Series of functions. Chapter lim sup and lim inf

Transcription

1 Chpter 4 Series of functions In this chpter we shll see how the theory in the previous chpters cn be used to study functions. We shll be prticulrly interested in how generl functions cn be written s sums of series of simple functions such s power functions nd trigonometric functions. This will tke us to the theories of power series nd Fourier series. 4. lim sup nd lim inf In this section we shll tke look t useful extension of the concept of limit. Mny sequences do not converge, but still hve rther regulr symptotic behvior s n goes to infinity they my, for instnce, oscillte between n upper set of vlues nd lower set. The notions of limit superior, lim sup, nd limit inferior, lim inf, re helpful to describe such behvior. They lso hve the dvntge tht they lwys exist (provided we llow them to tke the vlues ±). We strt with sequence { n } of rel numbers, nd define two new sequences {M n } nd {m n } by M n =sup{ k k n} nd m n =inf{ k k n} We llow tht M n = nd tht m n = s my well occur. Note tht the sequence {M n } is decresing (s we re tking suprem over smller nd smller sets), nd tht {m n } is incresing (s we re tking infim over incresingly smller sets). Since the sequences re monotone, the limits lim M n nd lim m n n! n! 93

2 94 CHAPTER 4. SERIES OF FUNCTIONS clerly exist, but they my be ±. Wenowdefinethelimit superior of the originl sequence { n } to be nd the limit inferior to be lim sup n = lim M n n! n! lim inf n! n = lim n! m n The intuitive ide is tht s n goes to infinity, the sequence { n } my oscillte nd not converge to limit, but the oscilltions will be symptoticlly bounded by lim sup n bove nd lim inf n below. The following reltionship should be no surprise: Proposition 4.. Let { n } be sequence of rel numbers. Then if nd only if lim n = b n! lim sup n =liminf n = b n! n! (we llow b to be rel number or ±.) Proof: Assume first tht lim sup n! n =liminf n! n = b. Since m n pple n pple M n, nd lim m n =liminf n = b, n! n! lim M n =limsup n = b, n! n! we clerly hve lim n! n = b by squeezing. We now ssume tht lim n! n = b where b 2 R (the cses b = ± re left to the reder). Given n >0, there exists n N 2 N such tht n b < for ll n N. In other words b < n <b+ for ll n N. Butthen nd b b pple m n <b+ <M n pple b + for n N. Since this holds for ll > 0, we hve lim sup n! n = lim inf n! n = b 2

3 4.2. INTEGRATING AND DIFFERENTIATING SEQUENCES 95 Exercises for section 4.. Let n =( ) n.findlimsup n! n nd lim inf n! n. 2. Let n = cos n 2.Findlimsup n! n nd lim inf n! n. 3. Let n = rctn(n)sin n 2.Findlimsup n! n nd lim inf n! n. 4. Complete the proof of Proposition 4.. for the cse b =. 5. Show tht nd lim sup n! ( n + b n ) pple lim sup n! n +limsupb n n! lim inf n! ( n + b n ) lim inf n! n +liminf n! b n nd find exmples which show tht we do not in generl hve equlity. Stte nd prove similr result for the product { n b n } of two positive sequences. 6. Assume tht the sequence { n } is nonnegtive nd converges to, nd tht b =limsupb n is finite nd positive. Show tht lim sup n! n b n = b (the result holds without the condition tht b is positive, but the proof becomes messy). Wht hppens if the sequence { n } is negtive? 7. We shll see how we cn define lim sup nd lim inf for functions f : R! R. Let 2 R, nd define M =sup{f(x) x 2 (, + )} m =inf{f(x) x 2 ( for >0 (we llow M = nd m = )., + )} ) Show tht M decreses nd m increses s! 0. b) Show tht lim sup x! f(x) =lim!0 + M nd lim inf x! f(x) =lim!0 + m exist (we llow ± s vlues). c) Show tht lim x! f(x) =b if nd only if lim sup x! f(x) =liminf x! f(x) = b d) Find lim inf x!0 sin x nd lim sup x!0 sin x 4.2 Integrting nd di erentiting sequences Assume tht we hve sequence of functions {f n } converging to limit function f. If we integrte the functions f n, will the integrls converge to the integrl of f? And if we di erentite the f n s, will the derivtives converge to f 0? In this section, we shll see tht without ny further restrictions, the nswer to both questions re no, but tht it is possible to put conditions on the sequences tht turn the nswers into yes. Let us strt with integrtion nd the following exmple. Exmple : Let f n :[0, ]! R be the function in the figure.

4 96 CHAPTER 4. SERIES OF FUNCTIONS n 6 E E E E E E E E E n It is given by the formul 8 >< f n (x) = >: Figure - 2n 2 x if 0 pple x< 2n 2n 2 x +2n if 2n pple x< n 0 if n pple x pple but it is much esier just to work from the picture. The sequence {f n } converges R pointwise to 0, but the integrls do not not converge to 0. In fct, 0 f n(x) dx = 2 since the vlue of the integrl equls the re under the function grph, i.e. the re of tringle with bse n nd height n. The exmple bove shows tht if the functions f n converge pointwise to function f on n intervl [, b], the integrls R b f n(x) dx need not converge to R b f(x) dx. The reson is tht with pointwise convergence, the di erence between f nd f n my be very lrge on smll sets so lrge tht the integrls of f n do not converge to the integrl of f. If the convergence is uniform, this cn not hppen (note tht the result below is ctully specil cse of Lemm 3.6.): Proposition 4.2. Assume tht {f n } is sequence of continuous functions converging uniformly to f on the intervl [, b]. Then the functions converge uniformly to on [, b]. F n (x) = F (x) = f n (t) dt f(t) dt Proof: We must show tht for given >0, we cn lwys find n N 2 N such tht F (x) F n (x) < for ll n N nd ll x 2 [, b]. Since {f n }

5 4.2. INTEGRATING AND DIFFERENTIATING SEQUENCES 97 converges uniformly to f, there is n N 2 N such tht f(t) f n (t) < b for ll t 2 [, b]. For n N, we then hve for ll x 2 [, b]: F (x) F n (x) = (f(t) f n (t)) dt pple f(t) f n (t) dt pple pple dt pple b Z b b dt = This shows tht {F n } converges uniformly to F on [, b]. 2 In pplictions it is often useful to hve the result bove with flexible lower limit. Corollry Assume tht {f n } is sequence of continuous functions converging uniformly to f on the intervl [, b]. For ny x 0 2 [, b], the functions converge uniformly to on [, b]. Proof: Recll tht f n (t) dt = F n (x) = F (x) = 0 x 0 f n (t) dt x 0 f(t) dt f n (t) dt + f n (t) dt x 0 regrdless of the order of the numbers, x 0,x, nd hence x 0 f n (t) dt = f n (t) dt 0 f n (t) dt The first integrl on the right converges uniformly to R x f(t) dt ccording to the proposition, nd the second integrl converges (s sequence of numbers) to R x 0 f(t) dt. Hence R x x 0 f n (t) dt converges uniformly to f(t) dt 0 f(t) dt = x 0 f(t) dt s ws to be proved. 2 Let us reformulte this result in terms of series. Recll tht series of functions P v n(x) converges pointwise/unifomly to function f on n intervl I if n only if the sequence {s n } of prtil sum s n (x) = P n k=0 v k(x) converges pointwise/uniformly to f on I.

6 98 CHAPTER 4. SERIES OF FUNCTIONS Corollry Assume tht {v n } is sequence of continuous functions such tht the series P v n(x) converges uniformly on the intervl [, b]. Then for ny x 0 2 [, b], the series P R x x 0 v n (t) dt converges uniformly nd v n (t) dt = v n (t) dt x 0 x 0 The corollry tell us tht if the series P v n(x) converges uniformly, we cn integrte it term by term to get v n (t) dt = x 0 x 0 v n (t) dt This formul my look obvious, but it does not in generl hold for series tht only converge pointwise. As we shll see lter, interchnging integrls nd infinite sums is quite tricky business. To use the corollry e ciently, we need to be ble to determine when series of functions converges uniformly. The following simple test is often helpful: Proposition (Weierstrss M-test) Let {v n } be sequence of continuous functions on the intervl [, b], nd ssume tht there is convergent series P M n of positive numbers such tht v n (x) pplem n for ll n 2 N nd ll x 2 [, b]. Then series P v n(x) converges uniformly on [, b]. Proof: Since (C([, b], R), ) is complete, we only need to check tht the prtil sums s n (x) = P n P k=0 v k(x) form Cuchy sequence. Since the series M n converges, we know tht its prtil sums S n = P n k=0 M k form Cuchy sequence. Since for ll x 2 [, b] nd ll m>n, s m (x) s n (x) = mx k=n+ v k (x) pple mx k=n+ v k (x) pple mx k=n+ M k = S m S n, this implies tht {s n } is Cuchy sequence. 2 Exmple : Consider the series P cos nx cos nx P n=. Since pple, nd n 2 n 2 n 2 converges, the originl series P cos nx n 2 n= converges uniformly to n 2 function f on ny closed nd bounded intervl [, b]. Hence we my intergrte termwise to get 0 f(t) dt = Z n= x cos nt n 2 dt = n= sin nx n 3

7 4.2. INTEGRATING AND DIFFERENTIATING SEQUENCES 99 Let us now turn to di erentition of sequences. This is much trickier business thn integrtion s integrtion often helps to smoothen functions while di erentition tends to mke them more irregulr. Here is simple exmple. sin nx Exmple 2: The sequence (not series!) { n } obviously converges uniformly to 0, but the sequence of derivtives {cos nx} does not converge t ll. The exmple shows tht even if sequence {f n } of di erentible functions converges uniformly to di erentible function f, the derivtives f 0 n need not converge to the derivtive f 0 of the limit function. If you drw the grphs of the functions f n, you will see why lthough they live in n incresingly nrrower strip round the x-xis, they ll wriggle eqully much, nd the derivtives do not converge. To get theorem tht works, we hve to put the conditions on the derivtives. The following result my look ugly nd unstisfctory, but it gives us the informtion we shll need. Proposition Let {f n } be sequence of di erentible functions on the intervl [, b]. Assume tht the derivtives f 0 n re continuous nd tht they converge uniformly to function g on [, b]. Assume lso tht there is point x 0 2 [, b] such tht the sequence {f(x 0 )} converges. Then the sequence {f n } converges uniformly on [, b] to di erentible function f such tht f 0 = g. Proof: The proposition is just Corollry in convenient disguise. If we pply tht proposition to the sequence {f 0 n}, we se tht the integrls R x x 0 f 0 n(t) dt converge uniformly to R x x 0 g(t) dt. By the Fundmentl Theorem of Clculus, we get f n (x) f n (x 0 )! x 0 g(t) dt uniformly on [, b] Since f n (x 0 ) converges to limit b, this mens tht f n (x) converges uniformly to the function f(x) =b + R x x g(t) dt. Using the Fundmentl Theorem of Clculus gin, we see tht f 0 (x) =g(x). 2 Also in this cse it is useful to hve reformultion in terms of series: Corollry Let P u n(x) be series where the functions u n re di erentible with continuous derivtives on the intervl [, b]. Assumetht the series of derivtives P u0 n(x) converges uniformly on [, b]. Assume lso tht there is point x 0 2 [, b] where the series P u n(x 0 ) converges.

8 00 CHAPTER 4. SERIES OF FUNCTIONS Then the series P u n(x) converges uniformly on [, b], nd 0 u n (x)! = u 0 n(x) The corollry tells us tht under rther strong conditions, we cn di erentite the series P u n(x) termbyterm. Exmple 3: Summing geometric series, we see tht e x = X e nx for x>0 (4.2.) If we cn di erentite term by term on the right hnd side, we shll get e x ( e x ) 2 = X n= ne nx for x>0 (4.2.2) To check tht this is correct, we must check the convergence of the differentited series (4.2.2). Choose n intervl [, b] where > 0, then ne nx pple ne n for ll x 2 [, b]. Using, e.g., the rtio test, it is esy to see tht the series P ne n converges, nd hence P ne nx converges uniformly on [, b] by Weierstrss M-test. The corollry now tells us tht the sum of the sequence (4.2.2) is the derivtive of the sum of the sequence (4.2.), i.e. e x ( e x ) 2 = X ne nx for x 2 [, b] n= Since [, b] is n rbitrry subintervl of (0, ), we hve e x ( e x ) 2 = X Exercises for Section 4.2 n= ne nx for ll x>0. Show tht P cos(nx) n 2 + converges uniformly on R. 2. Does the series P ne nx in Exmple 3 converge uniformly on (0, )? 3. Let f n :[0, ]! R be defined by f n (x) =nx( x 2 ) n. Show tht f n (x)! 0 for ll x 2 [0, ], but tht R 0 f n(x) dx! Explin in detil how Corollry follows from Corollry Explin in detil how Corollry follows from Proposition

9 4.3. POWER SERIES 0 6. ) Show tht series P cos x n n= n converges uniformly on R. 2 b) Show tht P sin x n n= n converges to continuous function f, nd tht 7. One cn show tht x = f 0 (x) = n= cos x n n 2 2( ) n+ sin(nx) for x 2 (, ) n n= If we di erentite term by term, we get = 2( ) n+ cos(nx) for x 2 (, ) n= Is this correct formul? 8. ) Show tht the sequence P [, ) where>. n= n x converges uniformly on ll intervls b) Let f(x) = P n= n x for x>. Show tht f 0 (x) = P n= ln x n x. 4.3 Power series Recll tht power series is function of the form f(x) = c n (x ) n where is rel number nd {c n } is sequence of rel numbers. It is defined for the x-vlues tht mke the series converge. We define the rdius of convergence of the series to be the number R such tht R =limsup p n cn n! with the interprettion tht R = 0 if the limit is infinite, nd R = if the limit is 0. To justify this terminology, we need the the following result. Proposition 4.3. If R is the rdius of convergence of the power series P c n(x ) n, the series converges for x < R nd diverges for x >R.If0 <r<r, the series converges uniformly on [ r, + r]. Proof: Let us first ssume tht x >R. This mens tht x < R, nd since lim sup p n n! c n = R, there must be rbitrrily lrge vlues of n such tht np c n [ > x. Hence c n(x ) n >, nd consequently the series must diverge s the terms do not decrese to zero.

10 02 CHAPTER 4. SERIES OF FUNCTIONS To prove the (uniform) convergence, ssume tht r is number between 0 nd R. Since r > R, we cn pick positive number b<such tht b r > R. Since limsup np n! c n = R, there must be n N 2 N such tht p n cn < b r when n N. This mens tht c nr n <b n for n N, nd hence tht c n (x ) n <b n for ll x 2 [ r, + r]. Since P n=n bn is P convergent, geometric series, Weierstrss M-test tells us tht the series n=n c n(x ) n converges uniformly on [ r, + r]. Since only the til of sequence counts for convergence, the full series P c n(x ) n lso converges uniformly on [ r, +r]. Since r is n rbitrry number less thn R, we see tht the series must converge on the open intervl ( R, + R), i.e. whenever x <R. 2 Remrk: When we wnt to find the rdius of convergence, it is occsionlly convenient to compute slightly di erent limit such s lim p n+ n! c n or lim p n n! c n insted of lim p n n! c n. This corresponds to finding the rdius of convergence of the power series we get by either multiplying or dividing the originl one by (x ), nd gives the correct nswer s multiplying or dividing series by non-zero number doesn t chnge its convergence properties. The proposition bove does not tell us wht hppens t the endpoints ± R of the intervl of convergence, but we know from clculus tht series my converge t both, one or neither endpoint. Although the convergence is uniform on ll subintervls [ r, + r], it is not in generl uniform on ( R, + R). Corollry Assume tht the power series f(x) = P c n(x ) n hs rdius of convergence R lrger thn 0. Then the function f is continuous nd di erentible on the open intervl ( R, + R) with f 0 (x) = nd nc n (x ) n = n= f(t) dt = (n+)c n+ (x ) n for x 2 ( R, +R) c n n + (x )n+ = n= c n n (x )n for x 2 ( R, +R) Proof: Since the power series converges uniformly on ech subintervl [ r, +r], the sum is continuous on ech such intervl ccording to Proposition Since ech x in ( R, + R) is contined in the interior of some of the subintervls [ r, + r], we see tht f must be continuous on the full intervl ( R, + R). The formul for the integrl follows immeditely by pplying Corollry on ech subintervl [ r, + r] in similr wy.

11 4.3. POWER SERIES 03 To get the formul for the derivtive, we shll pply Corollry To use this result, we need to know tht the di erentited series P n= (n + )c n+ (x ) n hs the sme rdius of convergence s the originl series; i.e. tht p n cn = R lim sup n! n+ p (n + )c n+ =limsup n! (note tht by the remrk bove, we my use the n + -st root on the left hnd side insted of the n-th root). Since lim p n+ n! (n + ) =, this is not hrd to show (see Exercise 6). Applying Corollry on ech subintervl [ r, + r], we now get the formul for the derivtive t ech point x 2 [ r, + r]. Since ech point in ( R, + R) belongs to the interior of some of the subintervls, the formul for the derivtive must hold t ll points x 2 ( R, + R). 2 A function tht is the sum of power series, is clled rel nlytic function. Such functions hve derivtives of ll orders. Corollry Let f(x) = P c n(x ) n for x 2 ( R, + R). Then f is k times di erentible in ( R, +R) for ny k 2 N, ndf (k) () =k!c k. Hence P c n(x ) n is the Tylor series f(x) = f (n) () (x ) n n! Proof: Using the previous corollry, we get by induction tht f (k) exists on ( R, + R) nd tht f (k) (x) = n(n )... (n k + )c n (x ) n k n=k Putting x =, we get f (k) () =k!c k, nd the corollry follows. 2 Exercises for Section 4.3. Find power series with rdius of convergence 0,, 2, nd. 2. Find power series with rdius of convergence tht converge t both, one nd neither of the endpoints. 3. Show tht for ny polynomil P,lim n! n p P (n) =. 4. Use the result in Exercise 3 to find the rdius of convergence: ) P 2n x n n 3 +

12 04 CHAPTER 4. SERIES OF FUNCTIONS b) P 2n2 +n 3n+4 x n c) P nx2n 5. ) Explin tht = P x 2 x2n for x <, 2x b) Show tht = P ( x 2 ) 2 2nx2n for x <. c) Show tht 2 +x ln x = P x2n+ 2n+ 6. Let P c n(x ) n be power series. for x <. ) Show tht the rdius of convergence is given by for ny integer k. R =limsup n! n+k p c n b) Show tht lim n! n+ p n +=(write n+p n +=(n + ) n+ ). c) Prove the formul lim sup n! n+ p (n + )c n+ =limsup n! in the proof of Corollry Abel s Theorem p n cn = R We hve seen tht the sum f(x) = P c n(x ) n of power series is continuous in the interior ( R, + R) of its intervl of convergence. But wht hppens if the series converges t n endpoint ± R? It turns out tht the sum is lso continuous t the endpoint, but tht this is surprisingly intricte to prove. Before we turn to the proof, we need lemm tht cn be thought of s discrete version of integrtion by prts. Lemm 4.4. (Abel s Summtion Formul) Let { n } nd {b n} be two sequences of rel numbers, nd let s n = P n k=0 k. Then NX NX n b n = s N b N + s n (b n b n+ ). If the series P n converges, nd b n! 0 s n!, then n b n = s n (b n b n+ ) in the sense tht either the two series both diverge or they converge to the sme limit.

13 4.4. ABEL S THEOREM 05 Proof: Note tht n = s n s n for n, nd tht this formul even holds for n =0ifwedefines = 0. Hence NX NX NX n b n = (s n s n )b n = s n b n X N s n b n Chnging P the index of summtion nd using tht s = 0, we see tht N s n b n = P N s nb n+. Putting this into the formul bove, we get NX n b n = NX NX s n b n NX s n b n+ = s N b N + s n (b n b n+ ) nd the first prt of the lemm is proved. The second follows by letting N!. 2 We re now redy to prove: Theorem P (Abel s Theorem) The sum of power series f(x) = c n(x ) n is continuous in its entire intervl of convergence. This mens in prticulr tht if R is the rdius of convergence, nd the power series converges t the right endpoint +R, then lim x"+r f(x) =f(+r), nd if the power series converges t the left endpoint R, then lim x# R f(x) = f( R). Proof: We lredy know tht f is continuous in the open intervl ( R, + R), nd tht we only need to check the endpoints. To keep the nottion simple, we shll ssume tht = 0 nd concentrte on the right endpoint R. Thus we wnt to prove tht lim x"r f(x) =f(r). Note tht f(x) = P c nr n x n R. If we ssume tht x <R,wemy pply the second version of Abel s summtion formul with n = c n R n nd n b n = to get x n f(x) = x n x n+ f n (R) = R R x R X x n f n (R) R where f n (R) = P n k=0 c kr k. Summing geometric series, we see tht we lso hve Hence f(r) = f(x) f(r) = x R X x n f(r) R x X x n (f n (R) f(r)) R R Iuselim x"b nd lim x#b for one-sided limits, lso denoted by lim x!b nd lim x!b +.

14 06 CHAPTER 4. SERIES OF FUNCTIONS Given n >0, we must find >0 such tht this quntity is less thn when R <x<r. This my seem obvious due to the fctor ( x/r), but the problem is tht the infinite series my go to infinity when x! R. Hence we need to control the til of the sequence before we exploit the fctor ( x/r). Fortuntely, this is not di cult: Since f n (R)! f(r), we first pick n N 2 N such tht f n (R) f(r) < 2 for n N. Then pple f(x) x R f(r) pple + N X = x R N X x n f n (R) f(r) + R x X x n f n (R) f(r) pple R R n=n x n f n (R) f(r) + R x R N X x n f n (R) f(r) + R 2 x X x n = R 2 R where we hve summed geometric series. Now the sum is finite, nd the first term clerly converges to 0 when x " R. Hence there is >0 such tht this term is less thn 2 when R <x<r, nd consequently f(x) f(r) < for such vlues of x. 2 Let us tke look t fmous exmple. Exmple : Summing geometric series, we clerly hve Integrting, we get +x 2 = X ( ) n x 2n for x < rctn x = ( ) n x2n+ 2n + for x < Using the Alternting Series Test, we see tht the series converges even for x =. By Abel s Theorem 4 = rctn = lim x" Hence we hve proved rctn x =lim x" ( ) n x2n+ 2n + = X ( ) n 2n + 4 =

15 4.4. ABEL S THEOREM 07 This is often clled Leibniz or Gregory s formul for, but it ws ctully first discovered by the Indin mthemticin Mdhv (c. 350 c. 425). This exmple is rther typicl; the most interesting informtion is often obtined t n endpoint, nd we need Abel s Theorem to secure it. It is nturl P to think tht Abel s Theorem must hve converse sying tht if lim x"+r c nx n exists, then the sequence converges t the right endpoint x = + R. This, however, is not true s the following simple exmple shows. Exmple 2: Summing geometric series, we hve +x = X ( x) n for x < P Obviously, lim x" ( x)n =lim x" +x = 2, but the series does not converge for x =. It is possible to put extr conditions on the coe ensure convergence t the endpoint, see Exercise 2. cients of the series to Exercises for Section 4.4. ) Explin why +x = P ( )n x n for x <. b) Show tht ln( + x) = P xn+ ( )n n+ c) Show tht ln 2 = P ( )n n+. for x <. 2. In this problem we shll prove the following prtil converse of Abel s Theorem: Tuber s Theorem Assume tht s(x) = P c nx n is power series with rdius of convergence. Assume tht s =lim x" P c nx n is finite. If in ddition lim n! nc n = 0, then the power series converges for x = nd s = s(). ) Explin tht if we cn prove tht the power series converges for x =, then the rest of the theorem will follow from Abel s Theorem. b) Show tht lim N! N P N n c n = 0. c) Let s N = P N c n. Explin tht s(x) s N = NX c n ( x n )+ c n x n n=n+ d) Show tht x n pple n( x) for x <.

16 08 CHAPTER 4. SERIES OF FUNCTIONS e) Let N x be the integer such tht N x pple x <N x + Show tht s x ". XN x XN x c n ( x n ) pple ( x) n c n pple XN x n c n!0 N x f) Show tht n=n x+ c n x n pple n=n x+ n c n xn n = d x N x x n where d x! 0 s x ". Show tht P n=n x+ c nx n! 0 s x ". g) Prove Tuber s theorem. 4.5 Normed spces In lter chpter we shll continue our study of how generl functions cn be expressed s series of simpler functions. This time the simple functions will be trigonometric functions nd not power functions, nd the series will be clled Fourier series nd not power series. Before we turn to Fourier series, we shll tke look t normed spces nd inner product spces. Strictly speking, it is not necessry to know bout such spces to study Fourier series, but bsic understnding will mke it much esier to pprecite the bsic ides nd put them into wider frmework. In Fourier nlysis, one studies vector spces of functions, nd let me begin by reminding you tht vector spce is just set where you cn dd elements nd multiply them by numbers in resonble wy. More precisely: Definition 4.5. Let K be either R or C, nd let V be nonempty set. Assume tht V is equipped with two opertions: Addition which to ny two elements u, v 2 V ssigns n element u + v 2 V. Sclr multipliction which to ny element u 2 V nd ny number 2 K ssigns n element u 2 V. We cll V vector spce over K if the following xioms re stisfied: (i) u + v = v + u for ll u, v 2 V. (ii) (u + v)+w = u +(v + w) for ll u, v, w 2 V. (iii) There is zero vector 0 2 V such tht u + 0 = u for ll u 2 V. (iv) For ech u 2 V, there is n element u 2 V such tht u +( u) =0.

17 4.5. NORMED SPACES 09 (v) (u + v) = u + v for ll u, v 2 V nd ll 2 K. (vi) ( + )u = u + u for ll u 2 V nd ll, 2 K: (vii) ( u) =( )u for ll u 2 V nd ll, 2 K: (viii) u = u for ll u 2 V. To mke it esier to distinguish, we sometimes refer to elements in V s vectors nd elements in K s sclrs. I ll ssume tht you re fmilr with the bsic consequences of these xioms s presented in course on liner lgebr. Recll in prticulr tht subsetu V is vector spce if it closed under ddition nd sclr multipliction, i.e. tht whenever u, v 2 U nd 2 K, thenu + v, u 2 U. To mesure the seize of n element in metric spce, we introduce norms: Definition If V is vector spce over K, norm on V is function : V! R such tht: (i) u 0 with equlity if nd only if u = 0. (ii) u = u for ll 2 K nd ll u 2 V. (iii) u + v pple u + v for ll u, v 2 V. Exmple : The clssicl exmple of norm on rel vector spce, is the eucliden norm on R n given by q x = x 2 + x x2 n where x =(x,x 2...,x n ). The corresponding norm on the complex vector spce C n is z = p z 2 + z z n 2 where z =(z,z 2...,z n ). The spces bove re the most common vector spces nd norms in liner lgebr. More relevnt for our purposes in this chpter re: Exmple 2: Let (X, d) be compct metric spce, nd let V = C(X, R) be the set of ll continuous, rel vlued functions on X. ThenV is vector spce over R nd f =sup{ f(x) x 2 X} is norm on V. To get complex exmple, let V = C(X, C) nd define the norm by the sme formul s before. From norm we cn lwys get metric in the following wy:

18 0 CHAPTER 4. SERIES OF FUNCTIONS Proposition Assume tht V is vector spce over K nd tht is norm on V. Then d(u, v) = u v is metric on V. Proof: We hve to check the three properties of metric: Positivity: Since d(u, v) = u v, we see from prt (i) of the definition bove tht d(u, v) 0 with equlity if nd only if u v = 0, i.e. if nd only if u = v. Symmetry: Since u v = ( )(v u) = ( ) v u = v u by prt (ii) of the definition bove, we see tht d(u, v) =d(v, u). Tringle inequlity: By prt (iii) of the definition bove, we see tht for ll u, v, w 2 V : d(u, v) = u v = (u w)+(w v) pple pple u w + w v = d(u, w)+d(w, v) Whenever we refer to notions such s convergence, continuity, openness, closedness, completeness, compctness etc. in normed vector spce, we shll be refering to these notions with respect to the metric defined by the norm. In prctice, this mens tht we continue s before, but write u v insted of d(u, v) for the distnce between the points u nd v. Remrk: The inverse tringle inequlity (recll Proposition 2..4) d(x, y) d(x, z) ppled(y, z) (4.5.) is useful tool in metric spces. In normed spces, it is most conveniently expressed s u v pple u v (4.5.2) (use formul (4.5.) with x = 0, y = u nd z = v). Note tht if {u n } n= is sequence of elements in normed vector spce, we define the infinite sum P P n= u n s the limit of the prtil sums s n = n k= u k provided this limit exists; i.e. u n = lim n= nx u k n! k= 2

19 4.5. NORMED SPACES When the limit exists, we sy tht the series converges. Remrk: The nottion u = P n= u n is rther trecherous it seems to be purely lgebric reltionship, but it does, in fct, depend on which norm we re using. If we hve two di erent norms nd 2 on the sme spce V,wemyhveu = P n= u n with respect to, but not with respect to 2, s u s n! 0 does not necesrily imply u s n 2! 0. This phenomenon is ctully quite common, nd we shll meet it on severl occsions lter in the book. Recll from liner lgebr tht t vector spce V is finite dimensionl if there is finite set e, e 2,...,e n of elements in V such tht ech element x 2 V cn be written s liner combintion x = e + 2 e n e n in unique wy. We cll e, e 2,...,e n bsis for V, nd sy tht V hs dimension n. A spce tht is not finite dimensionl is clled infinte dimensionl. Most of the spces we shll be working with re infinite dimensionl, nd we shll now extend the notion of bsis to (some) such spces. Definition Let {e n } n= be sequence of elements in normed vector spce V. We sy tht {e n } is bsis 2 for V if for ech x 2 V there is unique sequence { n } n= from K such tht x = n e n n= Not ll normed spces hve bsis; there re, e.g., spces so big tht not ll elements cn be reched from countble set of bsis elements. Let us tke look t n infinite dimensionl spce with bsis. Exmple 3: Let c 0 be the set of ll sequences x = {x n } n2n of rel numbers such tht lim n! x n = 0. It is not hrd to check tht {c 0 } is vector spce nd tht x =sup{ x n : n 2 N} is norm on c 0. Let e n =(0, 0,...,0,, 0,...) be the sequence tht is on element number n nd 0 elsewhere. Then {e n } n2n is bsis for c 0 with x = P n= x ne n. If normed vector spce is complete, we cll it Bnch spce. The next theorem provides n e cient method for checking tht normed spce 2 Strictly speking, there re two notions of bsis for n infinite dimensionl spce. The type we re introducing here is sometimes clled Schuder bsis nd only works in normed spces where we cn give mening to infinite sums. There i nother kind of bsis clled Hmel bsis which does not require the spce to be normed, but which is less prcticl for pplictions.

20 2 CHAPTER 4. SERIES OF FUNCTIONS is complete. We sy tht series P P n= u n in V converges bsolutely if n= u n converges (note tht P n= u n is series of positive numbers). Proposition A normed vector spce V is complete if nd only if every bsolutely convergent series converges. Proof: Assume first tht V is complete nd tht the series P u n converges bsolutely. We must show tht the series converges in the ordinry sense. Let S n = P n k=0 u k nd s n = P n k=0 u k be the prtil sums of the two series. Since the series converges bsolutely, the sequence {S n } is Cuchy sequence, nd given n >0, there must be n N 2 N such tht S n S m < when n, m N. Without loss of generlity, we my ssume tht m>n. By the tringle inequlity s m s n = mx k=n+ u k pple mx k=n+ u k = S m S n < when n, m N, nd hence {s n } is Cuchy sequence. Since V is complete, the series P u n converges. For the converse, ssume tht ll bsolutely convergent series converge, nd let {x n } be Cuchy sequence. We must show tht {x n } converges. Since {x n } is Cuchy sequence, we cn find n incresing sequence {n i } in N such tht x n x m < for ll n, m n 2 i i. In prticulr x ni+ x ni <, nd clerly P P2 i i= x ni+ x ni converges. This mens tht the series i= (x n i+ x ni ) converges bsolutely, nd by ssumption it converges in the ordinry sense to some element s 2 V. The prtil sums of this sequence re NX s N = (x ni+ x ni )=x nn+ x n i= (the sum is telescoping nd lmost ll terms cncel), nd s they converge to s, we see tht x nn+ must converge to s + x n. This mens tht subsequence of the Cuchy sequence {x n } converges, nd thus the sequence itself converges ccording to Lemm Exercises for Section 4.5. Check tht the norms in Exmple relly re norms (i.e. tht they stisfy the conditions in Definition 4.5.2). 2. Check tht the norms in Exmple 2 relly re norms (i.e. tht they stisfy the conditions in Definition 4.5.2). 3. Let V be normed vector spce over K. Assume tht {u n }, {v n } re sequences in V converging to u og v, respectively, nd tht { n }, { n } re sequences in K converging to og,respectively.

21 4.6. INNER PRODUCT SPACES 3 ) Show tht {u n + v n } converges to u + v. b) Show tht { n u n } converges to u c) Show tht { n u n + n v n } converges to u + v. 4. Let V be normed vector spce over K. ) Prove the inverse tringle inequlity u v pple u v for ll u, v 2 V. b) Assume tht {u n } is sequence in V converging to u. Show tht { u n } converges to u 5. Show tht is norm on C([0, ], R). f = Z 0 f(t) dt 6. Prove tht the set {e n } n2n in Exmple 3 relly is bsis for c Let V 6= {0} be vector spce, nd let d be the discrete metric on V.Show tht d is not generted by norm (i.e. there is no norm on V such tht d(x, y) = x y ). 8. Let V 6= {0} be normed vector spce. Show tht V is complete if nd only if the unit sphere S = {x 2 V : x =} is complete. 9. Show tht if normed vector spce V hs bsis (s defined in Definition 4.5.4), then it is seprble (i.e. it hs countble, dense subset). 0. l is the set of ll sequences x = {x n } n2n of rel numbers such tht P n= x n converges. ) Show tht is norm on l. x = x n n= b) Show tht the set {e n } n2n in Exmple 3 is bsis for l. c) Show tht l is complete. 4.6 Inner product spces The usul (eucliden) norm in R n cn be defined in terms of the sclr (dot) product: x = p x x This reltionship is extremely importnt s it connects length (defined by the norm) nd orthogonlity (defined by the sclr product), nd it is the key to mny generliztions of geometric rguments from R 2 nd R 3 to R n. In this section we shll see how we cn extend this generliztion to certin infinite dimensionl spces clled inner product spces.

22 4 CHAPTER 4. SERIES OF FUNCTIONS The bsic observtion is tht some norms on infinite dimensionl spces cn be defined in terms of n inner product just s the eucliden norm is defined in terms of the sclr product. Let us begin by tking look t such products. As in the previous section, we ssume tht ll vector spces re over K which is either R or C. As we shll be using complex spces in our study of Fourier series, it is importnt tht you don t neglect the complex cse. Definition 4.6. An inner product h, i on vector spce V over K is function h, i : V V! K such tht: (i) hu, vi = hv, ui for ll u, v 2 V (the br denotes complex conjugtion; if the vector spce is rel, we just hve hu, vi = hv, ui). (ii) hu + v, wi = hu, wi + hv, wi for ll u, v, w 2 V. (iii) h u, vi = hu, vi for ll 2 K, u, v 2 V. (iv) For ll u 2 V, hu, ui 0 with equlity if nd only if u = 0 (by (i), hu, ui is lwys rel number). 3 As immedite consequences of (i)-(iv), we hve (v) hu, v + wi = hu, vi + hu, wi for ll u, v, w 2 V. (vi) hu, vi = hu, vi for ll 2 K, u, v 2 V (note the complex conjugte!). (vii) h u, vi = 2 hu, vi (combine (i) nd(vi) nd recll tht for complex numbers 2 = ). Exmple : The clssicl exmples re the dot products in R n nd C n.if x =(x,x 2,...,x n ) nd y =(y,y 2,...,y n ) re two rel vectors, we define hx, yi = x y = x y + x 2 y x n y n If z =(z,z 2,...,z n ) nd w =(w,w 2,...,w n ) re two complex vectors, we define hz, wi = z w = z w + z 2 w z n w n Before we look t the next exmple, we need to extend integrtion to complex vlued functions. If, b 2 R, <b, nd f,g :[, b]! R re continuous functions, we get complex vlued function h :[, b]! C by letting h(t) =f(t)+ig(t) 3 Strictly speking, we re defining positive definite inner products, but they re the only inner products we hve use for.

23 4.6. INNER PRODUCT SPACES 5 We define the integrl of h in the nturl wy: Z b h(t) dt = Z b f(t) dt + i Z b g(t) dt i.e., we integrte the rel nd complex prts seprtely. Exmple 2: Agin we look t the rel nd complex cse seprtely. For the rel cse, let V be the set of ll continuous functions f :[, b]! R, nd define the inner product by hf,gi = Z b f(t)g(t) dt For the complex cse, let V be the set of ll continuous, complex vlued functions h :[, b]! C s descibed bove, nd define hh, ki = Z b h(t)k(t) dt Then h, i is n inner product on V. Note tht these inner products my be thought of s nturl extensions of the products in Exmple ; we hve just replced discrete sums by continuous products. Given n inner product h, i, wedefine : V! [0, ) by u = p hu, ui in nlogy with the norm nd the dot product in R n nd C n. For simplicity, we shll refer to s norm, lthough t this stge it is not t ll cler tht it is norm in the sense of Definition On our wy to proving tht relly is norm, we shll pick up few results of geometric nture tht will be useful lter. We begin by defining two vectors u, v 2 V to be orthogonl if hu, vi = 0. Note tht if this is the cse, we lso hve hv, ui =0sincehv, ui = hu, vi = 0 = 0. With these definitions, we cn prove the following generliztion of the Pythgoren theorem: Proposition (Pythgoren Theorem) For ll orthogonl u, u 2,..., u n in V, Proof: We hve u + u u n 2 = u 2 + u u n 2 u + u u n 2 = hu + u u n, u + u u n i =

24 6 CHAPTER 4. SERIES OF FUNCTIONS = X pplei,jpplen hu i, u j i = u 2 + u u n 2 where we hve used tht by orthogonlity, hu i, u j i =0wheneveri 6= j. 2 Two nonzero vectors u, v re sid to be prllel if there is number 2 K such tht u = v. As in R n,theprojection of u on v is the vector p prllel with v such tht u p is orthogonl to v. Figure shows the ide. u p @ p - Figure : The projection p of u on v Proposition Assume tht u nd v re two nonzero elements of V. Then the projection p of u on v is given by: p = hu, vi v 2 v The norm of the projection is p = hu,vi v Proof: Since p is prllel to v, it must be of the form p = v. Todetermine, we note tht in order for u p to be orthogonl to v, wemusthve hu p, vi = 0. Hence is determined by the eqution 0=hu v, vi = hu, vi h v, vi = hu, vi v 2 Solving for, we get = hu,vi v 2, nd hence p = hu,vi v 2 v. To clculte the norm, note tht p 2 = hp, pi = h v, vi = 2 hv, vi = hu, vi 2 hu, vi 2 v 4 hv, vi = v 2 (recll property (vi) just fter Definition 4.6.). 2 We cn now extend Cuchy-Schwrz inequlity to generl inner products:

25 4.6. INNER PRODUCT SPACES 7 Proposition (Cuchy-Schwrz inequlity) For ll u, v 2 V, hu, vi pple u v with equlity if nd only if u nd v re prllel or t lest one of them is zero. Proof: The proposition clerly holds with equlity if one of the vectors is zero. If they re both nonzero, we let p be the projection of u on v, nd note tht by the pythgoren theorem u 2 = u p 2 + p 2 p 2 with equlity only if u = p, i.e. whenu nd v re prllel. Since p = hu,vi v by Proposition 4.6.3, we hve u 2 hu, vi 2 v 2 nd the proposition follows. 2 We my now prove: Proposition (Tringle inequlity for inner products) For ll u, v 2 V u + v pple u + v Proof: We hve (recll tht Re(z) refers to the rel prt of complex number z = + ib): u + v 2 = hu + v, u + vi = hu, ui + hu, vi + hv, ui + hv, vi = = hu, ui + hu, vi + hu, vi + hv, vi = hu, ui + 2Re(hu, vi)+hv, vi pple pple u 2 +2 u v + v 2 =( u + v ) 2 where we hve used tht ccording to Cuchy-Schwrz inequlity, we hve Re(hu, vi) pple hu, vi pple u v. 2 We re now redy to prove tht relly is norm: Proposition If h, i is n inner product on vector spce V, then defines norm on V, i.e. u = p hu, ui (i) u 0 with equlity if nd only if u = 0.

26 8 CHAPTER 4. SERIES OF FUNCTIONS (ii) u = u for ll 2 C nd ll u 2 V. (iii) u + v pple u + v for ll u, v 2 V. Proof: (i) follows directly from the definition of inner products, nd (iii) is just the tringle inequlity. We hve ctully proved (ii) on our wy to Cuchy-Schrz inequlity, but let us repet the proof here: u 2 = h u, ui = 2 u 2 where we hve used property (vi) just fter Definition The proposition bove mens tht we cn think of n inner product spce s metric spce with metric defined by d(x, y) = x y = p hx y, x yi Exmple 3: Returning to Exmple 2, we see tht the metric in the rel s well s the complex cse is given by Z b d(f,g) = f(t) g(t) 2 2 dt The next proposition tells us tht we cn move limits nd infinite sums in nd out of inner products. Proposition Let V be n inner product spce. (i) If {u n } is sequence in V converging to u, then the sequence { u n } of norms converges to u. (ii) If the series P w n converges in V, then w n = lim N! NX w n (iii) If {u n } is sequence in V converging to u, then the sequence hu n, vi of inner products converges to hu, vi for ll v 2 V. In symbols, lim n! hu n, vi = hlim n! u n, vi for ll v 2 V. (iv) If the series P w n converges in V, then. h w n, vi = hw n, vi n= n=

27 4.6. INNER PRODUCT SPACES 9 Proof: (i) follows directly from the inverse tringle inequlity u u n pple u u n (ii) follows immeditely from (i) if we let u n = P n k=0 w k (iii) Assume tht u n! u. To show tht hu n, vi!hu, vi, issu ces to prove tht hu n, vi hu, vi = hu n u, vi!0. But by Cuchy-Schwrz inequlity hu n u, vi pple u n u v! 0 since u n u! 0 by ssumption. (iv) We use (iii) with u = P n= w n nd u n = P n k= w k.then nx h w n, vi = hu, vi = lim hu n, vi = lim h w k, vi = n! n! n= = lim n! k= nx hw k, vi = hw n, vi We shll now generlize some notions from liner lgebr to our new setting. If {u, u 2,...,u n } is finite set of elements in V,wedefinethe spn Sp{u, u 2,...,u n } n= k= of {u, u 2,...,u n } to be the set of ll liner combintions 2 u + 2 u n u n, where, 2,..., n 2 K AsetA V is sid to be orthonorml if it consists of orthogonl elements of length one, i.e. if for ll, b 2 A, wehve 8 < 0 if 6= b h, bi = : if = b If {e, e 2,...,e n } is n orthonorml set nd u 2 V,wedefinetheprojection of u on Sp{e, e 2,...,e n } by P e,e 2,...,e n (u) =hu, e ie + hu, e 2 ie hu, e n ie n This terminology is justified by the following result. Proposition Let {e, e 2,...,e n } be n orthonorml set in V. For every u 2 V, the projection P e,e 2,...,e n (u) is the element in Sp{e, e 2,...,e n } closest to u. Moreover, u P e,e 2,...,e n (u) is orthogonl to ll elements in Sp{e, e 2,...,e n }.

28 20 CHAPTER 4. SERIES OF FUNCTIONS Proof: We first prove the orthogonlity. It su ces to prove tht for ech i =, 2,...,n, s we then hve hu P e,e 2,...,e n (u), e i i = 0 (4.6.) hu P e,e 2,...,e n (u), e + + n e n i = = hu P e,e 2,...,e n (u), e i n hu P e,e 2,...,e n (u), e n i =0 for ll e + + n e n 2 Sp{e, e 2,...,e n }. To prove formul (4.6.), just observe tht for ech e i hu P e,e 2,...,e n (u), e i i = hu, e i i hp e,e 2,...,e n (u), e i i = hu, e i i hu, e i ihe, e i i + hu, e 2 ihe 2, e i i + + hu, e n ihe n, e i i = = hu, e i i hu, e i i =0 To prove tht the projection is the element in Sp{e, e 2,...,e n } closest to u, letw = e + 2 e n e n be nother element in Sp{e, e 2,...,e n }. Then P e,e 2,...,e n (u) w is in Sp{e, e 2,...,e n }, nd hence orthogonl to u P e,e 2,...,e n (u) by wht we hve just proved. By the Pythgoren theorem u w 2 = u P e,e 2,...,e n (u) 2 + P e,e 2,...,e n (u) w 2 > u P e,e 2,...,e n (u) 2 As n immedite consequence of the proposition bove, we get: Corollry (Bessel s inequlity) Let {e, e 2,...,e n,...} be n orthonorml sequence in V.Fornyu 2 V, hu, e i i 2 pple u 2 i= Proof: Since u P e,e 2,...,e n (u) is orthogonl to P e,e 2,...,e n (u), we get by the Pythgoren theorem tht for ny n u 2 = u P e,e 2,...,e n (u) 2 + P e,e 2,...,e n (u) 2 P e,e 2,...,e n (u) 2 Using the Pythgoren Theorem gin, we see tht P e,e 2,...,e n (u) 2 = hu, e ie + hu, e 2 ie hu, e n ie n 2 = = hu, e ie 2 + hu, e 2 ie hu, e n ie n 2 = = hu, e i 2 + hu, e 2 i hu, e n i 2 2

29 4.6. INNER PRODUCT SPACES 2 nd hence u 2 hu, e i 2 + hu, e 2 i hu, e n i 2 for ll n. Letting n!, the corollry follows. 2 We hve now reched the min result of this section. Recll from Definition tht {e i } is bsis for V if ny element u in V cn be written s liner combintion u = P i= ie i in unique wy. The theorem tells us tht if the bsis is orthonorml, the coe sients i re esy to find; they re simply given by i = hu, e i i. Theorem (Prsevl s Theorem) If {e, e 2,...,e n,...} is n orthonorml bsis for V, then for ll u 2 V, we hve u = P i= hu, e iie i nd u 2 = P i= hu, e ii 2. Proof: Since {e, e 2,...,e n,...} is bsis, we know tht there is unique sequence, 2,..., n,... from K such tht u = P n= ne n. This mens tht u P N n= ne n! 0 s N!. Since the projection P e,e 2,...,e N (u) = P N n= hu, e nie n is the element in Sp{e, e 2,...,e N } closest to u, wehve NX NX u hu, e n ie n pple u n e n! 0 s N! n= n= nd hence u = P n= hu, e nie n. To prove the second prt, observe tht since u = P n= hu, e P nie n =lim N N! n= hu, e nie n, we hve (recll Proposition 4.6.7(ii)) u 2 = lim N! NX hu, e n ie n 2 = lim n= N! n= NX hu, e n i 2 = hu, e n i 2 n= 2 The coe cients hu, e n i in the rguments bove re often clled (bstrct) Fourier coe cients. By Prsevl s theorem, they re squre summble in the sense tht P n= hu, e ni 2 <. A nturl question is whether we cn reverse this procedure: Given squre summble sequence { n } of elements in K, does there exist n element u in V with Fourier coe cients n,i.e. such tht hu, e n i = n for ll n? The nswer is rmtive provided V is complete. Proposition 4.6. Let V be complete inner product spce over K with n orthonorml bsis {e, e 2,...,e n,...}. Assume tht { n } n2n is sequence from K which is squre summble in the sense tht P n= n 2 converges. Then the series P n= ne n converges to n element u 2 V,nd hu, e n i = n for ll n 2 N.

30 22 CHAPTER 4. SERIES OF FUNCTIONS Proof: We must prove tht the prtil sums s n = P n k= ke k form Cuchy sequence. If m>n,wehve s m s n 2 = mx k=n+ n e n 2 = mx k=n+ n 2 Since P n= n 2 converges, we cn get this expression less thn ny >0 by choosing n, m lrge enough. Hence {s n } is Cuchy sequence, nd the series P n= ne n converges to some element u 2 V. By Proposition 4.6.7, hu, e i i = h n e n, e i i = h n e n, e i i = i n= n= 2 Completeness is necessry in the proposition bove if V is not complete, there will lwys be squre summble sequence { n } such tht P n= ne n does not converge (see exercise 3). A complete inner product spce is clled Hilbert spce. Exercises for Section 4.6. Show tht the inner products in Exmple relly re inner products (i.e. tht they stisfy Definition 4.6.). 2. Show tht the inner products in Exmple 2 relly re inner products. 3. Prove formul (v) just fter Definition Prove formul (vi) just fter Definition Prove formul (vii) just fter Definition Show tht if A is symmetric (rel) mtrix with strictly positive eigenvlues, then hu, vi =(Au) v is n inner product on R n. 7. If h(t) =f(t) +ig(t) is complex vlued function where f nd g re differentible, define h 0 (t) =f 0 (t)+ig 0 (t). Prove tht the integrtion by prts formul Z b pple b Z b u(t)v 0 (t) dt = u(t)v(t) u 0 (t)v(t) dt holds for complex vlued functions. 8. Assume tht {u n } nd {v n } re two sequences in n inner product spce converging to u nd v, respectively. Show tht hu n, v n i!hu, vi.

31 4.6. INNER PRODUCT SPACES Show tht if the norm is defined from n inner product by u = hu, ui 2, we hve the prllelogrm lw u + v 2 + u v 2 =2 u 2 +2 v 2 for ll u, v 2 V. Show tht the norms on R 2 defined by (x, y) = mx{ x, y } nd (x, y) = x + y do not come from inner products. 0. Let {e, e 2,...,e n } be n orthonorml set in n inner product spce V.Show tht the projection P =P e,e 2,...,e n is liner in the sense tht P ( u) = P (u) nd P (u + v) =P (u)+p (v) for ll u, v 2 V nd ll 2 K.. In this problem we prove the polriztion identities for rel nd complex inner products. These identities re useful s they express the inner product in terms of the norm. ) Show tht if V is n inner product spce over R, then hu, vi = 4 u + v 2 u v 2 b) Show tht if V is n inner product spce over C, then hu, vi = 4 u + v 2 u v 2 + i u + iv 2 i u iv 2 2. If S is nonempty subset of n inner product spce, let S? = {u 2 V : hu, si = 0 for ll s 2 S} ) Show tht S? is closed subspce of S. b) Show tht if S T,thenS? T?. 3. Let l 2 be the set of ll rel sequences x = {x n } n2n such tht P n= x2 n <. ) Show tht if x = {x n } n2n nd y = {y n } n2n re in l 2,thentheseries P n= x ny n converges. (Hint: For ech N, NX x n y n pple n= NX x 2 n n=! 2 N X yn 2 n=! 2 by Cuchy-Schwrz inequlity) b) Show tht l 2 is vector spce. c) Show tht hx, yi = P n= x ny n is n inner product on l 2. d) Show tht l 2 is complete. e) Let e n be the sequence where the n-th component is nd ll the other components re 0. Show tht {e n } n2n is n orthonorml bsis for l 2. f) Let V be n inner product spce with n orthonorml bsis {v, v 2,..., v n,...}. Assume tht for every squre summble sequence { n }, there is n element u 2 V such tht hu, v i i = i for ll i 2 N. Show tht V is complete.

32 24 CHAPTER 4. SERIES OF FUNCTIONS 4.7 Liner opertors In liner lgebr the importnt functions re the liner mps. The sme holds for infinitely dimensionl spces, but the mps re then usully referred to s liner opertors or liner trnsformtions: Definition 4.7. Assume tht V nd W re two vector spces over K. A function A : V! W is clled liner opertor if it stisfies: (i) A( u) = A(u) for ll 2 K nd u 2 V. (ii) A(u + v) =A(u)+A(v) for ll u, v 2 V. Combining (i) nd (ii), we see tht A( u + v) = A(u)+ A(v) Using induction, this cn be generlized to A( u + 2 u n u n )= A(u )+ 2 A(u 2 )+ + n A(u n ) (4.7.) It is lso useful to observe tht since A(0) = A(00) = 0A(0) = 0, we hve A(0) = 0 for ll liner opertors. As K my be regrded s vector spce over itself, the definition bove covers the cse where W = K. The mp is then usully referred to s (liner) functionl. Exmple : Let V = C([, b], R) be the spce of continuous functions from the intervl [, b] tor. The function A : V! R defined by A(u) = Z b u(x) dx is liner functionl, while the function B : V! V defined by B(u)(x) = u(t) dt is liner opertor. Exmple 2: Just s integrtion, di erentition is liner opertion, but s the derivtive of di erentible function is not necessrily di erentible, we hve to be creful which spces we work with. A function f :(, b)! R is sid to be infinitely di erentible if it hs derivtives of ll orders t ll points in (, b), i.e. if f (n) (x) exists for ll n 2 N nd ll x 2 (, b). Let U be the spce of ll infinitely di erentible functions, nd define D : U! U by Du(x) =u 0 (x). Then D is liner opertor. We shll minly be interested in liner opertors between normed spces, nd the following notion is of centrl importnce:

33 4.7. LINEAR OPERATORS 25 Definition Assume tht (V, v ) nd (W, W ) re two normed spces. A liner opertor A : V! W is bounded if there is constnt M 2 R such tht A(u) W pple M u V for ll u 2 V. Remrk: The terminology here is rther trecherous s bounded opertor is not bounded function in the sense of, e.g., the Extreml Vlue Theorem. To see this, note tht if A(u) 6= 0, we cn get A( u) W = A(u) W s lrge s we wnt by incresing the size of. The best (i.e. smllest) vlue of the constnt M in the definition bove is denoted by A nd is given by A(u) W A =sup : u 6= 0 u V An lterntive formultion (see Exercise 4) is A =sup{ A(u) W : u V =} (4.7.2) We cll A the opertor norm of A. The nme is justified in Exercise 7. It s instructive to tke new look t the opertors in Exmples nd 2: Exmple 3: The opertors A nd B in Exmple re bounded if we use the (usul) supremum norm on V. To see this for B, note tht B(u)(x) = u(t) dt pple u(t) dt pple u du = u (x ) pple u (b ) which implies tht B(u) pple (b ) u for ll u 2 V. Exmple 4: If we let U hve the supremum norm, the opertor D in Exmple 2 is not bounded. If we let u n =sinnx, wehve u n =, but D(u n ) = n cos nx! s n!. Tht D is n unbounded opertor is the source of lot of trouble, e,g. the rther unstisfctory conditions we hd to enforce in our tretment of di erentition of series in Proposition We shll end this section with brief study of the connection between boundedness nd continuity. One wy is esy: Lemm A bounded liner opertor A is uniformly continuous. Proof: If A = 0, A is constnt zero nd there is nothing to prove. If A 6= 0, we my for given >0, choose = A. For u v V <,we then hve A(u) A(v) W = A(u v) W pple A u v V < A A <