Let $X$ be a metric space and $B(X)$ be the space of bounded functions $f: X \to \mathbb{C}$. Consider the embedding $J : X \to B(X)$ defined via $y \mapsto J(x)(y) = d(x, y) − d(x_0, y)$ for some fixed $x_0 \in X$. Show that this embedding is isometric. Hence $\overline{J(X)}$ is another (equivalent) completion of $X$.
There was an exercise before that where we defined the metric for the space of bounded functions $B(X)$ as follows:
$$d(f,g) = \text{sup}_{x\in X}|f(x) - g(x)| \qquad f,g \in B(X)$$
I know that a function is isometric if it preserves distances between elements $x,y \in X$. So, I tried showing that we have $d_X(y,z) = d_B(J(x)(y), J(x)(z))$. However, I am struggling with that. Could anybody help me?
I'm going to slightly change the notation. We are given a map $J \colon X \to B(X)$ defined by $$J(x)(z) = d(x,z)-d(x_0,z)$$
To show that $J$ is an isometry, we need to show that for $x, y \in X$,
$$ d(x,y) = d_{B(X)}(J(x), J(y))$$ Since $B(X)$ has the sup-metric, that reduces to $$ d(x,y) = \sup_{z \in X} | J(x)(z) - J(y)(z)|$$ That reduces to $$d(x,y) = \sup_{z\in X}|(d(x,z)-d(y,z))|$$ By the triangle inequality, for any $z \in Z$, we have $d(y,z) \leq d(y,x)+d(x,z)$, implying $d(x,y) \geq d(y,z)- d(x,z)$. By a similar application of the triangle inequality, $d(x,z) \leq d(x,y)+d(y,z)$, so $d(x,y) \geq d(x,z)-d(y,z)$. Combining these two inequalities gives $$d(x,y) \geq |d(x,z) - d(y,z)|$$ Thus $d(x,y)$ is an upper bound of $\{|d(x,z)-d(y,z)| \mid z \in X\}$.
To complete the argument, we need to show that $d(x,y)$ is the least upper bound of that set. But that follows from the fact that $|d(x,z)-d(y,z)|$ attains the value $d(x,y)$ when $z=y$.