Let $(X,\mathcal{M},\mu)$ be a measure space, and suppose $g\in L^{\infty}(X)$. If for every function $f\in L^{2}(X)$, we always have $$\int |f\overline{g}|\,\mathrm{d}\mu <\infty,$$ (i.e., $f\overline{g}\in L^{1}(X)$) then does $g$ necessarily lie in $L^{2}(X)$?
This question is asking a partial converse of the general fact that $L^{2}(X)$ usually forms a Hilbert space with the inner product $\langle f, g\rangle = \int f\overline{g}\,\mathrm{d}\mu$, provided that $f,g\in L^{2}(X)$.
For any $f \in L^2(X)$, $\int |fg| d \mu = \int|\bar{f}\bar{g}| d\mu < \infty$ ( Using the fact that $\forall f \in L^2(X)$ we have that $f\bar{g} \in L^1(X)$ )
Hence, $\forall f \in L^2(X), |\int fg d\mu| \le \int|fg|d \mu < \infty$ , in other words, the sending $f \mapsto \int fg d\mu$ defines a bounded linear functional on $L^2(X)$ . Since, $L^2(X)^* \cong L^2(X)$ , it follows that $g \in L^2(X)$ .