Suppose \((f_n)\) is a sequence in \(L^2(\mathbb R)\) that converges weakly to \(f\in L^2(\mathbb R)\). Is it possible that \[ \|f_n\|_{L^2}\longrightarrow\infty? \]
Solution
1. The associated functionals have uniformly bounded operator norms.
For each \(n\), define the bounded linear functional \[ T_n(g)=\langle f_n,g\rangle_{L^2}, \qquad g\in L^2(\mathbb R). \] Weak convergence means that for every fixed \(g\in L^2\), \[ T_n(g)=\langle f_n,g\rangle\longrightarrow\langle f,g\rangle. \] Hence for every \(g\) the scalar sequence \((T_n(g))\) is bounded. By the uniform boundedness principle, \[ \sup_n\|T_n\|<\infty. \]
2. The functional norms are the \(L^2\) norms of the representing functions.
By the Riesz representation theorem for Hilbert spaces, \[ \|T_n\|=\|f_n\|_2. \] Therefore \[ \sup_n\|f_n\|_2<\infty. \]
3. Q.E.D.
By step 1, the norms \(\|T_n\|\) are uniformly bounded; step 2 therefore shows that \(\|f_n\|_2\) cannot tend to infinity.