Proof of Limits of a Uniformly Bounded Sequence of Real Functions with a Common Modulus of Continuity: Bounds, Extraction and Uniform Convergence on Sequentially Compact Sets

Each result cited below is universally quantified over the data in its own statement. We work in the setting of The Real Numbers: Standing Notation and Background, which carries the elementary arithmetic, order and absolute-value facts of $\mathbb{R}$ used below; natural numbers start at $1$, as in Natural Numbers.

Notation. For $x\in X$ and $k\in\mathbb{N}$ put $A_{k}(x)=\{u_{m}(x):m\in\mathbb{N},\ m\ge k\}$. Since $|u_{N}(x)|\le C<C+1$ for every $N$, the number $C+1>0$ witnesses that $(u_{N}(x))_{N\in\mathbb{N}}$ is a bounded sequence, and each $A_{k}(x)$ is nonempty and bounded above by $C$ and below by $-C$. By the definition of the limit superior and that of the limit inferior,

$$\bar{u}(x)=\inf\{\sup A_{k}(x):k\in\mathbb{N}\},\qquad \underline{u}(x)=\sup\{\inf A_{k}(x):k\in\mathbb{N}\}.$$

Step 0 (Shifting suprema and infima). Let $S\subseteq\mathbb{R}$ be nonempty, let $c\in\mathbb{R}$, and put $S+c=\{s+c:s\in S\}$. We show: if $S$ is bounded above, then $S+c$ is bounded above and $\sup(S+c)=\sup S+c$; if $S$ is bounded below, then $S+c$ is bounded below and $\inf(S+c)=\inf S+c$. Suppose $S$ is bounded above. For $s\in S$ we have $s\le\sup S$, hence $s+c\le\sup S+c$, so $\sup S+c$ is an upper bound of $S+c$. If $b$ is any upper bound of $S+c$, then $s\le b-c$ for every $s\in S$, so $b-c$ is an upper bound of $S$ and therefore $\sup S\le b-c$, that is, $\sup S+c\le b$. Thus $\sup S+c$ is a least upper bound of $S+c$ in the sense of Upper Bound and Least Upper Bound, and by the uniqueness recorded in The Real Numbers: Standing Notation and Background §bounds it equals $\sup(S+c)$. The statement for infima is proved in the same way with all inequalities reversed, using Lower Bound and Greatest Lower Bound in a Totally Ordered Set: $\inf S+c$ is a lower bound of $S+c$, and if $\ell$ is a lower bound of $S+c$ then $\ell-c$ is a lower bound of $S$, so $\ell-c\le\inf S$, that is, $\ell\le\inf S+c$.

Consequently, if $(b_{N})_{N\in\mathbb{N}}$ is a bounded sequence of real numbers, say $|b_{N}|\le M$ with $M>0$, and $c\in\mathbb{R}$, then $(b_{N}+c)_{N\in\mathbb{N}}$ is bounded, since $|b_{N}+c|\le M+|c|$ and $M+|c|>0$; and

$$\limsup_{N}(b_{N}+c)=\limsup_{N}b_{N}+c,\qquad \liminf_{N}(b_{N}+c)=\liminf_{N}b_{N}+c. \tag{0.1}$$

Indeed, writing $B_{k}=\{b_{m}:m\ge k\}$, the corresponding set for $(b_{N}+c)$ is $B_{k}+c$, so by the first part $\sup(B_{k}+c)=\sup B_{k}+c$ for every $k$; hence $\{\sup(B_{k}+c):k\in\mathbb{N}\}=\{\sup B_{k}:k\in\mathbb{N}\}+c$, and the second part gives that its infimum is $\limsup_{N}b_{N}+c$. The identity for the limit inferior follows in the same way, first shifting $\inf B_{k}$ and then the supremum of $\{\inf B_{k}:k\in\mathbb{N}\}$.

Step 1 (Clause 1, bounds). Fix $x\in X$. From $|u_{N}(x)|\le C$ we get $-C\le u_{N}(x)\le C$ for every $N$. By claim 2 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence, taken with $M=C$ for the first inequality and with $M=-C$ for the second, we get $\bar{u}(x)\le C$ and $-C\le\underline{u}(x)$, and by claim 1 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence we have $\underline{u}(x)\le\bar{u}(x)$. Hence $-C\le\underline{u}(x)\le\bar{u}(x)\le C$, and so $|\bar{u}(x)|\le C$ and $|\underline{u}(x)|\le C$.

Step 2 (Clause 2, modulus). Fix $x,y\in X$. By claim 1 of Metric Space, $0\le d(x,y)$, so $t=\omega(d(x,y))$ is defined. For every $N$ the hypothesis gives $u_{N}(x)-u_{N}(y)\le|u_{N}(x)-u_{N}(y)|\le t$, that is, $u_{N}(x)\le u_{N}(y)+t$. The sequence $(u_{N}(y)+t)_{N\in\mathbb{N}}$ is bounded by Step 0, so claim 2 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence applies to $a_{N}=u_{N}(x)$ and $b_{N}=u_{N}(y)+t$, and together with (0.1) it gives

$$\bar{u}(x)\le\limsup_{N}\bigl(u_{N}(y)+t\bigr)=\bar{u}(y)+t,\qquad \underline{u}(x)\le\liminf_{N}\bigl(u_{N}(y)+t\bigr)=\underline{u}(y)+t.$$

By claim 3 of Metric Space, $d(y,x)=d(x,y)$, so the same argument with $x$ and $y$ exchanged gives $\bar{u}(y)\le\bar{u}(x)+t$ and $\underline{u}(y)\le\underline{u}(x)+t$. Hence $|\bar{u}(x)-\bar{u}(y)|\le t$ and $|\underline{u}(x)-\underline{u}(y)|\le t$.

Step 3 (Clause 3, extraction). Fix $x\in X$ and write $a_{N}=u_{N}(x)$, $L=\bar{u}(x)$ and $L'=\underline{u}(x)$. For each $j\in\mathbb{N}$ the number $\tfrac{1}{j}$ is positive.

(a) By claim 3 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence, taken with $\varepsilon=\tfrac{1}{j}$, for every $j\in\mathbb{N}$ there is $P_{j}\in\mathbb{N}$ with $a_{m}<L+\tfrac{1}{j}$ for every $m\ge P_{j}$.

(b) Let $j,k\in\mathbb{N}$. Since $L$ is the infimum, and in particular a lower bound, of $\{\sup A_{k'}(x):k'\in\mathbb{N}\}$, we have $L\le\sup A_{k}(x)$, hence $L-\tfrac{1}{j}<\sup A_{k}(x)$. The set $A_{k}(x)$ is nonempty and bounded above, so Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ §strict-above, applied with $S=A_{k}(x)$ and $b=L-\tfrac{1}{j}$, gives an element of $A_{k}(x)$ exceeding $b$; that is, there is $m\in\mathbb{N}$ with $m\ge k$ and $L-\tfrac{1}{j}<a_{m}$.

We now choose $N_{1},N_{2},\dots$ recursively. Let $N_{1}$ be an index $m$ given by (b) for $j=1$ and $k=P_{1}$. Having chosen $N_{j}$, let $N_{j+1}$ be an index $m$ given by (b) for $j+1$ in place of $j$ and $k=\max\{P_{j+1},N_{j}+1\}$. Then $N_{j+1}\ge N_{j}+1>N_{j}$ for every $j$, and $N_{j}\ge P_{j}$, so (a) and (b) give $L-\tfrac{1}{j}<a_{N_{j}}<L+\tfrac{1}{j}$, that is, $|a_{N_{j}}-L|<\tfrac{1}{j}$ for every $j\in\mathbb{N}$. Given $\varepsilon>0$, claim 3 of The Archimedean Property of the Real Numbers gives $J\in\mathbb{N}$ with $\tfrac{1}{J}<\varepsilon$, and then $|a_{N_{j}}-L|<\tfrac{1}{j}\le\tfrac{1}{J}<\varepsilon$ for every $j\ge J$. By Limit of a Sequence of Real Numbers, $(u_{N_{j}}(x))_{j\in\mathbb{N}}$ converges to $\bar{u}(x)$.

For the limit inferior we argue symmetrically. (a') By claim 3 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence, for every $j$ there is $P'_{j}\in\mathbb{N}$ with $L'-\tfrac{1}{j}<a_{m}$ for every $m\ge P'_{j}$. (b') For $j,k\in\mathbb{N}$, $L'$ is the supremum, and in particular an upper bound, of $\{\inf A_{k'}(x):k'\in\mathbb{N}\}$, so $\inf A_{k}(x)\le L'<L'+\tfrac{1}{j}$; the set $A_{k}(x)$ is nonempty and bounded below, so Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ §strict-below, applied with $S=A_{k}(x)$ and $b=L'+\tfrac{1}{j}$, gives $m\ge k$ with $a_{m}<L'+\tfrac{1}{j}$. Choosing $N'_{1}$ by (b') with $j=1$ and $k=P'_{1}$, and $N'_{j+1}$ by (b') with $j+1$ and $k=\max\{P'_{j+1},N'_{j}+1\}$, we obtain $N'_{1}<N'_{2}<\cdots$ with $|a_{N'_{j}}-L'|<\tfrac{1}{j}$ for every $j$, and the same Archimedean argument shows that $(u_{N'_{j}}(x))_{j\in\mathbb{N}}$ converges to $\underline{u}(x)$.

Step 4 (Clause 4, uniform convergence). Let $K$ be as in clause 4 and suppose, for a contradiction, that the conclusion fails. Then there is a real $\varepsilon>0$ such that for every $N_{0}\in\mathbb{N}$ there exist $N\in\mathbb{N}$ with $N\ge N_{0}$ and $x\in K$ with $|u_{N}(x)-\bar{u}(x)|>\varepsilon$. Fix such an $\varepsilon$; all later choices depend on it.

Choose recursively $N_{1}\in\mathbb{N}$ and $x_{1}\in K$ for $N_{0}=1$, and, having chosen $N_{j}$, choose $N_{j+1}$ and $x_{j+1}\in K$ for $N_{0}=N_{j}+1$. Then $N_{1}<N_{2}<\cdots$, $x_{j}\in K$, and

$$|u_{N_{j}}(x_{j})-\bar{u}(x_{j})|>\varepsilon\qquad\text{for every }j\in\mathbb{N}. \tag{4.1}$$

Since $K$ is sequentially compact, there are $x^{*}\in K$ and a strictly increasing sequence $(j_{i})_{i\in\mathbb{N}}$ in $\mathbb{N}$ such that $(x_{j_{i}})_{i\in\mathbb{N}}$ converges to $x^{*}$ in $(X,d)$.

The remaining choices are made in the following order. First, claim 2 of Modulus of Continuity, applied to $\tfrac{\varepsilon}{3}>0$, gives $\delta>0$ such that $\omega(t)\le\tfrac{\varepsilon}{3}$ for every $t\ge0$ with $t\le\delta$. Second, by Convergent Sequence in a Metric Space there is $i_{1}\in\mathbb{N}$ with $d(x_{j_{i}},x^{*})<\delta$ for every $i\ge i_{1}$. Third, since $x^{*}\in K$ we have $\bar{u}(x^{*})=\underline{u}(x^{*})$, so claim 5 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence, taken with $a=\bar{u}(x^{*})$, shows that $(u_{N}(x^{*}))_{N\in\mathbb{N}}$ converges to $\bar{u}(x^{*})$; by Limit of a Sequence of Real Numbers there is $P\in\mathbb{N}$ with $|u_{N}(x^{*})-\bar{u}(x^{*})|<\tfrac{\varepsilon}{3}$ for every $N\ge P$. Finally, put $i=\max\{i_{1},P\}$, $z=x_{j_{i}}$ and $N=N_{j_{i}}$.

By Strictly Increasing Sequences of Natural Numbers Dominate Their Index, applied to the strictly increasing sequences $(N_{j})_{j\in\mathbb{N}}$ and $(j_{i})_{i\in\mathbb{N}}$, we have $N=N_{j_{i}}\ge j_{i}\ge i\ge P$. Since $i\ge i_{1}$, $d(z,x^{*})<\delta$, and $d(x^{*},z)=d(z,x^{*})$ by claim 3 of Metric Space; both are nonnegative by claim 1 of Metric Space. Hence the hypothesis on $(u_{N})$ gives $|u_{N}(z)-u_{N}(x^{*})|\le\omega(d(z,x^{*}))\le\tfrac{\varepsilon}{3}$; the choice of $P$ gives $|u_{N}(x^{*})-\bar{u}(x^{*})|<\tfrac{\varepsilon}{3}$; and Step 2 gives $|\bar{u}(x^{*})-\bar{u}(z)|\le\omega(d(x^{*},z))\le\tfrac{\varepsilon}{3}$. By the triangle inequality for the absolute value,

$$|u_{N}(z)-\bar{u}(z)|\le|u_{N}(z)-u_{N}(x^{*})|+|u_{N}(x^{*})-\bar{u}(x^{*})|+|\bar{u}(x^{*})-\bar{u}(z)|<\varepsilon .$$

But $z=x_{j_{i}}$ and $N=N_{j_{i}}$, so (4.1) with $j=j_{i}$ gives $|u_{N}(z)-\bar{u}(z)|>\varepsilon$, a contradiction. Hence for every $\varepsilon>0$ there is $N_{0}\in\mathbb{N}$ with $|u_{N}(x)-\bar{u}(x)|\le\varepsilon$ for every $N\ge N_{0}$ and every $x\in K$.