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Given a sequence $(f_k)_{k\in\mathbb{N}}\subset C^2(\mathbb{R}^2)$ of strictly positive functions $f_k\equiv f_k(x,y)$ with $\|f_k(x,\cdot)\|_{L^1}=1$ for all $x\in\mathbb{R}$ and such that for each $x\in\mathbb{R}$ we have

$$\tag{1} \lim_{k\rightarrow\infty} f_k(x,\cdot) = \delta_x \qquad\text{ in the distributional sense},$$

with $\delta_x$ the delta distribution at $x$.

(The 'heat kernel' $f_k:=\frac{k}{4\pi}e^{-k(x-y)^2/4}$ is an example of such a sequence.)

Question: Can we infer that for infinitely many $k\in\mathbb{N}$, the mixed derivatives $\partial_y\partial_x\log(f_k)$ vanish (almost) nowhere on the diagonal $\Delta:=\{(x,x)\mid x\in\mathbb{R}\}$ (or any compact subset thereof)?

(Vague) Intuition: Let $k\in\mathbb{N}$ and $x_0\in\mathbb{R}$ be fixed, and $\delta>0$ be small. Considering the restriction of $f_k$ to the square $\mathcal{R}:= B_\delta(x_0)\times B_\delta(x_0)$ with $y$-sections $\mathcal{R}_y:= B_\delta(x_0)\times\{y\}$, we by $(1)$ find the functions $\phi_k^y := \left.f_k\right|_{\mathcal{R}_y}$ to 'bulk' increasingly at the point $(y,y)$ and 'flatten sharply' on $\mathcal{R}_y\setminus\{(y,y)\}$ as $k$ increases. Consequently, the (monotonic) transformations $\varphi_k^y:=\log(\phi^y_k)$ show a 'rapid decay below zero' on $\mathcal{R}_y\setminus\{(y,y)\}$ (as $k\rightarrow\infty$).

This suggests that (for $k$ large enough) we have $\psi_k^y:= \left.\partial_x(\varphi_k^y)\right|_{x=x_0}>0$ for $y>x_0$, and $\psi_k^y<0$ for $y<x_0$, so that $\left.\partial_y[\partial_x\log(f_k)]\right|_{(x,y)=(x_0, x_0)} = \partial_y(\left.\psi_k^y)\right|_{y=x_0} > 0$, provided that $(1)$ guarantees that $$\tag{2}\lim_{h\rightarrow 0^+}\frac{\psi_k^{x_0+h} - \psi_k^{x_0-h}}{2h}>0.$$

Do you see a way to put this intuition into a rigorous proof? (Or is it wrong altogether and the claim doesn't hold?)

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    $\begingroup$ That would be way too much to ask for: take any $\psi$ with integral $1$ supported on $(-1,1)$ with integral $1$ and put $f_n(x,y)=n\psi(nx-ny+2)$, say (you do not need to be right in the middle to converge to the delta-measure; you can approach the diagonal "from the side" instead). Perhaps you will be happy with some weaker property? $\endgroup$
    – fedja
    Sep 18, 2020 at 1:55
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    $\begingroup$ (What I would really need is that: for every (non-empty) open (wrt. the subspace topology of $\mathbb{R}^2$) subset $U$ of $\Delta$ there exists a (non-empty) open subset $U'\subseteq U$ and an $n$ such that the restriction $\left.[\partial_y\partial_x\log(f_n)]\right|_{U'}$ is nowhere zero. [In my case, the$f_n$ are fundamental solutions to the parabolic (backward/forward) Kolmogorov equations.]) $\endgroup$
    – cts12
    Sep 18, 2020 at 11:11
  • $\begingroup$ (In the above, instead of one $n=n_{U'}$ I would in fact need to have infinitely many such $n$.) $\endgroup$
    – cts12
    Sep 18, 2020 at 12:06
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    $\begingroup$ Just make $\psi$ a positive constant on $(1,3)$ then and descend a bit later staying positive, if you wish.. $\endgroup$
    – fedja
    Sep 18, 2020 at 15:11
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    $\begingroup$ Alas, no. You will, probably, need to use the equation somehow, but how exactly to do it in general (i.e., without playing with the particular (assumptions about) coefficients) is unclear to me. $\endgroup$
    – fedja
    Sep 18, 2020 at 18:58

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