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Suppose $f: U \rightarrow R$ for some open subset $U$ of $R^2$ is continuous in each variable ie. $f(- , y)$ continuous for each fixed y, and $f(x , -)$ continuous for each fixed x.

I know the counterexample that $f = \frac{xy}{x^2 +y^2} $ for $(x,y) \neq (0,0)$ , $f = 0$ for $(x,y) = (0,0)$ is separately continuous but not continuous at the origin.

Where does the following proof that it should be continuous fail?

Suppose we try to show continuity at $(x_1 , y_1)$. Then for any $(x_2, y_2)$ in $U$, $|f(x_1 , y_1) -f(x_2 , y_2)| \leq |f(x_1 , y_1) -f(x_2 , y_1)| + |f(x_2 , y_1) -f(x_2 , y_2)|$ by the triangle inequality.

Fix $\epsilon > 0$.

Then $\exists a>0$ s.t $|x_1 - x_2| < a \implies |f(x_1 , y_1) -f(x_2 , y_1)| < \epsilon $.

Similarly $\exists b>0$ s.t $|y_1 - y_2| < b \implies |f(x_2 , y_1) -f(x_2 , y_2)| < \epsilon $.

Let $\delta = min(a,b)$, then for $|(x_1,y_1) - (x_2,y_2)|<\delta$, we have $|f(x_1,y_1) - f(x_2,y_2)|< 2\epsilon$. Done.

Is it because whilst it may work for that particular choice of $(x_2, y_2)$, there may be another choice, also within distance $\delta$ of $(x_1, y_1)$, such that $|f(x_1,y_1) - f(x_2,y_2)| > \epsilon$? If I add the condition that $f$ is Lipschitz in $y$, say, with Lipschitz constant independent of $y$, how is this sufficient for continuity?

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That proof fails because that $b$ depends on $x_2$. It would wourk if you could choose $a$ and $b$ depending on $x_1$ and on $y_1$ alone.

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    $\begingroup$ Ahh thank you, that's obvious now you say it. $\endgroup$ – Nev Nov 30 '18 at 13:52
  • $\begingroup$ @Nev If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Nov 30 '18 at 15:32

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