Let $f$ be a nonnegative Schwartz function on $n$-dimensional Euclidean space.
Then I wonder if it is possible to find a sequence of complex-valued Schwartz functions $f_n$ such that $\lvert f_n \rvert^2$ converges to $f$ in the Frechet topology of the Schwartz space.
I think it is plausible but cannot figure out how to prove / disprove.
Could anyone help me?
Start with $f\in \mathcal{S}(\mathbb{R}^n)$. For every $L>0$ let $\chi_1\in C_c^\infty(\mathbb{R})$ be the standard cut-off function such that $\varphi(t) =0$ for $\vert t \vert \geq 1$, $\varphi(0)=1$ and $\varphi^{(k)}(1)=0$ for all $k\geq 1$. Then we define for every $L\geq 0$ the cut-off function $$ \varphi_L: \mathbb{R} \rightarrow \mathbb{R}, \varphi_L(t) = \begin{cases} 0,& \vert t \vert\geq L+1, \\ 1,& \vert t \vert \leq L,\\ \varphi(t-\mathrm{sgn(t)L}),& L<\vert t \vert<L+1. \end{cases} $$ One readily checks that $\varphi_L\in C_c^\infty(\mathbb{R})$. We now define the cut-off function $$ \chi_L:\mathbb{R}^n \rightarrow \mathbb{R}, \chi_L(x) = \prod_{j=1}^n \varphi_L(x_j). $$ It is easy to check that for every $\alpha \in \mathbb{N}^n$ and all $L\geq 1$ we have $$ \sup_{x\in \mathbb{R}^n} \vert \partial^\alpha \chi_L(x) \vert = \sup_{x\in \mathbb{R}^n} \vert\partial^\alpha \chi_1(x) \vert < \infty. $$
The problem boils down to find a positive $g\in \mathcal{S}(\mathbb{R}^n)$ such that $\sqrt{g}\in \mathcal{S}(\mathbb{R}^n)$. Once we have found such a function, then we can define
$$ f_\varepsilon(x):=\sqrt{\chi_{1/\varepsilon}(x)f(x)+ \varepsilon g(x)}. $$
Define the semi-norms $$ \Vert f \Vert_{\alpha, \beta} := \sup_{x \in \mathbb{R}^n} \vert x^\beta \partial^\alpha f(x) \vert. $$
Since \begin{align*} \sup_{x\in \mathbb{R}^n} \vert \partial^\alpha \chi_{1/\epsilon}(x) \vert <\infty \end{align*} as noted above, we have \begin{align*} \lVert f_\epsilon^2 \rVert_{\alpha. \beta} <\infty \end{align*} for all $\alpha, \beta$, which implies that $f_\epsilon^2 \in \mathcal{S}(\mathbb{R}^n)$.
Now, consider another Schwartz function \begin{align*} F_\epsilon:=(1-\chi_{1/\epsilon}) f \end{align*} Since $F_\epsilon$ is supported on $[\frac{1}{\epsilon}, \infty)^n \sqcup (-\infty, -\frac{1}{\epsilon}]^n$, the estimate \begin{align*} \frac{1}{\epsilon^n} \lvert x^{\beta}\partial^\alpha F_\epsilon(x) \rvert \leq \Bigl \lvert \Bigl(\prod_{i=1}^n x_i \Bigr) x^{\beta}\partial^\alpha F_\epsilon(x) \Bigr \rvert=\lvert x^{\beta+(1,\cdots,1)}\partial^\alpha F_\epsilon(x) \rvert \end{align*} holds for $x \in \mathbb{R}^n$ so that \begin{align*} \frac{1}{\epsilon^n} \Vert F_\epsilon \Vert_{\alpha, \beta} \leq \Vert F_\epsilon \Vert_{\alpha, \beta+(1, \dots, 1)} \end{align*}
We also observed above that the supremum of $\partial^\alpha \chi_L$ does not depend on $L$. Hence $\Vert F_\epsilon \Vert_{\alpha, \beta+(1, \dots, 1)}$ does not depend on $\epsilon$, implying \begin{align*} \Vert f_\varepsilon^2 - f \Vert_{\alpha, \beta} \leq \Vert F_\epsilon \Vert_{\alpha, \beta} + \varepsilon \Vert g \Vert_{\alpha, \beta} \leq \epsilon^n \Vert F_\epsilon \Vert_{\alpha, \beta+(1, \dots, 1)} + \varepsilon \Vert g \Vert_{\alpha, \beta} \to 0^+ \text{ as } \epsilon \to 0^+ \end{align*} for all $\alpha,\beta$. That is, $f_\varepsilon^2 \to f$ in the Fréchet topology of $\mathcal{S}(\mathbb{R}^n)$.
Lastly, we need to check if $f_\epsilon \in \mathcal{S}(\mathbb{R}^n)$. However, because $g$ is assumed to be positive, it is clear that $f_\epsilon$ is smooth on whole $\mathbb{R}^n$. Moreover, $f_\varepsilon (x)=\sqrt{\varepsilon g(x)}$ for $\vert x \vert > 1/\varepsilon$ so that \begin{equation} \lvert x^\beta \partial^\alpha f_\epsilon(x) \rvert \leq \sup_{ \lvert x \rvert \leq \frac{1}{\epsilon} } \lvert x^\beta \partial^\alpha f_\epsilon(x) \rvert + \sqrt{\epsilon}\sup_{ \lvert x \rvert > \frac{1}{\epsilon} } \lvert x^\beta \partial^\alpha \sqrt{g}(x) \rvert \end{equation} for all $x \in \mathbb{R}^n$. Therefore, \begin{equation} \lVert f_\epsilon \rVert_{\alpha, \beta} < \infty \end{equation} for all $\alpha, \beta$, implying $f_\epsilon \in \mathcal{S}(\mathbb{R}^n)$.
Note that $g(x)=e^{-x^2}$ does the job as $\sqrt{g(x)}= e^{-x^2/2}$ is again Schwartz.
