I am interested in a particular instacne of the phenomena

"Partial derivatives + (A certain degree of) continuity" implies differentiablilty. My case assumes less regularity than usual:

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x_0,y_0) \in \mathbb{R}^2$.

Assume that the partial derivatives of $f$ with respect to $x$ and $y$ exist at the point $(x_0,y_0)$ and one of them exists and continuous w.r.t to the other variable (for instance the function $y \mapsto\frac{\partial{f}}{\partial x}(x_0,y)$ is continuous at the point $y_0$).

Is it true that $f$ is differentiable at $(x_0,y_0)$?

Note: It is known that if both partial derivatives exists, and one of them is continuous (as a functions of two variables) then $f$ is differentiable. (For a proof see here).

However, the proof uses:

1) The existence of $\frac{\partial{f}}{\partial x}$ on some ball around $(x_0,y_0)$

(I assume only $\frac{\partial{f}}{\partial x}$ exists on $\{x_0\} \times (y_0-\epsilon,y_0+\epsilon)$).

2) The continuity of $(x,y) \mapsto \frac{\partial{f}}{\partial x}(x,y)$ at $(0,0)$.

(I assume only continuity of $\frac{\partial{f}}{\partial x}(x_0,y)$).


1 Answer 1


This is false.

Counterexample, at the point $(x_0,y_0)=(0,0)$: $$f(x,y) = \begin{cases} x^3/y, & y\neq 0 ,\\ 0, & y=0 . \end{cases} $$ This function satisfies the hypotheses:

  • $f_y(0,0)=0$ (since $f$ is constant along the $y$ axis: $f(0,y)=0$ for all $y$), so in particular $f_y(0,0)$ exists.
  • $f_x(0,y) = 0$ for all $y$ (since the restriction of $f$ to a line $y=c$ is either identically zero or a constant times $x^3$, which has zero derivative at the origin), so the one-variable function $y \mapsto f_x(0,y)=0$ is continuous.

But $f$ isn't even continuous at the origin, hence not differentiable there.


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.