Consider $f \in L^{2}(0,1)$ and $g \in L^{\infty}(0,1)$ such that


1. $ \text{lim} ~g(x) = 0 \ \ \text{when} \ \ x \to 0^{+};$
2. $g(x) > 0 \ \forall x \in (0,1)$;
3. $\text{lim}~\dfrac{g(x)}{x^{\alpha}} = N > 0,$ when $x \to 0^{+}$, $0 < \alpha < 1$


Moreover, suppose
$$
\int\_{0}^{1}g(x)|f(x)|^{2} = M < \infty
$$
**Question:** Is it possible to get an estimate of the form
$$
\|f\|\_{L^{2}(0,1)}^{2} \leq C \|gf\|\_{L^{2}(0,1)}^{2} ?
$$


**My idea**
$$
\|gf\|\_{L^{2}(0,1)}^{2} \leq \|g\|\_{L^{\infty}(0,1)} \int\_{0}^{1}g(x)|f(x)|^{2} = M\|g\|\_{L^{\infty}(0,1)} 
$$
It is correct to write
$$
\int\_{0}^{1}|f(x)|^{2} = \dfrac{1}{|g(s)|}\int\_{0}^{1}|g(s)||f(x)|^{2} dx
$$
But I can't do anything more than that :(



Get an estimate on $L^{2}(0,1)$
No. Think of a sequence of $f\_n$ such that $\| f\_n\|\_2=1$ while $\|g f\|\_2\to 0$. And you might as well assume $g(x)= x^{\alpha}$.


It is equivalent to asking the same question with $f\_n\geq0$, $\| f\_n\|\_1 =1$, $\| x^{2\alpha}f\_n\|\_1\to0$ (just take $g\_n=\sqrt{f\_n}$ for the $L^2$ case).


Let us try $f\_n=\begin{cases} n & \textrm{when } x<\frac1n\\ 0 &\textrm{otherwise}\end{cases}.$ We check that it works:
$$\int f\_n \textrm{d} x =1,\quad \int\_0^1 x^{2\alpha} f\_n \textrm{d} x = \frac{1}{(2\alpha+1)n^{2\alpha}}\to0.$$


