Solution Set 3: Part B

Benjamin Bakker Mathematics 25a Prof. Berger

October 10, 2003

B.1. (1) Show that if f : ℝ_{≥0} → ℝ is continuous and f(x) ⟶ ℓ when x ⟶ +∞, then f is uniformly continuous.

Solution. ∀ϵ > 0, because f(x) ⟶ ℓ as x ⟶ +∞, there exists A ≥ 0 such that f(x) ∈ B(ℓ, ϵ/2) whenever x ≥ A. Consider the restriction of f to [0, A + 1] ⊂ ℝ_{≥0}. [0, A + 1] is compact, so the restriction is uniformly continuous, and therefore there exists δ′ > 0 such that for all x, x′ ∈ ℝ_{≥0}

|x − x′| < δ′ ⇒ |f(x) − f(x′)| < ϵ

The smaller of δ′ and 1 will do the same job, so we may as well let δ = min(δ′, 1). For any x, x′ ∈ [A, ∞), we already know that f(x), f(x′) ∈ B(ℓ, ϵ/2) ⇒ |f(x)−f(x′)| < ϵ. If |x−x′| < δ, then either x, x′ ∈ [0, A+1] or [A, ∞), so it follows that for any x, x′ ∈ ℝ_{≥0},

|x − x′| < δ ⇒ |f(x) − f(x′)| < ϵ

and f is uniformly continuous.

(2) Show that if f : ℝ_{≥0} → ℝ is uniformly continuous then there exist A, B ∈ ℝ such that |f(x)| ≤ Ax + B for all x ∈ ℝ_{≥0}.

Solution. Let ϵ = 1, and B ∈ ℝ such that B > |f(0)| + 1. There then exists δ > 0 such that |x − x′| < δ ⇒ |f(x) − f(x′)| < 1 for all x − x′. Let A = 2/δ. Then for any x, x′ ∈ [0, δ/2], |f(x) − f(0)| < 1 implies that |f(x)| < 1 + |f(0)| < B, so that |f(x)| < B ≤ Ax+B on [0, δ/2]. Now, assume |f(nδ/2)| < A((n−1)δ/2)+B = (n−1)+B. Then once again |f(x) − f(nδ/2)| < 1 on [nδ/2, (n + 1)δ/2], so |f(x)| < 1 + |f(nδ/2)| = n + B for all x ∈ [nδ/2, (n+1)δ/2], and therefore by induction |f(x)| < Ax+B on [0, (n+1)δ/2] for any n. Similarly, we may extend the above argument to negative x, and it follows that |f(x)| < Ax + B for all x.

(3) Which of the following functions ℝ_{≥0} → ℝ is uniformly continuous?

(a) x → e^{x} (b) x → \{ \begin{matrix}x sin \frac{1}{x} & if x ≠ 0 \\ 0 & if x = 0 \end{matrix} (c) x → \sqrt{x}

Solution. (a). If e^{x} was uniformly continuous, then given ϵ = 1 there exists δ > 0 such that for all x, x′ ∈ ℝ_{≥}0, |x − x′| < δ ⇒ |e^{x} − e^{x′}| < 1. If this were the case, then

for all x ∈ ℝ_{≥0}, e^{x+δ/2} − e^{x} = e^{x}(e^{δ/2} − 1) < 1 and therefore

e^{x} < \frac{1}{e^{δ/2} − 1}

e^{x} is not bounded, and thus the above cannot be true. (b). f is continuous at 0 since for any ϵ > 0, if x − x′ < ϵ/2, then |f(x)| ≤ ϵ/2 and

similarly for x′, so |f(x) − f(x′)| < ϵ; f is evidently continuous everywhere else as well. The limit of f as x ⟶ +∞ is 1 (by L’Hopital’s rule; alternatively, sin x is essentially defined by its power series, and x sin \frac{1}{x} = 1 − 1/6x^{2} + ⋯ ⟶ 1), so by (1) f is uniformly continuous.

(c). For all ϵ > 0, let δ = ϵ^{2}. For any x, x′ ∈ ℝ_{≥0}, by the triangle inequality

|\sqrt{x} − \sqrt{x′}| ≤ \sqrt{x} + \sqrt{x′}

and so

(\sqrt{x} − \sqrt{x′})^{2} ≤ |\sqrt{x} − \sqrt{x′}|(\sqrt{x} + \sqrt{x′}) = |x − x′| < ϵ^{2} ⇒ |\sqrt{x} − \sqrt{x′}| < ϵ

B.2. Show that if F ⊂ [0, 1] is closed then there exists a continuous function f : [0, 1] → ℝ such that f^{−1}(0) = F

Solution. Let f(x) = d(x, F) as defined in A.1. By A.1, f is continuous, and f(x) = 0 ⇔ x ∈ \overline{F} = F since F is closed.

B.3. Show that if f : ℝ → ℝ is bijective and continuous, then f is a homeomorphism.

Solution. We need to show that the inverse of f is continuous; since f is bijective, we must have that the image f(U) of any open U ⊂ ℝ is open. Let B = B(x, ϵ) ⊂ be any ball; B is contained in the compact set I = [x − ϵ, x + ϵ]. In lecture we showed that a continuous bijection from a compact space is a homeomorphism, so the restriction of f to I is a homeomorphism, and f(B) is open in f(I). f(I) is a connected compact subset of ℝ (the image of a connected (compact) space under a continuous map is connected (compact)) and is therefore an interval [a, b]; neither a nor b is in f(B), for if a were, then f(B) \ {a} would be connected but B \ {f^{−1}(a)} would not, and similarly for b. Thus, f(B) is open. . For any open set U ⊂ ℝ and any point x ∈ U, there must be an open ball B_{x} ⊂ U centered at x by definition. We then have ∪_{x∈U}B_{x} = U, and since f is a bijection, f(U) = ∪_{x∈U}f(B_{x}). By the above, each f(B_{x}) is open and therefore f(U) is open. f is then a homeomorphism.