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?

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    $\begingroup$ How about $f(x,y)=\cases{0&\text{if }x=0\vee y=0\\1&\text{otherwise}}$ $\endgroup$ Dec 2, 2011 at 9:51
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    $\begingroup$ 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/… $\endgroup$ Dec 2, 2011 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, 2011 at 11:27

2 Answers 2


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.


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, 2011 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, 2011 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, 2011 at 11:22

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