PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION

PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION SAMEER CHAVAN Abstrct. These re the lecture notes prepred for the prticipnts of IST to be conducted t BP, Pune from 3rd to 15th November, 2014. Contents 1. Pointwise nd Uniform Convergence 1 2. Lebesgue s Proof of Weierstrss Theorem 2 3. Bernstein s Theorem 3 4. Applictions of Weierstrss Theorem 4 5. Stone s Theorem nd its Consequences 5 6. A Proof of Stone s Theorem 7 7. Müntz-Szász Theorem 8 8. Nowhere Differentible Continuous Function 9 References 10 1. Pointwise nd Uniform Convergence Exercise 1.1 : Consider the function f n (x) = x n for x [0, 1]. Chec tht {f n } converges pointwise to f, where f(x) = 0 for x [0, 1) nd f(1) = 1. Exercise 1.2 : Consider the function f m (x) = lim n (cos(m!πx)) n for x R. Verify the following: (1) {f m } converges pointwise to f, where f(x) = 0 if x R \ Q, nd f(x) = 1 for x Q. (2) f is discontinuous everywhere, nd hence non-integrble. Remr 1.3 : If f is pointwise limit of sequence of continuous functions then the set of continuties of f is everywhere dense [1, Pg 115]. Consider the vector spce B[, b] of bounded function from [, b] into C. Let f n, f be such tht f n f B[, b]. A sequence {f n } converges uniformly to f if f n f := sup f n (x) f(x) 0 s n. x [,b] 1

2 SAMEER CHAVAN Theorem 1.4. Let {f n } be sequence of continuous functions. If {f n } converges uniformly to f on [, b] then f is continuous. Moreover, lim n b f n (x)dx = b f(x)dx. Proof. Let N 1 be such tht f N f < ɛ/3. Recll tht f N is uniformly continuous on [, b], tht is, for some δ, f N (x) f N (y) < ɛ/3 whenever x y < δ. Finlly, for ll x, y such tht x y < δ, f(x) f(y) f(x) f N (x) + f N (x) f N (y) + f N (y) f(y) The remining prt follows from b f n (x)dx Tht s the end of the proof. f N f + ɛ/3 + f N f ɛ/3 + ɛ/3 + ɛ/3. b f(x)dx f n f (b ). Corollry 1.5. Let P = {f B[, b] : {p n } such tht p n f 0}. Then P is contined in C[, b]. A remrble result of Weierstrss sserts indeed tht P = C[, b]. In prticulr, ny continuous function on [, b] cn be pproximted uniformly by sequence of polynomils. Theorem 1.6. For ny f C[, b], there exists sequence {p n } of polynomils such tht f p n 0 s n. There re severl proofs of different flvors of this gret theorem (prt from Weierstrss originl proof). Perhps the most elementry proof is due to Lebesgue (which relies on the conclusion of Weierstrss theorem for the function x ). A constructive proof is due to Bernstein (which relies on Chebyshev s version of Bernoulli s lw of lrge numbers). We will lso discuss two remrble generliztions of Weierstrss Theorem: Stone s Theorem nd Müntz-Szász s Theorem. Exercise 1.7 : Let {p n } be sequence of polynomils of degree d n. Suppose tht p n f 0 s n for some continuous function f C[, b]. If f is not polynomil then d n s n. Hint. The subspce of polynomils of degree t most d is finite-dimensionl, nd hence closed in C[, b]. 2. Lebesgue s Proof of Weierstrss Theorem Exercise 2.1 : Consider the function f n (x) = nx(1 x 2 ) n for x [0, 1]. Verify: (1) {f n } converges pointwise to f, where f(x) = 0 for ll x [0, 1]. 1 (2) lim n 0 f n(x)dx 1 0 f(x)dx, nd hence {f n} cn not converge uniformly to f. Here is prtil converse to Theorem 1.4.

UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 3 Theorem 2.2. Let {f n } be sequence in C[, b] converging pointwise to continuous function f. If {f n (x)} decresing for ll x [, b] then {f n } converges uniformly to f. Proof. Let g n := f n f 0. For ɛ > 0, consider the closed subset K n := {x [, b] : g n (x) ɛ} of [, b]. Since g n g n+1, K n+1 K n. In prticulr, finite intersection of K n s is non-empty if every K n is non-empty. If x [, b] then since g n (x) 0, x / K n for sufficiently lrge n. If ech K n is non-empty then we must hve n=1k n (Exercise), nd hence K N is empty for some N. Tht is, 0 g n (x) < ɛ for n N. Here is n importnt specil cse of Weierstrss Theorem. Corollry 2.3. Define sequence {p n } of polynomils by p 0 (x) = 0, nd p n+1 (x) := p n (x) + (x p n (x) 2 )/2 (n 0). If q n (x) := p n (x 2 ) then the {q n } converges uniformly to f(x) = x on [ 1, 1]. Proof. Let y = 1 x 2 (x [ 1, 1]) then f(x) = 1 y for y [0, 1]. Thus it suffices to chec tht {p n } converges uniformly to x on [0, 1]. One my verify inductively tht 0 p n (x) x for x [0, 1]. It follows tht p n (x) p n+1 (x) for ll n 0 nd ll x [0, 1]. In prticulr, {p n (x)} converges pointwise to x. Now pply Theorem 2.2 to f n := p n. Exercise 2.4 : Show tht the function g : R R given by g(x) = 0 for x 0, = x for x 0. Show tht for ny positive number α, g cn be uniformly pproximted by polynomils on [ α, α]. Hint. Note tht g(x) = 1 2 (x + x ). Lebesgue s Proof of Weierstrss Theorem. Since f is uniformly continuous on [, b], there exists positive integer N such tht f(x) f(y) < ɛ/2 whenever x y < 1/N. For x i := + (b )(i/n) (i = 0,, N), consider the function h(x) with grph polygon of vertices t (, f()), (x 1, f(x 1 )),, (x n 1, f(x n 1 )), (b, f(b)). Chec tht f g ɛ/2. On the other hnd, N 1 h(x) = f() + c i g(x x i ) (x [, b]) i=0 for some sclrs c 0,, c N 1. The result now follows from Exercise 3.2. 3. Bernstein s Theorem For f C[0, 1], define the nth Bernstein polynomil B n (f) by n ( ) n B n (f)(x) := f(/n) x (1 x) n (x [0, 1], n 1). =0 Remr 3.1 : B n (1) = 1 (Probbility distribution) nd B n (nt) = nx (Men of the distribution), nd B n (n 2 t 2 ) = n(n 1)x 2 + nx (Vrince of the distribution).

4 SAMEER CHAVAN Exercise 3.2 : If g n (x) := n =0 (x /n)2( n ) x (1 x) n (x [0, 1]) then verify tht g n 1/4n. Hint. Use Remr 3.1. Theorem 3.3. If f C[0, 1] then B n f f 0 s n. Remr 3.4 : Since φ(x) = (1 x) + xb is homeomorphism from [0, 1] onto [, b], Weierstrss Theorem follows from Bernstein s theorem. The proof of Bernstein s Theorem depends on the following vrint of Bernoulli s lw of lrge numbers due to Chebyshev. Lemm 3.5. Given δ > 0 nd x [0, 1], let F denote the set Then, for every x [0, 1], F := { N : 0 n such tht /n x δ}. F ( ) n x (1 x) n 1 4nδ 2. Proof. Note tht for ny x [0, 1] nd F, /n x 2 δ 1. It follows tht 2 ( ) n x (1 x) n 1 ( ) n δ 2 (/n x) 2 x (1 x) n F F 1 n ( ) n δ 2 (/n x) 2 x (1 x) n. =0 The desired estimte follows from Exercise 3.2. Proof of Bernstein s Theorem. Since f is uniformly continuous, there exists n integer N 1 such tht f(x) f(y) < ɛ/2 whenever x y < δ = 1/N. Let us estimte B n (f) f. Let F be s in Lemm 3.5, nd note tht n ( ) n B n (f)(x) f(x) = (f(/n) f(x)) x (1 x) n =0 ( n f(/n) f(x) F + ( n f(/n) f(x) / F ) x (1 x) n ) x (1 x) n 2 f 4nδ 2 + ɛ/2, where we used Remr 3.1. Choose now n integer n 1 so tht f nδ 2 note tht B n (f) f < ɛ. < ɛ, nd 4. Applictions of Weierstrss Theorem Exercise 4.1 : Use Weierstrss Theorem to show tht C[, b] is seprble.

UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 5 Exercise 4.2 : For 0 x < y 1, consider the indictor function χ [x,y] : [0, 1] R. Show tht there exists sequence {p n } of polynomils such tht 1 0 p n (x) χ [x,y] (x) dx 0 s n. Conclude tht finite liner combintion of indictor functions of subintervls of [0, 1] cn be pproximted uniformly by polynomils. Hint. Approximte χ [x,y] by continuous functions in the L 1 norm. Exercise 4.3 : Let f C[, b] be such tht b tn f(t)dt = 0 for ll non-negtive integers n. Show tht f(t) = 0 for every t [, b]. Hint. Get sequence {p n } such tht p n f 0, nd pply Theorem 1.4. Exercise 4.4 : Let f : [, b] R be continuouly differentible function. Show tht there exists sequence {r n } of polynomils such tht r n f 0 nd r n f 0 s n. Conclude tht C 1 [, b] is seprble with norm f := f + f. Hint. Let g(x) = f(x) f() nd note tht g = f. Find sequence {q n } of polynomils such tht q n g 0. Set p n (x) := x q n(t)dt. Note tht p n = q n, nd hence p n g 0. Also, x x p n (x) g(x) = q n (t)dt g (t)dt (b ) q n g. Let r n (x) := p n (x) + f(). Exercise 4.5 : Let f : [, b] R be Riemnn-integrble function. Show tht there exists sequence {p n } of polynomils such tht b p n (x) f(x) 2 dx 0 s n. Hint. Approximte ny Riemnn-integrble function by continuous function, nd then pply Weierstrss Theorem. Exercise 4.6 : Let K be compct subset of R n nd let A be sublgebr of C(K). If f A then show tht f A, where f (x) = f(x). Hint. Let := sup x K f(x) nd let ɛ > 0. Let p : [, ] R be polynomil such tht p t < ɛ. Since p(f) A, we must hve sup p(f)(x) f(x) < ɛ. x K 5. Stone s Theorem nd its Consequences Recll tht the Weierstrss Theorem sys tht P = C[, b]. It is interesting to note tht P is n lgebr which enjoys the following properties: If x y [, b] then trivilly id(x) id(y), nd if x [, b] then 1(x) 0, where id(x) = x nd 1(x) = 1. It turns out these properties of sublgebrs A of C[, b] ensure tht

6 SAMEER CHAVAN A = C[, b]. This remrble fct ws first discovered by Stone in much more generlity. Theorem 5.1. Let K be compct subset of K n, where K is either R or C. Let A be n lgebr of continuous functions f : K R with the following properties: (1) If x y K, then there exists f A such tht f(x) f(y). (2) For every x K, there exists f A such tht f(x) 0. Then A = C R (K). Before we present proof of Stone s Theorem, it is dvisble to understnd it through its wide rnge of pplictions/consequences. Exercise 5.2 : For positive integer n, consider the polynomil sublgebr A of C R [0, 1] generted by 1 nd x n. Show tht A = C R [0, 1]. Exercise 5.3 : Show tht the lgebr P of nlytic polynomils on the closed unit D stisfies ssumptions of Theorem 5.1, however, P C(D). Wht goes wrong? Hint. z belongs to C(D) but z / P. Corollry 5.4. Let K be compct subset of K n. Let A be n lgebr of continuous functions f : K C with the following properties: (1) If x y K, then there exists f A such tht f(x) f(y). (2) For every x K, there exists f A such tht f(x) 0. (3) For every f A, f A, where f(x) = f(x). Then A = C(K). Proof. Let A R denote the lgebr functions in A which re rel-vlued, nd let RA be the set of rel prts of functions in A. Since Rf := (f + f)/2 A for every f A, RA = A R. Chec tht A R stisfies (1) nd (2) of Theorem 5.1. Thus every rel-vlued continuous function on K cn be uniformly pproximted by polynomils. Since A is n lgebr, A = C(K). Exercise 5.5 : Show tht the trigonometric polynomils re dense in C(T), where T stnds for the unit circle. Exercise 5.6 : Let f : T R be continuous function such tht ˆf(n) := 2π 0 f(e iθ )e inθ dθ = 0 for ll n Z. Show tht f is identiclly 0. Hint. Use the lst exercise. Exercise 5.7 : For n integer n 1, consider the polynomil sublgebr A of C[, b] generted by x. If 0 / [, b] then show tht A = C[, b]. Corollry 5.8. Let A be s in Theorem 5.1 except tht (2) is not given. If f A whenever f A, then there exists K such tht A = {f C(K) : f() = 0}. Proof. If (2) does not hold true then there exists K such tht f() = 0 for every f A, tht is, A {f C(K) : f() = 0}. Since {f C(K) : f() = 0} is closed in C(K), we must hve A {f C(K) : f() = 0}. Now let f C(K) be such tht f() = 0. By Stone s Theorem, the lgebr B generted by A nd 1 is dense in C(K). Thus there exists sequence {f n } in B such tht f n f 0

UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 7 s n. Note tht f n = g n + α n for g n A. Also, f n () f() = 0, nd hence α n 0. It follows tht g n f f n f + α n 0 s n. Exercise 5.9 : Let A := {x 2 p(x) : p is polynomil}. Show tht A = {f C[0, 1] : f(0) = 0}. Exercise 5.10 : Show tht the sublgebr A := {p(x 2 ) : p is polynomil} is not dense in C[ 1, 1]. Hint. If possible then there exists sequence {p n } of polynomils such tht p n (x 2 ) x 0 s n. In prticulr, 2 1 0 p n (x 2 )dx = 1 1 p n (x 2 )dx 1 However, 1 0 p n(x 2 )dx 1 xdx = 1, which is bsurd. 0 1 xdx = 0. Exercise 5.11 : Show tht for ny > 0, A := {p( x ) : p is polynomil} is not dense in C[, ]. Corollry 5.12. Let f C(K) be such tht f(x) > 0 for ll x K. Let A be sublgebr of C(K) tht contins f. If f is injective then A = C(K). Proof. Note tht f A stisfies (1) nd (2) of Stone s Theorem. Remr 5.13 : Let X = [, b]. Consider the lgebr A of functions of the form p(f), where p is polynomil such tht p(0) = 0. The lst corollry is pplicble to f(x) = x α for α > 0, f(x) = e x, f(x) = 1/x if 0 / [, b]. 6. A Proof of Stone s Theorem We strt with some elementry properties of sublgebrs of C R (K). Exercise 6.1 : Let A be sublgebr of C R (K). If f 1,, f A then so re mx{f 1,, f } nd min{f 1,, f }. Hint. mx{f 1, f 2 } = f1+f2 induction. 2 + f1 f2 2 A by Exercise 4.6. Now pply finite Lemm 6.2. Under the ssumptions of Theorem 5.1, for distinct points x 1, x 2 of K nd (rel) sclrs c 1, c 2, there exists f A such tht f(x 1 ) = c 1 nd f(x 2 ) = c 2. Proof. By ssumptions there exist g, h, A such tht g(x 1 ) g(x 2 ) nd h(x 1 ) 0 h(x 2 ). Then u := (g g(x 1 )) nd v := (g g(x 2 )h belong to A, nd f := does the job. c 1v v(x + c2u 1) u(x 2) Proof of Stone s Theorem. Fix x K nd ɛ > 0. For x nd every y K, by the preceding lemm, there exists f y A such tht f y (x) = f(x) nd f y (y) = f(y). Note tht f y f is continuous such tht (f y f)(x) = 0. Thus there exists n open neighbourhood U y of y such tht f y (t) f(t) > ɛ for every t U y. Now {U y } is n open cover of K nd K is compct. Thus, for some y 1,, y K, K i=1 U y i. Also, for g x := mx{f y1,, f y } A (Exercise 6.1) nd (6.1) g x (t) > f(t) ɛ for every t K.

8 SAMEER CHAVAN Now we vry x. Note tht g x (x) = mx{f y1 (x),, f y (x)} = f(x). Hence, by the continuity of g x, there exists n open neighbourhood V x of x such tht g x (t) f(t) < ɛ for every t V x. Now {V x } is n open cover of K nd K is compct. Thus, for some x 1,, x l K, K l i=1 V x i. Also, for h := min{g x1,, g xl } A nd h(t) < f(t) + ɛ for every t K. Also, by (6.1), h(t) > f(t) + ɛ for every t K. Tht is, h f < ɛ. 7. Müntz-Szász Theorem In this section, we see nother remrble generliztion of Weierstrss Theorem, which reltes the topologicl property of density of polynomils with divergence of certin Hrmonic series of positive sclrs. Theorem 7.1. Suppose 0 < λ 1 < λ 2 <. Then the closed liner spn of {1, t λ1, t λ2, } equls C[0, 1] if nd only if =1 1 λ =. Remr 7.2 : The choice λ = gives the conclusion of Weierstrss Theorem. Note tht for ny integer l 1, the closed liner spn of {1, t λ l, t λ l+1, } equls C[0, 1] provided =1 1 λ =. Corollry 7.3. The closed liner spn of the constnt function 1 nd monomils {t p : p is prime number} equls C[0, 1]. Proof. This follows from the fct tht p 1/p =, where the sum is ten over the set of prime numbers. Corollry 7.4. Suppose 0 < λ 1 < λ 2 <. Then the closed liner spn of {t λ1, t λ2, } equls {f C[0, 1] : f(0) = 0} provided =1 1 λ =. Proof. Clerly, the closed liner spn of {t λ1, t λ2, } is contined in {f C[0, 1] : f(0) = 0}. Let f C[0, 1] be such tht f(0) = 0. Then, by Müntz-Szász Theorem, there exists sequence {p n } in the closed liner spn of {1, t λ1, t λ2, } such tht p n f 0 s n. But then p n (0) f(0) = 0, nd hence q n (t) := p n (t) p n (0) in the liner spn of {t λ1, t λ2, } converges uniformly to f. We will only setch the outline of the proof of the sufficiency prt. Let us first collect ll the ingredients required for proof of Müntz-Szász Theorem. Lemm 7.5. Let M be closed subspce of C[0, 1]. If M C[0, 1] then there exists non-zero bounded liner mp φ : C[0, 1] C such tht φ(f) = 0 for every f M. Proof. Let f C[0, 1] \ M nd let Y = M + {λf : λ C}. Define ψ(g + αf) = αd (f, M) for g M nd α C. Clerly, ψ = 0 on M nd φ(f) = d (f, M) > 0. The desired conclusion now follows from Hnh-Bnch Extension Theorem. Lemm 7.6. Every bounded liner mp φ : C[0, 1] C is given by φ(f) = f(t)dµ(t) for complex Borel mesure µ on [0, 1]. [0,1] Lemm 7.7. If f is bounded holomorphic function defined on the open unit disc D with zeros α 1, α 2, then n=1 (1 α n ) = implies tht f(z) = 0 for ll z D.

UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 9 Proof of Müntz-Szász Theorem. Suppose there exists bounded liner mp φ : C[0, 1] C such tht φ(f) = 0 for every f M. By Lemm 7.6, for some complex Borel mesure µ on [0, 1], φ(t λ ) = [0,1] tλ dµ(t) = 0 for every positive integer. In view of Lemm 7.5, it suffices to chec tht φ = 0. By Weierstrss Theorem, it suffices to chec tht φ(t ) = [0,1] t dµ(t) = 0 for every positive integer. We now define function g on the open right hlf plne H by setting g(z) = e z log t dµ(t) (z H). (0,1] Since e z log t = e log trz 1 for ll z H nd t (0, 1], by the dominted convergence theorem, g is continuous on H. By theorems of Fubini nd Morer, it is esily seen tht g is holomorphic in H such tht g(λ ) = 0 for every integer 1. Recll tht the zeros of ny non-zero holomorphic function on connected domin re isolted. If {λ } is bounded sequence then by the Identity Theorem, ( ) 1+z g = 0. So suppose tht λ s. Define h : D C by h(z) = g 1 z, ( ) nd note tht h is bounded holomorphic function such tht g λ 1 λ +1 = 0 for every integer 1. Also, since =0 1 λ =, ( ) n=1 1 λ 1 λ +1 = 2 1 n=1 λ +1 =. By Lemm 7.7, we must hve h = 0, nd hence g = 0. It follows tht g() = 0, nd therefore φ(t ) = 0 for every integer 1. We seprte out one importnt technique employed in the preceding proof, which cn be used to prove mny pproximtion results (e.g. Stone s Theorem). Exercise 7.8 : Let S be subset of C(K). Show tht S = C(K) if nd only if for ny complex Borel mesure, f(t)dµ(t) = 0 for every f S implies K f(t)dµ(t) = 0 for every f C(K). K Hint. Use Lemms 7.5 nd 7.6. 8. Nowhere Differentible Continuous Function By Weierstrss Theorem, ny continuous function on [0, 1] is uniform limit of infinitely differentible functions. It is quite striing tht there exists nowhere differentible continuous function on [0, 1]. In prticulr, uniform limit of infinitely rel differentible functions could be nowhere differentible. Theorem 8.1. Let φ(x) = x for x [ 1, 1], which is extended periodiclly (with period 2) to R by setting φ(x + 2) = φ(x) (x R). Then the function f : R R given by ( ) n 3 f(x) = φ(4 n x) 4 is continuous. However, for every x R, there exists sequence {δ m } converging to 0 such tht (f(x + δ m ) f(x))/δ m s m. Proof. Clerly, φ is continuous function such tht 0 φ(x) 1 (x R). Also, ( ) n 3 f(x) φ(4 x) n 4 ( ) n 3 ( ) n 3 φ(4 n x) 0. 4 4 n=+1 n=+1

10 SAMEER CHAVAN Thus, ( 3 4) n φ(4 n x) converges uniformly to f on R, nd hence f is continuous. Let x R nd let m be positive integer. Set δ m = ± 1 2 4 m, where the sign is so chosen tht no integer lies between 4 m x nd 4 m x + δ m. Define γ n := (φ(4 n (x + δ m )) φ(4 n x))/δ m. If n > m then φ(4 n (x + δ m )) = φ(4 n x + 4 n m /2) = φ(4 n x), nd hence γ n = 0. Note tht γ m = 4 m. When 0 n < m, γ n = 4n (x + δ m ) 4 n x δ m 4n δ m δ m = 4 n. Now we complete the rgument. Note tht ( ) n 3 (f(x + δ m ) f(x))/δ m = γ n 4 = m ( ) n 3 m 1 γ n 4 = ( ) n 3 γ n + 3 m 4 which blows up s m. m 1 3 m 3 n = 1 2 (3m + 1), References [1] R. Bos, A Primer of Rel Functions, The mthemticl ssocition of Americ, 1981. [2] J. Burill, Lectures on Approximtion by Polynomils, Lecture Notes, Tt Institute of Fundmentl Reserch, Bomby, 1959. [3] N. Crothers, Rel Anlysis, Cmbridge Univ. Press, 2000. [4] W. Rudin, Principles of Mthemticl Anlysis, McGrw-Hill, 1976. [5] W. Rudin, Rel nd Complex Anlysis, McGrw-Hill Boo Co. New Yor, 1987.