Proof of The Weighted Uniform Distance Is a Metric, and Is Complete When the Target Is Complete

Write $C=C([a,b],Y)$. By The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (i), the elements of $C$ are exactly the maps $[a,b]\to Y$ continuous on $[a,b]$, and by The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (ii), $d_{w}$ is a map $C\times C\to\mathbb{R}$. Throughout, The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (iii) is used in the form: for $u,v\in C$, $w(t)\,d_{Y}(u(t),v(t))\le d_{w}(u,v)$ for every $t\in[a,b]$, with equality at some $t_{0}\in[a,b]$.

The metric ((i)). Let $x,y,z\in C$.

Nonnegative. $a\in[a,b]$ by Basic Facts about Intervals of the Real Line (v), $w(a)>0$ and $d_{Y}(x(a),y(a))\ge0$ by Metric Space (ii); so $0\le w(a)\,d_{Y}(x(a),y(a))\le d_{w}(x,y)$.

Zero. If $x=y$, take $t_{0}$ with $d_{w}(x,x)=w(t_{0})\,d_{Y}(x(t_{0}),x(t_{0}))$; this is $0$, as $d_{Y}(x(t_{0}),x(t_{0}))=0$ by Metric Space (iii). Conversely, let $d_{w}(x,y)=0$ and $t\in[a,b]$. Then $0\le w(t)\,d_{Y}(x(t),y(t))\le d_{w}(x,y)=0$, so $w(t)\,d_{Y}(x(t),y(t))=0$; as $w(t)>0$, $d_{Y}(x(t),y(t))=0$, and $x(t)=y(t)$ by Metric Space (iii). Hence $x=y$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).

Symmetric. Take $t_{0}$ with $d_{w}(x,y)=w(t_{0})\,d_{Y}(x(t_{0}),y(t_{0}))$. By Metric Space (iv) this equals $w(t_{0})\,d_{Y}(y(t_{0}),x(t_{0}))\le d_{w}(y,x)$. Exchanging $x$ and $y$ gives $d_{w}(y,x)\le d_{w}(x,y)$, so $d_{w}(x,y)=d_{w}(y,x)$.

Triangle. Take $t_{0}$ with $d_{w}(x,z)=w(t_{0})\,d_{Y}(x(t_{0}),z(t_{0}))$. By Metric Space (v) and $w(t_{0})>0$,

$$d_{w}(x,z)\le w(t_{0})\,d_{Y}(x(t_{0}),y(t_{0}))+w(t_{0})\,d_{Y}(y(t_{0}),z(t_{0}))\le d_{w}(x,y)+d_{w}(y,z).$$

So $d_{w}$ is a metric on $C$ by Metric Space (i), and $(C,d_{w})$ is a metric space by Metric Space (vi).

Nonempty ((ii)). Let $c\in Y$ and let $h:[a,b]\to Y$ be the map with $h(s)=c$ for every $s\in[a,b]$, which exists by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v). For all $s,t\in[a,b]$, $d_{Y}(h(s),h(t))=d_{Y}(c,c)=0$ by Metric Space (iii). So for every $t\in[a,b]$ and every real $\varepsilon>0$, $\delta=1$ has the property required in Continuous Map Between Metric Spaces (i); hence $h$ is continuous on $[a,b]$ by Continuous Map Between Metric Spaces (ii), and $h\in C$.

Completeness ((iii)). Let $(Y,d_{Y})$ be complete, and let $(x_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $(C,d_{w})$.

Step 1: bounds for the weight. By Extreme Value Theorem on a Closed Real Interval (ii) applied to $w$, there are $p,q\in[a,b]$ with $w(p)\le w(t)\le w(q)$ for every $t\in[a,b]$. Put $m=w(p)$ and $M=w(q)$; then $m>0$ and $M>0$. For $u,v\in C$ and $t\in[a,b]$, $d_{Y}(u(t),v(t))\ge0$ by Metric Space (ii), so $m\,d_{Y}(u(t),v(t))\le w(t)\,d_{Y}(u(t),v(t))\le d_{w}(u,v)$, that is,

$$d_{Y}(u(t),v(t))\le d_{w}(u,v)/m.\qquad(1)$$

Step 2: the pointwise limit. As $C\subseteq Y^{[a,b]}$, the set of maps from $[a,b]$ to $Y$, $(x_{k})$ is a sequence in $Y^{[a,b]}$. Let $t\in[a,b]$. By Sections of a Sequence of Maps into a Metric Space, and the Map of Their Limits (i), with $[a,b]$ for its $D$, the map $k\mapsto x_{k}(t)$ is a sequence $(x_{k}(t))_{k}$ in $Y$. It is a Cauchy sequence in $(Y,d_{Y})$: given a real $\varepsilon>0$, the Cauchy property of $(x_{k})$ applied to $m\varepsilon>0$ gives $N\in\mathbb{N}$ with $d_{w}(x_{k},x_{l})<m\varepsilon$ for all $k,l\in\mathbb{N}$ with $k,l\ge N$, and then $d_{Y}(x_{k}(t),x_{l}(t))<\varepsilon$ by (1). As $(Y,d_{Y})$ is complete, $(x_{k}(t))_{k}$ is convergent in $(Y,d_{Y})$ by Complete Metric Space (i). By Sections of a Sequence of Maps into a Metric Space, and the Map of Their Limits (ii), with $[a,b]$ for its $D$ and $S$, there is exactly one map $x:[a,b]\to Y$ such that $(x_{k}(t))_{k}$ converges to $x(t)$ in $(Y,d_{Y})$ for every $t\in[a,b]$, and $x(t)$ is the limit $\lim_{k\to\infty}x_{k}(t)$.

Step 3: the convergence is uniform. We show: for every real $\varepsilon>0$ there is $N\in\mathbb{N}$ with

$$d_{Y}(x_{k}(t),x(t))<\varepsilon\qquad\text{for all }k\in\mathbb{N}\text{ with }k\ge N\text{ and all }t\in[a,b].\qquad(2)$$

The choices are made in this order. First $\varepsilon$ is given. (Here and in Steps 4 and 5, the quotients $m\varepsilon/2$, $\varepsilon/2$, $\varepsilon/3$ and $\varepsilon/(2M)$ are real numbers, positive as $m,M>0$, and are computed in the type $\mathrm{Real}$ of The Real Numbers as a Type (i), by Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (ii).) Then the Cauchy property of $(x_{k})$ applied to $m\varepsilon/2>0$ gives $N\in\mathbb{N}$ with $d_{w}(x_{k},x_{l})<m\varepsilon/2$ for all $k,l\ge N$, so that, by (1), $d_{Y}(x_{k}(t),x_{l}(t))<\varepsilon/2$ for all $k,l\ge N$ and all $t\in[a,b]$; this $N$ does not depend on $t$. Now let $k\ge N$ and $t\in[a,b]$. By Step 2 and Convergent Sequence in a Metric Space (i) there is $N'\in\mathbb{N}$, depending on $t$, with $d_{Y}(x_{l}(t),x(t))<\varepsilon/2$ for all $l\ge N'$. Let $l$ be the larger of $N$ and $N'$, which exists by Arithmetic and Order of the Natural Numbers (vii); then $l\in\mathbb{N}$, $N\le l$ and $N'\le l$ by Arithmetic and Order of the Natural Numbers (vi). By Metric Space (v),

$$d_{Y}(x_{k}(t),x(t))\le d_{Y}(x_{k}(t),x_{l}(t))+d_{Y}(x_{l}(t),x(t))<\varepsilon/2+\varepsilon/2=\varepsilon.$$

Step 4: the limit is continuous. Let $t\in[a,b]$ and let a real $\varepsilon>0$ be given. First, (2) applied to $\varepsilon/3$ gives $N\in\mathbb{N}$, and, with $k=N$ (as $N\le N$ by Arithmetic and Order of the Natural Numbers (vi)), $d_{Y}(x_{N}(s),x(s))<\varepsilon/3$ for every $s\in[a,b]$. Then, as $x_{N}\in C$ is continuous on $[a,b]$, it is continuous at $t$ relative to $[a,b]$ by Continuous Map Between Metric Spaces (ii), and Continuous Map Between Metric Spaces (i) gives a real $\delta>0$ with $d_{Y}(x_{N}(s),x_{N}(t))<\varepsilon/3$ for every $s\in[a,b]$ with $d_{\mathbb{R}}(t,s)<\delta$. For such $s$, by Metric Space (v) applied twice and Metric Space (iv),

$$d_{Y}(x(s),x(t))\le d_{Y}(x_{N}(s),x(s))+d_{Y}(x_{N}(s),x_{N}(t))+d_{Y}(x_{N}(t),x(t))<\varepsilon.$$

So $x$ is continuous at $t$ relative to $[a,b]$; as $t$ was arbitrary, $x$ is continuous on $[a,b]$ by Continuous Map Between Metric Spaces (ii), and $x\in C$.

Step 5: convergence in $d_{w}$. Let a real $\varepsilon>0$ be given, and let $N$ be given by (2) for $\varepsilon/(2M)$. Let $k\ge N$. For every $t\in[a,b]$, as $w(t)\le M$ and $d_{Y}(x_{k}(t),x(t))\ge0$,

$$w(t)\,d_{Y}(x_{k}(t),x(t))\le M\,d_{Y}(x_{k}(t),x(t))\le M\cdot\frac{\varepsilon}{2M}=\frac{\varepsilon}{2}.$$

Taking $t_{0}$ with $d_{w}(x_{k},x)=w(t_{0})\,d_{Y}(x_{k}(t_{0}),x(t_{0}))$ gives $d_{w}(x_{k},x)\le\varepsilon/2<\varepsilon$. Hence $(x_{k})$ converges to $x$ in $(C,d_{w})$ by Convergent Sequence in a Metric Space (i), and is convergent by Convergent Sequence in a Metric Space (ii).

As $(x_{k})$ was an arbitrary Cauchy sequence in the metric space $(C,d_{w})$ of (i), $(C,d_{w})$ is complete by Complete Metric Space (i).