I'm trying to see why this is true -- here $d$ is a distance metric: $$ \ \left| \inf_{y\in X} (d(x,y)-g(y)) - \inf_{y\in X} (d(x',y)-g(y)) \right| \leq \sup_{y\in X} \left| d(x,y)-d(x',y)\right| $$
I just don't see how this is true since we may have different optimal $y$'s for $d(x,y)-g(y)$ and $d(x',y)-g(y)$. Anyone have an idea? Thanks!
(I assume here that the two infima are finite.)
Let $\varepsilon > 0$. By definition of $\inf$, there exists $y'\in X$ such that $$ \inf_{y\in X} [d(x',y) - g(y)] \geq d(x', y') - g(y')-\varepsilon, $$ hence $$ \begin{split} \inf_{y\in X} [d(x,y) - g(y)] -\inf_{y\in X} [d(x',y) - g(y)] & \leq d(x,y') - g(y') - [d(x', y') - g(y') - \varepsilon] \\ & = d(x,y') - d(x',y') + \varepsilon \\ & \leq \sup_{y\in X} |d(x,y) - d(x',y)| + \varepsilon. \end{split} $$ Similarly, we can prove the symmetric inequality $$ \inf_{y\in X} [d(x',y) - g(y)] -\inf_{y\in X} [d(x,y) - g(y)] \leq \sup_{y\in X} |d(x,y) - d(x',y)| + \varepsilon. $$ Hence, $$ \left| \inf_{y\in X} [d(x,y) - g(y)] -\inf_{y\in X} [d(x',y) - g(y)] \right| \leq \sup_{y\in X} |d(x,y) - d(x',y)| + \varepsilon. $$