3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ +1. To me, trying to understand how standard results can be generalized and how is mathematical research. $\endgroup$ – Daniele Tampieri Nov 16 '19 at 8:32
  • $\begingroup$ Thank you @DanieleTampieri for your encourage! Honestly, I proved the theorem in case mixed partial derivatives are of order $2$ here. $\endgroup$ – LE Anh Dung Nov 16 '19 at 8:35
  • $\begingroup$ Doesn’t this follow from the fact that weak partial derivatives commute? $\endgroup$ – Deane Yang Nov 16 '19 at 19:09
2
$\begingroup$

Assume $f:X\to F$ has the partial derivatives $\partial_m\partial_{m-1}\dots\partial_1f(x)$ at any point $x$ of the open rectangle $X:=\prod_{1\le i\le n}]b_i,c_i[$.

For $u\in \mathbb{R}^n$, let $\delta_u $ denote the finite difference $\delta_uf(x):=f(x+u)-f(x)$, defined for $x$ and $x+u\in X$, and let $(e_i)_i$ be the standard basis of $\mathbb{R}^n$.

Iterating the Mean Value Theorem, we have, say with $|t_j|<r$ and $[x_j-r,x_j+r]\subset\; ]b_j,c_j[$ for $j=1\dots m$ $$\big\|\delta_{t_me_m}\dots \delta_{t_1e_1}f(x)\big\|_F\le |t_1\dots t_m| \sup_{|x_j-y_j| \le r\atop 1\le j\le m}\big\|\partial_m \dots\partial_1f(y) \big\|_F.$$

If $p\in F$ and we apply the latter to the function $g(x):=f(x)-px_1\dots x_m $ we also get $$\big\|\delta_{t_me_m}\dots \delta_{t_1e_1}f(x)-p\,t_1\dots t_m\big\|_F\le |t_1\dots t_m| \sup_{|x_j-y_j| \le r\atop 1\le j\le m}\big\|\partial_m \dots\partial_1f(y) -p\big\|_F.$$
In particular if we take $p:=\partial_m \dots\partial_1f(a)$, and we assume that $\partial_m \dots\partial_1f$ is continuous at $x=a$, we have

$$\big\|\frac{ \delta_{t_me_m}\dots \delta_{t_1e_1}f(a)} {t_1\dots t_m} -\partial_m \dots\partial_1f(a)\big\|_F=o(1) $$ for $\|t\|\to0$. Assume now there exists the $(m-1)$-order partial derivative $\partial_{\sigma_{m-1}}\dots\partial_{\sigma_1}f(x)$ for all $x\in X$ and for a given permutation $\sigma$ of $\{1,\dots,m\}$. Since all $\delta_ {t_j e_j}$ commute, taking the limit for $ t_{\sigma_{j}}\to0$ from $j=1$ to $j=m-1$, we arrive at
$$\big\|\frac{\partial_{\sigma_{m-1}}\dots\partial_{\sigma_1}f(a+t_{\sigma_m})-\partial_{\sigma_{m-1}}\dots\partial_{\sigma_1}f(a)} {t_{\sigma_m}} -\partial_m \dots\partial_1f(a)\big\|_F=o(1) $$ for $t_{\sigma_m}\to0,$ whence the thesis: there exists $\partial_{\sigma_{m}}\dots\partial_{\sigma_1}f$ at $x=a$ and coincides with $\partial_m \dots\partial_1f(a)$.

(Note: To simplify the notation I assumed $(j_1,\dots,j_m)$ are $(1,\dots,m)$; nothing changes in general even if some of the $j_k$ coincide, because any mixed derivative can be written $\partial_{j_m}\dots\partial_{j_1}f(a)=\frac{\partial }{\partial s_m}\dots\frac{\partial }{\partial s_1}f(a+s_1e_{j_1}+\dots+s_m e_{j_m})|_{s_1=0\dots s_m=0}$).

$\endgroup$
  • $\begingroup$ After long time digesting your proof using finite difference operator, I have combined it with my previous attempt to to give my it a try. I have posted my proof here. If you don't mind, please have a look at it. Thank you so much! By the way, I'm just exposed to Real Analysis, so your proof is quite advanced for me. $\endgroup$ – LE Anh Dung Nov 17 '19 at 15:44
  • $\begingroup$ Could you please explain how to go from $g(x):=f(x)-px_1\dots x_m$ to $\delta_{t_me_m}\dots \delta_{t_1e_1}g(x) = \delta_{t_me_m}\dots \delta_{t_1e_1}f(x)-p\,t_1\dots t_m$? $\endgroup$ – LE Anh Dung Nov 18 '19 at 0:32
  • $\begingroup$ The difference operator is linear, so you only need to observe how it acts on $px_1\dots x_m$; just consider the case m=1... $\endgroup$ – Pietro Majer Nov 18 '19 at 7:45
  • $\begingroup$ IHi, are $x_1,\ldots,x_m$ in the expression $g(x):=f(x)-px_1\dots x_m$ constant, or are they dependent on $x$? it seems to me that $x_1,\ldots,x_m \in \mathbb R$. $\endgroup$ – LE Anh Dung Nov 18 '19 at 10:33
2
$\begingroup$

Suppose $\partial_{j_1} \ldots \partial_{j_m} f$ exists in a neighbourhood of $a$ and it is continuous at $a$. Let $\Delta^h_k f(x) = f(x + h e_k) - f(x)$ be the difference operator, with $e_k$ the $k$-th vector in the standard basis. By Fubini's theorem (and induction), $$ \Delta_{j_1}^{h_1} \ldots \Delta_{j_m}^{h_m} f(a) = \idotsint\limits_{[0,h_1]\times\ldots\times[0,h_m]} \partial_{j_1} \ldots \partial_{j_m} f(a + y) dy_{j_1} \ldots dy_{j_m} ,$$ where we understand that $y_k = 0$ for $k \notin \{j_1, \ldots, j_m\}$. It follows that $$ \lim_{(h_1, \ldots, h_m) \to 0} \frac{\Delta_{j_1}^{h_1} \ldots \Delta_{j_m}^{h_m} f(a)}{h_1 \ldots h_m} = \partial_{j_1} \ldots \partial_{j_m} f(a) .$$ The difference operators commute, and therefore $$ \lim_{(h_{\sigma(1)}, \ldots, h_{\sigma(m)}) \to 0} \frac{\Delta_{j_{\sigma(1)}}^{h_{\sigma(1)}} \ldots \Delta_{j_{\sigma(m)}}^{h_{\sigma(m)}} f(a)}{h_{\sigma(1)} \ldots h_{\sigma(m)}} = \partial_{j_1} \ldots \partial_{j_m} f(a) .$$ In particular, $$ \lim_{h_{\sigma(1)} \to 0} \lim_{(h_{\sigma(2)}, \ldots, h_{\sigma(m)}) \to 0} \frac{\Delta_{j_{\sigma(1)}}^{h_{\sigma(1)}} \ldots \Delta_{j_{\sigma(m)}}^{h_{\sigma(m)}} f(a)}{h_{\sigma(1)} \ldots h_{\sigma(m)}} = \partial_{j_1} \ldots \partial_{j_m} f(a) ,$$ that is, $$ \lim_{h_{\sigma(1)} \to 0} \frac{1}{h_{\sigma(1)}} \Delta_{j_{\sigma(1)}}^{h_{\sigma(1)}} \biggl(\lim_{(h_{\sigma(2)}, \ldots, h_{\sigma(m)}) \to 0} \frac{\Delta_{j_{\sigma(2)}}^{h_{\sigma(2)}} \ldots \Delta_{j_{\sigma(m)}}^{h_{\sigma(m)}} f(\cdot)}{h_{\sigma(2)} \ldots h_{\sigma(m)}}\biggr)(a) = \partial_{j_1} \ldots \partial_{j_m} f(a) .$$ By assumption, the parenthesized expression is equal to $\partial_{j_{\sigma(2)}} \ldots \partial_{j_{\sigma(m)}} f(\cdot)$ in a neighbourhood of $a$, and so finally $$ \lim_{h_{\sigma(1)} \to 0} \frac{1}{h_{\sigma(1)}} \Delta_{j_{\sigma(1)}}^{h_{\sigma(1)}} \bigl(\partial_{j_{\sigma(2)}} \ldots \partial_{j_{\sigma(m)}} f\bigr)(a) = \partial_{j_1} \ldots \partial_{j_m} f(a) ,$$ that is, $$ \partial_{j_{\sigma(1)}} \partial_{j_{\sigma(2)}} \ldots \partial_{j_{\sigma(m)}} f(a) = \partial_{j_1} \ldots \partial_{j_m} f(a) ,$$ as desired.

$\endgroup$
  • $\begingroup$ It is a bit delicate the integral of $\partial_{j_1}\dots\partial_{j_m}f$, since a priori it is just an $F$-valued measurable function, but you want to use Fubini's thm and the fundamental theorem of calculus. $\endgroup$ – Pietro Majer Nov 16 '19 at 15:54
  • $\begingroup$ For instance, even in one variable, $f(t):=t^2\sin(t^{-2})$ for $t\neq0$ and $f(0)=0$, is everywhere differentiable but $f'$ is not integrable on $[-1,1]$ $\endgroup$ – Pietro Majer Nov 16 '19 at 17:10
  • $\begingroup$ @PietroMajer: Wait, here we assume that $\partial_{j_1}\ldots\partial_{j_m}f$ is continuous at $a$, and hence bounded (and in particular integrable) in a neighbourhood of $a$. Or am I missing something? $\endgroup$ – Mateusz Kwaśnicki Nov 16 '19 at 21:11
  • $\begingroup$ ok! (maybe it was better to state that the integral formula is meant for $h_j$ small) $\endgroup$ – Pietro Majer Nov 16 '19 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.