Let $\,\,\,f_h : [0,1] \rightarrow \mathbb{R}$ the functions defined by $f_h(x) = h$ if $0 \le x \le \frac{1}{h}$ and $0$ otherwise. Prove that there do not exist a subsequence $(f_{h_{k}})_k$ and $f \in L^1((0,1))$ such that $f_{h_{k}} \rightharpoonup f $ in $L^1((0,1))$.
I have no idea how to resolve this exercise, can anyone help me?
Suppose that there exists a subsequence $\left(f_{h_k}\right)_{k\geqslant 1}$ and a function $f$ such that $f_{h_k}\to f$ weakly in $\mathbb L^1$. For positive integers $M$ and $N$, define the function $$ g_{M,N}:=\mathbf 1\left\{x\in [0,1]\mid f(x)\gt \frac 1M\mbox{ and }x\gt \frac 1N\right\}. $$ Then for $k$ such that $h_k\gt N$, the function $f_{h_k} g_{M,N}$ is identically zero hence $$ 0=\int fg_{M,N}\geqslant \frac 1M \lambda\left\{x\in [0,1]\mid f(x)\gt \frac 1M\mbox{ and }x\gt \frac 1N\right\} $$ which shows that $f(x)\leqslant 1/M$ on $[1/N,1]$. Since this is valid for any $M$ and $N$, we derive that $f(x)\leqslant 0$ almost everywhere. Applying the definition of weak convergence to the function $g$ constant equal to $1$ leads to a contradiction.