Solution Set 5: Part B

Benjamin Bakker Mathematics 25a Prof. Berger

October 24, 2003

B.1. Simple Convergence and uniform convergence I. Let I = [a, b] and f_{n} : I → ℝ be a sequence of functions.

(1) Prove that if f_{n} converges uniformly to f : I → ℝ, then it converges simply.

Solution. This is a one-liner: for any y ∈ I, if we are given ϵ > 0, there exists N such that |f_{n}(y) − f(y)| ≤ sup_{x∈I} |f_{n}(x) − f(x)| < ϵ for all n > N, and thus f_{n}(y) ⟶ f(y). □

(2) Prove that the converse is not true by considering the sequence f_{n} : [0, 1] → ℝ,

f_{n}(x) = x^{n} and f(x) = \{ \begin{matrix}0 & x ≠ 1 \\ 1 & x = 1 \end{matrix}

Solution. First note that f_{n} converges simply to f; since f_{n}(1) = 1 and f_{n}(0) = 0 for all n, f_{n}(1) ⟶ f(1) and f_{n}(0) ⟶ 0, while for all x ∈ (0, 1), f_{n}(x) = x^{n} ⟶ 0 = f(x) as n ⟶ ∞. (For any ϵ > 0, if x ∈ (0, 1) we may take N = ⌈log_{x} ϵ⌉, in which case x^{n} < ϵ for all n > N.)

If we take ϵ = 1/4 and x_{n} = (1/2)^{1/n}, then for any N, |f_{N}(x_{N})−f(x_{N})| = 1/2 > ϵ, and f_{n} cannot converge uniformly to f. Indeed, for any n, sup_{x∈[0,1]} |f_{n}(x) − f(x)| = 1. □

(3) Prove that even if f is continuous then simple convergence does not necessarily imply uniform convergence by considering

f_{n}(x) = \{ \begin{matrix}nx & if 0 ≤ x ≤ 1/(2n) \\ 1 − nx & if 1/(2n) ≤ x ≤ 1/n \\ 0 & if 1/n ≤ x ≤ 1 \end{matrix}

Solution. f_{n} converges simply to f = 0; f_{n}(0) = 0 for all n, so f_{n}(0) ⟶ 0, and for all x ∈ (0, 1], there exists an N such that 1/n < x ⇒ f_{n}(x) = 0 for all n > N, and thus f_{n}(x) ⟶ 0.

The convergence is not uniform, however. If we take ϵ = 1/4, then for all N, |f_{N}(1/(2N))− f(1/(2N))| = 1/2, so that sup_{x∈[0,1]} |f_{n}(x) − f(x)| ≥ 1/2 for all n. (In fact the supremum is always equal to 1/2.) □

B.2. Simple Convergence and uniform convergence II. Assume f_{n} : I → ℝ is a sequence of continuous functions converging simply to a continuous function f : I → ℝ, such that f_{n+1}(x) ≥ f_{n}(x) for every n ≥ 1, x ∈ I.

(1) Show that y_{n} = sup_{x∈I} |f_{n}(x) − f(x)| is non-increasing.

Solution. For any fixed x, f_{n}(x) is a non-decreasing convergent sequence, and its limit f(x) must therefore by the supremum of {f_{n}(x)}_{n≥1} (which must exist): in particular, f(x) ≥ f_{n}(x) for all n. Thus, f(x) ≥ f_{n+1}(x) ≥ f_{n}(x) for all n ⇒ |f(x) − f_{n+1}(x)| ≤ |f(x) − f_{n}(x)| for all n and x ∈ I ⇒ y_{n+1} ≤ y_{n}. □

(2) Prove that f_{n} converges uniformly to f.

Solution. (adapted from Parker Meares’ solution.) y_{n} is a non-increasing sequence bounded from below, since y_{n} ≥ 0 for all n, so it converges to the infimum of {y_{n}}_{n≥1}. Suppose the limit is 2ϵ > 0. Let h_{n}(x) = |f_{n}(x) − f(x)|, and let A_{n} ⊂ I be the set of all x such that h_{n}(x) ≥ ϵ; A_{n} is nonempty since sup_{x∈I} h_{n}(x) > ϵ. h_{n} is continuous, so h_{n}^{−1}(B(0, ϵ)) is open; its complement, A_{n}, is a closed subset of the compact interval I and is itself compact. f_{n+1}(x) ≥ f_{n}(x) ⇒ h_{n+1}(x) ≥ h_{n}(x) for any n, so A_{n+1} ⊂ A_{n}. For any x ∈ I, h_{n}(x) ⟶ 0, so for some N we must have x ∉ A_{n} for n > N and therefore x ∉ ∩_{n≥1}A_{n}. It then follows that ∩_{n≥1}A_{n} = ∅, contradicting problem C.1 on problem set 3 (the intesection of a countable nested sequence of compact sets is nonempty). □

B.3. Simple Convergence and uniform convergence III. Assume f_{n} : I → ℝ is a sequence of functions converging simply to f : I → ℝ, such that for any x ∈ I and any sequence x_{n} ⟶ x, f_{n}(x_{n}) ⟶ f(x).

(1) Prove that for every ϵ > 0 and x ∈ I, there exists δ > 0 and N such that y ∈ B(x, δ) ⇒ f_{n}(y) ∈ B(f_{n}(x)) for all n > N.

Solution. Suppose not; assume that for some ϵ, given any δ and N there is some y ∈ B(x, δ) for which f_{n}(y) is not within ϵ of f_{n}(x) for some n > N. This implies that for all i there is some φ(i) > i and y_{φ(i)} ∈ B(x, 1/i) such that f_{φ(i)}(y_{φ(i)}) ∉ B(f_{φ(i)}(x), ϵ). Note that we can always choose φ(i) greater than all previously chosen φ(j), j < i. f_{n}(x) ⟶ f(x) by simple convergence, so for some N, f_{n}(x) ∈ B(f(x), ϵ/2) for all n > N. By the triangle inequality, for any z ∈ B(f(x), ϵ/2),

d(z, f_{n}(x)) ≤ d(z, f(x)) + d(f(x), f_{n}(x)) < ϵ ⇒ B(f(x), ϵ/2) ⊂ B(f_{n}(x), ϵ) ∀n > N

y_{φ(i)} converges to x by construction, but f_{φ(i)}(y_{φ(i)}) is not in B(f(x), ϵ/2) for any φ(i) > N, and therefore cannot converge to f(x), a contradiction. □

(2) Prove that f is continuous.

Solution. For all ϵ > 0, there exists δ > 0 and M such that for all x, y ∈ I, |x − y| < δ ⇒ |f_{n}(x)−f_{n}(y)| < ϵ/3 for all n > M by the above. By simple convergence, for each such pair x, y, there exists N, P such that |f_{n}(x) − f(x)| < ϵ/3 for n > N, and |f_{n}(y) − f(y)| < ϵ/3 for n > N, so that

|f(x) − f(y)| < |f(x) − f_{n}(x)| + |f_{n}(x) − f_{n}(y)| + |f_{n}(y) − f(y)| < ϵ

(for all n > max{M, N, P }, though we don’t need this any more). □

(3) Prove that f_{n} converges uniformly to f.

Solution. By the simple convergence of f_{n}, the continuity of f, and part (1), for any x ∈ I and ϵ > 0, there exists δ_{x} > 0 and N_{x} such that for any y ∈ B(x, δ) and n > N_{x}, |f_{n}(x) − f(x)| < ϵ/6, |f(x) − f(y)| < ϵ/6 and |f_{n}(x) − f_{n}(y)| < ϵ/6, where N_{x} is obtained by taking the maximum of the N’s guaranteed for each inequality, as in (2). Thus,

|f(y) − f_{n}(y)| < |f_{n}(x) − f(x)| + |f(x) − f(y)| + |f_{n}(x) − f_{n}(y)| < ϵ/2 ∀y ∈ B(x, δ_{x}), n > N_{x}

I is compact and therefore covered by finitely many balls of radius ϵ; take their centers to be x_{i}. Then if we let N = max N_{x_{i}} and δ = min{δ_{x_{i}}},

|f(y) − f_{n}(y)| < ϵ/2 ∀y ∈ B(x, δ), n > N

⇒ sup_{y∈I} |f(y) − f_{n}(y)| < ϵ ∀y ∈ B(x, δ), n > N

and f_{n} uniformly converges to f. □