Let $p\in (1,\infty)$.
To construct a counter example in some other context, I am seeking a bounded sequence $(A_{n})$ of operators in $B(\ell^{p})$ with the following property.
$\|A_{n}c_{n}\|_{\infty}$ is eventually bounded away from $0$, where $c_{n}\in \ell^{p}$ is defined by $c_{n}(k) = \begin{cases}\frac{1}{n^{\frac{1}{p}}} & k < n\\ \\ \frac{1}{k} & k \geq n \end{cases}$
I am not sure if such a sequence exists, but to prove that none exists seems impossible to me.
More specifically what is needed is a bounded sequence $(c_{n})$ in $\ell^{p}$ with $\displaystyle\lim\limits_{n\to\infty}\|c_{n}\|_{p} > 0$ and $\displaystyle\lim\limits_{n\to\infty}\|c_{n}\|_{\infty} = 0$ satisfying the above condition involving $(A_{n})$.
If anyone has any insight it would be greatly appreciated.
Let $0\ne c\in l^p$. By Hahn-Banach, there exists a linear $T:l^p\rightarrow\mathbb{R}$ such that $Tc = 1$ and $|Tv|\le\frac{\|v\|_{l^p}}{\|c\|_{l^p}}$ for all $v\in l^p$. Let $A:l^p\rightarrow l^p$ be defined by $$ Av = (Tv,0,0,\dots) $$ It follows that $\|Av\|_{l^p} = |Tv|\le\frac{\|v\|_{l^p}}{\|c\|_{l^p}}$, and hence $\|A\|_{l^p\rightarrow l^p}\le\frac{1}{\|c\|_{l^p}}$. On the other hand, $\|Ac\|_{l^{\infty}} = 1$.
Hence, given any sequence $(c_n)$ in $l^p$ with $\liminf\limits_{n\rightarrow\infty}{\|c_n\|_{l^p}}>0$ (no assumptions needed on the $l^{\infty}$ norms), we can define $A_n$ as above such that $\|A_nc_n\|_{l^{\infty}} = 1$ for all $n$, and $(A_n)$ is bounded in $B(l^p)$.