1

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?

Don John Prep
  • 500
  • 1
    This sequence does not converge weak* in $(\ell^\infty)^$! A cluster point of a sequence is (as far as I know) a limit point of a subsequence. BTW: Convergence in weak topology is just pointwise topology of the corresponding continuous linear functionals. – Zardo Jul 30 '15 at 19:27
  • 1
    Hint: You don't have to guess what "cluster point" means - it's in the index! (Not really the same as limit point, quite. If $x_j=0$ for all $j$ then $0$ is a cluster point of the sequence.) Hint: The sum converges because it's a finite sum. Hint: The limit you say you're having trouble with does not exist - nobody said it did. Hint: Banach-Alaoglu, – David C. Ullrich Jul 30 '15 at 19:29

1 Answers1

6

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 $$ \langle a,\delta_k\rangle=\sum_ja_j\delta_k(j)=a_k\ne0. $$

Martin Argerami
  • 217,281
  • 2
    Good catch David, thanks. – Martin Argerami Jul 30 '15 at 20:23
  • 2
    Do you mean "convergent subsequence" not "convergent subnet"? – Cameron L. Williams Jul 30 '15 at 20:26
  • 2
    I meant subnet, unless I'm missing something. As $\ell^\infty(\mathbb N)$ is not separable, I don't think that the weak$^*$-topology is metrizable, so I don't know if the unit ball is sequentially compact. – Martin Argerami Jul 30 '15 at 20:43
  • 2
    Oh you had just mixed up sequence and nets so I was a little lost but I get what you're saying now. – Cameron L. Williams Jul 30 '15 at 20:53
  • 1
    Since we have a sequence from the beginning, it is ok to talk about subsequence because any subnet for a sequence is in fact countable. – A.Γ. Jul 30 '15 at 22:14
  • 3
    @A.G.It doesn't follow that a convergent subsequence exists; see this example. Perhaps I'm misunderstanding what you're saying... Martin, $B(l_\infty^)$ is not weak sequentially compact (nor is $B(X^)$ if $X$ contains $\ell_\infty$ (or just $\ell_1(\Bbb R)$)). A proof of this can be found in Diestel's Sequences and Series in Banach Spaces*. – David Mitra Jul 30 '15 at 23:07