Let $f$ be defined as: $$ f(x,y)= \begin{cases} \frac{xy}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases} $$

What's the oscillation of $f$ at the point $(0,0)$?

Definition: oscillation of $f$ at the point $c$:

$$ O(f,c) = \inf_{c\in U}\sup_{x_1,x_2\in U}|f(x_1)-f(x_2)| $$, where $U$ is an open subset containing $c$.

$\lim_{x\rightarrow0}f(x,0) = 0 \neq 1/2 = \lim_{x \rightarrow 0}f(x,x)$

So, it's clear that $f$ isn't continuous at $(0,0)$, so $O(f,c) > 0$.

But I'm not really sure how to approach this. How to compute the values needed?


1 Answer 1


Making $y = \lambda x$ we have

$$ \frac{x y}{x^2+y^2} = \frac{\lambda}{1+\lambda^2} $$

and the oscillation is

$$ -\frac{1}{2} \le \lambda \le \frac{1}{2}$$

enter image description here


$$ \frac{\lambda}{1+\lambda^2} $$

enter image description here

  • $\begingroup$ Wait, what just happened there? How did you get those values and how can I be sure that's the oscillation? $\endgroup$
    – Collapse
    May 1, 2018 at 18:59
  • 2
    $\begingroup$ Going to the origin along the generic line $y = \lambda x$ $\endgroup$
    – Cesareo
    May 1, 2018 at 19:01
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    $\begingroup$ @CesarEo How did you bound the $\lambda$ with $-\frac{1}{2}$ and $\frac{1}{2}$? Can you argue why the oscillation can't be bigger? $\endgroup$ May 1, 2018 at 19:11
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    $\begingroup$ Included an additional plot. $\endgroup$
    – Cesareo
    May 1, 2018 at 19:46
  • $\begingroup$ @CesarEo Thank you. And what is the formal argument for the statement that the oscillation can't be bigger? $\endgroup$ May 1, 2018 at 19:48

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