Let $f:\mathbb R^2\rightarrow \mathbb R$ s.t. $f(x,0)=x^2, f(0,y)=y^2$ then $f$ is differentiable

Question

Let $f:\mathbb R^2\rightarrow \mathbb R$ s.t. $f(x,0)=x^2, f(0,y)=y^2$

Prove or disprove: $f$ is differentiable at $(0,0)$

I think that this is a simple disprove question, but couldn't find a counterexample.

Answer 1

It is indeed false. You can take, say$$f(x_1,\ldots,x_n)=x_1^{\,2}+x_2^{\,2}+\cdots+x_n^{\,2}+\begin{cases}1&\text{ if }x_1=x_2=\cdots=x_n\ne0\\0&\text{ otherwise.}\end{cases}$$It is discontinuous at $(0,0,\ldots,0)$.

Answer 2

On a more conceptual level, without giving an explicit counterexample, it's possible to see that this is false. The given conditions only determine how $f$ looks on the coordinate axes. Within any of the four quadrants, we are completely free to choose all kinds of horrible behavior for our function. We can make it be a constant in one quadrant, but a completely different constant in the next quadrant. We can make it wildly discontinuous and whatnot. Make it look like a sine wave along one diagonal, but like a cosine wave along the other diagonal. Be creative!

Anyway, this boatload of choices also comes with ones which make the function discontinuous at the origin, which makes differentiability impossible.