Questions: Uniform Continuity

Question 1

Consider f(x) = 1/x on the interval (0, 1). Why is f NOT uniformly continuous on this domain?

• A. f is not differentiable at x = 0, which lies in the closure of (0, 1)
• B. Near x = 0, the function grows arbitrarily steep, so any fixed δ eventually fails: for small enough x and y = x + δ/2, |f(x) − f(y)| can exceed any ε
• C. The interval (0, 1) is open, and no function on an open interval can be uniformly continuous
• D. f(x) = 1/x is discontinuous at x = 0, which makes the whole interval fail

Answer: (B)

Near x = 0, the slope of 1/x grows without bound. For any proposed δ > 0, we can find x small enough that |f(x) − f(x + δ/2)| is arbitrarily large — the same δ that works near x = 1/2 will catastrophically fail near x = 0.001. This is the essence of non-uniform continuity: the δ required to control |f(x) − f(y)| depends on x and cannot be bounded below by a single positive constant over the whole domain. Option C is wrong: f(x) = x is uniformly continuous on any open interval.

Question 2

A student argues: 'f(x) = x² is continuous everywhere on ℝ, so it must be uniformly continuous on ℝ.' This reasoning fails because:

• A. f(x) = x² is not actually continuous on all of ℝ
• B. Continuity guarantees a δ for each (ε, x) pair separately, but for x² the required δ shrinks to 0 as x → ∞ — no single δ covers all points simultaneously
• C. Uniform continuity and pointwise continuity are equivalent on all unbounded intervals
• D. Unbounded domains never support uniformly continuous functions

Answer: (B)

For f(x) = x², to control |x² − y²| = |x + y||x − y| < ε when |x − y| < δ, we need δ < ε / |x + y|. As x → ∞, this upper bound on δ goes to 0 — no single positive δ can work for all x ∈ ℝ. Pointwise continuity only requires a δ that works at each fixed x; uniform continuity requires one δ that works at all x simultaneously. Option D is wrong: f(x) = x is uniformly continuous on all of ℝ.

Question 3

In the definition of uniform continuity, the key distinction from pointwise continuity is that δ does not depend on ε — a fixed δ works regardless of what ε is.

• T. True
• F. False

Answer: False

This is a common confusion that reverses the key distinction. In BOTH uniform and pointwise continuity, δ may (and typically does) depend on ε — a smaller ε generally requires a smaller δ. The actual distinction is about dependence on the POINT x: in pointwise continuity, δ may also depend on x (shrinking to 0 as x changes); in uniform continuity, δ depends only on ε and works simultaneously for ALL pairs of points in the domain. Saying 'δ is constant' or 'δ independent of ε' both misstate the definition.

Question 4

Uniform continuity is a property of a function on a set, not at a point — it is a category error to ask whether f is 'uniformly continuous at x₀.'

• T. True
• F. False

Answer: True

The definition of uniform continuity quantifies over ALL pairs x, y in the domain: for every ε > 0, there exists δ > 0 such that |x − y| < δ implies |f(x) − f(y)| < ε for ALL x, y in S. There is no 'local' version — no way to say this property holds 'at' a particular point, since it is inherently about the behavior of the function across the entire domain at once. By contrast, pointwise continuity at x₀ is a local property.

Question 5

Explain what goes wrong when you try to prove that f(x) = x² is uniformly continuous on ℝ. Why does the argument that works for f(x) = x fail here?

Model answer: For f(x) = x, |f(x) − f(y)| = |x − y|, so choosing δ = ε works for ALL points simultaneously — the same δ covers the entire real line. For f(x) = x², |f(x) − f(y)| = |x² − y²| = |x + y| · |x − y|. To make this less than ε when |x − y| < δ, we need δ · |x + y| < ε, so δ < ε / |x + y|. The required δ depends on x: as x grows, |x + y| ≈ 2|x| grows without bound, forcing δ to shrink toward 0. No single δ can satisfy the constraint for all x ∈ ℝ simultaneously. The linear function had a bounded 'rate of variation'; the quadratic's rate grows unboundedly.

The key structural difference is that f(x) = x has constant derivative (slope always 1), while f(x) = x² has unbounded derivative (slope 2x). Lipschitz functions — those with bounded derivatives — are always uniformly continuous, because the Lipschitz constant gives a global bound on how much f can vary per unit of x. Uniform continuity fails exactly when this 'rate of change' is unbounded across the domain.