Is each mixed partial derivatives at one point independent of the order of differentiation ？ 
$\left(\textit{A.C.CLAIRAUT}\right)$
Suppose that $U$ is an open connected set in $\mathbf{R}^n,$ that $x_0\in U,$ and that $f：U\rightarrow \mathbf{R}.$Its mixed second partial derivatives $f_{ji},f_{ij} (1\leq i<j \leq n) $ exist on $U$. If $f_{ji},f_{ij}$ are continuous at  $x=x_0$, then$f_{ji}(x_0)=f_{ij}(x_0).$

By the Clairaut's theorem,the following proposition can be easily proved applying the induction method.

$\textbf{Proposition}$
$U$ is an open connected set in $\mathbf{R}^n.$ If $f：U\rightarrow \mathbf{R}$ all of whose partial derivatives up to $k$ are$\underline{\text{ defined and continuous in } U }$$(i.e.f\in C^{k}(U))$, then for any fixed $r (2\leq r\leq k),$the value $\partial_{i_1\cdots i_r}f(x)$ of the partial derivative remains the same for any permutation of the  indices $i_1\cdots i_r (1\leq i_{1}，\cdots，i_{r}\leq n).$

$\textbf{My Question:}$
Now we consider an elementary question,slightly modify the proposition'condition：replacing by " If $f：U\rightarrow \mathbf{R}$ all of whose partial derivatives up to order $k$ are$\underline{\text{defined in } U\text{ and continuous at } x_{0}\in U.}$ " Whether we also get that the value of $\partial_{i_1\cdots i_r}f(x)$ at $x=x_{0}$ is independent of the order $i_1\cdots i_r  (1\leq i_{1}，\cdots，i_{r}\leq n)$, for any fixed $r (2\leq r\leq k)$？When $k>2$,the conclusion will not holds (I think so).But I need some counterexamples to verify！
 A: The modified proposition is true.
To reduce the subscripts, fix vectors $v_1,\dots,v_k\in\mathbb R^n$ and let $\partial_i$ denote the partial derivative operator in the direction of $v_i,$ so $\partial_ig(x)=\lim_{t\to 0}\frac{g(x+tv_i)-g(x)}t.$

Theorem. Fix an open set $U,$ a point $x_0\in U,$ a function $f:U\to \mathbb R,$ and a permutation $\pi$ of $\{1,\dots,k\}.$ Assume:
  
  
*
  
*$(\partial_m \partial_{m+1}\dots\partial_kf)(x)$ exists for all $1\leq m\leq k$ and all $x\in U$
  
*$\partial_1 \partial_{2}\dots\partial_kf$ is continuous at $x_0$
  
*$(\partial_{\pi(m)}\partial_{\pi(m+1)}\dots\partial_{\pi(k)}f)(x)$ exists for all $2\leq m\leq k$ and all $x\in U$
Then $(\partial_{\pi(1)}\partial_{\pi(2)}\dots\partial_{\pi(k)}f)(x_0)$ exists and equals $(\partial_1 \partial_{2}\dots\partial_kf)(x_0).$

Define the forward quotient operators $$(D_i^tg)(x)=(g(x+tv_i)-g(x))/t.$$
I'll use the following weak form of the mean value theorem: if a function $g$ satisfies $C_1\leq\partial_i g(x+tv_i)\leq C_2$ for all $0<|t|<\delta,$ then it satisfies $C_1\leq(D_i^tg)(x)\leq C_2$ for all $0<|t|<\delta.$
Fix $\epsilon>0.$ Let $c=(\partial_1 \partial_{2}\dots\partial_kf)(x_0)$ and pick $\delta>0$ small enough that $$c-\epsilon\leq (\partial_1 \partial_{2}\dots\partial_kf)(x_0+\sum_{i=1}^kt_iv_i)\leq c+\epsilon\tag{1}$$ whenever $0<|t_1|,\dots,|t_k|< \delta.$ To prove the theorem we need to show that
$$c-\epsilon\leq (D_{\pi(1)}^{t} \partial_{\pi(2)}\dots\partial_{\pi(k)}f)(x_0)\leq c+\epsilon\tag{2}$$
whenever $0<|t|<\delta$ - this is just applying the definition of the derivative.
Applying the mean value theorem to (1) gives
$$c-\epsilon\leq (D^{t_1}_1 \partial_{2}\dots\partial_kf)(x_0+\sum_{i=2}^kt_iv_i)\leq c+\epsilon$$
whenever $0<|t_1|,\dots,|t_k|<\delta.$
Derivative operators commute with translations (as long as everything is properly defined), so we can commute the $D^{t_1}_1$ operator with the $\partial_i$ operators to give
$$c-\epsilon\leq (\partial_{2}\dots\partial_kD^{t_1}_1 f)(x_0+\sum_{i=2}^kt_iv_i)\leq c+\epsilon.$$
Applying the mean value theorem again, and continuing in this way, we eventually get
$$c-\epsilon\leq (D^{t_k}_kD^{t_{k-1}}_{k-1}\dots D^{t_1}_1 f)(x_0)\leq c+\epsilon.$$
Translations commute, so this is the the same as saying
$$c-\epsilon\leq (D^{t_{\pi(1)}}_{\pi(1)}\dots D^{t_{\pi(k)}}_{\pi(k)} f)(x_0)\leq c+\epsilon.$$
Taking the limit $t_{\pi(k)}\to 0$ gives
$$c-\epsilon\leq (D^{t_{\pi(1)}}_{\pi(1)}\dots D^{t_{\pi(k-1)}}_{\pi(k-1)}\partial_{\pi(k)} f)(x_0)\leq c+\epsilon$$
whenever $0<|t_{\pi(1)}|,\dots,|t_{\pi(k-1)}|<\delta.$
Continuing in this way, taking the limit as $t_{\pi(k-1)}\to 0$ and so on up to $t_{\pi(2)}\to 0,$ gives
$$c-\epsilon\leq (D^{t_{\pi(1)}}_{\pi(1)}\partial_{\pi(2)}\dots \partial_{\pi(k)} f)(x_0)\leq c+\epsilon$$
which gives (2) by setting $t_{\pi(1)}=t.$
