x) f (x) < ɛ. Remarks. (1) Prop 1. gives us a way to prove the o-uiform covergece. () For uiform covergece, we also have the Cauchy Criterio (8.1.10). Sometimes usig Cauchy Criterio is more coveiet, especially whe we do t kow the limit fuctio or we wat to defie some fuctio just as the limit fuctio. For example, i sectio 8.3 the defiitio of expoetial fuctio ad sectio 8.4 the defiitio of trigoometric fuctios usig power series. The we have to verify the covergece by usig Cauchy Criterio to see the limits really exist ad are still cotiuous. (3) Depedecy o the domai. Whe we talk about uiform covergece, we always eed to specify the domai o which fuctios f coverge uiformly. Example 3. (EXC 8.1.0) The sequece f (x) = x e x is uiformly coverget for x a with a > 0, but is ot uiformly coverget for x [0, ). Example 4. (EXC 8.1.1) The sequece f (x) = x is uiformly coverget for x a with a > 0, but is ot 1+ x uiformly coverget for x [0, ). Prop 3. (Subset Property) If f f o A, the f f o ay B if B is a subset of A. Coversely, if f is ot uiformly coverget to f o B, the for ay bigger iterval A, i.e. A B, we also have f ot uiformly coverget o A. This property is easy to prove. Example 1 (cot.) By this property, x is ot uiformly coverget o [0, ) or eve (, ) because we have show that it s ot uiformly coverget o [0, 1]. 3. Properties of uiform covergece. Prop 4. Uiformly coverget sequece preserves cotiuity, differetiatio, itegratio ad boudedess. Cocretely, Suppose f f. 4

5 (1) (8..)If f are all cotiuous, the the limit fuctio f is cotiuous. () (8..4)If f are all itegrable o [a, b], the f is also itegrable ad b b f (x)d x = lim a a f (x)d x. (3) (A weaker statemet of 8..3) If f are all differetiable o [a, b], ad additioally f g o [a, b], the f is differetiable ad f = g. (4) (EXC 8..7) If f are all bouded, the the limit fuctio f is also bouded. Importat Note. The first property is very importat because it tells us that if f are cotiuous ad uiformly coverget, the the limit fuctio must be also cotiuous. That is, if f are cotiuous but the limit fuctio f is cotiuous, we ca obtai that { f } is ot uiformly coverget. Example 1 (cot.) x o [0, 1] is ot uiformly coverget because the limit fuctio has a discotiuity at x = 1. Example 5. x 1+x o [0, ) is ot uiformly coverget because the limit fuctio has a discotiuity at x = 0. Prop 5. (Liearity, a simple geeralizatio of EXC 8.1.1) If f f ad g g o A, the a f + bg a f + bg o A for a, b. The proof of this property follows from the triagular iequality. Example 6. We ca easily see that f (x) = x+ 1 because g (x) = x ad h (x) = 1 are both uiformly coverget o [0, ) ad f = g + h. But {f } is ot uiformly coverget because f (x) = x + x + 1. Obviously {x } ad { 1 }. But { x } is ot uiformly coverget o which is show i the textbook (8.1.9(a)). Therefore f ca ot be uiformly coverget because otherwise by liearity { x } must be also uiformly coverget. 5