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Let $p\in(1,+\infty)$ and let $(f_n)_{n\in\mathbb N}$ be a sequence in $\mathcal L^1(E) \cap \mathcal L^p(E)$ and suppose the sequence is bounded in $\mathcal L^p(E)$ and has strong convergence in $\mathcal L^1(E)$ to a function $f\in \mathcal L^1(E) \cap \mathcal L^p(E)$. I want to prove that $f_n \rightharpoonup f$ in $\mathcal L^p(E)$.

My attempt: I divided my problem into three steps:

  1. I choose a good subsequence $(f_{n_k})$ of $(f_n)$ such that converges weakly i.e. such that $$\lim_{n \rightarrow \infty} ||\varphi(f_{n_k}) - \varphi(f')||_p$$ for some $f'\in \mathcal L^p$ and for every $\varphi\in (\mathcal L^p)^*$.

  2. I show that the weak limit of the subsequence is exactly $f$.

  3. I show that the sequence is Cauchy hence by completeness of $\mathcal L^p$ I can conclude that converges and that the limit is the same of its subsequence (since it is unique).

Could it be correct? If so, I could write down the detailed solution here.

Thank you|

Jack Meow Meow
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    I don't see how you would get (or need) 3) since you only require $f_n \rightharpoonup f$ weakly in $L^p$. Steps 1) and 2) work. I would say 2) is the most tricky one but it is generally true that if $f_n \rightharpoonup \tilde f$ weakly and $f_n \to f$ a.e. then $f=\tilde f$. So it suffices to assume $f_n \to f$ a.e. and not in $L^1$. – Evangelopoulos Foivos Dec 10 '24 at 11:20
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    My idea was that "convergence in L^1" implies that exists a subsequence of $(f_n)$ that converges a.e. to $f$. Hence if I prove that the subsequence converges weakly it converges to $f$, but then I need to show that the sequence converges too right? that's why I put the point 3 – Jack Meow Meow Dec 10 '24 at 11:46
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    It suffices to show that every subsequence of $(f_n)$ has a further subsequence that converges weakly to $f$ in $L^p$. This is done by 1) and 2) – Evangelopoulos Foivos Dec 10 '24 at 13:27
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    Ahhhh perfect !!! I didn’t understand! Thank you so much – Jack Meow Meow Dec 10 '24 at 17:39

1 Answers1

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There is no need to go to subsequences. Holder's inequality gives the result.

Replacing $f_n$ by $f_n-f$ we may suppose $f=0$.

Let $g \in L^{q}$ ($\frac 1 p+\frac 1 q=1$). $L^{1}$ convergence shows that $\int f_ng1_{|g|\le N} \to 0$ for each fixed $N$.

Now, $|\int f_ng1_{|g|> N}| \le (\sup_k \|f_k\|_p) \|g1_{|g|> N}\|_q<\epsilon$ if $N$ is large enough (not depending on $n$). Put these together to complete the proof.

Kavi Rama Murthy
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