PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION SAMEER CHAVAN Abstract. These are the lecture notes prepared for the participants of IST to be conducted at BP, Pune from 3rd to 15th November, 2014. Contents 1. Pointwise and Uniform Convergence 1 2. Lebesgue's Proof of Weierstrass' Theorem 2 3. Bernstein's Theorem 3 4. Applications of Weierstrass' Theorem 4 5. Stone's Theorem and its Consequences 5 6. A Proof of Stone's Theorem 7 7. M¨untz-Sz´aszTheorem 8 8. Nowhere Differentiable Continuous Function 9 References 10 1. Pointwise and Uniform Convergence n Exercise 1.1 : Consider the function fn(x) = x for x 2 [0; 1]: Check that ffng converges pointwise to f; where f(x) = 0 for x 2 [0; 1) and f(1) = 1: n Exercise 1.2 : Consider the function fm(x) = limn!1(cos(m!πx)) for x 2 R: Verify the following: (1) ffmg converges pointwise to f; where f(x) = 0 if x 2 R n Q; and f(x) = 1 for x 2 Q: (2) f is discontinuous everywhere, and hence non-integrable. Remark 1.3 : If f is pointwise limit of a sequence of continuous functions then the set of continuties of f is everywhere dense [1, Pg 115]. Consider the vector space B[a; b] of bounded function from [a; b] into C: Let fn; f be such that fn − f 2 B[a; b]. A sequence ffng converges uniformly to f if kfn − fk1 := sup jfn(x) − f(x)j ! 0 as n ! 1: x2[a;b] 1 2 SAMEER CHAVAN Theorem 1.4. Let ffng be a sequence of continuous functions. If ffng converges uniformly to f on [a; b] then f is continuous. Moreover, Z b Z b lim fn(x)dx = f(x)dx: n!1 a a Proof. Let N ≥ 1 be such that kfN − fk1 < /3: Recall that fN is uniformly continuous on [a; b]; that is, for some δ, jfN (x)−fN (y)j < /3 whenever jx−yj < δ: Finally, for all x; y such that jx − yj < δ; jf(x) − f(y)j ≤ jf(x) − fN (x)j + jfN (x) − fN (y)j + jfN (y) − f(y)j ≤ kfN − fk1 + /3 + kfN − fk1 ≤ /3 + /3 + /3: The remaining part follows from Z b Z b j fn(x)dx − f(x)dxj ≤ kfn − fk1(b − a): a a That's the end of the proof. Corollary 1.5. Let P = ff 2 B[a; b]: 9 fpng such that kpn − fk1 ! 0g: Then P is contained in C[a; b]: A remarkable result of Weierstrass asserts indeed that P = C[a; b]: In particular, any continuous function on [a; b] can be approximated uniformly by a sequence of polynomials. Theorem 1.6. For any f 2 C[a; b]; there exists a sequence fpng of polynomials such that kf − pnk1 ! 0 as n ! 1: There are several proofs of different flavors of this great theorem (apart from Weierstrass' original proof). Perhaps the most elementary proof is due to Lebesgue (which relies on the conclusion of Weierstrass' theorem for the function jxj). A constructive proof is due to Bernstein (which relies on a Chebyshev's version of Bernoulli's law of large numbers). We will also discuss two remarkable generaliza- tions of Weierstrass' Theorem: Stone's Theorem and M¨untz-Sz´asz'sTheorem. Exercise 1.7 : Let fpng be a sequence of polynomials of degree dn. Suppose that kpn − fk1 ! 0 as n ! 1 for some continuous function f 2 C[a; b]: If f is not a polynomial then dn ! 1 as n ! 1: Hint. The subspace of polynomials of degree at most d is finite-dimensional, and hence closed in C[a; b]: 2. Lebesgue's Proof of Weierstrass' Theorem 2 n Exercise 2.1 : Consider the function fn(x) = nx(1 − x ) for x 2 [0; 1]: Verify: (1) ffng converges pointwise to f; where f(x) = 0 for all x 2 [0; 1]: R 1 R 1 (2) limn!1 0 fn(x)dx 6= 0 f(x)dx; and hence ffng can not converge uni- formly to f: Here is a partial converse to Theorem 1.4. UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 3 Theorem 2.2. Let ffng be a sequence in C[a; b] converging pointwise to a continu- ous function f: If ffn(x)g decreasing for all x 2 [a; b] then ffng converges uniformly to f: Proof. Let gn := fn − f ≥ 0: For > 0; consider the closed subset Kn := fx 2 [a; b]: gn(x) ≥ g of [a; b]: Since gn ≥ gn+1;Kn+1 ⊆ Kn: In particular, finite intersection of Kn's is non-empty if every Kn is non-empty. If x 2 [a; b] then since gn(x) ! 0; x2 = Kn for sufficiently large n: If each Kn is non-empty then we 1 must have \n=1Kn 6= ; (Exercise), and hence KN is empty for some N: That is, 0 ≤ gn(x) < for n ≥ N: Here is an important special case of Weierstrass' Theorem. Corollary 2.3. Define a sequence fpng of polynomials by p0(x) = 0; and 2 pn+1(x) := pn(x) + (x − pn(x) )=2 (n ≥ 0): 2 If qn(x) := pn(x ) then the fqng converges uniformly to f(x) = jxj on [−1; 1]: 2 p Proof. Let y = 1−x (x 2 [−1; 1]) then f(xp) = 1 − y for y 2 [0; 1]: Thus it suffices to check that fpngpconverges uniformly to x on [0; 1]: One may verify inductively that 0 ≤ pn(x) ≤ x for x 2 [0; 1]: It follows that pn(x) ≤ pn+1(px) for all n ≥ 0 and all x 2 [0; 1]: In particular, fpn(x)g converges pointwise to x. Now apply Theorem 2.2 to fn := −pn: Exercise 2.4 : Show that the function g : R ! R given by g(x) = 0 for x ≤ 0; = x for x ≥ 0: Show that for any positive number α; g can be uniformly approximated by polyno- mials on [−α; α]: 1 Hint. Note that g(x) = 2 (x + jxj): Lebesgue's Proof of Weierstrass' Theorem. Since f is uniformly continuous on [a; b]; there exists a positive integer N such that jf(x) − f(y)j < /2 whenever jx − yj < 1=N: For xi := a + (b − a)(i=N)(i = 0; ··· ;N), consider the function h(x) with graph a polygon of vertices at (a; f(a)); (x1; f(x1)); ··· ; (xn−1; f(xn−1)); (b; f(b)): Check that kf − gk1 ≤ /2: On the other hand, N−1 X h(x) = f(a) + cig(x − xi)(x 2 [a; b]) i=0 for some scalars c0; ··· ; cN−1: The result now follows from Exercise 3.2. 3. Bernstein's Theorem For f 2 C[0; 1], define the nth Bernstein polynomial Bn(f) by n X n B (f)(x) := f(k=n) xk(1 − x)n−k (x 2 [0; 1]; n ≥ 1): n k k=0 Remark 3.1 : Bn(1) = 1 (Probability distribution) and Bn(nt) = nx (Mean of 2 2 2 the distribution), and Bn(n t ) = n(n − 1)x + nx (Variance of the distribution). 4 SAMEER CHAVAN Pn 2n k n−k Exercise 3.2 : If gn(x) := k=0(x − k=n) k x (1 − x) (x 2 [0; 1]) then verify that kgnk1 ≤ 1=4n: Hint. Use Remark 3.1. Theorem 3.3. If f 2 C[0; 1] then kBnf − fk1 ! 0 as n ! 1: Remark 3.4 : Since φ(x) = (1 − x)a + xb is a homeomorphism from [0; 1] onto [a; b]; Weierstrass' Theorem follows from Bernstein's theorem. The proof of Bernstein's Theorem depends on the following variant of Bernoulli's law of large numbers due to Chebyshev. Lemma 3.5. Given δ > 0 and x 2 [0; 1]; let F denote the set F := fk 2 N : 0 ≤ k ≤ n such that jk=n − xj ≥ δg: Then, for every x 2 [0; 1]; X n 1 xk(1 − x)n−k ≤ : k 4nδ2 k2F jk=n−xj2 Proof. Note that for any x 2 [0; 1] and k 2 F; δ2 ≥ 1: It follows that X n 1 X n xk(1 − x)n−k ≤ (k=n − x)2 xk(1 − x)n−k k δ2 k k2F k2F n 1 X n ≤ (k=n − x)2 xk(1 − x)n−k: δ2 k k=0 The desired estimate follows from Exercise 3.2. Proof of Bernstein's Theorem. Since f is uniformly continuous, there exists an in- teger N ≥ 1 such that jf(x) − f(y)j < /2 whenever jx − yj < δ = 1=N: Let us estimate kBn(f) − fk1. Let F be as in Lemma 3.5, and note that n X n jB (f)(x) − f(x)j = (f(k=n) − f(x)) xk(1 − x)n−k n k k=0 X n ≤ jf(k=n) − f(x)j xk(1 − x)n−k k k2F X n + jf(k=n) − f(x)j xk(1 − x)n−k k k2 =F 2kfk ≤ 1 + /2; 4nδ2 kfk1 where we used Remark 3.1. Choose now an integer n ≥ 1 so that nδ2 < , and note that kBn(f) − fk1 < . 4. Applications of Weierstrass' Theorem Exercise 4.1 : Use Weierstrass' Theorem to show that C[a; b] is separable. UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 5 Exercise 4.2 : For 0 ≤ x < y ≤ 1; consider the indicator function χ[x;y] : [0; 1] ! R: Show that there exists a sequence fpng of polynomials such that Z 1 jpn(x) − χ[x;y](x)jdx ! 0 as n ! 1: 0 Conclude that finite linear combination of indicator functions of subintervals of [0; 1] can be approximated uniformly by polynomials.