Given $f:\mathbb{R}^n\to\mathbb{R}$ so that all partial derivatives up to second order exist. I am tempted to call that twice partially (not necessarily continuously) differentiable.

$\textbf{Does this imply for $f$ to be (totally) differentiable}$ or even continuously partially differentiable?

My intuition is, that the latter is actually the same as to ask wether a partially differentiable function is autoatically continuous (which is not true) BUT, however, i am quite confused and would appreciate clarification.

P.S. I think the problem, which troubles me is essentially the following:

Since $\frac{\partial f}{\partial x_i}$ is partially differentiable , this means that it is continuous along lines parallel to the cooordinate-axes, but I don't expect them to be continuous as functions on $\mathbb{R}^n$.


Let $h(\theta)$ be a $\pi/2$-periodic $C^{\infty}$ function which is constantly equal to zero in a neighbourhood of $0$ but such that $h(\pi/4) = 1.$ Define $f(x,y) = h(\theta)$, where $\theta$ is the angular coordinate of $(x,y) \ne (0,0)$, and $f(0,0) = 0$.

Then $f$ is not continuous at $(0,0)$, but the partial derivatives of $f$ of all orders exist everywhere.

  • $\begingroup$ No, $f'_x(0,y)=1/y$ for $y \neq 0$, while $f'_x(0,0)=0$, so $f''_{xy}(0,0)$ doesn't exist. $\endgroup$ Jun 20 '17 at 11:45
  • $\begingroup$ @HansLundmark Thanks. $\endgroup$
    – user49640
    Jun 20 '17 at 11:45
  • $\begingroup$ @HansLundmark It should be okay now. $\endgroup$
    – user49640
    Jun 20 '17 at 11:59
  • $\begingroup$ Yes, that should work! +1. $\endgroup$ Jun 20 '17 at 12:23

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