I was wondering that: Suppose that $\{P_n\}$ is a sequence of polynomials with degree $\le p$ such that for every test function $\varphi\in C^\infty_c(\mathbb{R}^d)$, the sequence of integrals $\left\{\displaystyle\int_{\mathbb{R}^d}P_n(x)\varphi(x){\rm d}x\right\}$ converges. Must the coefficients of $\{P_n\}$ of each degree converge? Here $C^\infty_c(\mathbb{R}^d)$ is the space of smooth functions over $\mathbb{R}^d$ with compact support.
I would like to find a function $\varphi$ with compact support such that, for a given multiindex $|\alpha_0|\le p$, we have
$$
\int_{\mathbb{R}^d}x^{\alpha_0}\varphi(x){\rm d}x\neq 0,\quad\int_{\mathbb{R}^d}x^\alpha\varphi(x){\rm d}x=0,\quad\forall|\alpha|\le p,\alpha\neq\alpha_0,
$$
then plugging this function into the integral shows that the $\alpha$-coefficients of $\{P_n\}$ converge. Could you please give some strategies to construct such $\varphi$ if it does exist, or is there any more elegant way to think about this question?
Edit: The following result would be applicable.
(Banach-Steinhaus theorem of distributions) Suppose that $\{T_n\}$ is a sequence of distributions over an open set $\Omega\subset\mathbb{R}^d$, and that $\langle T_n,\varphi\rangle$ converges for every $\varphi\in C^\infty_c(\Omega)$, then $\{T_n\}$ converges in $D'(\Omega)$ (the space of distributions).
So we know that $\{P_n\}$ converges to some distribution $T$. Moreover,
$$
0=\partial^{\alpha}P_n\to\partial^{\alpha}T,\quad\forall|\alpha|=p+1,
$$
so $\partial^{\alpha}T=0$ for every $|\alpha|=p+1$. It is standard (can be shown by induction on $p$) that $T$ is itself defined by a polynomial $P$ with degree $\le p$. By subtracting $P$ we can suppose that
$$
\lim_{n\to\infty}\displaystyle\int_{\mathbb{R}^d}P_n(x)\varphi(x){\rm d}x=0,\quad\forall\varphi\in C^\infty_c(\mathbb{R}^d),
$$
and I want to show that the coefficients of $\{P_n\}$ of each degree converge to $0$.
Best Answer
Define the functionals $T_n: C^\infty_c(\mathbb{R}^d) \to \mathbb{R}$ by $T_n(\varphi) =\int_{\mathbb{R}^d}x^{n}\varphi(x)\;{\rm d}x$ for $n\in\mathbb{N}$. These operators are of course linear independent. Now, once you choose $n\in\mathbb{N}$, for each $α_0 \leq n$ we have that $\cap_{α\neq α_0} ker(T_α)$ is not a subset of $ker(T_{α_0})$. Otherwise, it would be a linear combination of them (there is a theorem that states that) and this will get you the $\varphi$ you need.
The theorem: If $f_i \in V^*$ let $E = \cap ker(f_i)$ and let $f\in V^*$ such that $f(E)=0$ then $f = \sum α_if_i$.
Proof of the theorem can be found here: Intersection of kernels and linear dependence of functionals