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Part CCSIR NET December 2024periodic-plus-continuous-gives-uniform

Periodic plus continuous gives uniform

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a continuous function such that f(x) = 0 for all x ≤ 0 and for all x ≥ 1. Define F(x)=∑(n=−∞..∞)f(x+n),x∈RF(x) = \sum(n=-\infty..\infty) f(x + n), x \in \mathbb{R}. Which of the following statements are true?

  1. A.F is bounded.✓ Correct
  2. B.F is continuous on R\mathbb{R}.✓ Correct
  3. C.F is uniformly continuous on R\mathbb{R}.✓ Correct
  4. D.F is not uniformly continuous on R\mathbb{R}.

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: Continuous on a bounded interval ⇒ bounded

More on this topic

The chapter behind this: Continuity vs uniform continuity vs Lipschitz — free to read

From Continuity and Differentiation › Continuity, uniform continuity, Lipschitz

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