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I need some help with the following problem. It says give an example of a function $f: [0, 1] \times [0, 1] \to \mathbb R$ such that for each fixed $x$, $y \mapsto f(x, y)$ is continuous and for each fixed $y$, $x \mapsto f(x, y)$ is continuous, but $f$ is not continuous. A hint would be much appreciated. I would also love to know if there is any trick to finding counterexamples, as I have always been weak with those kind of questions. Thank you!

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Do you mean not continuous at all $(x,y) \in [0,1]^2$? – copper.hat Feb 22 '13 at 6:12
up vote 1 down vote accepted

Consider$$ f(x,y) = \begin{cases} \frac{xy}{x^2+y^2}, & \text{if }(x,y)\ne (0,0) \\ 0, & \text{if }(x,y)= (0,0) \end{cases} $$ Then it is easy to see that $f(x,y)$ is separately continuous (i.e. continuous in each variable) but $f(x,y)$ is not continuous at $(0,0)$ you can see that by showing that $$\displaystyle \lim_{(x,y)\to (0,0)}f(x,y)\ne0$$ to do this consider the path along $x=y$ then the limit should equal to $1/2$ along this path, and clearly $1/2\ne 0$.

As for a tricks for finding counter examples, I'm not aware if there is any, usually finding counter examples is not an easy thing to do, to be able to find counter examples you have to see lots of examples and grow a feeling for the subject.

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thanks for the answer. ahh i should have been able to recall this example from my analysis class last year. guess i gave a lot of what i learned back to the prof. – Aden Dong Feb 22 '13 at 14:39

Take $$f(x,y)=\frac{xy}{\sqrt{x^4+y^4}},~~~(x,y)\neq(0,0) ~~~\text{and}~~~f(x,y)=0,~~~(x,y)=(0,0)$$ It can be shown that the function is not continuous at the origin.

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+1 new gravatar? – amWhy Feb 22 '13 at 13:42
Mornin. Yes, I love him so much Amy. – Babak S. Feb 22 '13 at 13:44

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