Let $f_h:[0,1]\to\mathbb{R}$ be defined as $f_h(x)=h$ if $0\le x\le 1$ and $f_h(x)=0$ otherwise. Prove that it does not exist a subsequence $(f_{h_k})_k$ weakly convergent in $L^1(0,1)$.
Attempt of solution. Let's proceed by contraposition. I have to find a functional $\Phi\in(L^1(0,1))'$ such that $<\Phi,f_{h_k}>$ does not converge (in $\mathbb{R}$) to $<\Phi,f>$. By Riesz's Representation theorem, I can "represent" the elements of $(L^1(0,1))'$ through functions in $L^{\infty}(0,1)$.
$1)$If I find $g\in L^{\infty}(0,1)$ such that $\int_0^1f_{h_k}gdx$ does not converge to $\int_0^1fgdx$ I am done.
My doubt is: our professor suggested us to look for a sequence $(g_i)_i\in L^{\infty}(0,1)$ such that $lim_{k\to\infty}\int_0^1f_{h_k}g_idx=1$ for every $i$ fixed and $lim_{i\to\infty}\int_0^1fg_idx=0$.
I can't understand why this hint is equivalent to $1)$