In Question on differentiability at a point, it is mentioned (and in Equivalent condition for differentiability on partial derivatives it is cited from Apostol) that for $f:\mathbb{R}^2\to\mathbb{R}$ to be differentiable at a point, one sufficient condition is that both partials exist at the point, and one of the partials is continuous throughout a neighborhood of the point.
I have seen counterexamples (e.g. the one given on Wikipedia) of functions which have partials defined everywhere, but fail to be differentiable at some point. But the examples I've seen aren't satisfying because both partials aren't continuous at the point in question (in fact, the partials fail to have a limit at that point at all), let alone in a neighborhood of the point.
Can someone direct me to a function that shows why the "throughout a neighborhood" condition is required for the sufficient condition mentioned in Apostol? I'm having trouble imagining a function where a partial is continuous at the point, but not throughout a neighborhood of the point. One thought I had was to use something whose cross section looked like Volterra's function (centered at a point not in the fat Cantor set), but I'm not certain that this would work, and even if it does, I wonder if there's a nicer example.
Answer
As @Nameless pointed out, that Apostol quote says something stronger: the nicer partial only needs to be defined in a neighborhood and continuous at the point. Therefore, there is no such function where both partials are defined at a point $P$ and one of them is defined in an open neighborhood of $P$ and continuous at $P$ and yet the function is somehow not differentiable at $P$.
I had a follow-up question before which has now been separated into Continuous partials at a point without being defined throughout a neighborhood and not differentiable there?.
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