Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Do you mean not continuous at all $(x,y) \in [0,1]^2$? –  copper.hat Feb 22 '13 at 6:12

2 Answers 2

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.

share|improve this answer
    
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.

share|improve this answer
    
+1 new gravatar? –  amWhy Feb 22 '13 at 13:42
    
Mornin. Yes, I love him so much Amy. –  B. S. Feb 22 '13 at 13:44

Your Answer

 
discard

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.