Weak convergence of any uniformly bounded sequence of functions in $L^p (\Omega)$ which converges to zero in $L^1$-norm.

Question

Let $(\Omega, \mathcal F, \mu)$ be a measure space and $\{f_n\}_{n \geq 1}$ be a uniformly bounded sequence of functions in $L^p (\Omega)$ $(1 < p < \infty).$ If $\|f_n\|_1 \to 0$ then $f_n \rightharpoonup 0.$

I have tried to use Eberlein–Šmulian theorem to conclude that $f_{n_k} \rightharpoonup \hat f.$ If $\hat f \neq 0$ a.e. then there exists a measurable subset $E \subseteq \Omega$ of finite positive measure such that $\hat f \neq 0$ on $E.$ If needed, by reversing the sign of $\hat f$ we may assume that $\hat f \geq \varepsilon$ on $E$ for some $\varepsilon > 0.$ Then first of all $$\tag {1}\left \lvert \int_{\Omega} f_n \chi_E\ d\mu \right \rvert \leq \int_{\Omega} \left \vert f_n \right \rvert \chi_E\ d\mu \leq \|f_n\|_1 \to 0.$$

On the other hand, $$\int_{\Omega} f_{n_k} \chi_E\ d\mu \geq \frac {\varepsilon} {2} \mu (E),$$

which contradicts $(1).$ Hence $\hat f = 0$ $\mu$-a.e. This shows that $f_{n_k} \rightharpoonup 0.$ But I need to show that $f_n \rightharpoonup 0.$ So I modify my argument by starting with a subsequence of $\{f_n\}_{n \geq 1}$ and then implementing the same idea as above we can show that it has a further subsequence which converges to zero weakly and hence so does the main sequence.

I think that my argument works well. Do anyone have any suggestion on it?

Thanks for reading.

Answer 1

Your argument works well but the beginning can be formulated in a better way by saying that each subsequence of $(f_n)$ has a further subsequence that convergences weakly to some $f$ and then use your argument in order to show that $f=0$.

Alternatively, one can proceed as follows. Let $q$ be the conjugate exponent of $p$ and let $g\in\mathbb L^q$. For each $R$, \begin{align} \left\lvert \int_{\Omega}f_ngd\mu\right\rvert&\leqslant \int_{\Omega} \left\lvert f_ng\right\rvert d\mu \\ &\leqslant \int_{\Omega}\left\lvert f_ng\right\rvert\mathbf{1}_{\lvert g\rvert\leqslant R} d\mu+\int_{\Omega}\left\lvert f_ng\right\rvert\mathbf{1}_{\lvert g\rvert\gt R} d\mu\\ &\leqslant R\lVert f_n\rVert_1+\lVert f_n\rVert_p\lVert g \mathbf{1}_{\lvert g\rvert\gt R}\rVert_q\\ &\leqslant R\lVert f_n\rVert_1+\sup_{\ell\geqslant 1}\lVert f_\ell\rVert_p\lVert g \mathbf{1}_{\lvert g\rvert\gt R}\rVert_q. \end{align} Since $\lVert f_n\rVert_1\to 0$, we derive that for each $R$, $\limsup_{n\to\infty}\left\lvert \int_{\Omega}f_ngd\mu\right\rvert\leqslant\sup_{\ell\geqslant 1}\lVert f_\ell\rVert_p\lVert g \mathbf{1}_{\lvert g\rvert\gt R}\rVert_q$ and by monotone convergence, $\lim_{R\to\infty} \lVert g \mathbf{1}_{\lvert g\rvert\gt R}\rVert_q=0$.