Co-locating slowly increasing smooth functions in two different ways

Question

This question is subsequent from my previous one.

I will write everything in detail for the sake of completeness.

Let $g_1$ and $g_2$ be smooth functions on $\mathbb{R}$, whose derivatives of all orders are polynomially bounded. They are tempered distributions on $\mathbb{R}$.

Then, the tensor product $(g_1 \otimes g_2)(x,y):=g_1(x)g_2(y)$ is clearly a tempered distribution on $\mathbb{R}^2$.

Now, let $\Delta_n$ be a sequence of Schwartz functions on $\mathbb{R}^2$ converging to $\delta(x-y)$ in the weak$^*$ topology of distributions. That is, for any Schwartz function $F$ on $\mathbb{R}^2$, we have $$ \lim\limits_{n \to \infty}\int_{\mathbb{R}^2} \Delta_n(x,y)F(x,y)dxdy = \int_{\mathbb{R}} F(x,x)dx $$

Now, I wonder if the following is true: $$ \lim\limits_{n \to \infty} \int_{\mathbb{R}^2} \Delta_n(x,y)g_1(x)g_2(y) f(x)dxdy = \int_{\mathbb{R}} g_1(x)g_2(x) f(x)dx $$ for any Schwartz function $f$ on $\mathbb{R}$?

Obviously, the co-located product $(g_1g_2)(x):=g_1(x)g_2(x)$ has all derivatives polynomially bounded and is a tempered distribution $\mathbb{R}$ and heuristically, the above formula I question looks correct.

However, the answer in my previous one says that such naive thoughts may not be correct in general. So, I would like to check in a more simplified setting.

Could anyone help me?

Answer 1

The following is a counterexample: $$\Delta_n(x,y)=\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}\,e^{-(x^2+y^2)/n} +e^{-x^2-(y-n)^2},$$ $$g_1(x)=g_2(x)=x^2,\quad f(x)=e^{-x^2/2}.$$

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Indeed, the term $\sqrt{\frac n{2\pi}}\,e^{-n(x-y)^2/2}$ in the above expression for $\Delta_n(x,y)$ is a standard weak* approximation of $\delta(x-y)$, and $e^{-(x^2+y^2)/n}\to1$ uniformly over $(x,y)$ in any given bounded set. The second summand in the expression for $\Delta_n(x,y)$ is a weak* approximation of $0$, by (say) dominated convergence.

As for your second displayed equality, one way to see that it fails to hold is to compute explicitly both integrals there, which is easy to do. In particular, the limit on the left-hand side of that equality is $\infty$.