I asked this question on MSE here. One person gave an answer but then he deleted it because my version of Clairaut-Schwarz theorem is stronger than his. I meant my version only requires the continuity of **one** mixed partial derivative while his may requires the continuity of **all** mixed partial derivatives.

It seems that this question will receive no answer in MSE, so I have no choice but to post it on *mathoverflow.net*.

I usually encounter Clairaut-Schwarz theorem where the mixed partial derivatives are of order $2$, i.e.

$\textbf{Clairaut-Schwarz Theorem:}$ Let $X$ be open in $\mathbb R^n$, $f:X \to F$, and $i, j \in\{1,\ldots,n\}$. Suppose that $\partial_j \partial_i f$ is continuous at $a$ and that $\partial_j f$ exists in a neighborhood of $a$. Then $\partial_i \partial_j f (a)$ exists and $$\partial_i \partial_j f (a) = \partial_j \partial_i f (a)$$

I would like to ask if Clairaut-Schwarz theorem holds in case the mixed partial derivatives are of arbitrary order $m$, i.e.

Let $X$ be open in $\mathbb R^n$, $f:X \to F$, and $m \in \mathbb N$. Suppose $j_1, j_2, \ldots, j_m \in\{1,\ldots,n\}$ and $\sigma$ is a permutation of $\{1, \ldots, m\}$. If $\partial_{j_1} \partial_{j_2} \cdots \partial_{j_m} f$ is continuous at $a$ and $\partial_{j_{\sigma(2)}} \cdots \partial_{j_{\sigma(m)}} f$ exists in a neighborhood of $a$, then $$\partial_{j_1} \partial_{j_2} \cdots \partial_{j_m} f (a)= \partial_{j_{\sigma(1)}} \partial_{j_{\sigma(2)}} \cdots \partial_{j_{\sigma(m)}} f(a)$$

Thank you so much for your help!