Lemma 40.8.2.label Let $(X, \cm, \mu)$ be a measure space, $f: X \to \complex$ be a measurable function, and $C \ge 0$ such that $\norm{fg}_{L^2(X; \complex)}\le C\norm{g}_{L^2(X; \complex)}$ for all $g \in L^{2}(X; \complex) \cap \Sigma(X; \complex)$, then $f \in L^{\infty}(X; \complex)$ with $\norm{f}_{L^\infty(X; \complex)}\le C$.
Proof. For any $h \in L^{1}(X; \complex) \cap L^{\infty}(X; \complex)$,
\begin{align*}\normn{f^2h}_{L^1(X; \complex)}&= \int_{X} |f^{2}| \cdot |h| = \int_{X} |f||h|^{1/2}\cdot |f||h|^{1/2}= \normn{f|h|^{1/2}}_{L^2(X; \complex)}^{2} \\&\le C^{2} \normn{|h|^{1/2}}_{L^2(X; \complex)}^{2} = C^{2} \norm{h}_{L^1(X; \complex)}\end{align*}
By Theorem 16.3.3, $f^{2} \in L^{\infty}(X; \complex)$ with $\normn{f^2}_{L^\infty(X; \complex)}\le C^{2}$. Hence $f \in L^{\infty}(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))}\le C$.$\square$
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