Suppose that $f \in L^2(\mathbb{R})$ and its $L^2$ derivative exists, call it, $f'$, i.e $$\lim_{y\to 0} \Bigg|\Bigg|\frac{f(x-y)-f(x)}{y}-f'(x)\Bigg|\Bigg|_{L^2} = 0$$ Then $$\Big|\Big|f\Big|\Big|_{L^2}^4 \leqslant 4\Big|\Big|xf(x)\Big|\Big|_{L^2}^2 \cdot\Big|\Big|f'(x)\Big|\Big|_{L^2}^2$$ This has been asked a few times here but I couldn't find a rigorous proof that shows the above for complex-valued-$L^2$ functions.
I am able to show this for Schwartz functions using integration by parts and tried using a density argument to extend it to all of $L^2$.
We can pick a sequence $\{f_n\}$ of Schwartz functions such that $f_n\to f$ and $f_n' \to f'$ in $L^2$. But to conclude I need to show $xf_n(x)\to xf(x)$ in $L^2$. How can I show this?
If $f$ has support in some large ball, say $(-R,R)$, this follows simply from
$$ \Vert x (f_n-f)\Vert_{L^2} \leq 1.000000001 R \Vert f_n -f\Vert_{L^2} \rightarrow 0 $$
for $n\rightarrow \infty$ (we need some coefficient larger than $1$ in front of $R$ as the support of $f_n$ might be slightly larger than the support of $f$, pick your favourite number larger than $1$. I did not want to go for $2$ as one might get the wrong idea that this factor $2$ has anything to do with the length of $(-R,R)$).
Now for a general $f\in L^2(\mathbb{R})$ with your conditions, pick $\chi_n\in C_c^\infty(\mathbb{R})$ such that $0\leq \chi_n \leq 1$, $\vert \chi'_n\vert \leq 2$, $\chi_n(x) = 0$ for $\vert x \vert \geq n+1$ and $\chi_n(x)=1$ for $\vert x \vert \leq n$ (i.e. pick a bump function). Then consider $f_n = \chi_n f$. Note that we know from the first step that your inequality holds true for $f_n$. Then $\Vert xf_n \Vert_{L^2} \rightarrow \Vert x f\Vert_{L^2}$ by monotone convergence (we do not need that it is finite). We have $f_n'=\chi_n' f + \chi_n f'$. By dominated convergence we have $\chi_n' f \rightarrow 0$ in $L^2$ and by monotone convergence we have $\Vert \chi_n f' \Vert_{L^2} \rightarrow \Vert f' \Vert_{L^2}$.