$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ implies the convergence of $f_n$ to $f$ in $\textit{L}^p$ with the right hypotesis? I think this might work, but how i don't know:
Let $p\in (1,+\infty)$ and $\{f_n\}$, $f\in \textit{L}^p(E)$, if we know that:

*

*$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ as $n\longrightarrow +\infty$;

*$\forall$ measurable $F\subseteq E$ with $|F| < +\infty \Longrightarrow \int_Ff_n\longrightarrow \int_Ff$, as $n\rightarrow +\infty$;

is it true that $f_n$ converges to $f$ in $\textit{L}^p(E)$ in norm (strong convergence)?
Any suggestions?
 A: Here is a sketch for a proof:

*

*The sequence is bounded, hence it has a weakly convergent subsequence.

*The second bullet gives that the weak limit is $f$.

*A subsequence-subsequence argument shows that the entire sequences converges weakly.

*Weak convergence and convergence of norms implies strong convergence in $L^p$, $1 < p <\infty$.

A: I'll procede as suggested by gerw.
For the first of the two hipothesys i have that
$$\forall \epsilon > 0 \mbox{, } \exists N\in \mathbb{N} \mbox{ : } \forall n \ge N \mbox{, } \bigl| ||f_n||_p-||f||_p \bigr|< \epsilon$$
so $\forall n\ge N$, $||f_n||_p <||f||_p + \epsilon$, now letting $M:=\max\{||f1||_p$, ...,$||f_{N-1}||_p$, $||f||_p+\epsilon\}$, we obtain $||F_n||_p \le M$, that is the boudedness of the sequence. With what we have we can apply:
Theorem 1.1. $(1 < p < +\infty)$ Let $\{f_n\}$ be a bounded sequence in $\textit{L}^p(U)$. Then there exists a subsequence $\{f_{n_k}\}$ which converges weakly to some $f\in \textit{L}^p(U)$. For a proof click here.
So given $\{f_{n_k}\}_{k\in \mathbb{N}}$, $\overline{f}\in \textit{L}^p(E)$ such that $f_{n_k} \rightharpoonup \overline{f}$ we have shown the first step.
Now, passing to the second one, by the definition of weak convergence:
$$\int_Egf_{n_k} \longrightarrow \int_Eg\overline{f} \mbox{ as } k\rightarrow +\infty \mbox{, } \forall g\in \textit{L}^p(E) \mbox{ (for Riesz's rep. theorem, with } \frac{1}{p}+\frac{1}{q}=1\mbox{).}$$
So for $g=\chi_F\in  \textit{L}^p(E)$ if $|F|<+\infty$, gives us
$$\int_Ff_{n_k} \longrightarrow \int_F\overline{f} \mbox { as } k\rightarrow +\infty \mbox{, } \forall \mbox{ measurable } F \mbox{ s.t. } |F| <+\infty \mbox.$$
Furthermore, by hipothesys
$$\int_Ff_n \longrightarrow \int_Ff \mbox { as } k\rightarrow +\infty \mbox{, } \forall \mbox{ measurable }F \mbox{ s.t. } |F| <+\infty$$
and since $\{\int_Ff_{n_k}\}_{k\in \mathbb{N}}\subseteq \{\int_Ff_n\}_{n\in \mathbb{N}}\subset \mathbb{R}$, also the real subsequence converges.
So
$$\forall F:|F|<+\infty \mbox{, } \int_Ff=\int_F\overline{f} \Longrightarrow f=g \mbox{ a.e. on these } F\mbox;$$
if we pick $F_k:=B(0,k)$, then $\forall k\in \mathbb{N}$, $f=\overline{f}$ on $F_k$ except on a $N_k$ such that $|N_k|=0$; so, observing that $E=\cup_{k\in \mathbb{N}}F_k$, we obtain that $f=\overline{f}$ on $E$ except on the measurable set $\cup_{k\in \mathbb{N}}N_k$, but $\bigl|\cup_{k\in \mathbb{N}}N_k\bigr|\le \sum_{k\in \mathbb{N}}|N_k|=0$, and we can deduce that $f=\overline{f}$ a.e., which means that they are the same element in $\textit{L}^p(E)$. This concludes the second point.
Now to show the third point we first suppose that $f_n \not \rightharpoonup f$, then sow it is absurd, implying it is not true, which means weak convergence is true.
So if $f_n \not \rightharpoonup f$, then
$$\exists \overline{g}\in \textit{L}^p(E) \mbox{ s.t. } \int_E\overline{g}f_n \longrightarrow \int_E\overline{g}f \mbox{ as } n\rightarrow +\infty \mbox{ is NOT true.}$$
This means that there exists a subsequence $\{f_{n_i}\}_{i\in \mathbb{N}}$ that doesn't converge weakly to $f$ and neither do all of it's subsequences, but by hipothesys $||f_n||_p \longrightarrow ||f||_p$ so that also $||f_{n_i}||_p \longrightarrow ||f||_p$ as $i\rightarrow +\infty$; also $||f_{n_i}||_p \le M$ defined as before.
So now we can apply to $\{f_{n_i}\}_{i\in \mathbb{N}}$ the first two points already proven to obtain that
$$\exists \{f_{n_{i_k}}\}_{k\in \mathbb{N}} \mbox{ subsequence, s.t. } \int_E\overline{g}f_{n_{i_k}} \longrightarrow \int_E\overline{g}f \mbox{ as } k\rightarrow +\infty \mbox,$$
which is not possible since we said the contrary before. So we deduce that $f_n \rightharpoonup f$, proving also the third point.
To finish the proof we just have to apply the Radon-Riez Theorem (Theorem 5 here)
In fact under the condition $f_n \rightharpoonup f$, the following are equivalent as $n \rightarrow +\infty$:
$$||f_n||_p \longrightarrow ||f||_p \iff f_n\longrightarrow f \mbox{ in norm }\textit{L}^p(E) \mbox{ (strong topology).}$$
Thanks to @gerw who gave me valuable hints.
