Let \((f_n)\) be a sequence in a Hilbert space \(X\) such that \(f_n\rightharpoonup f\), meaning \[ \langle f_n,g\rangle\to\langle f,g\rangle \qquad(g\in X). \]
Prove that \[ \|f\|\le \liminf_{n\to\infty}\|f_n\|, \] and give an example where the inequality is strict.
If in addition \(\|f_n\|\to\|f\|\), prove that \(f_n\to f\) in norm.
Solution
1. The norm is weakly lower semicontinuous.
If \(f=0\), the claimed inequality is immediate. Assume \(f\ne0\) and set \[ g=\frac{f}{\|f\|}. \] Then \(\|g\|=1\), so by Cauchy–Schwarz \[ |\langle f_n,g\rangle|\le\|f_n\|. \] Weak convergence gives \[ \langle f_n,g\rangle\longrightarrow\langle f,g\rangle=\|f\|. \] Therefore \[ \|f\| =\lim_{n\to\infty}|\langle f_n,g\rangle| \le\liminf_{n\to\infty}\|f_n\|. \]
2. Strict inequality can occur.
Take \(X=\ell^2(\mathbb N)\) and let \(e_n\) be the standard orthonormal basis. For every \(g=(g_k)\in\ell^2\), one has \(g_n\to0\), hence \[ \langle e_n,g\rangle=g_n\longrightarrow0. \] Thus \(e_n\rightharpoonup0\). But \[ \|e_n\|=1 \qquad\text{for all }n, \] so \[ 0=\|0\|<1=\liminf_n\|e_n\|. \]
3. Weak convergence plus convergence of norms implies norm convergence.
Since \(f_n\rightharpoonup f\), taking the test vector \(g=f\) gives \[ \langle f_n,f\rangle\longrightarrow\langle f,f\rangle=\|f\|^2. \] Using the Hilbert-space identity, \[ \|f_n-f\|^2 =\|f_n\|^2+\|f\|^2-2\operatorname{Re}\langle f_n,f\rangle. \] By hypothesis, \(\|f_n\|\to\|f\|\), so the right-hand side tends to \[ \|f\|^2+\|f\|^2-2\|f\|^2=0. \] Hence \[ \|f_n-f\|\longrightarrow0, \] which is exactly strong convergence.