Is this condition sufficient for differentiability of f?

Question

One of the partial derivatives of f:R^n --> R exists at a point a and the other n - 1 exist in an open ball centered at a and are continuous at a. I thought all partial derivatives needed to be continuous at a and exist in a neighborhood of a, but I was told this is not the case.
Thank you

Answer 1

Yes, this is, indeed, a sufficient condition for differentiability at a. I think this not very known and, in many books, maybe most, the proofs presume all partial derivatives are continuous at a. But in fact, if n-1 partial derivatives are continuous at a and exist in a neighborhood of a, then all that's required of the remaining derivative is that it exist at a. As stated, it doesn't need to be continuous at a and doesn't even need to exist in a neighborhood of a.

An excellent proof of this fact is given in Apostol book (Real Analysis). You "zig -zag" in R^n, in steps parallel to the axis, and apply the mean value theorem.

A poof of this can also be found in the google group sci.math. Some years ago, a friend of mine posted a message about this and she gave the same proof that's in Apostol book. I don't remember exactly when. But if you do a google search putting in the subject field something like condition for differentiability in R^n, you'll find her message with the proof. Her name is Amanda

Answer 2

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