Let \(f_j\in L^2(\mathbb R^n)\) with \(f_j\to f\) in \(L^2\). Suppose there exists \(M<\infty\) such that \[ \|f_j\|_{H^\sigma}\le M \qquad(j\ge1), \] where \[ \|g\|_{H^\sigma}^2=\int_{\mathbb R^n}(1+|\xi|^2)^\sigma|\widehat g(\xi)|^2\,d\xi. \] Must \(f\in H^\sigma\)? Prove your answer.
Solution
1. After passing to a subsequence, the Fourier transforms converge almost everywhere.
By the Plancherel theorem, \[ \widehat f_j\longrightarrow\widehat f \qquad\text{in }L^2(\mathbb R^n). \] Hence there is a subsequence \((j_k)\) such that \[ \widehat f_{j_k}(\xi)\longrightarrow\widehat f(\xi) \] for almost every \(\xi\).
2. The limit belongs to \(H^\sigma\) and satisfies \(\|f\|_{H^\sigma}\le M\).
Set \[ w(\xi)=(1+|\xi|^2)^\sigma \] which is positive and measurable for every real \(\sigma\). By step 1 and Fatou's lemma, \[ \begin{aligned} \|f\|_{H^\sigma}^2 &=\int w(\xi)|\widehat f(\xi)|^2\,d\xi\\ &\le\liminf_{k\to\infty}\int w(\xi)|\widehat f_{j_k}(\xi)|^2\,d\xi\\ &=\liminf_{k\to\infty}\|f_{j_k}\|_{H^\sigma}^2\le M^2. \end{aligned} \] The weighted integral is therefore finite, so \[ f\in H^\sigma(\mathbb R^n), \qquad \|f\|_{H^\sigma}\le M. \]
3. Q.E.D.
Step 2 proves that the uniform \(H^\sigma\) bound passes to the \(L^2\) limit.