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Suppose $f:[0, 1] \times [0,1] \to \mathbb R$ is a function continuous in each variable and that $\{x \mapsto f(x, y): y \in [0, 1]\}$ is equicontinuous. Prove that $f$ is continuous. So fix $(x, y) \in [0, 1] \times [0, 1]$ and let $\varepsilon > 0$. We need to show that there is a $\delta > 0$ such that $|f(s, t)-f(x, y)| < \varepsilon$ whenever $(s-x)^2+(t-y)^2 < \delta^2$. But \begin{equation*}|f(s, t)-f(x, y)| \leq |f(s, t)-f(s, y)| + |f(s, y) - f(x, y)|\end{equation*} By equicontinuity, the second term can be made arbitrarily small. However, I'm not sure if I can use the continuity of $f$ in the second variable to make the first term arbitrarily small. In fact, I'm not even sure if I'm on the right track. Can someone help me with this? Thank you!

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What is $q {}{}{}{}$? –  copper.hat Feb 22 '13 at 16:16
    
sorry. i've edited the question. –  Aden Dong Feb 22 '13 at 16:18

1 Answer 1

up vote 1 down vote accepted

Hint: You will be better off adding and subtracting $f(x,t)$ instead of $f(s,y)$ in your triangle inequality step. (It matters because the problem setup is not symmetric in the two variables.)

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oh yes, indeed. thank you for the very helpful hint. –  Aden Dong Feb 22 '13 at 16:29
    
You're welcome. –  Noah Stein Feb 22 '13 at 16:30

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