Is the following statement true? Bounded and continuous on ℝ ⇒ uniformly continuous

No. f(x) = sin(x²) xₙ = √(2πn), yₙ = √(2πn + π/2) satisfy |xₙ − yₙ| → 0 while |f(xₙ) − f(yₙ)| = 1.

Counterexample

Is this true?

Bounded and continuous on $\mathbb{R} \Rightarrow$ uniformly continuous

No — it is false.

The counterexample

$f(x) = \sin(x^{2})$

Why it fails

$x_{n} = \sqrt{2\pi{}n}, y_{n} = \sqrt{2\pi{}n + \pi/2}$ satisfy | $x_{n} - y_{n}$ | → 0 while | $f(x_{n}) - f(y_{n})$ | = 1.

Keep practicing

Pointwise vs uniform

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

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.