- Let \(g _ { n }\) be a sequence of functions in \(L ^ { 1 } ( S ^ { 1 } , d \theta )\) where \(S ^ { 1 }\) is the unit circle \(\{ e ^ { i \theta } : 0 \leq \theta \leq 2 \pi \}\) We say that \(g _ { n } \ \to \ 0\) weakly if \(\int_{S^1} g_n(e^{i\theta}) f(e^{i\theta})\,d\theta \to 0\) as \(n \to \infty\) for all \(f \in C ( S ^ { 1 } )\)
Question: Suppose that \(\left\{ g _ { n } \right\}\) is a sequence in \(L ^ { 1 } ( S ^ { 1 } , d \theta )\) and \(\int_{S^1} e^{ik\theta} g_n(e^{i\theta})\,d\theta \to 0\) as \(n \to \infty\) for all \(k \in \mathbb { Z }\) . Need \(g _ { n } \to 0\) weakly? Give either a proof or a counterexample.
Solution
No. Let \(g_n(e^{i\theta})\da ne^{in\theta}\).
1. For each fixed \(k\in\ZZ\), \(\int_{S^1}e^{ik\theta}g_n(e^{i\theta})\,d\theta\to0\).
The integral is \(n\int_0^{2\pi}e^{i(k+n)\theta}\,d\theta\), which is \(0\) whenever \(n>\abs k\).
2. \(g_n\) does not converge weakly to \(0\).
The functional \(\Lambda_n(f)\da\int_{S^1}g_nf\,d\theta\) on the Banach space \(C(S^1)\) has norm \(\norm{g_n}_{L^1}=2\pi n\). If \(\Lambda_n(f)\to0\) for every \(f\), the uniform boundedness principle would give \(\sup_n\norm{\Lambda_n}<\infty\), which is false.
3. Q.E.D.
Steps 1 and 2 show that the sequence \(g_n\) is a counterexample.