Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 3H (Analysis II)

A real-valued function $f$ on an interval $I$ is uniformly continuous whenThe essential point in uniform continuity is that $\delta$ is independent of the locations of $x$ and $y$. Uniform continuity on the whole real line does not imply boundedness: $f(x)=x$ is Lipschitz continuous with constant $1$, hence uniformly continuous, but is unbounded.

Take $x_n=2\pi n$ and $y_n=2\pi n+1/n$. Then $|y_n-x_n|=1/n\to 0$, whileThe sequential criterion for uniform continuity would force $f(y_n)-f(x_n)\to 0$. These pairs contradict it: for example, for all sufficiently large $n$ their output separation is greater than $\pi$, despite arbitrarily small input separation. Thus $x\sin x$ is not uniformly continuous on $\mathbb{R}$.

The derivative is $f^{\prime }(x)=-4x^3{e}^{-x^4}$. Its absolute value is continuous and tends to zero as $x\to \pm \infty$, so it is bounded. More explicitly,obtained by differentiating $4t^3{e}^{-t^4}$ for $t\ge 0$. The mean value theorem gives $|f(x)-f(y)|\le M|x-y|$ for every $x,y$. Thus $f$ is Lipschitz continuous, and choosing $\delta =\epsilon /M$ proves $e^{-x^4}$ is uniformly continuous on $\mathbb{R}$.