# 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 $$f_{ji},f_{ij} (1\leq i exist on $$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！

• Suggestion: Elementary algebra, the definition of partial derivatives and the mean-value theorem applied twice will give you two points P' and P" arbitrarily close to your point P such that one mixed partial at P' equals the other mixed partial at P".You should have no trouble taking it from there. Commented Jan 16, 2020 at 6:32
• @P.Lawrence: I am sorry to say that are you sure you have understand my question?
– Nemo
Commented Jan 16, 2020 at 6:51
• Sorry for the delay in replying. My understanding is that you are asking if existence of the mixed partials everywhere in U and continuity of the mixed partials at P is enough to guarantee equality of the mixed partials at P. Commented Jan 17, 2020 at 3:31
• @P.Lawrence :Don't mention it ^__^ I considered whether the argument fails or not,when $k\geq r>2$.Actually,you said is based on $k=r=2$,which holds according to Clairaut's theorem(the method of proof as you suggested).
– Nemo
Commented Jan 22, 2020 at 10:15
• Please, avoid making several edits. Commented Feb 12, 2020 at 18:36

## 1 Answer

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 $$t_{\pi(1)}=t.$$

• I really appreciate your help！But I want to confirm whether those partial derivatives $\partial_{1},\cdots,\partial_{k}$(or those fixed vectors $v_{1},\cdots,v_{k}$) are distinctive from each other in the theorem that you referred to？If we need not such requirement , my modified proposition will be true.
– Nemo
Commented Jan 22, 2020 at 12:28
• @Nemo: they aren't required to be distinct, so you can use it for things like $f_{xxy}=f_{xyx}.$
– Dap
Commented Jan 22, 2020 at 12:36
• I see.This theorem can be used on more generalized scope than the condition which the modified propostion gives.Thanks for your elegant proof！
– Nemo
Commented Jan 22, 2020 at 12:48