Let $f_n$ be a sequence of functions in $L^1(K, m ; \mathbb{C})$, $K$ metric compact and $m$ a Radon measure on $K$. Assume that $\| f_n \|_1 \leq 1$.
From what I understand, there is a subsequence converging weakly in $L^1$ if and only if the $f_n$ are equi integrable, which means :
$$\forall \epsilon>0 \exists \delta >0 \forall A, m(A) \leq \delta \forall n \int_A |f_n| dm \leq \epsilon $$
Now remark that the measures $f_n m$ have bounded total variation, and therefore there exists weakly-* converging subsequences. Would I be correct to assume that there are no weakly converging subsequences in $L^1(m)$ if and only if all weak-* adherence values of $f_n m$ are singular with respect to $m$ ?
I'm pretty sure of the implication "no weak $L^1$ adherence values $\Rightarrow$ all adherence values of $f_n m$ must be singular". The other implication seems clearly true but I have no argument.
All help appreciated !
Best Answer
Not true as stated. One can interlace a convergent sequence with a non-equi-integrable sequence; the result will still have a convergent subsequence. What is true (and what you probably meant) is that equi-integrability is equivalent to precompactness in the weak topology.
No. Let $K=[0,1]$ and consider the sequence $$f_n = 2^n \chi_{[2^{-n-1},2^{-n}]}-2^{-n-1}\chi_{[2^{-n},2^{-n+1}]}$$ It satisfies $\|f_n\|_1=1$ and has no weakly convergent subsequence in $L^1$ (test the convergence with $$\phi = \sum_{n \text{ even }} \chi_{[2^{-n},2^{-n+1}]}$$ to see that it fails.) Yet, the measures $f_n \,dm$ converge to $0$ in the weak* topology of $C(K)^*$.
On the other hand, it is true that having a subsequence that weakly converges in $L^1$ implies having a non-singular weak* cluster point. This is simply because weak $L^1$ convergence implies weak* convergence of associated measures.