Skip to content
Part CCSIR NET June 2025uniform-convergence-carries-uniform-continuity-and-none-of-the-others

Uniform convergence carries uniform continuity and none of the others

Let (fn)n≥1(f_{n})_{n}\ge_{1} be a sequence of real-valued functions on R\mathbb{R}. Which of the following statements are true?

  1. A.If each fnf_{n} is uniformly continuous and (fn)n≥1(f_{n})_{n}\ge_{1} converges to f uniformly, then f is uniformly continuous.✓ Correct
  2. B.If each fnf_{n} is bounded and (fn)n≥1(f_{n})_{n}\ge_{1} converges to f pointwise, then f is bounded.
  3. C.If each fnf_{n} is bounded and continuous, (fn)n≥1(f_{n})_{n}\ge_{1} converges pointwise to a bounded and continuous function f, then the convergence is uniform.
  4. D.If each fnf_{n} is differentiable and (fn)n≥1(f_{n})_{n}\ge_{1} converges to f uniformly, then f is differentiable.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests pointwise vs uniform.

See pricing

50 are analysed free — try those first.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Drill statements like this

Related counterexample: fₙ → f uniformly ⇒ fₙ′ → f′

More on this topic

The chapter behind this: Uniform convergence — what it buys and how to test it — free to read

From Sequences and Series of Functions › Pointwise vs uniform convergence, M-test, Dini

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram