Recall that a Hausdorf topological space $X$ is called compactly generated if any set whose intersections with compacts are compact is closed. Locally compact and first countable spaces are compactly generated.



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> Let $E$ be a Banach space with the norm $\|\cdot\|$ and the unit ball $B\_E$. Let $|||\cdot|||\le \|\cdot\|$ be another norm, and let $\tau$ be a linear (or even locally convex) topology which is stronger than the $|||\cdot|||$-topology, but weaker than the $\|\cdot\|$-topology on $B\_E$. Does it follows that $B\_E$ with $\tau$ is compactly generated?
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Is a topology sandwiched between two norms compactly generated?
Let $\tau$ be the weak topology on the Banach space $\ell\_1$. It is known that each weakly convergent sequence in $\ell\_1$ is norm convergent (i.e., $\ell\_1$ has the Shur property). This property implies that $\tau$ is not compactly generated (otherwise it would be equal to the norm topology). Now consider the norm
$$|||(x\_n)\_{n\in\omega}|||=\sum\_{n=0}^\infty\frac{|x\_n|}{2^n}$$and observe that the topology $\tau$ is stronger that the topology generated by the norm $|||\cdot|||$ on the unit ball of $\ell\_1$.


If we want the topology $\tau$ to be stronger that the topology generated by the norm $|||\cdot|||$ we can replace $\tau$ by the supremum of the weak topology and the topology generated by the norm $|||\cdot|||$. This modified topology still will not be compactly generated (because of the same Shur property).


