Assume \(f, f_j \subset L^2(\mathbb{R}^n)\) for \(j = 1, 2, \ldots\), \(f_j \to f\) a.e. and \(\int f_j^2 \, dx \to \int f^2 \, dx\). Prove \(\int |f_j - f|^2 \, dx \to 0\).
Solution
1. The sequence is uniformly bounded in \(L^2\).
By hypothesis, \[ \|f_j\|_2^2\longrightarrow\|f\|_2^2. \] Hence there is \(M<\infty\) such that \[ \sup_j\|f_j\|_2\le M. \]
2. Show that \(f_j\rightharpoonup f\) weakly in \(L^2\).
Fix \(h\in L^2(\mathbb R^n)\) and \(\varepsilon>0\). Choose a bounded compactly supported function \(\phi\) such that \[ \|h-\phi\|_2<\varepsilon. \] Then \[ \left|\int (f_j-f)(h-\phi)\right| \le (M+\|f\|_2)\varepsilon. \]
It remains to prove \[ \int (f_j-f)\phi\to0. \] Let \(K\) be a finite-measure set containing the support of \(\phi\). Since \(f_j\to f\) almost everywhere on \(K\), Egorov’s theorem gives, for every \(\delta>0\), a measurable set \(A\subset K\) with \(m(A)<\delta\) such that \(f_j\to f\) uniformly on \(K\setminus A\).
On \(K\setminus A\), \[ \left|\int_{K\setminus A}(f_j-f)\phi\right| \le \|\phi\|_\infty m(K) \sup_{K\setminus A}|f_j-f|\longrightarrow0. \] On \(A\), Cauchy–Schwarz gives \[ \left|\int_A(f_j-f)\phi\right| \le \|\phi\|_\infty (\|f_j\|_2+\|f\|_2)m(A)^{1/2} \le \|\phi\|_\infty(M+\|f\|_2)\delta^{1/2}. \] First choose \(\delta\) small and then \(j\) large. Thus \[ \int (f_j-f)\phi\to0. \] Since \(\varepsilon\) was arbitrary, \[ \int f_jh\to\int fh \] for every \(h\in L^2\), so \(f_j\rightharpoonup f\).
3. Combine weak convergence with convergence of norms.
Using the Hilbert-space identity, \[ \|f_j-f\|_2^2 =\|f_j\|_2^2+\|f\|_2^2 -2\operatorname{Re}\langle f_j,f\rangle. \] By Step 2, \[ \langle f_j,f\rangle\to\|f\|_2^2, \] while the hypothesis gives \[ \|f_j\|_2^2\to\|f\|_2^2. \] Therefore \[ \boxed{\|f_j-f\|_2^2\longrightarrow0.} \]