uniform convergence in individual variables implies uniform convergence in general? I have a family of functions $K_n : [0,1]^2 \to R$ that are continuous and symmetric in the sense that $K_n(s,t) = K_n(t,s)$ for all $s,t \in [0,1]$. I know that, for fixed $s$, $\sum_{n=1}^\infty K_n(s,\cdot)$ converges uniformly to some $K(s,\cdot)$; the same property holds for fixed $t$.
This means that $\sum_{n=1}^\infty K_n$ converges pointwise to $K$ on $[0,1]^2$. But does it mean that $\sum_{n=1}^\infty K_n$ must converge uniformly to $K$ on $[0,1]^2$? If so, why; if not, are there additional properties the $K_n$ must satisfy for this to hold?
 A: It does not hold in general. I cannot construct additional properties that are non-trivial for which it holds though. Before I give a counter-example lets shift the goalposts a bit to get a nicer situation.
Consider any sequence of functions $f_n$ that converge uniformly. Define $K_1:=f_1$, $K_n:= f_{n+1}-f_n$ for $n>1$, so $\sum_n^NK_n=f_{N+1}$ and the question of uniform convergence of a series is equivalent to uniform convergence of a sequence.
The condition of symmetry can also be thrown out of the window by using a rescaling $g: I\times I\to[0,\frac{1}{4}]\times[\frac{3}{4},1]$. Since $g$ is uniformly continuous $f_n \circ g$ converges uniformly to $f \circ g$. In this way you can get any function on the square.
So the situation we have is a sequence of continuous functions $f_n(t,s)$ on $I\times I$ and a function $f(t,s)$ continuous in both variables so that $\sup_s (f_n(t,s)-f(t,s) )$ and $\sup_t (f_n(t,s)-f(t,s))$ converge pointwise to zero.
The counter-example is then:
$$f_n(t,s) := \frac{nts}{1+n^2(t^4+s^4)}$$
These functions are continuous. They converge to zero pointwise and in this case they are also symmetric. So we need only prove $\sup_s f_n(t,s)$ converge to zero for any $t$. But $0≤f_n(t,s)≤\frac{nt}{1+n^2t^4}$, so $\sup_s f_n(t,s)$ does converge to zero for all $t$.
However $f_n(\frac{1}{\sqrt n},\frac{1}{\sqrt n})=\frac{1}{3}$ does not converge to zero as $n\to\infty$, so the sequence cannot converge uniformly to 0.
