Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 11G (Analysis II)

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.

True. If $|f^{\prime }(x)|\le K$ everywhere, the mean value theorem applied between any two distinct points givesThus $f$ is Lipschitz continuous and hence uniformly continuous. No continuity of $f^{\prime }$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the mean value theorem there.

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.