Given any metric space $(X,d)$ there is an embedding from $X$ to $C_b(X,\mathbb R)$, which is the set of bounded continuous functions from X to $\mathbb R$, with the $\ell_{\infty}$ norm. Given an $x_0\in X$ let a function $G$ be defined where $x\in X$ goes to the function $f_x$ defined as $f_x(y)= d(y,x)-d(y,x_0)$. Showing that $G$ is a function and that for all $x$, $G(x)=f_x$ is a continuous bounded function from $X$ to $\mathbb R$ is easy. What I don't understand is how $G$ is an isometric embeding. My approach was given two $f_x$ and $f_z$ their distance is clearly less or equal to $d(x,z)$. To show they are greater or equal I note that $d(f_x, f_z)$ (defined by the $\ell_\infty$ norm) must be greater than $d(f_x(x),f_z(x))$ (this is in the metric $d$ of $X$) which in turn is equal to $d(x,z)$, Is this correct? Am I missing something?
Proof that a metric space $X$ isometrically embeds to the set of bounded continuous functions from $X$ to $\mathbb R$
functionsisometrymetric-spaces
Related Solutions
$(i)$ Take a Cauchy sequence $(f_{n})_{n=1}^{\infty}\subseteq C_{b}(X;\mathbb{R})$ in the sup-norm $\|\cdot\|_{\infty}$ and $\varepsilon>0$. Hence there exists $n_{\varepsilon}\in\mathbb{N}$ so that \begin{equation*} |f_{n}(x)-f_{m}(x)|\leq \|f_{n}-f_{m}\|_{\infty}<\frac{\varepsilon}{3} \end{equation*} for all $n,m\geq n_{\varepsilon}$, which shows that $(f_{n}(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$ for every $x\in X$. Since $\mathbb{R}$ is complete, for every $x\in X$ there exists $f(x)\in \mathbb{R}$ so that $f_{n}(x)\to f(x)$. Since limits are unique in metric spaces, you may define a function $f:X\to\mathbb{R}$ so that $x\mapsto f(x)$. Now $f$ is the point-wise limit of $f_{n}$. We show that $\|f_{n}-f \|_{\infty}\to 0$ and $f\in C_{b}(X;\mathbb{R})$. Fix $x\in X$. Now there exists $n_{0}\in \mathbb{N}$ so that $|f_{n}(x)-f(x)|<\frac{\varepsilon}{3}$ for all $n\geq n_{0}$.. Thus for every $n\geq n_{1}:=\max\{n_{\varepsilon},n_{0}\}$ we have \begin{equation*} |f_{n}(x)-f(x)|\leq |f_{n}(x)-f_{n_{0}}(x)|+|f_{n_{0}}(x)-f(x)|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\frac{2\varepsilon}{3}. \end{equation*} Hence by taking supremum over all $x\in X$, we have $\|f_{n}-f\|_{\infty}\leq \frac{2\varepsilon}{3}<\varepsilon$ for all $n\geq n_{1}$. Hence $\|f_{n}-f\|_{\infty}\to 0$. You may now either conclude that $f\in C_{b}(X;\mathbb{R})$ since $f$ is the uniform limit of continuous bounded functions, or prove it as follows. Let $x\in X$ and everything else as above remain fixed. Since $f_{n_{1}}$ is continuous, there exists $\delta>0$ so that $f_{n_{1}}B(x,\delta)\subseteq B(f_{n_{1}}(x),\frac{\varepsilon}{3})$. Now for all $y\in B(x,\delta)$ we have \begin{align*} |f(x)-f(y)| &\leq |f(x)-f_{n_{1}}(x)|+|f_{n_{1}}(x)-f_{n_{1}}(y)|+|f_{n_{1}}(y)-f(y)| \\ &\leq 2\|f_{n_{1}}-f\|_{\infty}+|f_{n_{1}}(x)-f_{n_{1}}(y)| \\ &<3\frac{\varepsilon}{3}=\varepsilon, \end{align*} and thus $fB(x,\delta)\subseteq B(f(x),\varepsilon)$. Hence $f$ is continuous. Finally, since \begin{equation*} \|f\|_{\infty}\leq \|f-f_{n_{1}}\|_{\infty}+\|f_{n_{1}}\|_{\infty}<\frac{\varepsilon}{3}+\|f_{n_{1}}\|_{\infty}<\infty, \end{equation*} then $f$ is bounded. Hence $f\in C_{b}(X;\mathbb{R})$, and since $\|f_{n}-f\|_{\infty}\to 0$, we have proven that $C_{b}(X;\mathbb{R})$ is complete.
$(ii)$ Note that $|O(x)(y)|=|d(y,x)-d(y,x_{0})|\leq d(x,x_{0})$ for all $y\in X$ by the reverse triangle-inequality. Hence $\|O(x)\|_{\infty}\leq d(x,x_{0})$, which shows that $O(x)$ is bounded for every $x\in X$. The functions $y\mapsto d(y,x)$ and $y\mapsto -d(y,x_{0})$ are continuous, so each $O(x)$ is continuous as a sum of two continuous functions. Hence $O(x)\in C_{b}(X;\mathbb{R})$ for every $x\in X$.
$(iii)$ Note at first, that again by the reverse triangle-inequality it follows that for all $x,y\in X$: \begin{align*} \|O(x)-O(y)\|_{\infty}&= \sup_{z\in X}\|O(x)(z)-O(y)(z)\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,x_{0})-d(z,y)+d(z,x_{0})\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,y)\|_{\infty} \\ &\leq d(x,y). \end{align*} On the other hand, \begin{align*} 0\leq d(x,y)&=d(x,y)-d(x,x_{0})+d(x,x_{0})-d(x,x) \\ &=O(y)(x)-O(x)(x) \\ &= |O(y)(x)-O(x)(x)| \\ &\leq \|O(y)-O(x)\|_{\infty}. \end{align*} Hence $\|O(y)-O(x)\|_{\infty}=d(x,y)$ for all $x,y\in X$, which shows that $x\mapsto O(x)$ is an isometric embedding $X\to C_{b}(X;\mathbb{R})$.
Let $\|\mu\|$ denote the total variation norm of $\mu$.
$T$ is well defined because $\|\sum \alpha_n \delta_{x_n}\| \leq \sum |\alpha_n|=\|(\alpha_n)\|$ since $\|\delta_{x_n}\|=1$. [In any Banach space $\sum \|y_n|| <\infty$ implies that the series $\sum y_n$ converges].
Also $\|\sum \alpha_n \delta_{x_n}\|\geq \sum_k | (\sum_n\alpha_n \delta_{x_n}) (E_k)|$ where $E_k=\{x_k\}$ so $\|T(\alpha_n)\|=\sum |\alpha _n|=\|(\alpha_n)\|$ and $T$ is an isometry.
The fact that $\ell^{1}$ is isometrically isomorphic to a subspace of $C_b(X)$ implies that $C_b(X)$ is not reflexive. Hence $C_b(X)$ is not reflexive whenever $X$ is an infinite set. If $X$ is finite it is clear that $C_b(X)$ is reflexive.
Best Answer
As in the comment by @PaulSinclair we have $d(f_x,f_z)=\sup_{t\in X}|d(x,t)-d(z,t)|.$
The "subtraction version of the $\triangle$ inequality" is $|d(x,t)-d(y,t)|\le d(x,z)\; (See \;Footnote).$ Therefore $$d(f_x,f_z)\le d(x,z).$$
For any $t\in X$ we have $d(f_x,f_z)\ge |d(x,t)-d(z,t)|.$ In particular when $t=z$ we have $$d(f_x,f_z)\ge |d(x,z)-d(z,z)|=d(x,z).$$
Therefore $d(f_x,f_z)=d(x,z).$
$Footnote.$ $d(x,t)-d(z,t)\le d(x,z)$ because $d(x,t)\le d(x,z)+d(z,t).$ And $d(z,t)-d(x,t)\le d(x,z)$ because $d(z,t)\le d(z,x)+d(x,t)=d(x,z)+d(x,t).$ Therefore $|d(x,t)-d(z,t)|\le d(x,z).$