$L^2$ function in Schwartz space?

Question

Let $f:\mathbb R^n \rightarrow \mathbb R$ be a smooth function whose derivatives are all polynomially bounded and $f \in L^{\infty}.$

Such a function has the property that when multiplied with any Schwartz function $\varphi \in \mathcal S$ the product $f\varphi $ is again a Schwartz function.

Now consider the following linear and bounded operator $T:C([0,T];L^2) \rightarrow C([0,T];L^2)$ for some $\alpha \in L^1((0,T),\mathbb R)$ and $\varphi_0 \in \mathcal S:$

$(T\varphi)(t) =\varphi_0+ \int_0^t \alpha(s) f \varphi(s) \ ds.$ Banach's fixed point theorem (apply it iteratively on small time-intervals) implies the existence of a function $\varphi \in C([0,T];L^2) $ such that

$$(T\varphi)(t) =\varphi_0+ \int_0^t \alpha(s) f \varphi(s) \ ds = \varphi(t)$$

I would like to understand whether this fixed point is actually also in $\mathcal S$, i.e. $\varphi(s) \in \mathcal S$ or whether $L^2$ is the best we can do?

The reason I think it could be true is because everything makes perfect sense if $\varphi$ was Schwartz I think. In particular $T$ maps Schwartz functions to Schwartz functions. However, I fail to verify the contraction property for $T$ in $\mathcal S.$

Answer 1

This is quite ad hoc, but your integral equation can be reverse engineered to a differential equation (Maybe this is what you really want to solve?)

$$ \partial_t\varphi=\alpha(t)f\varphi,\ \varphi(0)=\varphi_0, $$

which has an explicit solution

$$ \varphi(x,t)=\varphi_0(x)\exp\left( \int_0^t \alpha(s)f(x)ds \right). $$

From here it is easy to see that $\varphi(\cdot,t)\in\mathcal S$.