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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    $\begingroup$ @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 (equality of iterated integrals) for continuous functions. $\endgroup$
    – t.b.
    Commented Jun 17, 2011 at 10:00
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    $\begingroup$ @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. $\endgroup$ Commented Jun 18, 2011 at 21:43
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    $\begingroup$ @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. $\endgroup$ Commented Jun 21, 2011 at 19:59
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    $\begingroup$ 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. $\endgroup$ Commented Jun 21, 2011 at 20:20
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    $\begingroup$ reposting from mathkb.com/Uwe/Forum.aspx/math/32761/… 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] $\endgroup$
    – yoyo
    Commented Jun 26, 2011 at 23:17

2 Answers 2


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 $f_{xy}$.

  • $\begingroup$ It is getting late here: Why is it that continuity on horizontal and vertical lines gives measurability? Apart from that I have no objections. $\endgroup$
    – t.b.
    Commented Jun 27, 2011 at 1:19
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    $\begingroup$ 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 $A$? $\endgroup$ Commented Jun 27, 2011 at 1:41
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    $\begingroup$ @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. $\endgroup$ Commented Jun 27, 2011 at 1:43
  • $\begingroup$ @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. $\endgroup$ Commented Jun 27, 2011 at 1:49
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    $\begingroup$ 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). $\endgroup$
    – t.b.
    Commented Jun 27, 2011 at 10:12

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.

  • $\begingroup$ 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. $\endgroup$ Commented Jun 26, 2011 at 22:10
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Jun 26, 2011 at 22:11
  • $\begingroup$ @George: I couldn't access them either. I just added a reference to the English translation of the first paper to the answer. $\endgroup$
    – t.b.
    Commented Jun 26, 2011 at 22:14
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    $\begingroup$ @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.) $\endgroup$
    – Grigory M
    Commented Jun 26, 2011 at 22:35
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    $\begingroup$ Thanks a lot for the references! And taking to time to outline the argument in the second paper. $\endgroup$ Commented Jun 29, 2011 at 23:38

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