Questions: Equicontinuity and the Arzelà-Ascoli Theorem
4 questions to test your understanding
Score: 0 / 4
Question 1 Multiple Choice
Which family of functions on [0, 1] is equicontinuous?
Afₙ(x) = xⁿ for n ≥ 1
Bfₙ(x) = sin(nx) for n ≥ 1
CAll differentiable f with |f'(x)| ≤ 5 on [0, 1]
DAll continuous f with |f(x)| ≤ 5 on [0, 1]
By the Mean Value Theorem, |f(x) - f(y)| ≤ 5|x - y| for every such f, so δ = ε/5 works for the whole family. The functions xⁿ become steep near x = 1, sin(nx) oscillates faster as n grows, and a bound on |f| alone does not control how fast f changes.
Question 2 True / False
Every sequence of continuous functions on [0, 1] satisfying |fₙ(x)| ≤ 1 has a uniformly convergent subsequence.
TTrue
FFalse
Answer: False
fₙ(x) = xⁿ satisfies the bound, but any subsequence converges pointwise to the function that is 0 on [0, 1) and 1 at x = 1. That limit is discontinuous, and a uniform limit of continuous functions must be continuous, so no subsequence converges uniformly. The missing hypothesis is equicontinuity.
Question 3 Short Answer
State the definition of equicontinuity for a family F of functions on a metric space K.
Think about your answer, then reveal below.
Model answer: For every ε > 0 there is a δ > 0 such that |f(x) - f(y)| < ε for every f in F and every x, y in K with d(x, y) < δ.
The order of quantifiers is the whole point: δ depends only on ε, not on the particular function f. For a single continuous function on a compact set, this is uniform continuity.
Question 4 Short Answer
In the proof of the Arzelà-Ascoli theorem, what role does the diagonal argument play?
Think about your answer, then reveal below.
Model answer: It extracts a subsequence that converges at every point of a countable dense subset of K.
Pointwise boundedness gives a convergent subsequence at each point of the dense set, one point at a time, and the diagonal subsequence converges at all of them. Equicontinuity and compactness then show this subsequence is uniformly Cauchy on all of K.