Show that $\|\phi_x - \phi_y \|=d(x,y)$ for all $x,y \in X$.

Question

Let $(X,d)$ be a metric space.Let $B(X)$ denote the real vector space of bounded real valued funtions on $X$.For $f \in B(X)$ define $\|f\|=sup\{|f(x)|:x \in X \}$. Fix $x_0 \in X$.For each $x\in X$,let $\phi_x:X \to \Bbb{R}$ be the function $$\phi_x(y)=d(x,y)-d(x_0,y)$$ for all $y\in X$.

(a)Show that $\phi_x \in B(X)$ for all $x\in X$.

Show that $\|\phi_x - \phi_y \|=d(x,y)$ for all $x,y \in X$.

Solution: (a)$|\phi_x(y)|=|d(x,y)-d(x_0,y)| \le d(x,x_0)$ by reverse triangle inequality.

(b)$|\phi_x(t)-\phi_y(t)|=|d(x,t)-d(y,t)| \le d(x,y)$ for all $t \in X$.Hence $\| \phi_x - \phi_y \| \le d(x,y)$.How do I show the other direction?

Answer 1

$\phi_x(y)-\phi_y(y)=d(x,y)$, so $\sup_t |\phi_x(t)-\phi_y(t)| \geq d(x,y)$.

Answer 2

$$\sup_z |\phi_x(z) - \phi_y(z)| = \sup_z |d(x,z)-d(y,z)|$$ and $|d(x,z)-d(y,z)|\leq d(x,y)$ due to the triangle inequality.