= Block averaging compactness counterexample
{title2=$(Tx)_k=m_k^{-1/q}\sum_{j\in B_k}x_j$}
For $1<p<\infty$, let $q=p/(p-1)$ and partition the coordinates into consecutive finite blocks $B_k$ of lengths $m_k\to\infty$. The displayed map defines a norm-one operator on the <l-p sequence space>. By <Hölder's inequality>, $|(Tx)_k|^p\leq\sum_{j\in B_k}|x_j|^p$, giving the upper <norm> bound. For the unit <vector> $e_j$ with $j\in B_k$, $Te_j=m_k^{-1/q}e_k$, so its <norm> tends to zero as $j\to\infty$.
Nevertheless, $x^{(k)}=m_k^{-1/p}\mathbf1_{B_k}$ has <norm> one and maps to $e_k$. Those images are pairwise distance $2^{1/p}$ apart. Thus $T$ is not a <compact operator>, and has <norm> exactly one. Vanishing images of a chosen basis do not alone test <compactness> on $\ell^p$ for $p>1$. The map is the adjoint of the isometric block embedding $(Uy)_j=m_k^{-1/q}y_k$, $j\in B_k$, on $\ell^q$.
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