Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 1A (Analysis II)

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe crucial point is that $\delta$ works throughout the interval, not just near one chosen point.

Yes, it is bounded on the closed unit interval. Choose $\delta$ for $\epsilon =1$, and choose finitely many grid points $t_j$ such that every $x\in [0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\le 1+{\max }_j|f(t_j)|$, a finite bound. Equivalently, uniform continuity implies continuity, and a continuous real function on a compact set is bounded.

Yes, a bounded derivative gives uniform continuity on the half-line. If $|f^{\prime }|\le K$, the mean value theorem gives $|f(x)-f(y)|\le K|x-y|$. This is a Lipschitz condition, so for $K>0$ use $\delta =\epsilon /K$. If $K=0$, the function is constant. As usual for a derivative on this closed half-line, continuity at its endpoint is included.

Not uniformly continuous. Take $x_n=n$, $y_n=n+1/n$. Then $|x_n-y_n|\to 0$, butThis contradicts the defining implication for uniform continuity, for example with $\epsilon =1$. The sequential criterion for uniform continuity makes explicit why arbitrarily close arguments at large distances from the origin matter.

Not uniformly continuous. ChooseTheir separation is $\pi /(x_n+y_n)\to 0$, while $\sin (x_n^2)=1$ and $\sin (y_n^2)=-1$. Thus the function values stay distance two apart, contradicting uniform continuity. Boundedness of the function does not prevent increasingly rapid oscillations.

Uniformly continuous. For $x\ge 1$, the derivative satisfiesThe mean value theorem therefore gives a Lipschitz condition with constant two, hence uniform continuity.