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I am trying to prove this statement: given a sequence $\{f_n | f_n > 0, c_1 \leq \int_{\Omega} f_n \log{f_n} \leq c_2 \}$, here $\Omega$ is a bounded domain; can we prove the $L^1$ strong convergence of $f_n$?

With the given information, now I can only say that $f_n \rightharpoonup f$ in $L^1(\Omega)$ weakly, and $ \int_{\Omega} f \log f \leq \lim \inf \int_{\Omega} f_n \log f_n \leq c_2$ ... But this does not give me $L^1$ strong convergence.

Calvin Khor
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Ziyao Yu
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  • Hello, and welcome to math.stackexchange. Please be a little more precise when formulating your question. Is this $L^1(\Omega)$ where $\Omega$ is a general set with a measure? Or a bounded subset of $\mathbb{R}^n$? Or a general subset of $\mathbb{R}^n$? And why do you think this might be true? Where does the problem come from? – Hans Engler Dec 12 '21 at 23:15
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    Thannk you! In my question, the $\Omega$ is bounded. I am currently learning about Fokker Planck equation and related Wasserstein gradient flow. I have learned that the strong $L^1$ convergence for $p$ in Fokker Planck equation from the Felix Otto's paper, and I am wondering why that is true... Now I think that is from the PDE structure. – Ziyao Yu Dec 13 '21 at 03:12
  • @ZiyaoYu Welcome to math.stackexchange! Otto's work on the Fokker-Planck equation is a beautiful piece of mathematics. I hope you enjoy it! And I'd like to see more questions from this direction. ;) – MaoWao Dec 13 '21 at 19:37

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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 $\|f_n - f\|_{L^1} = 4 = const.$

Hans Engler
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  • Thank you! Can you be more specific about your reasoning? – Ziyao Yu Dec 13 '21 at 03:23
  • What does "converges weak-∗ in 1 (not weakly!)" mean? (Weak-$*$ topology with respect to which pairing?) –  Dec 13 '21 at 17:02
  • Weak-$\ast$ in the space of Radon measures (dual space of continuous functions). The limiting measure then is actually an $L^1$ function since the $L \log L$ bound implies equi-integrability. – Hans Engler Dec 13 '21 at 17:13
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    @HansEngler The convergence is actually weak convergence. By the Dunford-Pettis theorem, a bounded equi-integrable subset of $L^1$ is weakly relatively compact. – MaoWao Dec 13 '21 at 19:39
  • Thanks, I'll put this comment in the answer text for completeness. – Hans Engler Dec 13 '21 at 20:17