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Part CCSIR NET December 2024shifting-is-tame-and-rescaling-is-not

Shifting is tame and rescaling is not

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a uniformly continuous function. For each positive integer n and x∈Rx \in \mathbb{R}, let gng_{n} and hnh_{n} be defined by gn(x)=f(x+1/n)g_{n}(x) = f(x + 1/n) and hn(x)=f(h_{n}(x) = f(nx). Which of the following statements are necessarily true?

  1. A.(gn)n≥1(g_{n})_{n}\ge_{1} converges uniformly on any compact subset of R\mathbb{R} but not on R\mathbb{R}.
  2. B.(gn)n≥1(g_{n})_{n}\ge_{1} converges uniformly on R\mathbb{R}.✓ Correct
  3. C.There exists a subsequence of (hn)n≥1(h_{n})_{n}\ge_{1} that converges uniformly on R\mathbb{R}.
  4. D.For every compact subset K⊆RK \subseteq \mathbb{R}, there exists a subsequence of (hn)n≥1(h_{n})_{n}\ge_{1} that converges uniformly on K.

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

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