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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 jointly continuous, but this is not what I am looking for.

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    $\begingroup$ 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
    Commented Feb 1, 2018 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
    Commented Feb 2, 2018 at 7:15
  • $\begingroup$ Unfortunately, in my setting there are no additional algebraic conditions $\endgroup$
    – user493456
    Commented Feb 2, 2018 at 11:19

3 Answers 3

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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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This paper is too good to pass up; it generalizes MasleniZZa's answer to general topological spaces:

Yaroslav I. Grushka, "On monotonous separately continuous functions", (English), Applied General Topology 20, No. 1, 75-79 (2019), MR3938648, Zbl 1414.54005. (Also available here in preprint form).

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It is a well-known result of Functional Analysis/Topological Vector Spaces [1] that every linear map from a finite dimensional normed space to any topological vector space $E$ is continuous (=jointly continuous).

[1]: Rudin, W. (1991). Functional analysis, mcgrawhill. Inc, New York.

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    $\begingroup$ This seems unlikely to be helpful in the setting which the original questioner was asking about $\endgroup$
    – Yemon Choi
    Commented Nov 22, 2020 at 18:18

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