This is a question I come across when reading my notes.
Find a sequence in the unit ball of $L^1$([0,1]) that has no convergent subsequence in the weak topology.
By a corollary of Banach-Alaoglu, we know that
If Banach space X is reflexive, the closed unit ball of X is weakly compact.
While, although generally $L^1$(μ) spaces is not reflexive, I think $L^1$([0,1]) is reflexive because [0,1] is of finite measure. Hence [0,1] should be compact?
I know sequentially compactness and compactness are not totally same, is this the problem here? Better, can someone give a sequence as an example?
Thanks in advance~
$L^1(0,1)$ is not reflexive, it is also not the dual space of another normed space. Hence boundedness of a sequence in $L^1$ does not mean much.
The prime example is the sequence $$ u_n(x):= n\chi_{[0,\frac1n]}(x) \quad x\in (0,1), $$ where $\chi_M$ is the characteristic function of the set $M$. The sequence satisfies $$ \|u_n\|_{L^1}=1, $$ but has no weakly converging subsequence. There is also no weak-star converging subsequence due to the lack of a predual space.
To see that $(u_n)$ does not converge weakly, note first that $$ \int_0^1 u_n \phi dx \to \phi(0) $$ for all continuous functions $\phi\in C([0,1])$. Thus, the functionals $f_n \in C([0,1])^*$ defined by $$ f_n(\phi) := \int_0^1 u_n \phi dx $$ converge weak-star in $C([0,1])^*$ to the Dirac delta functional. As this functional cannot be represented by a $L^1$-function, the sequence $(u_n)$ cannot have a weakly converging subsequence.