Let $X \subset \mathbb R^n$, $f:X\to\mathbb R^m$, $x_0\in X$

Assumption: All partial derivatives of f at $x_0$ exist and are continuous

$\Rightarrow$ f is differentiable at $x_0$.

$\Rightarrow D_vf(x_0)=\nabla f(x_0)\cdot v$ (assuming $m=1$ for simplicity)

Which means that all directional derivatives of f at $x_0$ can be expressed as linear combination of the (continuous) partial derivatives of f at $x_0$

Therefore these directional derivatives also have to be continuous. (*)

Is the conclusion correct? (Or why not?) And if yes, is my proof correct? (Or why not?)

(*) I'm implicitly assuming that differentiability at $x_0$ implies differentiability at all points around $x_0$ if they are close enough. Only with this assumption I can conclude that the partial derivatives are defined around $x_0$ and therefore ask if they are continuous around $x_0$ or not.

I hope you can follow my thoughts. Else just ask for clarifications, it's my first question. Thank you for your help :)

  • $\begingroup$ I believe continuity of partial derivatives implies existence of the derivative and hence all directional derivatives. You can see Spivak's 'Calculus on Manifolds' theorem 2.8 $\endgroup$ – NL1992 Apr 11 at 19:50
  • $\begingroup$ Yes, I understand that: continuous partial derivatives => differentiability <=> differential exists => directional derivatives exist. But are the directional derivatives continuous? $\endgroup$ – Stefan Apr 11 at 19:52
  • $\begingroup$ If you assume f has continuous partial derivatives at A open, then it has a continuous differential in A, hence the directional derivatives are also continuous. $\endgroup$ – NL1992 Apr 11 at 20:08
  • $\begingroup$ In your proof, to be safe, I would've used the fact f is continuously differentiable. $\endgroup$ – NL1992 Apr 11 at 20:10
  • $\begingroup$ You need to tell us more about $X.$ $\endgroup$ – zhw. Apr 11 at 20:53

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