An inequality of integrals Let $f \in L^{2}(\mathbb{R})$ be continuously differentiable on $\mathbb{R}$. I am trying to show the following:
$( \int |f|^{2} dx)^{2} \leq 4 ( \int |xf(x)|^{2} dx) ( \int |f'|^{2} dx))$. 
My first thought is to think about this inequality as $ ||f||^{4}_{2} \leq 4 ||xf(x)||^{2}_{2} ||f'||^{2}_{2} $ and apply Holder's inequality to get $ 4 (\int |xf(x)f'(x)| dx)^{2} \leq 4 ||xf(x)||^{2}_{2} ||f'||^{2}_{2}$ but beyond this I have no intuition, especially what to do with the continuously differentiable assumption. Could someone lend me a hint?
 A: Hints:
The claim holds trivially true if the right-hand side equals $+\infty$, therefore we may assume without loss of generality that
$$\int x^2 f(x)^2 \, dx < \infty. \tag{1} $$
Let us first assume that $f$ has bounded support.


*

*By the integration by parts formula, we have $$\int_{-R}^R f(x)^2 \, dx = \big[ x f(x)^2 \big]_{x=-R}^R -2 \int_{-R}^R x f(x) f'(x) \, dx$$ for all $R>0$. Since $f$ has bounded support, we can let $R \to \infty$ and obtain $$\int f(x)^2 \, dx = -2 \int x f(x) f'(x) \, dx. \tag{1}$$


For the general case:


*Choose a smooth cut-off function $\chi$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $|x| \leq 1$ and $\chi(x) = 0$ for $|x| \geq 2$. Set $\chi_R(x) := \chi(x/R)$.

*Applying step 1 to $f_R(x) := f(x) \chi_R(x)$ gives $$\int \chi_R(x)^2 f(x)^2 \, dx \leq 2 \int \left|x f(x) \frac{d}{dx}(f(x) \chi_R(x)) \right| \, dx. \tag{3}$$

*Show that by the product rule and the definition of $\chi_R$, $$\begin{align*}\left| \frac{d}{dx} (f(x) \chi_R(x)) \right| &\leq |f'(x)| + |f(x)| R^{-1} \|\chi'\|_{\infty}. \end{align*}$$

*Use the finiteness of the integral $(1)$ and the monotone convergence theorem to let $R \to \infty$ in $(3)$ and conclude $$\int f(x)^2 \, dx \leq 2 \int |x f(x) f'(x)| \, dx.$$

*Apply the Cauchy-Schwarz inequality.

