Suppose that \(f_j \in L^2(\mathbb{R}^n)\), \(j = 1, 2, 3, \ldots\) and that \(f_j \to f\) in \(L^2\). Suppose further that there is a constant \(M < \infty\) so that
\[\int e^{100|x|^2} |f_j(x)|^2 \, dx \leq M, \quad j = 1, 2, 3, \ldots.\]
Is it true that \(\int e^{99|x|^2} |f(x)|^2 \, dx < \infty\)? Give a proof or counterexample.
Solution
Yes. In fact the stronger estimate \[ \int_{\mathbb R^n} e^{100|x|^2}|f(x)|^2\,dx\le M \] holds.
1. Extract a subsequence converging almost everywhere.
Since \(f_j\to f\) in \(L^2(\mathbb R^n)\), there is a subsequence \(f_{j_k}\) such that \[ f_{j_k}(x)\longrightarrow f(x) \] for almost every \(x\in\mathbb R^n\). For instance, choose the subsequence so that \[ \|f_{j_k}-f\|_2^2\le 2^{-k}, \] and apply Tonelli to \[ \sum_{k=1}^\infty |f_{j_k}-f|^2. \] Its integral is finite, so the series is finite almost everywhere; hence its terms tend to zero almost everywhere [@Fol13].
2. Apply Fatou’s lemma with the Gaussian weight.
For almost every \(x\), \[ e^{100|x|^2}|f_{j_k}(x)|^2 \longrightarrow e^{100|x|^2}|f(x)|^2. \] These functions are nonnegative, so Fatou’s lemma gives [@Fol13] \[ \int_{\mathbb R^n} e^{100|x|^2}|f(x)|^2\,dx \le \liminf_{k\to\infty} \int_{\mathbb R^n} e^{100|x|^2}|f_{j_k}(x)|^2\,dx \le M. \] Since \(e^{99|x|^2}\le e^{100|x|^2}\), \[ \int_{\mathbb R^n} e^{99|x|^2}|f(x)|^2\,dx\le M<\infty. \]