I am covering now Lp spaces in my summer real analysis course and this problem from Folland related to the dual of Lp stumped me hard, it is problem 19 chapter 6 reads as follows:
We define $ \phi_n \in (l_\infty)^* $ by $ \phi_n(f) = n^{-1}(\sum\limits_{1}^n f(j)) $. I am asked to show this sequence $ \phi_n $ has a weak * cluster point $\phi$ and $\phi$ is an element of $ (l^\infty)^* $ that does not arise from an element of $ l^1 $.
I figured by cluster point they mean a limit point of the sequence in the weak * topology (which I still do not understand completely) but I have no idea how to show this sum is convergent as needed let alone showing its limit arises not from $ l^1 $. I am trying hard but cannot solve this. Could I please have some help on this?
First note that the sequence is bounded: $$ |\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty. $$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it admits a convergent subnet. Let $\phi$ be the (weak$^*$) limit of such a subnet.
So $\phi$ is a pointwise limit of a net $\{\phi_{n_\alpha}\}$. Now consider the elements $\delta_k\in\ell^\infty$, i.e. $$\delta_k(j)=\begin{cases}1,&\ j=k,\\ 0,&\ j\ne k\end{cases}$$ Note that $\phi(1)=1$, since $\phi_n(1)=1$ for all $n$. We have, for any $k\in \mathbb N$ and $n\geq k$, $$ \phi_n(\delta_k)=\frac1n, $$ so for all $k$ we have $\phi(\delta_k)=0$.
On the other hand if $\phi$ were given by some $a\in\ell^1$, then there exists $k$ with $a_k\ne0$ (otherwise we would have $\phi=0$) and this would imply $$ \langle a,\delta_k\rangle=\sum_ja_j\delta_k(j)=a_k\ne0. $$