I have a Schwartz function $f$ and I wish to know if $(.)^2*f$ is square integrable. I have the following heuristics: Using Plancherel we can move to the Fourier side and then by the convolution theorem the integrand is the product of the Fourier transforms. Now to find $\mathscr{F}(t^2)$ we observe that (ignoring the factors of $2\pi i) \;$$\mathscr{F}(t^2)=\mathscr{F}(t.t)=D(\mathscr{F}(t.1))=D^2(\mathscr{F}(1))=D^2(\delta)$ where $\delta$ is the Dirac delta. We are interested in the integral $\int |D^2(\delta).\hat{f}|^2.$ I'm not sure what to say about this other than the fact that $\int D^2(\delta).\hat{f}=(\hat{f})''(0)$ is finite.
Just knowing whether this claim is true or false (i.e., without proof) would be helpful as well. Although a proof is always appreciated!
Suppose that $f\geq0$ and $f\geq1$ on $[0,1]$. Then, with $g(x)=x^2$, $$ g*f(x)=\int_{-\infty}^\infty t^2\,f(x-t)\,dt\geq\int_{-\infty}^\infty t^2\,1_{[0,1]}(x-t)\,dt =\int_{x-1}^x t^2\,dt=\frac{x^2-x+1}3. $$ Then $$ \int_{-\infty}^\infty |g*f(x)|^2\,dt \geq\int_{-\infty}^\infty \Big(\frac{x^2-x+1}3\Big)^2\,dx=\infty. $$