Uniform convergence of multivariate continuous function on compact set Let $f:A\times B\to C$ be a continuous map from the product of topological spaces $A,B$ to some metric space $C$. We assume $B$ to be compact and denote $f_{a}:=f(a,\cdot):B\to C$ for $a\in A$.
Suppose that we have a convergent sequence $a_n\to a \in A$ my question is whether $f_{a_n} \to f_a$ uniformly on $B$.
I hoped that this was the case, but so far I couldn't find a proof.
There in a related case in which the convergence is not uniform. However this problem is different, because we are dealing with a continuous two-parameter map instead of just a sequence of one-parameter maps.
 A: Here is a proof that assumes that both $A$ and $B$ are topological sequential spaces and $B$ is also Hausdorff, hence continuity is equivalentent to sequential continuity and $B$ is sequentially compact:
First observe that the function $g:B\rightarrow\mathbb{R}$ defined by $$g(b)=d(f(a_n,b),f(a,b))$$
is continuous, since it is a composition of continuoues function. It follows from the compactness of $B$ that there exist $b_n\in B$ such that
$$\sup_{b\in B}g(b)=g(b_n),$$
or
$$\sup_{b\in B} d(f(a_n,b),f(a,b))=d(f(a_n,b_n),f(a,b_n)).$$
Now let's assume by contradiction that
$$\lim_{n\rightarrow \infty} \sup_{b\in B} d(f(a_n,b_n),f(a,b_n))=\lim_{n\rightarrow \infty}d(f(a_n,b_n),f(a,b_n)) \neq 0,$$
that is, there are $\varepsilon>0$ and a subsequence $n_k$ such that
$$d(f(a_{n_k},b_{n_k}),f(a,b_{n_k}))>\varepsilon.$$
Now, by the compactness of $B$, there exists a subsequence ${n_{k_j}}$ such that $b_{n_{k_j}}\rightarrow b$ for some $b\in B$, but in such case using the continuity of the metric and $f$
$$\lim_{j\rightarrow \infty}d(f(a_{n_{k_j}},b_{n_{k_j}}),f(a,b_{n_{k_j}}))=d(f(a,b),f(a,b))=0,$$
which contradicts that $d(f(a_{n_k},b_{n_k}),f(a,b_{n_k}))>\varepsilon.$
Conuterexample if $B$ is not compact:
Consider $A\times B= [0,1]\times (0,1)$, $C=\mathbb{R}$ and $f:A\times B \rightarrow C$ given by $f(x,y)=y^x$. Now observe that $a_n=1/n\rightarrow 0$, but
$$ \sup_{y\in B}|f_{a_n}(y)-f_0(y)|=\sup_{y\in (0,1)}|y^{1/n}-1|=1,$$
which proves that $f_{a_n}$ does not converges uniformly to $f_0$.
