Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 1 / a / ii / 3 / Solution

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.