Solution Set 7: Part A

Benjamin Bakker Mathematics 25a Prof. Berger

November 7, 2003

A.1. Prove that the function

f(x) = \{ \begin{matrix}0 & if x = 0 \\ x^{2} sin(1/x) & if x ≠ 0 \end{matrix}

is differentiable everywhere on ℝ but that f is not C^{1}.

Solution. x^{2}, sin x, and 1/x are all C^{1} on ℝ \ {0}, so f is C^{1} on ℝ \ {0} with derivative given by

f′(x) = 2x sin(1/x) − cos(1/x)

Clearly, f is C^{0} (although this isn’t really necessary), since

lim_{x→0} |f(x)| = lim_{x→0} |x^{2} sin(1/x)| ≤ lim_{x→0} x^{2} = 0 = f(0)

We claim that the derivative f′(0) of f at 0 exists and that f′(0) = 0. By similar reasoning,

lim_{h→0} |h sin(1/h)| ≤ lim_{h→0} h = 0

and setting ϕ(h) = h sin(1/h), we may write

f(h) = hϕ(h) = f(0) + hϕ(h) ⇒ f′(0) = 0

since ϕ(h) → 0. Thus, f is differentiable everywhere. However,

lim_{x→0} f′(x) = lim_{x→0} (2x sin(1/x) − cos(1/x)) = − lim_{x→0} cos(1/x)

does not exist, and therefore f′(x) is not continuous on ℝ. To see this explicitly, we could look at the sequence x_{n} = (2πn)^{−1} → 0; if f′ were continuous, we would have f′(x_{n}) → f′(0) = 0, but for all n

f′(x_{n}) = \frac{1}{πn} sin(2πn) − cos(2πn) = −1

so f′ is not continuous at 0. The fact that lim_{x→0} f′(x) doesn’t exist means that we couldn’t make f′ continuous even by redefining f′ at 0 (to be −1, for instance). □

A.2. Suppose that {f_{n}}_{n≥1} is a sequence of C^{1} functions f_{n} : [0, 1] → ℝ, and that f, g : [0, 1] → ℝ are two functions such that: f_{n} → f pointwise, and f′_{n} → g uniformly on [0, 1]. Prove that f is C^{1}, that f′ = g, and that f_{n} → f uniformly.

Solution. (due to Allan Chu) Note that g is continuous since it is the uniform limit of continuous functions f′_{n}. Indeed, for any ϵ > 0, there exist N such that for all n > N,

|f′_{n}(x) − g(x)| < \frac{ϵ}{3} and |f′_{n}(y) − g(y)| < \frac{ϵ}{3}

For any particular n > N, ∃δ such that for all x, y ∈ [0, 1] with |x − y| < δ,

|f′_{n}(x) − f′_{n}(y)| < \frac{ϵ}{3}

⇒ |g(x) − g(y)| ≤ |g(x) − f′_{n}(x)| + |f′_{n}(x) − f′_{n}(y)| + |f′_{n}(y) − g(y)| < ϵ

Given ϵ > 0, since f′_{n} → g uniformly, there exists N such that for all n > N, x ∈ [0, 1]

|∫_{0}^{x}(f′_{n} − g)| ≤ ∫_{0}^{x} |f′_{n} − g| ≤ ∫_{0}^{x} ϵ ≤ ϵ

f_{n} is C^{1}, so ∫_{0}^{x} f′_{n} = f_{n}(x) − f_{n}(0), and the above says that f_{n}(x) − f_{n}(0) converges uniformly to G(x) = ∫_{0}^{x} g. But we already know that f_{n}(x) − f_{n}(0) converges pointwise to f(x) − f(0), so f(x)−f(0) = G(x). f_{n}(x)−f_{n}(0) converges to f(x)−f(0) uniformly and f_{n}(0) → f(0) ⇒ f_{n} → f uniformly. g is continuous, so by the fundamental theorem of calculus G(x) = f(x) − f(0) is C^{1}, and therefore f is C^{1} with derivative f′ = G′ = g. □