Extract a subsequence converging a.e.

Question

Let $T>0$. Let $(g_n)_n\subseteq L^\infty([0, T])$ be uniformly bounded and $g\in L^\infty([0, T])$, such that for all $f\in L^1([0, T])$, $$\int_0^T f(t)g_n(t)dt \underset{n\rightarrow \infty}{\longrightarrow} \int_0^T f(t)g(t)dt.$$ Is it possible to extract a subsequence $(g_{n_k})_k$ such that $g_{n_k}\underset{k\rightarrow \infty}{\longrightarrow} g$ a.e.?

Answer 1

No, it's not true. Take $T=2\pi$ and consider

$$g_n(t) := \cos(nt).$$

Clearly, $\|g_n\|_{\infty} \leq 1$ for all $n \in \mathbb{N}$. Moreover, it follows from the Riemann-Lebesgue lemma that

$$\lim_{n \to \infty} \int (f(t) 1_{[0,2\pi]}(t)) \cos(nt) \, dt = 0$$

for all $f \in L^1$. Hence, $$\lim_{n \to \infty} \int_0^{2\pi} f(t) g_n(t) \, dt = \int_0^{2\pi} f(t) g(t) \, dt, \quad f \in L^1,$$

holds for $g:=0$. On the other hand, there cannot exist a subsequence $g_{n_k}$ which converges almost surely to $g=0$. Otherwise, it would follow from the dominated convergence theorem that

$$\lim_{k \to \infty} \int_0^{2\pi} g_{n_k}^2(t) \, dt = 0,$$

contradicting

$$\int_0^{2\pi} g_n(t)^2 \,d t =\pi, \qquad n \in \mathbb{N},$$

(which can be easily checked using the definition of $g_n$).