Why is there no subsequence of $\frac{1}{n}\boldsymbol 1_{[0,n]}(x)$ has no weak subsequence that converge weakly in $L^1$?

Question

Let $f_n(x)=\frac{1}{n}\boldsymbol 1_{[0,n]}(x)$. This sequence is bounded in $L^1(\mathbb R)$ since $\|f_n\|_{L^1}=1$. But why is there no subsequence that convergent weakly ? I know that if such subsequence exist (still denote $f_n$), then $\|f_n\|_{L^1}=1$ Let denote $f$ it's limit. Then, since $f_n\to 0$ pointwise then $f=0$ because $$\lim_{n\to \infty }\int_{\mathbb R}f_n\varphi=\int_{\mathbb R}\lim_{n\to \infty }f_n\varphi=0.$$

Therefore $\|f\|=0$. Now, for me there is no reason to have $$\lim_{n\to \infty }\|f_n\|=0.$$ Indeed, using semi lower continuity of the norm, we have that $$1=\liminf_{n\to \infty }\|f_n\|\geq \|f\|=0,$$ and thus, there is no contradiction. But I know that I should have a contradiction, but where ?

Answer 1

As you noticed, if a subsequence $\left\{f_n\right\}$ of $\frac{1}{n}\mathbf{1}_{[0,n]}$ converges, it must converge to $0$. But for a sequence $\left\{f_n\right\}\subset L^1(\mathbb{R})$ we have $f_n\to 0$ weakly in $L^1(\mathbb{R})$ if and only if $$\int_{-\infty}^{+\infty} f_n\varphi \to 0,\qquad \forall \varphi \in L^{\infty}(\mathbb{R}) $$ Which fails, for instance, with $\varphi=1$.

Answer 2

Let $\{n_k\}_{k \geq 1}$ be any subsequence. Then we readily compute $$\lim_{k \to \infty} \int_{\mathbb{R}} 1 \cdot \left(f_{n_k}(x) - 0\right)dx = \lim_{k \to \infty} 1 = 1 \neq 0.$$