Is $f(x)\exp(-x^2)$ summable if $f$ is square summable?

Question

Suppose that $f \in L^2(\mathbb R)$; i.e. $$ \int_{- \infty}^\infty \vert f(x) \vert^2 dx < \infty. $$ Can we from this infer that $$ \int_{- \infty}^\infty \vert f(x)\vert e^{-x^2} \, dx < \infty\text{ ?} $$ Intuitively this is clear since $\exp(-x^2)$ decays at least as good as $f$, however how to prove it? Can we represent $f$ as some worst case function when it comes to decay, and prove it from there? For instance, can we say that there exists a $\Delta$ such that $e^{-x^2} \leq \vert f(x) \vert$ for $x \geq \Delta$? If we can then we would be done: $$ \int_{- \infty}^\infty \vert f(x) \vert e^{-x^2} dx \leq \int_{\vert x \vert \leq \Delta} \vert f(x) \vert e^{- x^2} dx + \int_{\vert x \vert \geq \Delta} \vert f(x) \vert^2 \, dx < \infty. $$

Answer 1

The essential fact is not something about $e^{-x^2}$ decaying fast, but is not unrelated to that either. The essential fact is this: $$ \int_{-\infty}^\infty e^{-x^2}\,dx < \infty. $$

The point is that if $(X,\mathcal F,\mu)$ is a measure space and $\mu(X)<\infty$ then $L^2(X)\subseteq L^1(X)$. So let $X=\mathbb R$ and let $\mu$ be this measure: $$ e^{-x^2}\,dx. $$

Answer 2

Big Hint :

By Cauchy-Schwarz,

$$\int|f(x)e^{-x^2}| \, dx\leq \sqrt{\int|f(x)|^2 \, dx}\sqrt{\int e^{-2x^2} \, dx}.$$