Linear functional over complex sequences 
My try:
At part (a) I can show that M is non-empty (giving explicit example) and convex (by definition). Problems appear when i try show it is bounded (i have no idea at all) and weak*-closed. At part (b) instructor said to use Markov-Kakutani fixed point theorem but i can't check all the conditions of it. 

I don't know how to prove (c).
If part (c) is proved i can prove (d).
Any help is appreciated. Thank you!
 A: a) Let $m$ be a mean. Every element  $a \in l_{\infty}(\mathbb{Z})$ with $\|a\| \leq 1$ can be decomposed as $a = a_1 - a_2 +i(a_3 - a_4)$ with $a_i \in l_{\infty}(\mathbb{Z}), a_i \geq 0 \text{ and } \| a\| \leq 1 $ for all $1\leq i\leq 4$. Just take $a_1 = $ positive real part of $a$, $a_2 = $ negative real part of $a$ and similarly for $a_3$ and $a_4$. Now $a_i\geq 0 $ and $a_i \leq 1 \implies 0\leq m(a_i)\leq 1$. Adding them up we get $|m(a)| \leq 4$ which shows $\|m\| \leq 4$. 
c) It is enough to show for $A = \{1\}$. Let $B = \{1,2,..,n \}$. Now if $m$ is an invariant mean then $m(1_B) = n*m(1_A)$ . But $\|1_B\| \leq 1$ and so $|m(1_B)| \leq 4$. Hence $|m(1_A)| \leq 4/n$ for all $n$. Hence $m(1_A) = 0$. For odd integers, note that a translate of them is even integers and the fact that $m(1) = 1$. Hence you would get $1/2$. Using a similar logic you would get that $m(S) = 0$, by using the fact that the squares get sparse as $n$ increases and using $m(1_A) = 0$ for any finte set $A$. 
