Let $(X,d)$ be a metric space. Take a fixed $a \in X$. Consider for every $x \in X$ the function $f_{x}:X \rightarrow \mathbb{R}:y \mapsto f_{x}(x)=d(x,y)-d(a,y)$.
Now it is clear that $f_{x}$ is continuous and bounded for every $x \in X$. I have to show that $d_{\infty}(f_{x},f_{x'})=d(x,x')$, where $d_{\infty}$ denotes the metric $d_{\infty}(f,g)= \sup \{|f(x)-g(x)|$ where $x \in X \}$ (where $f$ and $g$ must be continuous and bounded).
So far I have: $d_{\infty}(f_{x},f_{x'})= \sup \{|f_{x}(y)-f_{x'}(y)|$ where $y \in X\}= \sup \{|(d(x,y)-d(a,y))-(d(x',y)-d(a,y))| \}=\sup \{|d(x,y)-d(x',y)| \}=...$ $=d(x,x')$. I guess I can't just skip the dots because I feel something is missing in the argument. I don't think the triangle inequality on itself is sufficient? Any help? Thank you.
Fix $x,x'$ we have
$$ d_\infty (f_x,f_x')=sup\{|d(x,y)-d(a,y)-d(x',y)+d(a,y)|y\in X\}=sup\{|d(x,y)-d(x',y)|y\in X\}$$ Now by the triangle inequality $d(x,x')\geq |d(x,y)-d(x',y)|$ (1) we conclude that $d_\infty(f_x,f_x')\leq d(x,x')$ (2) But for $y=x$ we have $d(x,x')=|d(x,x)-d(x',x)|$ (3) and so we have $d_\infty(f_x,f_x')\geq d(x,x')$ which completes the proof.
(1) by the triangle inequality $d(x,x')+d(x',y)\geq d(x,y)$ and $d(x,x')+d(x,y)\geq d(x',y)$ this two gives $d(x,x')\geq |d(x,y)-d(x',y)|$
(2) here I use the fact that if every element of the set is bounded (in our case by $d(x,x')$) then so is the supremum.
(3) If one of the elements is bigger than $d(x,x')$ then so is the supremum.