Are there any functions, f:U⊂Rn→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 D2f, then D2f is symmetric almost everywhere.

**Answer**

I can give a proof of the following statement.

Let U⊆R2 be open, and f:U→R be such that fxx, fxy, fyx and fyy are well defined on some Lebesgue measurable A⊆U. Then, fxy=fyx 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

fxy=fyx=limh→01h2(f(x+h,y+h)+f(x,y)−f(x+h,y)−f(x,y+h)) (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∈S:|g(h)(x)−g(x)|>ϵ} tends to zero as h→0, for each ϵ>0 and S⊆A of finite measure).

First, there are some technical issues regarding measurability. However, as fx and fy 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 ϵ>0. Then, for any δ>0, let Aδ be the set of (x,y)∈A such that

- |fyy(x+h,y)−fyy(x,y)|≤ϵ for all |h|≤δ with (x+h,y)∈A.
- |fy(x+h,y)−fy(x,y)−fyx(x,y)h|≤ϵ|h| for all |h|≤δ with (x+h,y)∈A.
- |f(x,y+h)−f(x,y)−fy(x,y)h−12fyy(x,y)h2|≤ϵh2 for all |h|≤δ with (x,y+h)∈A.

This is Lebesgue measurable and existence and continuity of the partial derivatives restricted to A implies that Aδ increases to A as δ decreases to zero. By monotone convergence, the measure of A∖Aδ decreases to zero.

Now, choose nonzero |h|≤δ. If (x,y), (x+h,y), (x,y+h), (x+h,y+h) are all in Aδ then,

f(x+h,y+h)−f(x+h,y)−fy(x+h,y)h−12fyy(x+h,y)h2

−f(x,y+h)+f(x,y)+fy(x,y)h+12fyy(x,y)h2

12fyy(x+h,y)h2−12fyy(x,y)h2

fy(x+h,y)h−fy(x,y)h−fyx(x,y)h2

are all bounded by ϵh2. Adding them together gives

|f(x+h,y+h)+f(x,y)−f(x+h,y)−f(x,y+h)−fyx(x,y)h2|≤4ϵh2. (2)

Now, choose a sequence of nonzero real numbers hn→0. It is standard that, for any integrable g:R2→R then g(x+hn,y), g(x,y+hn) and g(x+hn,y+hn) all tend to g(x,y) in L1 (this is easy for continuous functions of compact support, and extends to all integrable functions as these are dense in L1). Applying this where g is the indicator of Aδ shows that the set of (x,y)∈Aδ for which one of (x+hn,y), (x,y+hn) or (x+hn,y+hn) is not in Aδ has measure decreasing to zero. So, for |h| chosen arbitrarily small, inequality (2) applies everywhere on Aδ outside of a set of arbitrarily small measure. Letting δ decrease to zero, (2) applies everywhere on A outside of a set of arbitrarily small measure, for small |h|. As ϵ>0 is arbitrary, this is equivalent to the limit in (1) holding in measure and equalling fyx almost everywhere on A. Finally, exchanging x and y in the above argument shows that the limit in (1) is also equal to fxy.

**Attribution***Source : Link , Question Author : AppliedSide , Answer Author : George Lowther*