Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 / A / 1

A subset is relatively compact in $M$ when its closure in $M$ is compact.

A subset is totally bounded if for every $\epsilon >0$ it has a finite cover by balls of radius $\epsilon$. If its closure is compact, the cover by all such balls has a finite subcover, proving total boundedness.

Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as ${\sigma }_r$. These sets are infinite and nested, and two terms indexed by ${\sigma }_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in {\sigma }_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the closure as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the closure has the same limit. Sequential compactness therefore proves total boundedness implies relative compactness in a complete ambient space.

The Arzelà-Ascoli theorem says a subset of $C[0,1]$ is relatively compact exactly when it is uniformly bounded and equicontinuous. Pointwise boundedness together with equicontinuity is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\le |t-s|$, uniformly over the family, while $\Vert f{\Vert }_{\infty }\le 1$. Thus this family is relatively compact.

For $V$, the image of the unit ball has sup norm at most one and derivative $f$ of sup norm at most one; it lies in that family, so $V$ is compact.

For $W$, take $f_n=n{1}_{[0,1/n]}$, each of $L^1$ norm one. Then $W{f}_n(t)=\min (nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus $W$ is not compact. The same integration formula has different compactness properties because the domain unit balls impose different controls.