Suppose \(f_j\rightharpoonup f\) weakly in \(L^2(\RR^n)\).
Prove \[ \norm{f}_2\le\liminf_j\norm{f_j}_2, \] and give an example of strict inequality.
If \(\norm{f_j}_2\to\norm{f}_2\), prove \(\norm{f_j-f}_2\to0\).
Solution
1. Weak convergence implies\[ \norm{f}_2\le\liminf_j\norm{f_j}_2. \]
If \(f=0\), the inequality is immediate. If \(f\ne0\), set \(g=f/\norm{f}_2\). Then \(\norm{g}_2=1\), and weak convergence gives \[ \norm{f}_2 =\abs{\inner{f}{g}} =\lim_j\abs{\inner{f_j}{g}} \le\liminf_j\norm{f_j}_2. \]
2. Strict inequality can occur.
Let \((e_j)\) be an orthonormal sequence in \(L^2(\RR^n)\). Bessel’s inequality gives \(e_j\rightharpoonup0\), whereas \(\norm{e_j}_2=1\) for every \(j\). Thus \[ \norm{0}_2=0<1=\liminf_j\norm{e_j}_2. \]
3. If \(\norm{f_j}_2\to\norm{f}_2\), then\[ \boxed{\norm{f_j-f}_2\to0}. \]
Weak convergence gives \[ \inner{f_j}{f}\to\inner{f}{f}=\norm{f}_2^2. \] Therefore \[ \begin{aligned} \norm{f_j-f}_2^2 &=\norm{f_j}_2^2+\norm{f}_2^2-2\operatorname{Re}\inner{f_j}{f}\\ &\longrightarrow 0. \end{aligned} \]
4. Q.E.D.
Step 1 proves the lower-semicontinuity assertion in part (a), step 2 gives the requested strict example, and step 3 proves part (b).