Suppose \(f_n\in L^2([0,1])\) converges weakly to \(f\in L^2([0,1])\). Prove that \(\limsup_{n\to\infty}\|f_n\|_2<\infty\), or give a counterexample.
Solution
Let \(H=L^2([0,1])\) with \(\langle g,h\rangle=\int_0^1g\bar h\), and for each \(n\) let \(\phi_n(g)\da\langle g,f_n\rangle\). The statement holds: \(\sup_n\norm{f_n}_2<\infty\).
1. Each \(\phi_n\) is a bounded linear functional with \(\norm{\phi_n}=\norm{f_n}_2\).
The Cauchy–Schwarz inequality gives \(\abs{\phi_n(g)}\le\norm g_2\norm{f_n}_2\). If \(f_n\neq0\), the unit vector \(g=f_n/\norm{f_n}_2\) attains the bound; if \(f_n=0\), both sides are \(0\).
2. For each \(g\in H\), \(\sup_n\abs{\phi_n(g)}<\infty\).
Weak convergence gives \(\langle f_n,g\rangle\to\langle f,g\rangle\); conjugating, \(\phi_n(g)\to\langle g,f\rangle\), and a convergent sequence of complex numbers is bounded.
3. Q.E.D.
\(H\) is complete, so by steps 1 and 2 the uniform boundedness principle [@Fol13] gives \(\sup_n\norm{\phi_n}<\infty\). By step 1 this is \(\sup_n\norm{f_n}_2\), which bounds \(\limsup_n\norm{f_n}_2\).