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.

  • 1
    $\begingroup$ If you want to prove this directly, you need to be more precise what you mean by a norm that "sees" the diagonal. Otherwise the sup-norm has this property. I think that such a norm is not allowed. $\endgroup$ Jan 15 '20 at 11:53
  • $\begingroup$ @DieterKadelka well, I specified the $H^1$ norm in this post. For the sup norm the question is of course v easy. $\endgroup$
    – Xin Wang
    Jan 15 '20 at 11:55
  • $\begingroup$ @GeraldEdgar sorry, a typo $\endgroup$
    – Xin Wang
    Jan 15 '20 at 13:06

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. $\Box$


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