UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE


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1 UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence (f n ) converges pointwise on E if there is function f : E Y such tht f n (p) f(p) for every p E. Clerly, such function f is unique nd it is clled the pointwise limit of (f n ) on E. We then write f n f on E. For simplicity, we shll ssume Y = R with the usul metric. Let f n f on E. We sk the following questions: (i) If ech f n is bounded on E, must f be bounded on E? If so, must sup p E f n (p) sup p E f(p)? (ii) If E is metric spce nd ech f n is continuous on E, must f be continuous on E? (iii) If E is n intervl in R nd ech f n is differentible on E, must f be differentible on E? If so, must f n f on E? (iv) If E = [, b] nd ech f n is Riemnn integrble on E, must f be Riemnn integrble on E? If so, must f n(x)dx f(x)dx? These questions involve interchnge of two processes (one of which is tking the limit s n ) s shown below. (i) lim sup f n (p) = sup lim f n(p). n p E p E n (ii) For p E nd p k p in E, lim lim f n(p k ) = lim lim f n(p k ). k n n k d (iii) lim n dx (f n) = d ( ) lim dx f n. n (iv) lim n f n (x)dx = ( lim f n(x) n ) dx. Answers to these questions re ll negtive. Exmples 1.1. (i) Let E := (0, 1] nd define f n : E R by { 0 if 0 < x 1/n, f n (x) := 1/x if 1/n x 1. Then f n (x) n for ll x E, f n f on E, where f(x) := 1/x. Thus ech f n is bounded on E, but f is not bounded on E. (ii) Let E := [0, 1] nd define f n : E R by f n (x) := 1/(nx + 1). Then ech f n is continuous on E, f n f on E, where f(0) := 1 nd f(x) := 0 if 0 < x 1. Clerly, f is not continuous on E. (iii) () Let E := ( 1, 1) nd define f n : E R by f n (x) := 1/(nx 2 + 1). Then ech f n is differentible on E nd f n f on ( 1, 1), 1
2 2 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE (iv) where f(0) := 1 nd f(x) := 0 if 0 < x < 1. Clerly, f is not differentible on E. (b) Let E := R nd define f n : R R by f n (x) := (sin nx)/n. Then ech f n is differentible, f n f on R, where f 0. But f n (x) = cos nx for x R, nd (f n ) does not converge pointwise on R. For exmple, (f n (π)) is not convergent sequence. (c) Let E := ( 1, 1) nd define f n (x) := { [2 (1 + x) n ]/n if 1 < x < 0, (1 x) n /n if 0 x < 1. Then ech f n is differentible on E. (In prticulr, we hve f n (0) = 1 by L Hôpitl s Rule.) Also, f n f on ( 1, 1), where f 0. Also, { f n(x) (1 + x) n 1 if 1 < x < 0, = (1 x) n 1 if 0 x < 1. Further, f n g on ( 1, 1), where g(0) := 1 nd g(x) := 0 for 0 < x < 1. Clerly, f g. () Let E := [0, 1] nd define f n : [0, 1] R by { 1 if x = 0, 1/n!, 2/n!,..., n!/n! = 1, f n (x) := 0 otherwise. Then ech f n is Riemnn integrble on [0, 1] since it is discontinuous only t finite number of points. { 1 if x is rtionl, Also, f n f on [0, 1], where f(x) := 0 if x is irrtionl. For if x = p/q with p {0, 1, 2,..., q} N, then for ll n q, we hve n!x {0, 1, 2,...} nd so f n (x) = 1, while if x is n irrtionl number, then f n (x) = 0 for ll n N. We hve seen tht the Dirichlet function f is not Riemnn integrble. (b) Let E := [0, 1], nd define f n : [0, 1] R by f n (x) := n 3 xe nx. Then ech f n is Riemnn integrble nd f n f on [0, 1], where f 0. (Use L Hôpitl s Rule repetedly to show tht lim t t 3 /e st = 0 for ny s R with s > 0.) However, using Integrtion by Prts, we hve 1 0 xe nx dx = 1 n 2 1 n 2 e n 1 ne n for ech n N, so tht 1 0 f n(x)dx = n (n/e n ) (n 2 /e n ). (c) Let E := [0, 1] nd define f n : [0, 1] R by f n (x) := n 2 xe nx. As bove, ech f n is Riemnn integrble, nd f n f on [0, 1], where f 0, nd 1 0 f n(x)dx = 1 (1/e n ) (n/e n ) 1, which is not equl to 1 0 f(x)dx = Uniform Convergence of Sequence In n ttempt to obtin ffirmtive nswers to the questions posed t the beginning of the previous section, we introduce stronger concept of convergence.
3 UNIFORM CONVERGENCE 3 Let E be set nd consider functions f n : E R for n = 1, 2,.... We sy tht the sequence (f n ) of functions converges uniformly on E if there is function f : E R such tht for every ɛ > 0, there is n 0 N stisfying n n 0, p E = f n (p) f(p) < ɛ. Note tht the nturl number n 0 mentioned in the bove definition my depend upon the given sequence (f n ) of functions nd on the given positive number ɛ, but it is independent of p E. Clerly, such function f is unique nd it is clled the uniform limit of (f n ) on E. We then write f n f on E. Obviously, f n f on E = f n f on E, but the converse is not true : Let E := (0, 1] nd define f n (x) := 1/(nx + 1) for 0 < x 1. If f(x) := 0 for x (0, 1], then f n f on (0, 1], but f n f on (0, 1]. To see this, let ɛ := 1/2, note tht there is no n 0 N stisfying 1 f n (x) f(x) = nx + 1 < 1 2 for ll n n 0 nd for ll x (0, 1], since 1/(nx + 1) = 1/2 when x = 1/n, n N. A sequence (f n ) of relvlued functions defined on set E is sid to be uniformly Cuchy on E if for every ɛ > 0, there is n 0 N stisfying m, n n 0, p E = f m (p) f n (p) < ɛ. Proposition 2.1. (Cuchy Criterion for Uniform Convergence of Sequence) Let (f n ) be sequence of relvlued functions defined on set E. Then (f n ) is uniformly convergent on E if nd only if (f n ) is uniformly Cuchy on E. Proof. = ) Let f n f. For ll m, n N nd p E, we hve f m (p) f n (p) f m (p) f(p) + f(p) f n (p). =) For ech p E, (f n (p)) is Cuchy sequence in R, nd so it converges to rel number which we denote by f(p). Let ɛ > 0. There is n 0 N stisfying m, n n 0, p E = f m (p) f n (p) < ɛ. For ny m n 0 nd p E, letting n, we hve f m (p) f(p) ɛ. We hve the following useful test for checking the uniform convergence of (f n ) when its pointwise limit is known. Proposition 2.2. (Test for Uniform Convergence of Sequence) Let f n nd f be relvlued functions defined on set E. If f n f on E, nd if there is sequence ( n ) of rel numbers such tht n 0 nd f n (p) f(p) n for ll p E, then f n f on E. Proof. Let ɛ > 0. Since n 0, there is n 0 N such tht n n 0 = n < ɛ, nd so f n (p) f(p) < ɛ for ll p E. Exmple 2.3. Let 0 < r < 1 nd f n (x) := x n for x [ r, r]. Thenf n (x) 0 for ech x [ r, r]. Since r n 0 nd f n (x) 0 r n for ll x [ r, r], (f n ) is uniformly convergent on [ r, r]. Let us now pose the four questions stted in the lst section with convergence replced by uniform convergence. We shll nswer them one by one, but not necessrily in the sme order.
4 4 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Uniform Convergence nd Boundedness. Proposition 2.4. Let f n nd f be relvlued functions defined on set E. If f n f on E nd ech f n is bounded on E, then f bounded on E. Proof. There is n 0 N such tht n n 0, p E = f n (p) f(p) < 1. Also, since f n0 is bounded on E, there is α 0 such tht p E = f n0 (p) α 0. Hence p E = f(p) f(p) f n0 (p) + f n0 (p) < 1 + α 0. The converse of the bove result is not true, tht is, ech f n s well s f bounded on E nd f n f f n f on E. For exmple, let E := (0, 1], f n (x) := 1/(nx + 1) nd f 0. Given set E, let B(E) denote the set of ll relvlued bounded functions defined on E. For f, g in B(E), define d(f, g) := sup{ f(p) g(p) : p E}. Then it is esy to see tht d is metric on B(E), known s the supmetric on B(E). Also, by Proposition 2.2, for f n nd f in B(E), we hve f n f on E if nd only if d(f n, f) 0, tht is, (f n ) converges to f in the supmetric on B(E). Similrly, (f n ) is uniformly Cuchy on E if nd only if d(f n, f m ) 0 s n, m, tht is, (f n ) is Cuchy sequence in the supmetric on B(E). Thus Propositions 2.1 nd 2.4 show tht B(E) is complete metric spce. Also, under the hypotheses of Proposition 2.4, we hve sup p E f n (p) sup p E nd so, sup p E f n (p) sup p E f(p). Uniform Convergence nd Integrtion. f(p) d(f n, f) 0, Proposition 2.5. Let (f n ) be sequence of relvlued functions defined on [, b]. If f n f on [, b] nd ech f n is Riemnn integrble on [, b], then f is Riemnn integrble on [, b] nd f n(x)dx f(x)dx. Proof. Since f n f nd ech f n is bounded, we see tht f is bounded on [, b] by Proposition 2.4. For n N, let α n := d(f n, f), where d denotes the supmetric on B([, b]). For ech n N nd x [, b], we hve f n (x) f(x) α n, tht is, f n (x) α n f(x) f n (x) + α n, nd so L(f n ) α n (b ) L(f) U(f) U(f n ) + α n (b ). But since f n is Riemnn integrble, we hve L(f n ) = U(f n ), nd hence 0 U(f) L(f) 2α n (b ) 0 s n. Thus L(f) = U(f), tht is, f is Riemnn integrble on [, b]. Also, tht is, f n (x)dx α n (b ) f(x)dx f n (x)dx + α n (b ), f n(x)dx f(x)dx αn (b ) 0 s n. The converse of the bove result is not true, tht is, ech f n s well s f Riemnn integrble on [, b], f n f on [, b] nd f n(x)dx f(x)dx f n f. For exmple, if f n (x) := 1/(nx + 1) for x [0, 1],
5 0 UNIFORM CONVERGENCE 5 f(0) := 1, f(x) := 0 for x (0, 1], then f n f on [0, 1], but f is integrble nd 1 ln(nx + 1) f n (x)dx = 1 ln(1 + n) 1 = 0 = f(x)dx. n 0 n Uniform Convergence nd Continuity. Proposition 2.6. Let (f n ) be sequence of relvlued functions defined on metric spce E. If f n f on E nd ech f n is continuous on E, then f is continuous on E. Proof. Let ɛ > 0. There is n 0 N such tht p E = f n0 (p) f(p) < ɛ/3. Consider p 0 E. Since f n0 is continuous t p 0, there is δ > 0 such tht p E, d(p, p 0 ) < δ = f n0 (p) f n0 (p 0 ) < ɛ/3. nd hence f(p) f(p 0 ) f(p) f n0 (p) + f n0 (p) f n0 (p 0 ) + f n0 (p 0 ) f(p 0 ) < ɛ, estblishing the continuity of f t p 0 E. The converse of the bove result is not true, tht is, ech f n s well s f continuous on metric spce E, f n f on E f n f. For exmple, let f n (x) := nxe nx nd f(x) := 0 for x [0, 1]. Since f n (0) = 0 nd for x (0, 1], f n (x) 0 s n by L Hôpitl s Rule, we see tht f n f. But there is no n 0 N such tht n n 0, x [0, 1] = nxe nx 0 < 1, since nxe nx = e 1 for x = 1/n, n N. However, the following prtil converse holds. Proposition 2.7. (Dini s Theorem) Let (f n ) be sequence of relvlued functions defined on compct metric spce E. If f n f on E, ech f n nd f re continuous on E, nd (f n ) is monotonic sequence (tht is, f n f n+1 for ll n N, or f n f n+1 for ll n N), then f n f on E. For proof, see Theorem 7.13 of [3]. The following exmples show tht neither the compctness of the metric spce E nor the continuity of the function f cn be dropped from Dini s Theorem: (i) E := (0, 1] nd f n (x) := 1/(nx + 1), x E, (ii) E := [0, 1] nd f n (x) := x n, x E. Uniform Convergence nd Differentition. Answers to the questions regrding differentition posed in the lst section re not ffirmtive even when f n f on n intervl of R. () Let f n (x) := x 2 + (1/n 2 ) nd f(x) := x for x [ 1, 1]. Since f n (x) f(x) = x 2 + (1/n 2 ) x 2 x 2 + (1/n 2 ) x 2 = 1 n for ll n N nd x [ 1, 1], Proposition 2.2 shows tht f n f on [ 1, 1]. Although ech f n is differentible on [ 1, 1], the limit function f is not. (b) In Exmple 1.1 (iii) (b), f n f on R, ech f n differentible, but (f n ) does not converge pointwise. (c) In Exmple 1.1 (iii) (c), f n f on ( 1, 1) nd ech f n s well s f is differentible on ( 1, 1), nd f n g on ( 1, 1), where g f. However, if we ssume the uniform convergence of the derived sequence (f n) long with the convergence of the sequence (f n ) t only one point of the intervl, we hve stisfctory nswer. 0
6 6 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Proposition 2.8. Let (f n ) be sequence of relvlued functions defined on [, b]. If (f n ) converges t one point of [, b], ech f n is continuously differentible on [, b] nd (f n) converges uniformly on [, b], then there is f : [, b] R such tht f is continuously differentible on [, b], f n f on [, b] nd in fct, f n f on [, b]. Proof. Let x 0 [, b] nd c 0 R be such tht f n (x 0 ) c 0. Also, let ech f n be continuously differentible nd g : [, b] R be such tht f n g on [, b]. By Proposition 2.6, the function g is continuous on [, b]. Define f : [, b] R by x f(x) := c 0 + g(t)dt for x [, b]. x 0 By prt (ii) of the Fundmentl Theorem of Clculus (FTC), f exists on [, b] nd f (x) = g(x) for x [, b]. Thus f is continuously differentible on [, b] nd f n g = f. Also, by prt (i) of the FTC, we hve f n (x) = f n (x 0 ) + x x 0 f n (t)dt for x [, b]. Hence for n N nd x [, b], we obtin x ( f n (x) f(x) f n (x 0 ) c 0 + f n (t) g(t) ) dt x 0 f n (x 0 ) c 0 + x x 0 sup f n (t) g(t) t [,b] Thus f n f on [, b] by Proposition 2.2. f n (x 0 ) c 0 + (b )d(f n, g). The converse of the bove result is not true, tht is, ech f n s well s f continuously differentible on [, b], f n f on [, b], f n f on [, b] f n f. For exmple, let f n (x) := (nx + 1)e nx /n nd f(x) := 0 for x [0, 1]. Since f n (x) = nxe nx for ech n N nd ll x [0, 1], ech f n is monotoniclly decresing on [0, 1]. As f n (0) = 1/n, we obtin f n (x) f(x) 1/n for ll x [0, 1], nd so f n f on [0, 1]. Also, we hve seen fter the proof of Proposition 2.6 tht f n f, but f n f on [0, 1]. Remrk 2.9. Proposition 2.8 holds if we drop the word continuously ppering (two times) in its sttement, but then the proof is much more involved. See Theorem 7.17 of [3]. The results in Propositions 2.4, 2.5, 2.6 nd 2.8 re summrized in the following theorem. Theorem (i) The uniform limit of sequence of relvlued bounded functions defined on set is bounded. (ii) The uniform limit of sequence of Riemnn integrble functions defined on [, b] is Riemnn integrble, nd its Riemnn integrl is the limit of the sequence of termwise Riemnn integrls, tht is, if (f n ) is uniformly convergent to f on [, b] nd ech f n is Riemnn integrble on [, b], then the function f is Riemnn integrble on [, b] nd f(x)dx = lim n f n(x)dx.
7 UNIFORM CONVERGENCE 7 (iii) The uniform limit of sequence of continuous functions defined on metric spce is continuous. (vi) If sequence of continuously differentible functions defined on [, b] is convergent t one point of [, b] nd if the derived sequence is uniformly convergent on [, b], then the given sequence converges uniformly on [, b], the uniform limit is continuously differentible on [, b] nd its derivtive is the limit of the sequence of termwise derivtives, tht is, if (f n ) converges t one point of [, b], ech f n is continuously differentible on [, b] nd (f n ) is uniformly convergent on [, b], then (f n ) converges uniformly to continuously differentible function f on [, b], nd f (x) = lim n f n(x) for ll x [, b]. 3. Uniform Convergence of series The reder is ssumed to be fmilir with the elementry theory of series of rel numbers. (See, for exmple, Chpter 9 of [1], or Chpter 3 of [3].) Let (f k ) be sequence of relvlued functions defined on set E. Consider the sequence (s n ) of relvlued functions on E defined by s n := f f n = n f k. Note: Just s the sequence (f k ) determines the sequence (s n ), so does (s n ) determine (f k ): If we let s 0 = 0, then we hve f k = s k s k 1 for ll k N. We sy tht the series f k converges pointwise on E if the sequence (s n ) converges pointwise on E, nd we sy tht the series f k converges uniformly on E if the sequence (s n ) converges uniformly on E. For n N, the function s n is clled the nth prtil sum of the series f k nd if s n s, then the function s is clled its sum. Results bout convergence / uniform convergence of sequences of functions crry over to corresponding results bout convergence / uniform convergence of series of functions. Proposition 3.1. (Cuchy Criterion for Uniform Convergence of Series) Let (f k ) be sequence of relvlued functions defined on set E. Then the series f k converges uniformly on E if nd only if for every ɛ > 0, there is n 0 N such tht m n n 0, p E = m f k (p) < ɛ. k=n Proof. Use Proposition 2.1 for the sequence (s n ) of prtil sums. Proposition 3.2. (Weierstrss MTest for Uniform Absolute Convergence of Series) Let (f k ) be sequence of relvlued functions defined on set E. Suppose there is sequence (M k ) in R such tht f k (p) M k for ll k N nd ll p E. If M k is convergent, then f k converges uniformly nd bsolutely on E.
8 8 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Proof. Note tht m f k (p) k=n nd use Proposition 3.1. m f k (p) k=n m M k for ll m n, Exmples 3.3. (i) Consider the series k=0 xk, where x ( 1, 1). If r < 1, then the series converges uniformly on {x R : x r} since the series k=0 M k is convergent, where M k := r k for k = 0, 1,... (ii) For k N, let f k (x) := ( 1) k (x + k)/k 2, where x [0, 1]. We show tht the series f k converges uniformly on [0, 1]. For k N, let g k (x) := ( 1) k x/k 2, where x [0, 1]. Letting M k := 1/k 2 for k N, we observe tht the series k=0 M k is convergent. Hence the series g k(x) converges uniformly on [0, 1]. Also, the series ( 1)k /k converges uniformly on [0, 1], being convergent series of constnts. Since f k (x) = g k (x) + ( 1) k /k for k N nd x [0, 1], the series f k(x) converges uniformly on [0, 1]. This exmple lso shows tht the converse of Weierstrss Mtest does not hold: If M k := sup x [0,1] f k (x) = (1 + k)/k 2 for k N, then M k does not converge, since 1/k2 converges, but 1/k diverges. Proposition 3.4. (Dirichlet s Test for Uniform Conditionl Convergence of Series) Let (f k ) be monotonic sequence of relvlued functions defined on set E such tht f k 0 on E. If (g k ) is sequence of relvlued functions defined on E such tht the prtil sums of the series g k re uniformly bounded on E, then the series f kg k converges uniformly on E. In prticulr, the series ( 1)k f k converges uniformly on E. Proof. For ech p E, the series f k(p)g k (p) converges in R by Dirichlet s Test for conditionl convergence of series of rel numbers. (See, for exmple, Proposition 9.20 of [1], or Theorem 3.42 of [3].) For p E, let H(p) := f k(p)g k (p). Also, for n N, let G n := n g k nd H n := n f kg k. Further, let β R be such tht G n (p) β for ll n N nd ll p E. Then by using the prtil summtion formul n n 1 f k g k = (f k f k+1 )G k + f n G n for ll n 2, we hve H(p) H n (p) 2β f n+1 (p) for ll p E. Since f n+1 0 on E, it follows tht H n H on E, tht is, the series f kg k converges uniformly on E. In prticulr, letting g k (p) := ( 1) k for ll k N nd p E, nd noting tht G n (p) 1 for ll n N nd ll p E, we obtin the uniform convergence of the series ( 1)k f k on E. Exmple 3.5. Let E := [0, 1] nd f k (x) := x k /k for k N nd x [0, 1]. Then (f k ) is momotoniclly decresing sequence nd since f k (x) 1/k for k N nd x [0, 1], we see tht f k 0 on [0, 1] by Proposition 2.2. Hence the series ( 1)k x k /k converges uniformly on [0, 1]. k=n
9 UNIFORM CONVERGENCE 9 Results regrding the boundedness, Riemnn integrbility, continuity nd differentibility of the sum function of convergent series of functions cn be esily deduced from the corresponding results for the sequence of its prtil sums. Theorem 3.6. (i) The sum function of uniformly convergent series of relvlued bounded functions defined on set is bounded. (ii) The sum function of uniformly convergent series of Riemnn integrble functions defined on [, b] is Riemnn integrble, nd the series cn be integrted term by term, tht is, if f k is uniformly convergent on [, b] nd ech f k is Riemnn integrble on [, b], then the function f k is Riemnn integrble on [, b] nd ( ) f k (x) dx = f k (x)dx. (iii) The sum function of uniformly convergent series of relvlued continuous functions defined on metric spce is continuous. (vi) If series of continuously differentible functions defined on [, b] is convergent t one point of [, b] nd if the derived series is uniformly convergent on [, b], then the given series converges uniformly on [, b], the sum function is continuously differentible on [, b] nd the series cn be differentited term by term, tht is, if f k converges t one point of [, b], ech f k is continuously differentible on [, b] nd f k is uniformly convergent on [, b], then f k converges uniformly to continuously differentible function, nd ( ) f k (x) = f k (x) for ll x [, b]. Proof. The results follow by pplying Theorem 2.10 to the sequence of prtil sums of the given series. 4. Two Celebrted Theorems on Uniform Approximtion We hve seen in Proposition 2.6 tht uniform limit of sequence of continuous functions on metric spce is continuous. In this section, we reverse the procedure nd sk whether every continuous function on closed nd bounded intervl of R is the uniform limit of sequence of some specil continuous functions. For function f : [0, 1] R nd n N, we define the nth Bernstein polynomil of f by n ( ) ( n k B n (f) := f x k n) k (1 x) n k. k=0 Theorem 4.1. (Polynomil Approximtion Theorem of Weierstrss) If f : [0, 1] R is continuous, then B n (f) f on [0, 1]. Consequently, every relvlued continuous function on [0, 1] is the uniform limit of sequence of relvlued polynomil functions.
10 10 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE For proof, see Theorem 7.26 of [3], or Corollry 3.12 of [2]. Remrk 4.2. Theorem 4.1 cn be used to prove tht every relvlued continuous function on ny closed nd bounded intervl [, b] is the uniform limit of sequence of relvlued polynomil functions. Let φ : [0, 1] [, b] be defined by φ(x) := (1 x) + xb for x [0, 1]. Then φ is bijective continuous function nd its continuous inverse φ 1 : [, b] [0, 1] is given by φ 1 (t) = (t )/(b ) for t [, b]. Given continuous relvlued function g on [, b], consider the continuous function f := g φ defined on [0, 1]. If (P n ) is sequence of polynomil functions such tht P n f on [0, 1], nd if we let Q n := P n φ 1, then since Q n (t) = P n ( (t )/(b ) ) for t [, b], ech Q n is polynomil function, nd Q n f φ 1 = g on [, b]. Insted of polynomils, let us now consider trigonometric polynomils for pproximting function. They re given by n 0 + ( k cos kx + b k sin kx) for n N, where 0, 1, 2,..., b 1, b 2,... re rel numbers. For Riemnn integrble function f on [ π, π], we define the Fourier coefficients of f by k (f) := 1 π π 0 (f) := 1 2π π π π f(t) cos kt dt, b k (f) := 1 π f(t)dt, nd for k N, π π f(t) sin kt dt. The series 0 (f) + ( k (f) cos kx + b k (f) sin kx ) of functions defined on [ π, π] is clled the Fourier series of the function f. For n = 0, 1, 2,..., let s n (f) denote the nth prtil sum of this series, nd consider the rithmetic mens of these prtil sums given by σ n (f) := s 0(f) + s 1 (f) + s n (f) n + 1 for n = 0, 1, 2... Theorem 4.3. (Trigonometric Polynomil Approximtion Theorem of Fejér) If f : [ π, π] R is continuous nd f( π) = f(π), then σ n (f) f on [ π, π]. Consequently, every relvlued continuous function on [ π, π] hving the sme vlue t π nd π is the uniform limit of sequence of relvlued trigonometric polynomil functions. For proof, see Theorem 8.15 nd Exercise 8.15 of [3], or Theorem 3.13 of [2]. References [1] S. R. Ghorpde nd B. V. Limye, A Course in Clculus nd Rel Anlysis, Springer Interntionl Ed., New Delhi, [2] B. V. Limye, Functionl Anlysis, New Age Interntionl, 2nd Ed., New Delhi, [3] W. Rudin, Principles of Mthemticl Anlysis, 3rd Ed., McGrw Hill, New Delhi, 1976.
Lecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
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Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
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