Let $X,Y$ be compact metric spaces and consider $f:X\times Y\rightarrow X$ a separately continuous function.

I am wondering if there could be some additional conditions on $f$ (for example $f(\cdot,y):X\rightarrow X$ being surjective or injective for every $y\in Y$) which would grant the joint continuity of $f:X\times Y\rightarrow X$.

The strongest result I have found is contained in this article by Namioka and states that in this case there is a dense subset $A\subset Y$ such that $f:X\times A\rightarrow X$ is joint continuous, but this is not what I am looking for.

  • $\begingroup$ Since $X$ and $Y$ are compact, a necessary ans sufficient (and a bit trivial) condition is that all $f(\cdot,y)$ are continuous, and all $f(x,\cdot)$ are equicontinuous (or viceversa). $\endgroup$ – Pietro Majer Feb 1 '18 at 18:36
  • $\begingroup$ Maybe you have some algebraic conditions on $f$? Like $X,Y$ are compact topological groups and $f(\cdot,y)$ is a homomorphism for every $y$. Then you can try to derive the joint continuity from the existence of many continuity points (given by the Namioka Theorem)? $\endgroup$ – Taras Banakh Feb 2 '18 at 7:15
  • $\begingroup$ Unfortunately, in my setting there are no additional algebraic conditions $\endgroup$ – user493456 Feb 2 '18 at 11:19

Young [1] proved that for $X=\mathbb R$ (or $X=[0,1]$) the monotonicity of $f(\cdot,y)$ implies the joint continuity.

[1] W. Young, A note on monotone functions, The Quarterly Journal of Pure and Applied Mathematics (Oxford Ser.) 41 (1910), 79–87.


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