Let $(\mathcal{C},d_\infty)$ be a compact metric space and let $C\in\mathcal{C}$ be equicontinuous and defined as $C_A(x,y)=(xy)^{A\left(\frac{\ln(x)}{\ln(xy)}\right)}$ with $x,y\in(0,1)$ and let $A:[0,1]\to[0,1]$ be a convex function and let $\Phi:(\mathcal{A},\lVert \cdot \rVert_\infty)\to(\mathcal{C},d_\infty)$ with $\Phi(A)=C_A$.
I have to show that: $$ A_n \to A \text{ uniform } \iff C_n\to C \text{ uniform } $$
"$\Rightarrow$" i think this direction is clear for me $$ \lim_{n\to\infty} C_{A_n}(x,y)=\lim_{n\to\infty}(xy)^{A_n\left(\frac{\ln(x)}{\ln(xy)}\right)}=(xy)^{A\left(\frac{\ln(x)}{\ln(xy)}\right)}=C_A(x,y) $$ and because the $C's$ are equicontinuous and $\mathcal{C}$ is compact, pointwise convergence equals uniform convergence (Corollar of Arzelà-Ascoli). (can someone tell me if this is right?)
"$\Leftarrow$" i have problems proving this direction. I think i have to use $\Phi$ because it's in the introduction of the problem. But i don't know which properties i could use. I only see that $\Phi$ has to be continuous, since $$ \lim_{n\to\infty}\Phi (A_n)=\lim_{n\to\infty} C_{A_n}=\lim_{n\to\infty}(xy)^{A_n\left(\frac{\ln(x)}{\ln(xy)}\right)}=(xy)^{A\left(\frac{\ln(x)}{\ln(xy)}\right)} =C_A=\Phi(A). $$ Having this how can i show that $A_n\to A$ uniform?
If $f_n$ converges uniformly to $f$, then for any uniformly continuous function (say $g$), $g(f_n)$ converges to $g(f)$ uniformly. (why ?)
Also $f_n(g)$ converges uniformly to $f(g)$ for any continue function $g$. (why?)
Therefore, $ln(C_n) = A_n(\frac{ln(x)}{ln(xy)}) ln(xy)$ converges uniformly to $ln(C)$.
Now use product of uniformly continuously functions is uniformly countinous, to show $A_n(\frac{ln(x)}{ln(xy)}) \ ln(xy)$ converges uniformly to $\frac{ln(C)}{ln(xy} = A(\frac{ln(x)}{ln(xy)})$