Bound a norm $L^{2}$

Question

Let $f \in L^{2}(-1,1)$ and $g \in L^{\infty}(-1,1)$ such that $g(x) \geq 0$, $\forall x \in [-1,1]$ and $g (x) > 0$, $\forall x \in [0,1]$.

If $$ \|g^{\frac{1}{2}}f\|_{L^{2}(-1,1)} \leq C_{1} $$ where $C_{1} >0$. Can I limit the norm $L^{2}(-1,1)$ of the function $f$? That is, I can show that there is $C_{2} > 0$ so that $$ \|f\|_{L^{2}(-1,1)} \leq C_{2} ? $$

Answer 1

No. Let $g(x)=|x|$ if $x \neq 0$ and $g(0)=1$. Let $f=g^{-1/2}$. Can you check that this is a counter-example?