Let \((f_n)\) be a sequence in \(L^4([0,1])\) such that \[ \int_0^1|f_n|^4\le1 \qquad\text{for all }n, \] and \[ \int_0^1|f_n|\longrightarrow0. \] Show that \(f_n\rightharpoonup0\) weakly in \(L^4([0,1])\).
Solution
1. The part of the pairing where \(g\) is bounded tends to zero.
Let \(g\in L^{4/3}([0,1])\) and fix \(\varepsilon>0\). Since \[ |g|^{4/3}\in L^1([0,1]), \] choose \(M>0\) so large that \[ \|g\mathbf 1_{\{|g|>M\}}\|_{4/3}<\frac\varepsilon2. \] This is possible by the dominated convergence theorem applied as \(M\to\infty\).
Split \[ \int_0^1 f_n g = \int_{\{|g|\le M\}}f_ng + \int_{\{|g|>M\}}f_ng. \] For the bounded part, \[ \left|\int_{\{|g|\le M\}}f_ng\right| \le M\|f_n\|_1\longrightarrow0. \]
2. The tail of the pairing is uniformly small.
The Hölder inequality and the uniform \(L^4\) bound give \[ \left|\int_{\{|g|>M\}}f_ng\right| \le \|f_n\|_4\,\|g\mathbf 1_{\{|g|>M\}}\|_{4/3} \le \frac\varepsilon2. \]
3. The sequence converges weakly to zero.
Hence for all sufficiently large \(n\), \[ \left|\int_0^1 f_n g\right|<\varepsilon. \] Since \(g\in L^{4/3}\) was arbitrary, the duality of $L^p$ spaces shows that this is exactly \[ f_n\rightharpoonup0\quad\text{in }L^4([0,1]). \]
4. Q.E.D.
Steps <1>1–<1>3 prove convergence against every element of the dual space \(L^{4/3}([0,1])\).