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Part CCSIR NET December 2025check-x-equals-0-first-fn-of-0-equals-n-breaks-boundedness-and-pointwise-convergence-in-one-stroke

Check x equals 0 first fn of 0 equals n breaks boundedness and pointwise convergence in one stroke

Let N\mathbb{N} denote the set of all positive integers. For n∈Nn\in\mathbb{N}, let fn:[0,1]→Rf_{n}:[0,1]\to\mathbb{R} be given by fn(x)=n(1−f_{n}(x)=n(1-nx) if 0≤x≤1/n, 0 if x>1/n. Which of the following statements are necessarily true?

  1. A.The sequence of functions {fnf_{n}}n≥1_{n}\ge1 is uniformly bounded.
  2. B.The sequence of functions {fnf_{n}}n≥1_{n}\ge1 does not converge pointwise.✓
  3. C.The sequence of functions {fnf_{n}}n≥1_{n}\ge1 converges uniformly to the constant function 0.
  4. D.The set {fn:n∈Nf_{n} : n\in\mathbb{N}} is compact in C[0,1], where C[0,1] denotes the space of all real valued continuous functions on [0,1] equipped with the supremum norm.

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.

See pricing

50 are analysed free — try those first.

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

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