MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and

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1 MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f 1, f 2,, f, : E R (where E is a subset of R is said to be coverges poitwise o E to fuctio f: E R if ad oly if f (x) = f x for all x E Similarly a series of fuctio oly if k=1 f k (x) coverges poitwise to S(x) o E if ad k=1 f k x = S x for all x E Example 1 (Poitwise Covergece of sequece of fuctios) Let two sequeces of fuctios f, g : [0,1] R be defied as f x = e x ad g x = x Show that f, g both coverge poitwise o [0,1] Solutio: For f x, ote that f (x) = e x = e 0 = 1 for x [0,1] Thus f (x) coverges poitwise to f x = 1 For g(x) By takig it, we have g (x) = x = x = 0 if x [0,1) 1 = 1 = 1 if x = 1 Thus g (x) coverges poitwise to g(x) = 1 if x = 1 0 if x [0,1)

2 Example 2 (Poitwise Covergece of series of fuctios) Discuss the poitwise covergece of series of fuctios 2 x o (0, ) Solutio: By applyig root test (or ratio test if you wish), we have a 2 x = 2 = x = 1 x for x (0, ) We see the series coverges (absolutely) if 1 x < 1 x > 1 ad diverges if 1 x > 1 x < 1. Now it remais to check the case 1 x = 1 x = 1 (which root test does ot give ay coclusio). For x = 1, the series become 2 Which clearly diverges by term test (as 2 = +, thus whe x > 1 (or x (1, )), the series coverges poitwise. (*Note: I the above example (1, ) is also called domai of covergece) Example Discuss the poitwise covergece of series of fuctios e x 1 Solutio: By applyig root test (or ratio test) agai, we have a = e x 1 e 2 = 1 + x 1 = e x 1 for x (0, ) Hece the series coverges if e 2 1+ x 1 < 1 x 1 > e2 1 x > e 2 or x < 2 e 2 ad diverges if e 2 1+ x 1 > 1 x 1 < e2 1 2 e 2 < x < e 2 It remais to check the case e 2 1+ x 1 = 1 x = 2 e2 ad x = e 2 For x = 2 e 2, both series become 1 which clearly diverges. For x = e 2, both series become 1 which clearly diverges. Hece the domai of covergece of this series is x < 2 e 2 ad x > e 2 (or x, 2 e 2 (e 2, ))

3 I case whe the series is a power series (i.e. a R. The oe may have the followig fact: a x c for some costat Give a power series =0 a x c, the domai of covergece of the series is a o-empty iterval (E) which E [c R + c + R] where R is so called radius of covergece of the series Example 4 Fid the domai of covergece ad radius of covergece of the power series Solutio: We ca apply root test a The series coverges whe = x 1 (x 1) x 1 = < 1 x 1 < 2 < x < 4 x 1 = x 1 The series diverges whe x 1 > 1 x 1 > x < 2 ad x > 4 At x = 2, the series become ( ) = 1 which diverges At x = 4, the series become () = which diverges Hece the domai of covergece is ( 2,4) which c = 1, the radius of covergece R =. Example 5 Fid the domai of covergece ad radius of covergece of the followig power series Solutio: 2! x Sice the terms ivolves factorial, istead of usig root test, it may better for us to use ratio test a +1 a = 2 +1 x ! 2 x! 2x = + 1 = 0 < 1 So the series coverges for all x R, the domai of covergece is (, ) ad R =

4 Uiform Covergece Besides poitwise covergece, ext we would like to itroduce aother type of covergece which is called uiform covergece. It allows us to do some operatios like x a a b d dx f (x) = f (x) x a f (x) dx = f x dx f (x) = Defiitio: (Uiform Covergece of Fuctio) a b d dx f (x) Give a sequece of fuctio f : E R, we say f coverges uiformly to f iff sup x E f x f x = 0 I other word, sup x E f x f x 0 as (Note: sup x E g(x) is called sup-orm of g(x) o E) Defiitio: (Uiform Covergece of Series of Fuctio) Let g : E R be a sequece of fuctios, we say the series uiformly to fuctio S(x) o E iff the partial sum S x = uiformly to S(x) o E k=1 g k (x) coverges k=1 g k (x) coverges Example 6 (Example 1 revisited) Discuss the uiform covergece of f x = e x ad g x = x o [0,1] Solutio: For f (x) (Step 1: Fid the Limit) From example 1, we see the it is f x = f (x) = 1 (Step 2: Compute sup-orm) The sup f x f x = sup x [0,1] x 0,1 e x 1 = sup (e x 1) = e 1 1 x 0,1 sup x E f x f x Thus the fuctio is uiformly coverget o [0,1] = (e 1 1) = 0

5 For g (x) (Step 1: Fid the it) From example 1, we see that the it is g x = g (x) = 1 if x = 1 0 if x [0,1) (Step 2: Compute sup-orm) Note that g x g x = x 1 = 1 1 = 0 for x = 1 x 0 = x for x [0,1) The sup x 0,1 g x g x = sup sup x 0,1 x, 0 = sup 1,0 = 1 sup x E f x f x The fuctio is ot uiformly coverget o [0,1] = 1 = 1 0 Remark: (Poitwise Covergece v.s. Uiformly Covergece) These two covergeces looks similar but there is a big fudametal differece: Poitwise covergece oly require for every x E, f x coverges to f(x), but the speed of covergece ca be varied amog differet poits. Some poits may coverge faster ad some poits may coverge slower. But uiform covergece also requires the speed of covergece is similar amog all poits besides poitwise coverget so that the property of fuctios ca be preserved whe takig it. (For example: it of cotiuous fuctio is cotiuous, it of itegrable fuctio is itegrable) Let cosider the two fuctios i Example 6, by plottig their graphs out, we ca see 1. f x = e x f x = f (x) = 1 f 1 = e x f 2 = e x 2 f = e x Cotiuous Cotiuous Fig 1: Graph of f (x) (coverges uiformly o [0,1]

6 2. g x = x g x = g (x) = 1 if x = 1 0 if x [0,1) g 1 = x 1 g 2 = x 2 g = x Cotiuous Not Cotiuous Fig 2: Graph of g (x) (does ot coverges uiformly o [0,1] From the above two examples, we see if f (x) coverges i similar speed, the the cotiuity of f ca be preserved (i (1)). However if g (x) coverges very fast at some poits ad coverges very slow i some poits (say (2)). The the property of g (x) may be destroyed at. L-test (for sequece of fuctios) Let f : E R be sequeces of fuctios o set E, suppose 1) f (x) = f(x) (Poitwise Limit) 2) For each = 1,2,, there is costat L such that f x f x L for all x E ) L = 0 The f (x) coverges uiformly to f(x) Example 7 Show that the followig sequece of fuctios f x = six 1 + x coverges uiformly o [c, ) where c is a positive umber. (Step 1: Fid the it first) For ay x [c, ), we have si (x) f x = f (x) = = 0 (Sice 1 + x ) 1 + x (Step 2: Compute the sup-orm) 0 f x f x = six 1 + x x c = L... for x [c, ) Note that 1 1+c = 0, By L-test, f (x) coverges uiformly o [c, ).

7 I the followig, there are some suggested exercises, you should try to do them i order to uderstad the material. If you have ay questios about them, you are welcome to fid me durig office hours. You are also welcome to submit your work (complete or icomplete) to me ad I ca give some commets to your work. Exercise 0 Determie whether the followig statemets are true or ot. Give a brief explaatio. (e.g. proof, provide couter-examples etc.) *I the followig problems, let f (x) be a sequece of fuctios o E a) If f (x) coverges uiformly to f(x) o E, the f (x) coverges poitwise to f(x) o E b) If f (x) coverges poitwise to f(x) o E, the f (x) coverges uiformly o E. c) Give a series of fuctios f (x), the its domai of covergece MUST be a iterval. d) The domai of covergece of power series is always a iterval WITH BOTH ENDPOINTS. e) The domai of covergece of power series is always a iterval WITHOUT BOTH ENDPOINTS. f) Give a sequece of cotiuous fuctios f (x) o E, the f x = f (x) is also cotiuous o E g) Give a sequece of itegrable fuctios g (x) o E, the g x = g (x) is also itegrable o E. Aswer: a) is true ad b), c), d), e), f), g) are all false. (Hit: To disprove g), cosider the followig couter-example: Q (set of ratioal umbers) is coutable, let Q = {r 1, r 2, r,. }, costruct g (x) as g x = 1 if x = r 1, r 2, r 0 otherwise the g x = g (x) = 1 if x Q 0 otherwise Oe ca show g is Riema itegrable ad g(x) is ot itegrable, if you have forgotte how to prove them, look at Tutorial Note #19 i MATH202 which is available at Exercise 1 Fid the domai of covergece of followig series of fuctios a) 1 xe x b) 1 1+ x 1 + x 2

8 Exercise 2 Fid the domai of covergece ad radius of covergece for the followig power series a) x b) 1 1 x c) 2 2 x + 1 Exercise a) Show that the sequece of fuctios f x = 1 x +1 coverge uiformly o (0,1). b) Show that the sequece of fuctios g x = x x +1 coverges poitwise but ot coverges uiformly o (0,1) Exercise 4 Show that if f ad g coverges uiformly to f, g respectively o a set E, the f + g coverges uiformly o E. Exercise 5 a) Show that if f ad g are bouded ad coverges uiformly to f, g respectively o a set E, the f g coverges uiformly o E. (Hit: Note that f g fg = f g f g + f g fg) b) Is the statemet still true if the coditio f ad g are bouded is omitted. Exercise 6 Show that the followig sequeces of fuctios o idicated itervals. a) f x = x 1+ x 1 + x 2 b) f x = 2 x 1+ x 2 o x R (Hit: Use Calculus!) o x [1, )