$f_n\underset{n}{\to} f\Leftrightarrow f_n \overset{\sigma (\mathcal{L}^1,\mathcal{L}^{\infty})}{\underset{n}{\rightharpoonup}} f$? Let :
$$\ell^1=\{\{x_n\}\in \mathbb{R}^{\mathbb{N}}~:~\sum_{n=0}^{\infty}{|x_n|<\infty}\}
$$

Schur's theorem
Let $\{x_n\}\subset \ell^1$ and $x\in \ell^1$. Then :$$x_n\underset{n}{\to} x\Leftrightarrow x_n \overset{\sigma (\ell^1,\ell^{\infty})}{\underset{n}{\rightharpoonup}} x$$

Let $(E,\mathcal{A},\mu)$ be a finite measure space. The set of all integrable real-valued function on $E$ is denoted bu $\mathcal{L}^{1}$.
Let $\{f_n\}\subset \mathcal{L}^{1} $ be a sequence and $f\in\mathcal{L}^{1} $.
Can we say that:
$$f_n\underset{n}{\to} f\Leftrightarrow f_n \overset{\sigma (\mathcal{L}^1,\mathcal{L}^{\infty})}{\underset{n}{\rightharpoonup}} f$$
An idea please .
 A: You cannot say this.
For example, let $(E,\mathcal{A},\mu)$ be a probability space supporting a sequence $(X_k)_{k \geq 1}$ of iid Bernoulli random variables (with $\mathbb{P}(X_k = 1) = \frac12$).
By Khintchine's inequality, there is $c > 0$ such that for any $x_1, \dots, x_n$ we have that
$$c^{-1}\sum_{j=1}^n x_j^2 \leq \mathbb{E}\bigg[\bigg|\sum_{j=1}^n X_j x_j\bigg|\bigg] \leq c \sum_{j=1}^n x_j^2.$$
Let $c_{00}$ be the space of sequences which are eventually $0$ equipped with the $\ell^2$-norm (notice this is not the usual choice of norm for this space). Notice that $c_{00}$ is dense in $\ell^2$ and the above inequality says that $c_{00}$ is isomorphically embedded in $L^1(E, \mathcal{A}, \mu)$ via the map $x \mapsto (X_j x_j)_{j \geq 1}$. In particular, this map extends to an isomorphic embedding of $\ell^2$ into $L^1(E, \mathcal{A}, \mu)$. However it is well known that $\ell^2$ does not have the schur property and so $L^1(E, \mathcal{A},\mu)$ also does not have the schur property.
A: Here is a simple counter-example: Consider $(0,1)$ with Lebesgue measure. Let $f_n(x)=\sin (2\pi n x)$ and $f(x)=0$ for all $x$. By Riemann - Lebesgue Lemma $f_n \to f$ weakly in $L^{1} [0,1]$. Howvever $f_n(x)$ does not tend to $0$ in $L^{1} [0,1]$.
