Math 321 Assignment 5 — some solutions 1. Let pn be a polynomial of degree mn and suppose that pn → f uniformly on [a, b] and that f is not a polynomial. Show that mn → ∞. Solution sketch: Assume by contradiction that mn does not tend to inﬁnity. Then one can extract a subsequence mnk that is bounded. Consider the polynomials of bounded degree pnk , and note that ||pnk ||∞ is a bounded sequence. Conclude from that that the coeﬃcients of pnk are bounded and hence one can extract further subsequence in which the coeﬃcients converge. The limiting polynomial must be the function f , hence f is a polynomial. 2. Prove that there exists a sequence of polynomials pn such that pn → 0 pointwise on [0, 1] but ∫1 0 pn (x)dx = 3. Solution sketch: ∫ 1 Construct a sequence of continuous functions fn which converge pointwise to [0, 1] and 0 fn = 3. Now for each n, apply Weierstrass’ theorem and ﬁnd a polynomial p∗n ∫1 such that ||fn − p∗n || ≤ 1/n and set pn = p∗n + 0 (3 − p∗n ). The last integral is at most 1/n, and so pn satisﬁes the required conditions (verify!). 5. Let C(R) denote the set of continuous functions f : R → R. Deﬁne a metric d on it by d(f, g) = ∞ ∑ 2−n n=1 dn (f, g) 1 + dn (f, g) where dn (f, g) = max |f (t) − g(t)| . |t|≤n Convince yourself that d is a metric (but you do not have to submit the proof for it). (a) Prove that fn converges to f in the above metric of C(R) if and only if fn converges uniformly to f on every compact subset of R. For this reason, convergence in C(R) is sometimes called uniform convergence on compact sets. (b) Show that C(R) is complete. Solution for (a): Assume that fn , g are such that d(fn , g) → 0. We need to prove that for any J we have dJ (fn , g) → 0. We ﬁrst note that for any two functions f, g the number dJ (f, g) increases in J. Secondly, if ϵ ∈ (0, 1/2), then the inequality x/(1 + x) ≤ ϵ implies that x ≤ 2ϵ. Fix J and ϵ ∈ (0, 1/2) and let N be such that for all n ≥ N ∞ ∑ k=1 2−k dk (fn , g) ≤ ϵ2−J . 1 + dk (fn , g) Looking at the Jth term in this sum we get that dJ (fn , g) ≤ 2ϵ for n ≥ N , as required. For the assume that for all J we have dJ (fn , g) → 0. Let ϵ > 0 and choose J such ∑∞converse, −k that k=J+1 2 ≤ ϵ. Let N be such that for all n ≥ N we have dJ (fn , g) ≤ ϵ, then ∞ ∑ k=1 2 −k J ∞ ∑ ∑ dk (fn , g) −k ≤ 2 ϵ+ 2−k ≤ 2ϵ . 1 + dk (fn , g) k=1 k=J+1