# Does existence of partial derivatives implies continuity at a point $(x_0,y_0)$?

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?

• How about $f(x,y)=\cases{0&\text{if }x=0\vee y=0\\1&\text{otherwise}}$ Dec 2, 2011 at 9:51
• 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/… Dec 2, 2011 at 10:40
• 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. Dec 2, 2011 at 11:27

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?

Atajh.

• 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. Dec 2, 2011 at 11:07
• 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). Dec 2, 2011 at 11:13
• 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 Dec 2, 2011 at 11:22