Problem: If $d$ and $d'$ are equivalent metrics on a Polish space $Y$, then show that the metrics $\delta$ and $\delta '$ are equivalent where $$\delta(f, g)= \sup_{x \in X} d\left(f (x), g (x)\right)$$ and $\delta' $ defined analogously on $C (X, Y)$, space of all continuous functions from $X$ to $Y$ where $X$ is a compact metrizable space.
I am trying to show that $\delta (f_n, f) \to 0$ implies $\delta ' (f_n, f) \to 0$. But the former is implying only that $f_n$ tends to $f$ pointwise (not uniformly) with $\delta '$.
Please help.
Suppose not. Then there is a subsequence $\left(g_j\right)_{j\geqslant 1} =\left(f_{n_j} \right)_{j\geqslant 1} $ and $\varepsilon_0$ such that $\delta'\left(g_j,f\right)\gt 3\varepsilon_0$ for all $j$. By definition of the supremum, for each $j$, there exists $x_j\in X$ such that $$d'\left(g_j\left(x_j\right),x_j\right)\gt 2\varepsilon_0.$$ Since $X$ is compact and metrizable, we can extract a convergent subsequence from the sequence $\left(x_j\right)_{j\geqslant 1}$, denoted $\left(x_{j_l}\right)_{l\geqslant 1}$, to some $x$. Therefore, writing $h_l=g_{j_l}$, the following inequality takes place for $l$ large enough: $$d'\left(h_l\left(x_l\right),x\right)\gt\varepsilon_0.$$ But using the convergence for $\delta$, we derive that $d\left(h_l\left(x_l\right),x\right)\to 0$, which contradict the equivalence between $d$ and $d'$.