Let $f\_n: \mathbb R^2 \rightarrow \mathbb R$ be a family of probability distributions with the property that they vanish on the diagonal $f\_n(x,x)=0.$


I would like to know: Can we show that a function like this can never converge to a standard Gaussian $f(x,y) = \frac{1}{2\pi} e^{- \frac{\vert x \vert^2+ \vert y \vert^2}{2}}?$


Of course, one has to measure non-convergence in a norm that "sees" the diagonal. Since the Fourier transform might be useful, I was thinking about showing 


$$\Vert \sqrt{f\_n}-\sqrt{f} \Vert\_{H^1} > \varepsilon$$


for $\varepsilon>0$ independent of $f\_n$ where $H^1$ is the Sobolev space. I take square roots in order to give $f$ and $f\_n$ unit mass in the $L^2$ sense. 


EDIT: I assume it to be true, as $H^1$ decomposes into the direct sum $H^1\_0$ and the harmonic functions on the zero set (which is in our case the diagonal). But I am wondering whether there is a very direct way of showing this.



Non-convergence to a Gaussian
Your conjecture is true.


Indeed, let $g\_n:=\sqrt{f\_n}$ and $g:=\sqrt{f}$. Let 
$$v:=\|g\|,
$$
where $\|h\|:=\|\,h|\_J\,\|\_{L^2(J)}$ for $h\in L^2(\mathbb R^2)$, $J:=I^2$, $I:=[-u,u]$, and $u\in(0,1/20)$ is small enough so that 
$$v>u/10;
$$
such a number $u$ exists, because $g(0,0)^2=1/(2\pi)>1/400$ and $g$ is continuous. For instance, one may take $u=1/21$, and then 
$$v>0.037[>u/10].$$
(The bounds below are numerically very loose, so that the above lower bound on $v$ is easy to significantly improve.)


One of the following two cases must occur. 


*Case 1: $\|g\_n\|\le\|g\|/2$*. Then 
$$\|\sqrt{f\_n}-\sqrt f\,\|\_{H^1}=\|g\_n-g\|\_{H^1}\ge\|g\_n-g\|\ge\|g\|/2=v/2. 
$$
So, Case 1 is good.


*Case 2: $\|g\_n\|>v/2$*. In this case, use the condition $g\_n(x,x)\equiv0$ to note that for all $x$ and $y$ in $I$ we have 
$g\_n(x,y)=\int\_x^y(D\_2g\_n)(x,z)\,dz$, where $D\_2$ is the partial derivative wrt the second argument and $\int\_x^y:=-\int\_y^x$ if $y<x$, whence 
$$g\_n(x,y)^2\le\Big(\int\_I|(D\_2g\_n)(x,z)|\,dz\Big)^2
\le\int\_I(D\_2g\_n)(x,z)^2\,dz. 
$$
So, 
$$\frac{v^2}4<\|g\_n\|^2=\int\_{I^2}g\_n^2
\le\int\_{I^3}dx\,dy\,dz\,(D\_2g\_n)(x,z)^2\le\|D\_2g\_n\|^2, 
$$
so that 
$$\|D\_2g\_n\|>v/2.
$$
On the other hand,
$$\|D\_2g\|^2\le\frac1{2\pi}\,\int\_{I^2}y^2\,dx\,dy<u^4/4<(u/40)^2<(v/4)^2,
$$
whence $\|D\_2g\|<v/4$. So,
$$\|\sqrt{f\_n}-\sqrt f\,\|\_{H^1}=\|g\_n-g\|\_{H^1}\ge\|D\_2g\_n-D\_2g\|\ge v/2-v/4=v/4. 
$$
So, Case 2 is good as well. $\Box$


