 Solution

The inequalities implyThus the two metrics induce the same open sets and are equivalent.

The converse fails. On $\mathbb{R}$, letThe metrics are equivalent because $\arctan :\mathbb{R}\to (-\pi /2,\pi /2)$ is a homeomorphism, butso no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take $X=\mathbb{N}$, $Y=\mathbb{R}$, and use the two metrics above. DefineThenso $f_n\to f$ uniformly for $d_2$, whereasso convergence is not uniform for $d_1$. This is the equivalent codomain metrics need not preserve uniform convergence phenomenon.