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 $f(x,y) = \frac{1}{2\pi} e^{- \frac{\Vert x \Vert^2}{2}}?$

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.

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.

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 $f(x,y) = \frac{1}{2\pi} e^{- \frac{\Vert x \Vert^2}{2}}?$

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.

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 $f(x,y) = \frac{1}{2\pi} e^{- \frac{\Vert x \Vert^2}{2}}?$

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.