$f^{(\ell)}(x) = P_{\ell}(x) e^{-x^2}$ and $|P_{\ell}(x)|$ is bounded by $C_{\ell}|x|^d$, where $C_{\ell}$ is some constant and $d \geq$ degree of $P_{\ell}$. Let $d$ be even for convenience (so we can divide by $2$).
We can write $e^{x^2} = 1 + x^2 + (x^2)^2/2! + \dots$ so that $e^{x^2} \geq 1 + (x^2)^{d/2}/(d/2)!$ (all terms are positive) and $e^{-x^2} \leq \frac{1}{1+x^d/(d/2)!} \implies |e^{-x^2}| \leq |\frac{1}{1+x^d/(d/2)!}|$.
So we have $\sup_{x \in \Bbb{R}}|x|^k|f^{\ell}(x)| \leq \sup C_{\ell}\left | \frac{x^d}{1 + \frac{x^{d}}{(d/2)!}} \right | \leq C_{\ell}\left | \frac{x^d}{x^d / (d/2)!} \right | \leq C_{\ell}(d/2)!$
Thus $f(x) = e^{-x^2}$ and all of its derivatives are rapidly decreasing in the Schwartz space sense and so $f$ belongs to $\mathcal{S}(\Bbb{R})$.
Proposition. Let $f\in C^\infty(\mathbb R^n)$. The following statements are equivalent.
(a) $\displaystyle\|f\|_{\alpha,\beta}:=\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|<\infty$ for all $\alpha,\beta\in\mathbb{N}^n$.
(b) $\displaystyle\sup_{x\in\mathbb{R}^n}\|x\|^k|D^\beta f(x)|<\infty$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(c) $\displaystyle\lim_{|x|\to\infty}\|x\|^kD^\beta f(x)=0$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(d) $\displaystyle\lim_{|x|\to\infty} x^\alpha D^\beta f(x)=0$ for all $\alpha,\beta\in\mathbb{N}^n$.
As a consequence we have the following
Corollary. Let $f\in C^\infty(\mathbb R^n)$. Then, $f$ is a Schwartz function if and only if $f$ is a rapidly decreasing function.
Proof of the Corollary: To say that $f$ is in the Schwartz space means that $f$ satisfies the condition $(a)$ of the proposition. To say that $f$ is rapidly decreasing means that each $D^\beta f$ tends to $0$ faster than $\|x\|^{-k}$ for all $k\geq 0$ as $|x|\to\infty$ which is precisely the condition $(c)$ of the proposition. $\blacksquare$
Proof of the Proposition:
$\text{(a)}\Rightarrow\text{(b)}$ Let $e_1,...,e_n$ be the canonical basis of $\mathbb{R}^n$. Then
$$\|x\|^k=\left(\sum_{j=1}^n x_j^2 \right)^{k/2}\leq C_{k,n}\sum_{j=1}^n(x_j^2)^{k/2}=C_{k,n}\sum_{j=1}^n|x_j|^k=C_{k,n}\sum_{j=1}^n|x^{ke_j}|$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$, where $C_{k,n}$ is a positive constant that depends on $k$ and $n$. Thus,
$$\begin{align}
\sup_{x\in\mathbb R^n}\|x\|^k|D^\beta f(x)|&\leq\sup_{x\in\mathbb R^n}\left(C_{k,n}\sum_{j=1}^n|x^{ke_j}|\right)|D^\beta f(x)|=C_{k,n}\sup_{x\in\mathbb R^n}\sum_{j=1}^n|x^{ke_j}D^\beta f(x)|\\
&\leq C_{k,n}\sum_{j=1}^n\|f\|_{ke_j,\beta}<\infty.
\end{align}$$
$\text{(b)}\Rightarrow\text{(c)}$ There exists a constant $M$ such that $\|x\|^{k+1}|D^{\beta}f(x)|\leq M$ for all $x\in\mathbb R^n$ and thus
$$\left|\|x\|^{k}D^{\beta}f(x)\right|\leq\frac{M}{|x|},\qquad\forall\ x \neq 0.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(c)}\Rightarrow\text{(d)}$ Let $\alpha=(\alpha_1,...,\alpha_n)$ be a multi-index and $k=\alpha_1+\cdots+\alpha_n$. Then
$$|x^\alpha|=\prod_{j=1}^n|x_j^{\alpha_j}|=\prod_{j=1}^n(x_j^{2})^{\alpha_j/2}\leq \prod_{j=1}^n\left(\sum_{r=1}^nx_r^2\right)^{\alpha_j/2}=\|x\|^{k}$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$. Thus,
$$|x^\alpha D^\beta f(x)|\leq\left|\|x\|^kD^\beta f(x)\right|,\qquad\forall \ x \in\mathbb R^n.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(d)}\Rightarrow\text{(a)}$ There exists a constant $M$ such that
$$|x^\alpha D^\beta f(x)|\leq 1,\quad\forall\ x\notin \overline{B(0,M)}$$
and thus
$$\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|\leq\left\{1,\max_{x\in \overline{B(0,M)}}|x^\alpha D^\beta f(x)|\right\}<\infty.\blacksquare$$
Best Answer
If $f\ge 0$ then $f^{1/2}= \lim_{n\to \infty} P_n(x)e^{-x/2}$ and $f =\lim_{n\to \infty} P_n(x)^2 e^{-x}$
If you want it to be strictly positive then add $e^{-x}/n$.
If you meant that $f\ge 0 \implies \sum_{n=1}^N L_n\frac{\langle f,L_n\rangle}{\langle L_n,L_n\rangle} \ge 0$ then no, $L_2-L_1$ is bounded so $(c+L_2-L_1)e^{-x/2}\ge 0$ but $(c-L_1)e^{-x/2}$ is not $\ge 0$.