Get an estimate on $L^{2}(0,1)$ [closed]

Question

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 :(

Answer 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: $$\int f_n \textrm{d} x =1,\quad \int_0^1 x^{2\alpha} f_n \textrm{d} x = \frac{1}{(2\alpha+1)n^{2\alpha}}\to0.$$