This is a follow-up to Continuous partials at a point but not differentiable there?, but I'll make this question self-contained. Throughout, $f$ will denote a function $\mathbb{R}^2\to\mathbb{R}$. An answer to Equivalent condition for differentiability on partial derivatives cites a theorem implying "If $f_x$ and $f_y$ exist at $P$ and $f_x$ is defined throughout a neighborhood of $P$ and $f_x$ is continuous at $P$, then $f$ is differentiable at $P$.", which is stronger than what you find in many calculus texts.

First note that this can't possibly be weakened to something like "exists in almost all of a neighborhood and is continuous along paths within that neighborhood of all but one limiting direction". One counterexample would be $f(x,y)=\sqrt[3]{xy}$, which has partials that both exist at the origin and only blow up along some axis (with the origin deleted) so that limits of a partial along paths whose limiting direction is not the direction of the problem-axis all exist and agree. But $f$ is not differentiable at the origin.

I wonder "is the fact that the partials blow up the problem?" Or is mere non-existence of the partials arbitrarily close to the point enough to give the function a chance to be non-differentiable? More formally...

My question: Is there a function $f:\mathbb{R}^2\to\mathbb{R}$ such that

  1. $f_x$ and $f_y$ exist at $P$.
  2. $f_x$ is defined in a neighborhood of $P$ minus an open ray (or a line with $P$ deleted, if it makes things easier).
  3. $f_x$ is continuous at $P$ in the sense that for all $Q$ in the domain of $f_x$ and for all $\varepsilon>0$ there exists $\delta>0$ so that $\Vert P-Q\Vert<\delta\Rightarrow\Vert f_x(Q)-f_x(P)\Vert<\varepsilon$.
  4. $f$ is not differentiable at $P$.

It may be the case that the answer is "no, because no function satisfies 1. through 3. without $f_x$ being defined in a neighborhood of $P$ so that 4. is impossible by the theorem in Apostol's book", but even if that's the case, I'd like to know. The reason I worry about that possibility is that I know that in the 1-d case a derivative can't have jump discontinuities, and this seems like a similar issue (something reminiscent of branch cuts). Additionally, naive functions built out of $x^2\sin\left(\frac{1}{x}\right)$ tend to be differentiable.

  • $\begingroup$ Thanks for the bounty, @postnobills! $\endgroup$
    – Mark S.
    Dec 27, 2013 at 3:53
  • $\begingroup$ Post no bills, but do post rewards :) $\endgroup$ Dec 27, 2013 at 5:22

1 Answer 1


If I understood correctly, an example would be the following. Define $f\colon \mathbb{R}^2 \to \mathbb{R}$ as $$f(x,y) = \begin{cases}y, & \text{if }y \ge 0 \text{ or }x \le 0,\\-y,&\text{otherwise}\end{cases}$$ and take $P = (0,0)$. Then $f_x = 0$ outside $\{{0\}} \times (-\infty,0)$, $f_y(0,0) = 1$, but $f$ is not differentiable at $(0,0)$.

  • $\begingroup$ That...wow. That looks very simple, but it certainly is an example of the type I asked for. I guess I should have asked for continuity... $\endgroup$
    – Mark S.
    Dec 31, 2013 at 2:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .