18.100B Problem Set 9 Solutions - DSpace@MIT

More documents

Recommendations

Info

3) So we have f n −→ f uniformly for f n bounded. Then for ε = 1, ∃N such that ∀n > N, x ∈ E, |f n (x)−f(x)| < 1. Thus if f n is bounded by B n , ∀x ∈ E, |f(x)| < 1+B N which implies that ∀n > N, x ∈ E, f n (x) < B N + 2. Thus (f n ) is uniformly bounded by max{B 1 , B 2 , ..., B N−1 , B N + 2}. If the f n are converging pointwise, f need not be bounded. For example, on (0, 1), if 1 f n (x) = , x + 1 n 1 1 then f n −→ pointwise, and each f n is bounded by n, but of course is unbounded on (0, 1). x x 4) We have f n −→ f and g n −→ g uniformly. ε a) Given ε > 0, ∃M, N ∈ N such that for any m > M, n > N, |f m − f| < 2 and |g n − g| < ε 2 . So for n, m > max{M, N} ε ε |(f n + g n ) − (f + g)| = |(f n − f) + (g n − g)| ≤ |f n − f| + |g n − g| < + = ε. 2 2 So (f n + g n ) −→ (f + g) uniformly. b) If each f n and g n is bounded on E, by the previous problem they are uniformly bounded, say by A and B. Say without loss of generality A ≥ B. Then for ε > 0, choose N such that if ε n > N, |f n − f| < 2 A and |g ε n − g| < 2 A . Then ε ε |f n g n − fg| = |f n g n − fg n + fg n − fg| ≤ |f n − f||g n | + |f||g n − g| < A + A = ε. 2A 2A So f n g n converges to fg uniformly. 5) Define 0 if x /∈ Q f (x) = x, g (x) = q if, in lowest terms, x = p/q so that 1 1 f n (x) = f (x) 1 + , and g n (x) = g (x) + . n n On any interval [a, b], with M = max (|a|, |b|) we have |f n (x) − f (x)| = |x| M , = 1 n ≤ |g n (x) − g (x)| n n so that f n → f and g n → g uniformly. On the other hand, with m = min (|a|, |b|) we have |f n (x) g n (x) − f (x) g (x)| = |(f n (x) − f (x)) g n (x) + f (x) (g n (x) − g (x))| f (x) n + 1 m = g (x) + ≥ g (x) . n n n Now notice that if L ∈ N is larger than b − a then there is an integer k such that k L ∈ [a, b], n choosing L ∈ N larger than m and larger than b − a we get m k m f n g n − fg ≥ g = L > 1 n L n and hence f n g n does not converge to fg uniformly.
6) We want to show g ◦ f n −→ g ◦ f uniformly, for g continuous on [−M, M]. Since g is continuous on a compact set, it is uniformly continuous. So given ε > 0, ∃δ > 0 such that if |x − y| < δ, |g(x) − g(y)| < ε. Now since f n −→ f uniformly, ∃N ∈ N such that for any n > N, x ∈ E, |f n (x) − f(x)| < δ. Thus |g(f n (x)) − g(f(x))| < ε, so |g ◦ f n (x) − g ◦ f(x)| < ε. Thus we have shown (g ◦ f n ) −→ g ◦ f uniformly on E. 7) a) First we claim that, for every x ∈ [0, 1], 0 ≤ P n (x) ≤ P n+1 (x) ≤ √ x. This is clearly true for n = 0, so assume inductively that it is true for n = k and notice that √ x − Pk+1 (x) = √ x − P k (x) + 1 x − Pk (x) 2 2 = √ x − P k (x) − 1 √ x − Pk (x) √ x + P k (x) 2 = √ x − P k (x) 1 − 1 √ x + Pk (x) ≥ 0. 2 It follows that P k+1 (x) ≤ √ x, and then 1 P k+1 (x) − P k (x) = x − P k (x) 2 ≥ 0 2 shows that P k (x) ≤ P k+1 (x), and proves the claim. Notice that for every fixed x, the sequence (P n (x)) is monotone increasing and bounded above (by √ x), it follows that this sequence converges, to say f (x). This function f (x) is non-negative and satisfies 1 f(x) = lim P n+1 (x) = lim P n (x) + (x − P n (x) 2 ) = f(x) + (x − f(x) 2 ) n−→∞ n−→∞ 2 which implies f (x) = √ x, and hence the polynomials converge pointwise to √ x on [0, 1]. Since they are continuous and converge monotonically to a continuous function on [0, 1], a compact set, they converge uniformly (by Dini’s theorem). b) Here we use a something similar to problem 6 along with the above work to show that P n (x 2 ) −→ |x| on [−1, 1]. The difference between this and problem 6 is we need (f n ◦ g) −→ (f ◦ g), but this is easier. Given ε > 0, ∃N ∈ N with ∀x ∈ [0, 1], n > N, |P n (x) − √ x| < ε. So for all x ∈ [−1, 1], n > N, |P 2 n (x 2 ) − √ x | < ε. Since |x| = √ x 2 , we have show that the polynomials P n (x 2 ) converge uniformly to |x| on [−1, 1].
Page 1: 18.100B Problem Set 9 Solutions Saw

18.100B Problem Set 9 Solutions - DSpace@MIT

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?