I have a smooth function $f(x)$ which is of rapid decay (actually I know it's a Schwartz function), and I have another function $g(x)$ which behaves like a polynomial as $x \to \pm \infty$; that is $$ \lim_{|x| \to \infty} |g(x)| = \lim_{|x| \to \infty} |x|^c $$ for some constant $c$. Is it true that $$ \int_{-\infty}^\infty |f(x)||g(x)| dx < \infty; $$ that is, is $fg \in L^1(\mathbb{R})$?
My feeling is that the rapid decay of $f$ and the fact that $g$ behaves like a polynomial should guarantee the $L^1$-convergence, but I'm having trouble proving this fact.
I am interpreting your statement "decays like a polynomial" to mean something a bit different than what you wrote ($|x|^c$ goes to infinity with $x$ when $c > 0$, but I don't think you just meant that $g$ blows up at infinity), instead that there exists $M$ such that for $|x| \geq M$ $g(x) \leq C|x|^c$ for some $C, c$. Does that capture your intention?
I think in this case for the tail integrals you could estimate $$\int_M^{\infty}|fg| \, dx \leq C \int_M^{\infty} |f||x|^c = C \int_M^{\infty} |f| |x|^{c+2} \frac{1}{x^2} \leq C \sup{|fx^{c+2}|}\int_M^{\infty} \frac{1}{x^2}$$
And now use the rapid decrease of $f$ to conclude this quantity is finite. Of course using an exponent of $2$ here is not critical, you could use any integrable exponent.