Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 11F (Analysis and Topology)

A function $f:(X,d_X)\to (Y,d_Y)$ is uniformly continuous if for every $\epsilon >0$ there is $\delta >0$ such thatfor all $x,x^{\prime }\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\epsilon >0$, choose $N$ such thatfor every $x$. Uniform continuity of $f_N$ supplies $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ impliesThe triangle inequality then gives $d_Y(f(x),f(x^{\prime }))<\epsilon$. Thus the uniform limit theorem for uniformly continuous functions proves that $f$ is uniformly continuous.

Pointwise convergence is insufficient. On $[0,1]$, the uniformly continuous functions $f_n(x)=x^n$ converge pointwise towhich is discontinuous and therefore not uniformly continuous.

The inequalities implyThus the two metrics induce the same open sets and are equivalent.

The converse fails. On $\mathbb{R}$, letThe metrics are equivalent because $\arctan :\mathbb{R}\to (-\pi /2,\pi /2)$ is a homeomorphism, butso no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take $X=\mathbb{N}$, $Y=\mathbb{R}$, and use the two metrics above. DefineThenso $f_n\to f$ uniformly for $d_2$, whereasso convergence is not uniform for $d_1$. This is the equivalent codomain metrics need not preserve uniform convergence phenomenon.