Let $c=\{a:\mathbb{N}\rightarrow \mathbb{C}: \exists \alpha\in \mathbb{C} \textrm{ so that } \lim\limits_{n\rightarrow \infty} a(n)=\alpha=:L_c(a)\}\subset \ell^\infty$, where $\ell^\infty$ is the Banach space of all bounded sequences with the supremum norm $\|a\|_\infty=\sup\limits_{n\in \mathbb{N}}|a(n)|$. Then the limit-operator $L_c:c\rightarrow \mathbb{C}$ is a continuous linear map ($c$ has the induced topology from $\ell^\infty$). Let $S:\ell^\infty\rightarrow\ell^\infty$ be the left-shift operator, $(Sa)(n)=a(n+1)$, $n\in \mathbb{N}$, $a\in \ell^\infty$. Clearly $L_c S=L_c$ on $c$. Using the Hahn-Banach theorem, it can be shown that there exists a continuous linear map $L:\ell^\infty\rightarrow \mathbb{C}$ such that
- $L|_c=L_c$, and
- $LS=L$ on $\ell^\infty$.
Question: Does there exist an $L:\ell^\infty\rightarrow \mathbb{C}$, which, besides 1. and 2., also satisfies:
- For all $k\in \mathbb{N}$, $LP_k =L$ on $\ell^\infty$, where for $k\in \mathbb{N}$, $P_k$ is the following `projection map': $(P_ka)(n)=a(n^k)$, $n\in \mathbb{N}$, $a\in \ell^\infty$.
