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Let $\{f_n\}_{n \in \mathbb{N}} \subset L^1$ be such that $f_n \rightarrow f$ in $L^1$.
And suppose that $\forall n \, \, \Vert f_n \Vert_p \le 1$ (with the usual $L^p$-norm) for some $p>1$.

Prove that $\Vert f \Vert_p \le 1$.

The hint given by the exercise is about using Fatou's Lemma, but I can't see useful ways to apply it. This is my attempt: $$ \int \underline{\lim} \vert f_n \vert^p \le \underline{\lim} \int \vert f_n \vert^p = \underline{\lim} \Vert f_n \Vert^p\le 1 $$ But I'm stuck here and cannot reconduct to $f$ (I cannot deduce any pointwise convergence).

SigmaBoy
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    Let $(X,||\cdot||)$ be a normed linear space and ${x_n}$ converges to $x$ w.r.t. $||\cdot||$, then $$||x_n||\to ||x||.$$ – Sumanta Sep 10 '20 at 12:22
  • @Sumanta But he have different norms here. We know that $f_n\to f$ in $L^1$, not in $L^p$. – Mark Sep 10 '20 at 12:28
  • Another thing that you can use $f_n\xrightarrow{\text{measure}}f$, so we have subsequence $f_{n_k}\xrightarrow{\text{a.e.}}f$. – Sumanta Sep 10 '20 at 12:28
  • $$ \int|f|^p=\int \underline{\lim} \vert f_{n_k} \vert^p \le \underline{\lim} \int \vert f_{n_k} \vert^p = \underline{\lim} \Vert f_{n_k} \Vert^p\le 1 $$ – Sumanta Sep 10 '20 at 12:30
  • See here $L^1$ convergence implies convergence in measure. https://math.stackexchange.com/questions/1281421/convergence-in-lp-implies-convergence-in-measure – Sumanta Sep 10 '20 at 12:31
  • And measure convergence implies a convergence of a subsequence almost everywhere https://math.stackexchange.com/questions/1006091/convergence-in-measure-implies-convergence-almost-everywhere-of-a-subsequence – Sumanta Sep 10 '20 at 12:33

2 Answers2

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$\|f_n-f\|_1 \to 0$ implies there is a subsequnce $f_{n_k}$ converging a.e. to $f$. So $\int |f|^{p} \leq \lim \inf \int |f_{n_k}|^{p} \leq 1$.

Kavi Rama Murthy
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If $f_n \to f$ in $L^1$ then $f_n \to f$ in measure. Then use this Exercise on convergence in measure (Folland, Real Analysis)