I have a lack of understanding regarding this proof, and since the proof is not in English, I will simply write it down up to the point where I can't go further:


Assume $U \subset \Bbb R^n$ is open and $f: U \rightarrow \Bbb R$ is a function that is partial differentiable in $U$. Let all partial derivatives $D_kf$ be continiuous in $x \in U.$ Then $f$ is total differentiable in $x.$


Since $U$ is open, there is a $\delta \gt 0$ such that the open ball around $x$ with radius $\delta$ lies completely in $U.$ Let $\xi := (\xi_1, ..., \xi_n) \in \Bbb R^n$ be a vector with $||\xi|| \le \delta.$ We define the points

$$z^{(k)} := x + \sum_{i=1}^k \xi_i e_i, i = 0, 1, ..., n$$


$z^{(0)} = x$ and $z^{(n)} = x + \xi.$ Applying the mean value theorem, there must be a $\lambda \in [0, 1]$ such that

$f(z^{(k)}) - f(z^{(k-1)})$ $=$ $D_kf(y^{(k)})\xi_k$ with $y^{(k)}$ $:=$ $z^{(k-1)} + \lambda_k \xi_k e_k.$

This is where I can't go further. I don't understand where the $\xi_k$ on the right side of the last equation comes from. I also consulted another book, but it didn't explain it either. Does someone have an "expanded" version of this step?


The mean value theorem tells us that there exists a point $y^{(k)} = (y_1,\ldots, y_k)$ on the segment joining $z^{(k)}$ to $z^{(k-1)}$, i.e. $y^{(k)} = z^{(k-1)} + \lambda_k (z^{(k)} - z^{(k-1)})$, hence $y^{(k)} = z^{(k-1)} + \lambda_k \xi_k e_k$, for some $\lambda_k \in [0,1]$, such that:

$$f(z^{(k)}) - f(z^{(k-1)}) = \nabla f(y^{(k)}) (z^{(k)} - z^{(k-1)})$$


$$f(z^{(k)}) - f(z^{(k-1)}) = \nabla f(y^{(k)}) (\xi_k e_k)$$


$$\nabla f(y^{(k)})(\xi_k e_k) =D_k f(y^{(k)})\xi_k $$

  • $\begingroup$ It took me a while, but I understood it now. :-) Thank you very much! $\endgroup$ – Julian Jun 22 '16 at 8:57

A full and detailed proof, with a step by step $\epsilon$ approximation step, is given here. You're welcome.

  • $\begingroup$ Thank you very much for your answer, but I honsetly don't see how this helps me with my question above. This proof seems to work completely differently. $\endgroup$ – Julian Jun 21 '16 at 21:18

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