I know that in functions of more than one variable, the existence of partial derivatives does not guarantee that the function will be continuous. However, can the reverse be stated? i.e. that the continuity of the function implies that the partial derivatives exist?

I have a feeling that the answer is no, just like in functions of one variable where e.g. $\mid x\mid$ is continuous at $x=0$ but does not have a derivative.

  • 2
    $\begingroup$ You just provided an example yourself. Extrapolate to whatever dimension you want. $\endgroup$ – Git Gud Jun 13 '16 at 21:27
  • 3
    $\begingroup$ so how about $f(x,y)=|x|+|y|$? $\endgroup$ – Alex R. Jun 13 '16 at 21:32

Nope. Consider $f\colon \mathbb R^2\rightarrow \mathbb R^2$ where $f(x,y) = |x + y|$ at $(x,y) = (0.0)$. It's not hard to show by a similar argument to the one for continuity of $f(x) = |x|$ doesn't imply differentiability that the partials don't exist.

  • $\begingroup$ I think that he meant $f(x,y) = |x + y|$. $\endgroup$ – Steven Harding Jun 13 '16 at 21:57
  • $\begingroup$ @StevenHarding Yes,totally my bad,was in a hurry. Fixed it and checked the limits when I got back,should be fine now. $\endgroup$ – Mathemagician1234 Jun 14 '16 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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