Let \((f_j)\) be functions in \(L^2([0,1])\) satisfying \[ \int_0^1 |f_j|\le \frac1j, \qquad \int_0^1 |f_j|^2\le1. \] Prove that \(f_j\rightharpoonup0\) weakly in \(L^2([0,1])\).
Solution
Fix \(g\in L^2([0,1])\) and \(\varepsilon>0\). For \(M>0\), define the truncation \[ g_M=g\mathbf1_{\{|g|\le M\}}. \] Then \(g_M\in L^\infty\) and \[ \|g-g_M\|_2\longrightarrow0 \qquad(M\to\infty). \] Choose \(M\) so large that \[ \|g-g_M\|_2<\frac{\varepsilon}{2}. \] For every \(j\), Cauchy–Schwarz and the uniform \(L^2\) bound give \[ \left|\int_0^1 f_j(g-g_M)\right| \le \|f_j\|_2\|g-g_M\|_2 <\frac{\varepsilon}{2}. \] On the bounded part, \[ \left|\int_0^1 f_jg_M\right| \le \|g_M\|_\infty\|f_j\|_1 \le \frac{M}{j}. \] Hence for all sufficiently large \(j\), this is \(<\varepsilon/2\). Therefore \[ \left|\int_0^1 f_j g\right|<\varepsilon \] for all sufficiently large \(j\). Since \(g\in L^2([0,1])\) was arbitrary, \[ f_j\rightharpoonup0 \qquad\text{in }L^2([0,1]). \]