I think the following is true and was hoping someone could help me think up a proof or (hopefully not) a counterexample.
Let $X$ and $Y$ be compact Hausdorff. If $f :X \times Y \to \mathbb{R}$ is separately continuous then $f$ is jointly continuous.
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$\begingroup$ Does anyone know under what additional conditions on $f$, $X,\,Y,\,Z$ does the statement hold? I would be happy with $X$ and $Y$ being cubes in $\mathbb{R}^n$, $Z=\mathbb{R}^m$ and $f$ being separately analytic (even $m=1$). $\endgroup$– SobolevJan 28, 2019 at 20:02
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1 Answer
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Unfortunately,
$$f(x,y) = \begin{cases}\qquad 0 &, x = y = 0\\ \dfrac{xy}{x^2+y^2} &, x^2+y^2 > 0\end{cases}$$
is separately continuous on $[-1,1]^2$.
answered Jun 25, 2014 at 12:46