Let ($E_i,d_i)$ be a metric space, $\overline d_3:=\min(1,d_3)$ and $$\overline\rho(f,g):=\sup_{x_2\in E_2}\overline d_3(f(x_2),g(x_2))\;\;\;\text{for }f,g\in E_3^{E_2}.$$
Now let $f:E_1\times E_2\to E_3$ and $$g:E_1\to E_3^{E_2}\;,\;\;\;x_1\mapsto f(x_1,\;\cdot\;).$$
I'm interested in the relations between the following claims:
- $f$ is continuous.$^1$
- $g(x_1)$ is continuous for all $x_1\in E_1$.
- $g$ is continuous.$^2$
(1.) clearly implies (2.), (3.) trivially implies (2.) and we should be able to show that (3.) implies (1.).
Now I wonder when does (1.) imply (3.). Can we show that (1.) implies (3.) when $E_1$ or $E_2$ is compact?
Let $x_1\in E_1$ and $\varepsilon>0$. Assuming $E_2$ is compact, $g(x_1)$ is uniformly continuous and hence there is a $\delta_1>0$ with $$\forall x_2,y_2\in E_2:d_2(x_2,y_2)<\delta_1\Rightarrow d_3(f(x_1,x_2),f(x_1,y_2))<\varepsilon\tag1.$$ Now we somehow need to utilize that $f$ is (jointly) continuous ...
$^1$ $E_1\times E_2$ is equipped with the product topology generated by $((x_1,y_1),(x_2,y_2))\mapsto\max(d_1(x_1,y_1),d_2(x_2,y_2))$.
$^2$ $C(E_2,E_2)$ is equipped with $\overline\rho$.
(1) implies (3) if $f$ is uniformly continuous (and this is the case if both $E_1$ and $E_2$ are compact): Suppose $f$ is uniformly continuous and let $x\in E_1$. We want to show that $g$ is continuous. Now, as $f$ is assumed to bei uniformly continuous, given $\varepsilon >0$ there exists $\delta>0$ such that $d_3(f(x_1,y_1), f(x_2,y_2))<\varepsilon$ whenever $d_1(x_1,y_1)<\delta$ and $d_2(x_2,y_2)<\delta$. In particular,
$\rho(g(x), g(x_1))=\sup_{y\in E_2}d_3 (f(x,y), f(x_1,y))\leq \varepsilon$ whenever $d_1(x,x_1)<\delta$.
This shows that $g$ is continuous.
As mentioned above, $f$ is uniformly continuous if $E_1$ and $E_2$ are compact. In general, the compactness of these spaces are necessary for the conclusion $(1)\Rightarrow (3)$ to hold (it is not hard to find counterexamples).