Assume \(f : [0,1] \to \mathbb{R}\) is a measurable function such that \(fg \in L^1([0,1])\) for all \(g \in L^2([0,1])\). Prove \(f \in L^2([0,1])\).
Solution
1. Truncate \(f\) to obtain bounded \(L^2\) functions.
For \(n\ge1\), define \[ f_n:=f\,\mathbf1_{\{|f|\le n\}}. \] Because \([0,1]\) has finite measure and \(|f_n|\le n\), we have \(f_n\in L^2([0,1])\).
Define \[ T_n:L^2([0,1])\to\mathbb R, \qquad T_n(g):=\int_0^1 f_n g. \] By Cauchy–Schwarz, \(T_n\) is a bounded linear functional and \[ \|T_n\|=\|f_n\|_2. \]
2. Apply the Uniform Boundedness Principle.
Fix \(g\in L^2([0,1])\). By hypothesis, \(fg\in L^1([0,1])\). Since \[ |f_ng|\le |fg|, \] we have \[ |T_n(g)| \le \int_0^1|f_ng| \le \int_0^1|fg|<\infty \] for every \(n\). Hence \[ \sup_n|T_n(g)|<\infty \qquad\text{for every }g\in L^2([0,1]). \] Because \(L^2([0,1])\) is Banach, the Uniform Boundedness Principle gives \[ \sup_n\|T_n\|<\infty. \] Thus there is \(C<\infty\) such that \[ \|f_n\|_2\le C \qquad\text{for all }n. \]
3. Pass to the limit.
The nonnegative functions \(|f_n|^2\) increase pointwise to \(|f|^2\). Therefore the Monotone Convergence Theorem gives \[ \int_0^1|f|^2 =\lim_{n\to\infty}\int_0^1|f_n|^2 \le C^2<\infty. \] Hence \[ \boxed{f\in L^2([0,1]).} \]