Locally uniform extraction for X_1-bounded, X_0-Lipschitz families with compact X_1-bounded subsequences

(The countable set of test points.) Put $D:=\mathbb{Q}\cap (-T,T)$; then $D$ is at most countable by [L3], and it is infinite: choosing an integer $m\ge 1$ with $m>1\mathrm{/}T$ [L4], the rationals $1\mathrm{/}(m+n)$, $n\ge 0$, are pairwise distinct and lie in $(0,T)\subseteq D$. Hence $D$ is countably infinite and we fix an enumeration $D=\{q_1,q_2,\dots \}$ with pairwise distinct $q_i$ [L3].

(Finite covers of bounded sequence ranges.) Let $(x_j)$ be any $X_1$-bounded sequence and fix $\epsilon >0$. Its range can be covered by finitely many open $X_0$-balls of radius $\epsilon$ centred at its terms. Otherwise define $j_1=1$ and recursively let $j_{l+1}$ be the least index whose term lies outside the union of the radius-$\epsilon$ balls centred at $x_{j_1},\dots ,x_{j_l}$. The assumed failure of every finite cover ensures this index exists. These canonically defined terms have pairwise distances at least $\epsilon$, so no subsequence converges in $X_0$, contrary to (ii). The same conclusion applies to every subsequence indexed by an infinite subset $S\subseteq \mathbb{N}$, enumerated in increasing order.

(Deterministic diagonal extraction.) Put $S_0=\mathbb{N}$. For $n\ge 1$, start with $S_{n-1}$ and refine its infinite index set successively at the rational times $q_1,\dots ,q_n$. At the refinement for $q_i$, let $S$ be the current infinite set. By step 1.2, finitely many radius-$2^{-n}$ balls centred at $f_j(q_i)$ with $j\in S$ cover the sequence range indexed by $S$. At least one such ball contains $f_j(q_i)$ for infinitely many $j\in S$, since a finite union of finite index sets is finite. Let $\mathrm{\ell }$ be the least index in $S$ whose centred ball has this property, and replace $S$ by $\{j\in S:\mathrm{\parallel }{f}_j(q_i)-f_{\mathrm{\ell }}(q_i){\mathrm{\parallel }}_0<2^{-n}\}$. This is an infinite subset of the preceding set. After the finitely many refinements define it to be $S_n$. All selections are least natural-number indices, so the recursive construction is determined without dependent choice. Define $k_n$ to be the least element of $S_n$ greater than $k_{n-1}$, with $k_0=0$. The $S_n$ are infinite and nested, so the $k_n$ exist and increase strictly. For fixed $i$ and $n\ge i$, any two terms $f_j(q_i),f_l(q_i)$ with $j,l\in S_n$ have distance less than $2^{1-n}$, by the refinement at $q_i$ and subsequent subset restrictions. Since $k_m\in S_m\subseteq S_n$ for $m\ge n$, the sequence $(f_{k_m}(q_i){)}_m$ is Cauchy and converges by [L1].

(Uniform Cauchy on a compact subinterval.) Fix $0<T^{\mathrm{\prime }}<T$ and $\epsilon >0$ and put $\delta :=\epsilon \mathrm{/}(3(C+1))$. Choose an integer $M\ge 1$ with $2T^{\mathrm{\prime }}\mathrm{/}M<\delta$ [L4]. For $i=0,\dots ,M-1$ the interval $(-T^{\mathrm{\prime }}+2T^{\mathrm{\prime }}i\mathrm{/}M,-T^{\mathrm{\prime }}+2T^{\mathrm{\prime }}(i+1)\mathrm{/}M)$ is a nonempty open interval, so it contains a rational $q_i^{\mathrm{\prime }}$ (finitely many choices [L3]), and each $q_i^{\mathrm{\prime }}\in D$. By step 2.1, for each $i$ the sequence $n\mapsto f_{k_n}(q_i^{\mathrm{\prime }})$ is Cauchy in $X_0$, so there is $J_i$ with $\mathrm{\parallel }{f}_{k_n}(q_i^{\mathrm{\prime }})-f_{k_m}(q_i^{\mathrm{\prime }}){\mathrm{\parallel }}_0<\epsilon \mathrm{/}3$ for $n,m\ge J_i$; put $J:={\max }_i{J}_i$. Let $t\in [-T^{\mathrm{\prime }},T^{\mathrm{\prime }}]$ and choose one of the closed partition intervals $[-T^{\mathrm{\prime }}+2T^{\mathrm{\prime }}i\mathrm{/}M,-T^{\mathrm{\prime }}+2T^{\mathrm{\prime }}(i+1)\mathrm{/}M]$ containing $t$. Such an interval exists also at both endpoints. Since $q_i^{\mathrm{\prime }}$ lies strictly inside its corresponding interval, $\mathrm{\mid }t-q_i^{\mathrm{\prime }}\mathrm{\mid }<2T^{\mathrm{\prime }}\mathrm{/}M<\delta$. Then for $n,m\ge J$ the triangle inequality gives $\mathrm{\parallel }{f}_{k_n}(t)-f_{k_m}(t){\mathrm{\parallel }}_0\le \mathrm{\parallel }{f}_{k_n}(t)-f_{k_n}(q_i^{\mathrm{\prime }}){\mathrm{\parallel }}_0+\mathrm{\parallel }{f}_{k_n}(q_i^{\mathrm{\prime }})-f_{k_m}(q_i^{\mathrm{\prime }}){\mathrm{\parallel }}_0+\mathrm{\parallel }{f}_{k_m}(q_i^{\mathrm{\prime }})-f_{k_m}(t){\mathrm{\parallel }}_0<2C\delta +\epsilon \mathrm{/}3<\epsilon$, using the Lipschitz hypothesis and $2C\delta \le 2C\epsilon \mathrm{/}(3(C+1))<2\epsilon \mathrm{/}3$; hence $(f_{k_n})$ is uniformly Cauchy on $[-T^{\mathrm{\prime }},T^{\mathrm{\prime }}]$.

(The limit function.) Let $t\in (-T,T)$ and choose $T^{\mathrm{\prime }}$ with $\mathrm{\mid }t\mathrm{\mid }<T^{\mathrm{\prime }}<T$. By step 3.1 the sequence $(f_{k_n}(t){)}_n$ is Cauchy in $X_0$, so by completeness it has a limit $f(t)\in X_0$, uniquely determined [L1]; passing to the limit $m\to \mathrm{\infty }$ in the estimate of step 3.1 at fixed $n\ge J$ gives $\mathrm{\parallel }{f}_{k_n}(t)-f(t){\mathrm{\parallel }}_0\le \epsilon$ for every $t\in [-T^{\mathrm{\prime }},T^{\mathrm{\prime }}]$, so $f_{k_n}\to f$ uniformly on $[-T^{\mathrm{\prime }},T^{\mathrm{\prime }}]$. Every compact subinterval $[a,b]$ of $(-T,T)$ lies in some $[-T^{\mathrm{\prime }},T^{\mathrm{\prime }}]$ with $T^{\mathrm{\prime }}<T$, namely for any $T^{\mathrm{\prime }}$ with $\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })<T^{\mathrm{\prime }}<T$, so the convergence is uniform on every compact subinterval.

(Continuity.) Let $t,t^{\mathrm{\prime }}\in (-T,T)$ and choose $T^{\mathrm{\prime }}>\max (\mathrm{\mid }t\mathrm{\mid },\mathrm{\mid }{t}^{\mathrm{\prime }}\mathrm{\mid })$ with $T^{\mathrm{\prime }}<T$. Norms of a convergent sequence converge to the norm of the limit (reverse triangle inequality), so by step 4.1 and the hypothesis $\mathrm{\parallel }f(t)-f(t^{\mathrm{\prime }}){\mathrm{\parallel }}_0={\lim }_{n\to \mathrm{\infty }}\mathrm{\parallel }{f}_{k_n}(t)-f_{k_n}(t^{\mathrm{\prime }}){\mathrm{\parallel }}_0\le C\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$; hence $f$ is Lipschitz with constant $C$ and therefore continuous (Continuity of a map of topological spaces at a point and globally), which with step 4.1 is the assertion.