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Are there any functions, $f:U\subset \mathbb{R}^n \to \mathbb{R}$, with Hessian matrix which is asymmetric on a large set (say with positive measure)?

I'm familiar with examples of functions with mixed partials not equal at a point, and I also know that if $f$ is lucky enough to have a weak second derivative $D^2f$, then $D^2 f$ is symmetric almost everywhere.

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    @Jonas: This doesn't immediately yield an answer (if at all) but it reinforces the observation that an example of this must be *very* ugly. One can show that (for $n = 2$) the existence of $f_{xy}(p)$ implies the existence of $f_{yx}(p)$ and equality $f_{xy}(p) = f_{yx}(p)$ provided $f_{xy}$ is continuous in a neighborhood of $p$. In fact, this statement is [equivalent to Fubini](http://www.cmc.edu/math/publications/aksoy/Mixed_Partials.pdf) (equality of iterated integrals) for continuous functions.2011-06-17
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    @Theo: The identity $f_{xy}=f_{yx}$ is always true *in the sense of distributions*, whether $f_{xy}$ is continuous or not. So, any counterexample will have the 'pointwise' definition of $f_{xy}$ disagreeing with the definition in the sense of distributions on a set of positive measure. That does seem hard to arrange, if its possible at all.2011-06-18
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    @George: Thank you, I have forgotten about that. Do you see any way of approaching this problem?2011-06-21
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    @Theo: I don't know whether or not it is possible. I'm torn between trying to show that$$f_{xy}=f_{yx}=\lim_{h\to0}\frac{1}{h^2}\left(f(x+h,y+h)+f(x,y)-f(x+h,y)-f(x,y+h)\right)$$almost everywhere and trying to construct a counterexample for which $f_{xy}\not=f_{yx}$ on something like a product of fat Cantor sets.2011-06-21
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    And, I'm not 100% sure what the hypotheses are. Should we assume that $f_{xx},f_{yy}$ also exist?2011-06-21
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    @George: Thanks! The first approach looks interesting because the second one is what I tried. I read the question as addressing the asymmetry of the classical Hessian $\begin{bmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{bmatrix}$ so I assumed that $f_{xx}$ and $f_{yy}$ are supposed to exist. But personally, I'd already be happy with an example for which $f_{xy} \neq f_{yx}$ on a set of positive measure regardless of the existence $f_{xx}$ and $f_{yy}$.2011-06-21
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    I'd be happy to see a result regarding $f_{xy}\neq f_{yx}$ on a large set even if $f_{xx}$ and $f_{yy}$ don't exist on that set, or an answer in the negative if $f_{xx}$ and $f_{yy}$ are required to exist. But I'll defer to @Jonas, since it's his bounty.2011-06-21
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    I'd also be happy to see what Nick would be happy to see.2011-06-22
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    @Jonas: Wow, you are really keen on getting an answer to this one! Either that or you really want to shed some points. I did think about it for a while, and the best I could think of was $f\colon\mathbb{R}^2\to\mathbb{R}$ with $f_{xy}\not=f_{yx}$ on an uncountable set, but not of positive measure.2011-06-26
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    @George: There was a bug that made the original bounty end early. That bounty was stricken from the record (see [here](http://meta.math.stackexchange.com/questions/2408/bounty-ended-early)), so I started a new one. I appreciate your work; it looks like the bounty's going to Grigory M, but I'm curious about your construction for an uncountable set.2011-06-26
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    @Jonas: It is not too hard to construct a function $f\colon\mathbb{R}^2\to\mathbb{R}$ with $f_{xy}=0$ and $f_{yx}=1$ on $S\times\{0\}$, with $S$ a fat Cantor set. The construction is a bit wasteful though, and could maybe be improved to $S\times U$ for some $U$ of positive measure.2011-06-26
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    http://www.pnas.org/content/18/7/517.full.pdf+html states a few theorems but doesn't prove anything; it also has a couple of references.2011-06-26
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    reposting from http://www.mathkb.com/Uwe/Forum.aspx/math/32761/second-partial-derivatives-commute-Clairaut-s-Thm Tolstov, "On partial derivatives" (Russian), Izvestiya Akad. Nauk SSSR. Ser. Mat. 13 (1949), 425-446. [MR 11,167b; Zbl 38.04003] Tolstov, "On partial derivatives", American Mathematical Society Translation 1952 (1952), no. 69, 30 pages. [MR 13,926a] Tolstov, "On partial derivatives", in "Translations, Series 1, Volume 10: Functional Analysis and Measure Theory", American Mathematical Society, 1962. [MR 38 #1985]2011-06-26
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    @Jonas Thank you. (If there were silver/gold version of Altruist badge, you would certainly qualify :-)2011-07-01
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    @Jonas: I definitely agree with the bounty going to Grigory. They were very interesting references which would seem to give a fairly complete answer to the question. And I only came up with my answer after seeing his, so I had a good idea what was reasonable to prove. But putting up *another* 500 bounty! That's almost too generous. Thanks for that.2011-07-01
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    @Jonas Thanks for bringing interest to this question. I'm also very happy with the outcome. I hope it's understood that I accepted George's submission since it explicitly answered the question, although it is clear that Grigory provided crucial information.2011-07-01

2 Answers 2

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I can give a proof of the following statement.

Let $U\subseteq \mathbb{R}^2$ be open, and $f\colon U\to\mathbb{R}$ be such that $f_{xx}$, $f_{xy}$, $f_{yx}$ and $f_{yy}$ are well defined on some Lebesgue measurable $A\subseteq U$. Then, $f_{xy}=f_{yx}$ almost-everywhere on $A$.

[Note: This is after seeing Grigory's answer. The statement here is a bit stronger than statement (1) due to Tolstov in his answer. I haven't, as yet, been able to see the translation of that paper, so I'm not sure if his argument actually gives the same thing.]

In fact, we can show that $$ f_{xy}=f_{yx}=\lim_{h\to0}\frac{1}{h^2}\left(f(x+h,y+h)+f(x,y)-f(x+h,y)-f(x,y+h)\right)\ \ {\rm(1)} $$ almost everywhere on $A$, where the limit is understood in the sense of local convergence in measure (functions $g_{(h)}$ tend to a limit $g$ locally in measure if the measure of $\{x\in S\colon\vert g_{(h)}(x)-g(x)\vert > \epsilon\}$ tends to zero as $h\to0$, for each $\epsilon > 0$ and $S\subseteq A$ of finite measure).

First, there are some technical issues regarding measurability. However, as $f_x$ and $f_y$ are assumed to exist on $A$, then $f$ is continuous along the intersection of $A$ with horizontal and vertical lines, which implies that its restriction to $A$ is Lebesgue measurable. Then, all the partial derivatives must also be measurable when restricted to $A$. By Lusin's theorem, we can reduce to the case where all the partial derivatives are continuous when restricted to $A$. Also, without loss of generality, take $A$ to be bounded.

Fix an $\epsilon > 0$. Then, for any $\delta > 0$, let $A_\delta$ be the set of $(x,y)\in A$ such that

  • $\left\vert f_{yy}(x+h,y)-f_{yy}(x,y)\right\vert\le\epsilon$ for all $\vert h\vert \le\delta$ with $(x+h,y)\in A$.
  • $\left\vert f_y(x+h,y)-f_y(x,y)-f_{yx}(x,y)h\right\vert\le\epsilon\vert h\vert$ for all $\vert h\vert\le\delta$ with $(x+h,y)\in A$.
  • $\left\vert f(x,y+h)-f(x,y)-f_y(x,y)h-\frac12f_{yy}(x,y)h^2\right\vert\le\epsilon h^2$ for all $\vert h\vert\le\delta$ with $(x,y+h)\in A$.

This is Lebesgue measurable and existence and continuity of the partial derivatives restricted to $A$ implies that $A_\delta$ increases to $A$ as $\delta$ decreases to zero. By monotone convergence, the measure of $A\setminus A_\delta$ decreases to zero.

Now, choose nonzero $\vert h\vert\le\delta$. If $(x,y)$, $(x+h,y)$, $(x,y+h)$, $(x+h,y+h)$ are all in $A_\delta$ then,

$$f(x+h,y+h)-f(x+h,y)-f_y(x+h,y)h-\frac12f_{yy}(x+h,y)h^2$$ $$-f(x,y+h)+f(x,y)+f_y(x,y)h+\frac12f_{yy}(x,y)h^2$$ $$\frac12f_{yy}(x+h,y)h^2-\frac12f_{yy}(x,y)h^2$$ $$f_y(x+h,y)h-f_y(x,y)h-f_{yx}(x,y)h^2$$

are all bounded by $\epsilon h^2$. Adding them together gives $$ \left\vert f(x+h,y+h)+f(x,y)-f(x+h,y)-f(x,y+h)-f_{yx}(x,y)h^2\right\vert\le4\epsilon h^2.\ \ {\rm(2)} $$

Now, choose a sequence of nonzero real numbers $h_n\to0$. It is standard that, for any integrable $g\colon\mathbb{R}^2\to\mathbb{R}$ then $g(x+h_n,y)$, $g(x,y+h_n)$ and $g(x+h_n,y+h_n)$ all tend to $g(x,y)$ in $L^1$ (this is easy for continuous functions of compact support, and extends to all integrable functions as these are dense in $L^1$). Applying this where $g$ is the indicator of $A_\delta$ shows that the set of $(x,y)\in A_\delta$ for which one of $(x+h_n,y)$, $(x,y+h_n)$ or $(x+h_n,y+h_n)$ is not in $A_\delta$ has measure decreasing to zero. So, for $\vert h\vert$ chosen arbitrarily small, inequality (2) applies everywhere on $A_\delta$ outside of a set of arbitrarily small measure. Letting $\delta$ decrease to zero, (2) applies everywhere on $A$ outside of a set of arbitrarily small measure, for small $\vert h\vert$. As $\epsilon > 0$ is arbitrary, this is equivalent to the limit in (1) holding in measure and equalling $f_{yx}$ almost everywhere on $A$. Finally, exchanging $x$ and $y$ in the above argument shows that the limit in (1) is also equal to $f_{xy}$.

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    It is getting late here: Why is it that continuity on horizontal and vertical lines gives measurability? Apart from that I have no objections.2011-06-27
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    This is great. I also have a question. How do we know that for a.e. $(x,y)\in A$ there is a sequence of $h_n\to 0$ such that $(x,y),(x+h_n,y),(x,y+h_n),$ and $(x+h_n,y+h_n)$ are in $A$?2011-06-27
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    @Theo: It's late here too...and that argument is a bit tricky. I didn't want to make the post much longer and consist of mainly of messing about with measurability arguments. One way is to choose a dense subset $S\subseteq\mathbb{R}$ such that almost every $(x,y)\in A$ is a limit of $(s,y)$ for $s\in S$. Then, by choosing finite subsets $S_n\subseteq S$, you can write $f$ as a limit of functions $f_n$ which are piecewise-contant w.r.t $x$ (with jumps on $S_n\times\mathbb{R}$) any measurable on horizontal lines. Then, $f_n$ are easily seen to be measurable.2011-06-27
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    @Nick: You don't need to know that. It drops out of the proof. I chose any sequence of nonzero $h_n\to0$ and then showed that the set of $(x,y)\in A$ for which $(x+h_n,y)$, $(x,y+h_n)$, $(x+h_n,y+h_n)$ are all in $A$ has measure increasing to that of $A$. This does imply that these are all in $A$ infinitely often (for almost every $(x,y)\in A$). If you like, you can pass to a subsequence and use Borel-Cantelli to get that they are all eventually in $A$ for almost every $(x,y)\in A$, although that is not needed for the proof.2011-06-27
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    @Theo: "dense subset $S\subseteq\mathbb{R}$" above should read "dense *countable* subset $S\subseteq\mathbb{R}$".2011-06-27
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    Ah yes! I misread the final paragraph. This certainly answers the question I originally posed, where I tacitly assume that all of the second partials exist. It's too bad that the remaining issue is solved in an untranslated paper.2011-06-27
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    Thank you for this explanation, this is really nice. The problems on the mathnet.ru-server seem to have been resolved, that is, I could download the papers (be sure to click on the full text link at the bottom of the pages, not the pdf-version link at the top).2011-06-27
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    @Theo: I modified the argument a bit to use convergence in measure. I'm not sure if that's any clearer, but you do get expression (1) for the mixed derivatives, which is quite neat.2011-06-27
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    The previous argument was clear enough for me, but having $(1)$ explicitly is even nicer. I like this argument a lot, thanks, that's brilliant!2011-06-28
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Looks like the problem was solved by G.P.Tolstov in 1949.

  1. If $f\colon\mathbb R^2\to\mathbb R$ has all mixed derivatives of second order everywhere, then $f_{xy}=f_{yx}$ almost everywhere. Reference: G.P.Tolstov, “On partial derivatives”, Izv. Akad. Nauk SSSR Ser. Mat., 13:5 (1949), 425–446 (MR0031544).

    English translation available as Amer. Math. Soc. Translation no. 69 (1952), 30pp. (MR0047758).

  2. There exists a function $f\colon\mathbb [0,1]^2\to\mathbb R$ s.t. $f_{xy}$ and $f_{yx}$ exist everywhere but $f_{xy}-f_{yx}$ is the characteristic function of a set of positive measure (proposition I); there also exists a function $f$ as above s.t. $f_{xy}\ne f_{yx}$ almost everywhere (proposition II). Reference: G.P.Tolstov, “On the mixed second derivative”, Mat. Sb. (N.S.), 24(66):1 (1949), 27–51 (MR0029971).


(Very rough) summary of the proof of proposition I from the second paper.

  1. Choose some variant of (thick) Cantor set $P=\bigcap\limits_n P_n$ (each $P_n$ is a union of $2^n$ intervals).

  2. Let $f_n$ be a sequence of continuous PL-functions s.t. $f'_n|_{P_n}=1$ and $f'_n\le 0$ on $I\setminus P_n$.

  3. Choose some $C^1$-smoothing $\phi_n$ of $f_n$ s.t. $\phi'_n|_P=1$, $|\phi_n|<2^{1-n}$ (+some other bounds from page 31).

  4. Define $\psi_0=x-\phi_0$, $\psi_n=\phi_n-\phi_{n-1}$. Note that $\psi'_n|_P=0$ (+some other bounds (11) from page 32).

  5. Define $F(x,y)=\sum \phi_n(x)\psi_n(y)$.

  6. $F_x=\sum\phi'(x)\phi(y)$ (since convergence is uniform — which relies on delicate choice of smoothing, AFAICS). So $F_x|_{I\times P}=\sum\psi(y)=y$ and $F_{xy}|_{P\times P}=1$.

  7. $F_y=\sum\phi(x)\phi'(y)$ (-//-). So $F_y|_{P\times I}=0$ and $F_{yx}|_{P\times P}=0$.

  8. $F_{xy}$ and $F_{yx}$ exist on the whole $I^2$... for some reason.

Something like that.

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    This sounds very interesting. I can't see the papers from those links, and I'd love to see these constructions/proofs or a brief sketch in English if anyone can see them.2011-06-26
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    That's great! Thank you. I added a link to the English translation since deciphering Russian is extremely time-consuming for me. I hope you don't mind. I couldn't locate a translation of the second article.2011-06-26
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    @George: I couldn't access them either. I just added a reference to the English translation of the first paper to the answer.2011-06-26
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    @Theo Of course, thank you. AFAIK there is no translation of the second one, I'm afraid. I shall try to read it and post some sketch later. (And, indeed, there is some technical problem with mathnet.ru at the moment. If it's not resolved till tomorrow, I'll add alternative link.)2011-06-26
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    Now I have an idea what to aim for, I think I can show that if all the second order derivatives exist on a set, then $f_{xy}=f_{yx}$ almost everywhere on that set. I don't know how close this is to Tolstov's proof of (1) above, as I can't see that paper.2011-06-26
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    (Re: summary of the construction) I'm not sure which details are important, but just to have something to begin with.2011-06-27
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    Thanks a lot for the references! And taking to time to outline the argument in the second paper.2011-06-29