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I have a question that requires a little introduction:

I am dealing with the Banach space $\ell_\infty(\mathbb{Z})$ (all bounded functions $f:\mathbb{Z} \rightarrow \mathbb{C}$) with the supremum norm and I have a definition of a mean on $\ell_\infty(\mathbb{Z})$ that is a linear functional $m \in\ell_\infty(\mathbb{Z})^*$ such that $m(f)\ge 0$ for all positive $f$ and $m(1)=1$. Furthermore, a mean is invariant if $m\circ \tau_n = m$ for all $n\in\mathbb{Z}$ where $\tau_n:\ell_\infty(\mathbb{Z})\rightarrow \ell_\infty(\mathbb{Z})$, $\tau_n(f) = f(k-n)$ for $k\in\mathbb{Z}$. I have shown that invariant means do exists and that for an invariant mean, $m$, $m(1_A) = 0$ where $A$ is a finite set.

What I need to show is that $m(1_B)=0$ where $B=\{n^2\mid n\in\mathbb{N}\}$. I have been told to look at the set $B_N = \{n^2\mid n\ge N\}$ for some natural number $N$, and that I should remember that the sets $S_N, S_{N}+1, \ldots, S_{N} + 2N$ are disjoint.

I know $m(1_{B\backslash B_{N}}) = 0$ because $B\backslash B_N = \{n^2\mid n< N\}$ is a finite set. I also know that $1-1_{B_N}$ is positive and in $\ell_\infty(\mathbb{Z})$, so $m(1-1_{B_n}) \ge 0$ i.e. $m(1) - m(1_{B_N}) = 1-m(1_{B_N}) \ge 0$, so $m(1_{B_N}) \le 1$. But I don't really know where to go from here. I am not sure how to use the hints. Anyone with any ideas?

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1 Answer 1

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Suppose that $m(1_B)>0$. Then $m(1_{B_N})=m(1_B)>0$ for all $N$. For $k=0,\dots,2N$ let $f_k=1_{B_N+k}$, and let $C=\bigcup_{k=0}^{2N}(B_N+k)$; then $$m(1_C)=\sum_{k=0}^{2N}m(f_k)=(2N+1)m(1_{B_N})\;.$$

Do you see the contradiction looming as $N$ increases?

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I see that if $N\rightarrow \infty$ then $(2N + 1)\rightarrow \infty$ and I know that $m(1_C) \le 1$. So because $m(1_{B_N})$ is a fixed number, say $\alpha >0$ (because $m(1_{B_N}) = m(1_B)$) then I have a contradiction, because then $m(1_C)\rightarrow \infty$. Thanks a lot! –  Helen Oct 31 '12 at 9:30
    
@Helen: You’re welcome! –  Brian M. Scott Oct 31 '12 at 15:52

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