If $F: \mathbb R^2 \to \mathbb R$ and $F_x$ (partial derivative of $F$ wrt $x$) and $F_y$ exist at $(x_0,y_0)$ then the function is continuous at that point. Is this true? If not what could be a counter-example?

  • 5
    $\begingroup$ How about $f(x,y)=\cases{0&\text{if }x=0\vee y=0\\1&\text{otherwise}}$ $\endgroup$ – hmakholm left over Monica Dec 2 '11 at 9:51
  • 1
    $\begingroup$ I've tried to edit the title to be more descriptive - so that the users of the site know what the question is about without needing to view it. I hope you don't mind - if you can come up with a better title, there's still the edit button.\\ You may also notice from my edits that using nice-formatted math is easy - for more about this see e.g. here: meta.math.stackexchange.com/questions/1773/… $\endgroup$ – Martin Sleziak Dec 2 '11 at 10:40
  • $\begingroup$ Thanks Martin. I'll look at the typesetting methods. I'm new here, but I love this place! I hope to learn and contribute as much as possible. $\endgroup$ – hargun3045 Dec 2 '11 at 11:27

The reason why such a statement cannot be true is, that e.g. the partial derivative $f_x$ at $(x_0,y_0)$ contains only Information of $f$ along the slice $\{(x_0+t,y_0)\mid t\in\mathbb{R}\}$.

A more interesting question therefore is, if $f$ would be continuous at $(x_0,y_0)$ if the directional derivatives $D_vf(x_0,y_0):=\left.\frac{d}{dt}\right|_{t=0}f((x_0+tv_1,y_0+tv_2)$ in every direction $v\in\mathbb{R}^2$ exist. Can you answer this?


  • $\begingroup$ But doesn't the same example as above by David Mitra disprove this? Directional derivative exists and equals zero at (x,y) = (0,0) but function is not continuous. $\endgroup$ – hargun3045 Dec 2 '11 at 11:07
  • $\begingroup$ Right! Actually i didn't take a closer look at David's example, as i only wanted to give you a more geometric view on this problem. I'm pretty much sure that Henning Makholm had something like this in mind, when we wrote down his counterexample (without any sweat). $\endgroup$ – atajh Dec 2 '11 at 11:13
  • $\begingroup$ Yeah! I got your point. I'm new to multi-variable, and the generic case of directional derivative for all approaches to a point (not just x or y slices) is giving me a better picture. Thank you $\endgroup$ – hargun3045 Dec 2 '11 at 11:22

Not true. Look at $f(x,y)= \cases{ {xy\over x^2+y^2},&$(x,y) \ne (0,0)$\cr 0,&$(x,y)=(0,0)$}$.

Here $f_x(0,0)=0=f_y(0,0)$ (as seen by applying the definitions; note, $f(0,y)$ and $f(x,0)$ are identically $0$).

But $f$ is not continuous at $(0,0)$ since the limit of $f$ as $(0,y)$ approaches $(0,0)$ is 0, but the limit as $(k,k)$ approaches 0 of $f$ is $1/2$.

See Henning's comment for a simpler, and better, example.


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.