Tensor product of a slowly increasing smooth function and a tempered distribution converging to a co-located product

Question

Let $T$ be a tempered distribution on $\mathbb{R}$ and $g$ be a smooth function on $\mathbb{R}$ whose derivatives of all orders are all polynomially bounded (a.k.a. slowly increasing).

For any pair of Schwartz functions $f_1, f_2$ on $\mathbb{R}$, let us define a separately continuous bilinear functional as $$ (f_1,f_2) \to T(f_1)\int_{\mathbb{R}} f_2 g $$

Then, by the nuclear theorem, there exists a tempered distribution $\Phi$ on $\mathbb{R}^2$ such that $$ \Phi(f_1 \otimes f_2)=T(f_1)\int_{\mathbb{R}} f_2 g $$

Now, let $\Delta_n$ be a sequence of Schwartz functions on $\mathbb{R}^2$ that converges to $\delta(x-y)$ in the weak$^*$ topology. That is, $$ \int_{\mathbb{R}^2} \Delta_n(x,y)F(x,y)dxdy \to \int_{\mathbb{R}} F(x,x) dx \text{ as } n \to \infty $$ for any Schwartz function $F$ on $\mathbb{R}^2$.

Now, my question is as follows:

For any Schwartz function $f$ on $\mathbb{R}$, does the following limit hold? $$ \Phi( f \Delta_n) \to T(gf) \text{ as } n \to \infty $$ Here, we understand the arguments as $(f \Delta_n)(x,y):=f(x)\Delta_n(x,y)$ and $(fg)(x):=f(x)g(x)$.

I think this is a plausible result because everything is well-defined and the computation is heuristcally correct as well. However, I cannot find a way to prove or disprove. Could anyone help me?

Add) It is well-known that there exists a sequence $\{ T_m \}$ of Schwartz functions on $\mathbb{R}$ that converges in the weak$^*$ topology to $T$. That is, $\lim\limits_{m \to \infty} \int_{\mathbb{R}} T_m h = T(h)$ for any Schwartz function $h$ on $\mathbb{R}$. Then, at least, we have the following: $$ \lim\limits_{m \to \infty} \lim\limits_{n \to \infty} \int_{\mathbb{R}^2} T_m(x)g(y)f(y)\Delta_n(x,y)dxdy = \lim\limits_{m \to \infty}\int_{\mathbb{R}} T_m(x)f(x)g(x)dx = T(fg) $$ as $T_m(x)g(y)f(y)$ is a Schwartz function on $\mathbb{R}^2$ for each $m$ and $fg$ is a Schwartz function on $\mathbb{R}$. Then, the issue will be on interchange of the two limits.

Add 2) The notion of wave front set and results of Las Hormander seem to be needed. It looks optimistic for now, as $g$ has empty wave front set.

Answer 1

The answer is that it doesn't work, as explained below.

For $n\in\mathbb{N}=\{0,1,2,\ldots\}$, recall that the $n$-th Hermite function is given by $$ h_n(x)=\pi^{-\frac{1}{4}}2^{-\frac{n}{2}}n!^{-\frac{1}{2}} \sum_{\ell=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{(-1)^{\ell}n!}{\ell!(n-2\ell)!}(2x)^{n-2\ell}\ . $$ The quantities $a_p:=h_p(0)$ vanish for $p$ odd and give $$ a_p=\pi^{-\frac{1}{4}}(-2)^{-\frac{p}{2}} \times\frac{\sqrt{p!}}{\left(\frac{p}{2}\right)!} $$ when $p$ is even. Defining the Fourier transform without the $2\pi$'s by $$ \mathcal{F}[f](\xi)=\int_{\mathbb{R}}e^{-i\xi x}f(x)\ dx\ , $$ we have $\mathcal{F}[h_n]=\sqrt{2\pi}\ (-i)^n h_n$. From this we find that the quantities $b_p:=\int_{\mathbb{R}}h_p(x)\ dx$ vanish for $p$ odd and give $$ b_p=\sqrt{2\pi}\times \pi^{-\frac{1}{4}}2^{-\frac{p}{2}} \times\frac{\sqrt{p!}}{\left(\frac{p}{2}\right)!} $$ when $p$ is even.

It is easy to show that $b_p$, for $p$ even, is decreasing and, by Stirling's formula, has the asymptotic $b_p\sim 2^{\frac{3}{4}} p^{-\frac{1}{4}}$. Hence, the series $$ \sum_{p=0}^{\infty}a_p b_p $$ is conditionally convergent. By the Riemann series rearrangement theorem, there exists a bijection $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ such that $$ \sum_{p=0}^{\infty}a_{\sigma(p)}b_{\sigma(p)}=0\ . $$ Now define, for $n,p,q\in\mathbb{N}$, $$ c_{n,p,q}=\mathbf{1}\{p=q, p\in\sigma(\{0,1,\ldots,n\})\}\ , $$ where $\mathbf{1}\{\cdots\}$ is the indicator function of the condition between braces. Recall that $h_p(x)h_q(y)$, with $(p,q)\in\mathbb{N}^2$, is a Schauder basis for $S(\mathbb{R}^2)$ as well as $S'(\mathbb{R}^2)$. Using the notation from my previous MO answer

Can distribution theory be developed Riemann-free?

we have $$ \langle\delta(x-y),h_p(x)h_q(y)\rangle_{x,y}=\int_{\mathbb{R}}h_p(x)h_q(x)\ dx=\delta_{p,q}\ , $$ by orthogonality.

Now define $$ \Delta_n(x,y)=\sum_{(p,q)\in\mathbb{N}^2}c_{n,p,q} h_p(x)h_q(y) $$ which is in $S(\mathbb{R}^2)$. Since the $c_{n,p,q}$ are bounded, and for each $(p,q)\in\mathbb{N}^2$, $$ \lim_{n\rightarrow\infty}c_{n,p,q}=\delta_{p,q}\ , $$ we have that $\Delta_n(x,y)$ converges to $\delta(x-y)$ in $S'(\mathbb{R}^2)$.

However, $$ \int_{\mathbb{R}}\Delta_n(0,y)\ dy=\sum_{(p,q)\in\mathbb{N}^2}c_{n,p,q}a_pb_q =\sum_{p=0}^{n}a_{\sigma(p)}b_{\sigma(p)}\rightarrow 0 $$ when $n\rightarrow \infty$.

Finally, take for $T$ the delta at the origin, and let $g$ be the constant functions equal to $1$.

Then, $\Phi(f\Delta_n)=f(0)\int_{\mathbb{R}}\Delta_n(0,y)\ dy$ goes to zero, while $T(fg)=f(0)$ is nonzero in general.