I'm working on this problem
"Let $X=[0,1]$ with the usual measure. For $n=1,2,...$, let $f_n=n^{1/p}\chi_{[0,1/n]}$. Prove that $f_n$ does not converge weakly to $0$ in $L^1([0,1])$."
Yes, this problem does not clarify what $p$ is.
My approach: We know that $(L^1)^*$ is isometric to $L^{\infty}$. So $f_n\to 0$ weakly in equivalent to $\int_0^1 f_ng \to 0$ for all $g$ s.t. $\sup_x |g(x)|<\infty$. We need to find a $g$ s.t. the limit is not 0. This is the tricky part for me.
The interval $[1/n,1]$ does not matter, so let's say $g=0$ there. But since $g$ is bounded, say, by $M$. Then $\int_0^1 f_ng\leq M\int_0^1 f_n = M \frac{1}{n} n^{1/p}$.
So what I just proved, is that $f_n\to 0$ weakly when $p>1$ but I have no conclusion for $p\leq 1$. This is weird since the problem tells me to prove $f_n$ DOES NOT CONVERGES.
WHere was I wrong in the $p>1$ part? And how should I deal with this problem?