Will $L^1\log L^1$ bound gives strong $L^1$ convergence?

Solution 1:

The answer is no, not even for a subsequence. If this conclusion were correct, then the embedding $L^2(\Omega) \to L^1(\Omega)$ would be completely continuous for bounded $\Omega$. And that is not the case.

A subsequence of $f_n$ converges weakly in $L^1$, since the $L \log L$ bound implies equi-integrability, and by the Dunford-Pettis theorem, a bounded equi-integrable subset of $L^1$ is weakly relatively compact. Thank you to @MaoWao for clarifying this.

For a concrete counterexample, consider $$ \Omega = [0,2\pi], \; f_n(x) = 2 + \sin 2 n x \, . $$ Then all $f_n$ are bounded in $L^\infty$ and thus in $L \log L$, and their weak-$\ast$ limit in $L^1$ is $f(x) = 2$. But the convergence is not strong, since $\|f_n - f\|_{L^1} = 4 = const.$