Weak convergence of $f_n$ in $L^p$ implies that of $|f_n|$? I am wondering that if $f_n$ converge weakly to $f$ in $L^p(\mathbb R^d)$, for $1<p<\infty$ then also $|f_n|$ converge weakly to $|f|$ in $L^p(\mathbb R^d)$?
I think it is true but I do not know how to prove it.
 A: No, it's not true in general. 
Consider $d=1$, $p=2$, $f_n(x)=\sin(nx)\chi_{(-\pi,\pi)}$, and $f(x)\equiv 0$. Riemann-Lebesgue Lemma implies $f_n \rightharpoonup f$. However, if we consider $g(x)=\chi_{(-\pi,\pi)}$, then one can compute that 
\begin{equation}
\int^\infty_{-\infty} |f_n|g \,dx= \int^\pi_{-\pi} |\sin(nx)|\,dx=4\not\to 0 = \int^\infty_{-\infty} |f|g\, dx.
\end{equation}
A: Consider the Hibert space $l^2(\mathbb{R})$ of complex square summable real sequences. For each $n \in \mathbb{N}$ let $(e^{(n)}_i)_{i=0}^{\infty} \in \ell^2$ be defined by $e^{(n)}_i = 1$ when $i = n$ and $e^{(n)}_i = 0$ otherwise. Then $(e^{(n)})_{n \in \mathbb{N}}$ is a sequence in $\ell^2$ which converges weakly to $0$. 
proof: Let $y:= (y_i)_{i=0}^{\infty} \in \ell^2$ be arbitrary, then $\langle y, e^{(n)} \rangle = y_n $ which converges to $0$ since $y$ is square summable. 
However, for every $n \in \mathbb{N}$ we have $\Vert e^{(n)} \Vert = 1$. 
A: Let $$f_n:=\sum_{k=1}^{2^n}(-1)^k\mathbf{1}_{\left[(k-1)2^{-n},k2^{-n}\right)}.$$
Using boundedness in $\mathbb L^p$ of $(f_n)_{n\geqslant 1}$ for $1\lt p\lt\infty$ and the fact that linear combinations of indicator functions of intervals of the form $((\ell-1) 2^{-N},\ell 2^{-N})$ are dense in $\mathbb L^{p'}$, with $1/p+1/p'=1$, we can see that $f_n\to 0$ weakly in $\mathbb L^p$, see here for more details.
However, since the interval $\left[(k-1)2^{-n},k2^{-n}\right)$, $1\leqslant k\leqslant 2^n$ are disjoint, $\lvert f_n\rvert=\mathbf 1_{[0,1)}$ which does not converge weakly to $0$.
