0

Wikipedia defines differentiability of a multivariate function as follows:

A function of several real variables $f: \Bbb{R}^m \rightarrow \Bbb{R}^n$ is said to be differentiable at a point $x_0$ if there exists a linear map $J: \Bbb{R}^m \rightarrow \Bbb{R}^n$ such that

$\lim_{h \rightarrow 0}{\frac{\|f(x_0 + h) - f(x_0) - J(h)\|_{\Bbb{R}^n}}{\|h\|_{\Bbb{R}^m}}} = 0$

They don't mention it outright, but I assume that there are no restrictions implied on how $h$ approaches 0 (i.e. from what direction, or indeed, whether it approaches zero along a straight path or not).

Later, they claim:

If all the partial derivatives of a function exist in a neighborhood of a point $x_0$ and are continuous at the point $x_0$, then the function is differentiable at that point $x_0$.

Does anyone know of a good proof of this claim? What bothers me is, how do you extrapolate that the limit exists from any direction, when all you know is the partial derivatives in the specific basis directions?

Or, said another way, why does knowing just the $n$ partial derivatives being continuous imply that the function is locally linear?

Community
  • 1
Henry Bigelow
  • 203
  • This question is basically a duplicate: https://math.stackexchange.com/questions/1007709/proof-that-continuous-partial-derivatives-implies-differentiability – symplectomorphic Oct 16 '18 at 19:10
  • You should note that the assumption of continuous partials is stronger than the mere assumption of the existence of partials. So your intuition that we are “merely” guaranteed the limit as $h\to0$ only along the coordinate axes is not quite right: the extra regularity assumption of continuity of the partials is what ensures the limit exists. – symplectomorphic Oct 16 '18 at 19:13
  • @symplectomorphic it seems like in copper.hat 's answer, he assumes that the partials are continuous not merely at x0 but in some neighborhood around x0. is wikipedia's statement that the partials only need to be continuous at x0 wrong? – Henry Bigelow Oct 16 '18 at 19:57
  • where do you think copper.hat uses that assumption? In my cursory glance, all I see is that he uses the assumption of continuity at $a$. – symplectomorphic Oct 16 '18 at 20:01
  • More or less the same proof: https://math.stackexchange.com/questions/1026114/why-does-continuous-partial-differentiability-imply-total-differentiability – symplectomorphic Oct 16 '18 at 20:03
  • This is a very fundamental theorem which is proved in any decent multivariable calculus textbook... – Hans Lundmark Oct 16 '18 at 20:07
  • And actually you can weaken the assumptions a little bit: https://math.stackexchange.com/a/53711/1242 – Hans Lundmark Oct 16 '18 at 20:14
  • @symplectomorphic I am studying copper.hat's answer. In it, he uses the mean value theorem, using $f(a + \phi_k(h))$ and $f(a + \phi_{k-1}(h))$ points. AFAIK the mean value theorem requires $f$ to be differentiable on an interval, but this was supposed to be a proof that only assumes $f$ is "differentiable" (i.e. in the n basis directions) at a single point. – Henry Bigelow Oct 17 '18 at 22:15
  • @HansLundmark I looked at Spivak's Theorem 2.8, but I don't think it's correct, because, like copper.hat, he uses the mean value theorem, which assumes differentiability in a neighborhood - a stronger assumption than differentiability at a point. This video looks more promising: https://www.youtube.com/watch?v=BYfTnrZXz3M. I'd be interested to know your thoughts. – Henry Bigelow Oct 17 '18 at 23:42
  • @HenryBigelow: If all you know is that the partial derivatives exist at a point, then the function need not even be continuous at that point, much less differentiable! So you need to assume that the partials (or all but one) exist in a neighbourhood of the point. (But they only need to be continuous at the point. The mean value theorem only requires the derivative to exist, not to be continuous.) In the video (at 7:00) he is using an even stronger assumption, namely that the derivatives are continuous in a neighbourhood of the point, so I don't know why you think that looks more promising? – Hans Lundmark Oct 18 '18 at 07:19
  • @HenryBigelow: The passage you quoted from Wikipedia also clearly states the assumption that the partials exist in a neighbourhood, so when you say that this was supposed to be a proof assuming only that the partials exist at the point, you are either misreading Wikipedia or talking about something that you must have gotten from some other (incorrect) source. – Hans Lundmark Oct 18 '18 at 07:29

0 Answers0