Let $g\in L^2[0,1)$. If for all $f\in L^2[0,1)$, the product $fg\in L^2[0,1)$, then $g\in L^{\infty}[0,1)$.
Suppose false. In other words, assume that $g\in L^2[0,1)\setminus L^{\infty}[0,1)$. My idea is to build up a function that is highly concentrated near the "singularity" of $g$. However, I'm not quite there yet. Here is my failed attempted at a counterexample. Define $A_k = \{t\in [0,1): \, |g(t)|\geq k\}$. Moreover, let $$ f_k(t) = \frac{k^2\chi_{A_k}}{|A_k|^{1/2}}, $$ where $|\cdot|$ denotes the measure. It follows that $$ \int_{0}^1|f_k(t)g(t)|^2\,dt = \int_0^1\frac{k^4\chi_{A_k}}{|A_k|}|g(t)|^2\,dt \geq \int_{A_k}\frac{k^2}{|A_k|}\,dt = k^2. $$ The issue is I didn't really show anything. We just showed that the sequence $\{f_kg\}_{k\in \Bbb{N}}$ has the property that $\lim_{k\to \infty}f_kg\notin L^2[0,1)$.
Additionally, we cannot think of the product as a linear functional regarding the $L^2$-norm as it is not linear. Any advice would be appreciated.
