Non-convergence to a Gaussian

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.

• 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. Jan 15 '20 at 11:53
• @DieterKadelka well, I specified the $H^1$ norm in this post. For the sup norm the question is of course v easy. Jan 15 '20 at 11:55
• @GeraldEdgar sorry, a typo Jan 15 '20 at 13:06

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.)
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 $$y, whence $$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 whence $$\|D_2g\|. So, $$\|\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$$