Questions: Uniform Convergence

Question 1

The sequence fₙ(x) = xⁿ on [0, 1) converges pointwise to 0 but fails to converge uniformly. What is the essential reason for this failure?

• A. The limit function is zero, and uniform convergence requires convergence to a nonzero function
• B. Near x = 1, fₙ(x) stays close to 1 for arbitrarily large n, so no single N works for all x simultaneously
• C. The functions fₙ are continuous, but the pointwise limit is not, which rules out uniform convergence on any interval
• D. Uniform convergence cannot occur on open intervals — it requires a closed, bounded domain

Answer: (B)

Uniform convergence requires one N (depending only on ε) such that |fₙ(x) − f(x)| < ε for *all* x simultaneously. For fₙ(x) = xⁿ with limit f = 0, we need xⁿ < ε for all x ∈ [0, 1). But for any n, choosing x close enough to 1 makes xⁿ as large as we like below 1. The 'tube' around 0 can never contain the entire graph of fₙ near x = 1. Option C describes a consequence (discontinuous limits), not the reason for this specific failure.

Question 2

Which of the following is guaranteed by uniform convergence but cannot be guaranteed from pointwise convergence alone?

• A. The limit function f exists at every point in the domain
• B. The limit of a sequence of continuous functions is continuous
• C. The sequence eventually reaches f exactly at each point
• D. All functions in the sequence share the same maximum value

Answer: (B)

The Uniform Limit Theorem states that if fₙ are continuous and fₙ → f uniformly, then f is continuous. This fails spectacularly under mere pointwise convergence: fₙ(x) = xⁿ on [0, 1] consists entirely of continuous functions yet converges pointwise to a discontinuous limit (0 on [0,1) and 1 at x=1). Uniform convergence is precisely the condition that prevents this kind of 'convergence to a worse function,' and it also enables the interchange of limit and integral.

Question 3

Uniform convergence requires that for every ε > 0 there exists N depending only on ε — not on x — such that |fₙ(x) − f(x)| < ε holds for all x in the domain whenever n > N.

• T. True
• F. False

Answer: True

This is the defining difference between uniform and pointwise convergence. In pointwise convergence, the required N can depend on both ε and the specific point x — different points may need different N values. Uniform convergence demands a single N that works everywhere simultaneously. Geometrically: the entire graph of fₙ must fit inside the ε-tube around the graph of f for all sufficiently large n.

Question 4

If fₙ → f pointwise and each fₙ is continuous, then f should also be continuous.

• T. True
• F. False

Answer: False

Pointwise convergence does not preserve continuity. The standard counterexample is fₙ(x) = xⁿ on [0, 1]: each fₙ is continuous, but the pointwise limit is 0 on [0, 1) and 1 at x = 1, which is discontinuous. Continuity is preserved by uniform convergence (the Uniform Limit Theorem), but not by pointwise convergence. This is precisely why the stronger condition of uniform convergence matters in analysis.

Question 5

What is the key difference between pointwise and uniform convergence, and why does it matter for preserving the continuity of limit functions?

Model answer: In pointwise convergence, each point x gets its own N(ε, x): the speed of convergence can vary arbitrarily across the domain. In uniform convergence, a single N(ε) works for all x simultaneously — convergence is equally fast across the entire domain. Continuity is preserved under uniform convergence because you can use the uniform N to control |f(x) − fₙ(x)| and |fₙ(y) − f(y)| uniformly near any point, then invoke continuity of fₙ for the middle term in an ε/3 argument. Pointwise convergence allows convergence to stall near some points, which can create jumps in the limit function.

The ε/3 proof of the Uniform Limit Theorem splits |f(x) − f(y)| ≤ |f(x) − fₙ(x)| + |fₙ(x) − fₙ(y)| + |fₙ(y) − f(y)|. The first and third terms are controlled by uniform convergence (they are both less than ε/3 for large enough n, independently of x and y). The middle term is controlled by continuity of fₙ. Without uniform convergence, the first and third terms depend on x and y, so you cannot choose n before choosing the nearby point y.