Recall that the inner product on \(L^2(\mathbb{R}^d)\) is given by
\[(f, g) = \int_{\mathbb{R}^d} f(x) \overline{g(x)} \, dx, \quad \text{for } f, g \in L^2(\mathbb{R}^d),\]
which induces the \(L^2\)-norm
\[\|f\|_{L^2} = (f, f)^{1/2}.\]
- If the sequence of functions \(\{f_n\}_{n=1}^\infty\) in \(L^2(\mathbb{R}^d)\) satisfy that \(\|f_n\|_{L^2} = 1\), show that there exists a subsequence of functions \(\{f_{n_j}\}_{j=1}^\infty\) such that \(f_{n_j}\) converges weakly to some function \(f\) in \(L^2(\mathbb{R}^d)\), i.e.,
\[(f_{n_j}, g) \to (f, g) \quad \text{for all } g \in L^2(\mathbb{R}^d).\]
- If \(f_n \rightharpoonup f\) weakly in \(L^2(\mathbb{R}^d)\) and \(\|f_n\|_{L^2} \to \|f\|_{L^2}\) as \(n \to \infty\), show that \(\|f_n - f\|_{L^2} \to 0\) as \(n \to \infty\).
Solution
1. Prove part (a).
The space \(L^2(\mathbb R^d)\) is a Hilbert space, hence reflexive. Therefore its closed unit ball is weakly compact. By the Eberlein–Smulian theorem, weak compactness in a Banach space is equivalent to weak sequential compactness.
Since every \(f_n\) lies in the unit sphere, in particular in the closed unit ball, there is a subsequence \((f_{n_j})\) and some \(f\in L^2(\mathbb R^d)\) such that \[ f_{n_j}\rightharpoonup f. \] Equivalently, \[ (f_{n_j},g)\longrightarrow(f,g) \qquad\text{for every }g\in L^2(\mathbb R^d). \]
2. Prove part (b).
Weak convergence gives \[ (f_n,f)\longrightarrow(f,f)=\|f\|_2^2. \] Using the Hilbert-space norm identity, \[ \begin{aligned} \|f_n-f\|_2^2 &=\|f_n\|_2^2+\|f\|_2^2-2\operatorname{Re}(f_n,f). \end{aligned} \] By hypothesis, \[ \|f_n\|_2^2\longrightarrow\|f\|_2^2, \] and the weak-convergence term tends to \(\|f\|_2^2\). Hence \[ \|f_n-f\|_2^2\longrightarrow \|f\|_2^2+\|f\|_2^2-2\|f\|_2^2=0. \] Therefore \[ \boxed{\|f_n-f\|_2\to0.} \]
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P-JHUU67RA2Weak convergence plus norm convergence implies strong convergence in $L^2$variant-of