Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.