# Is a topology sandwiched between two norms compactly generated?

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.

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?

• The discrete topology is compactly generated, so you're really asking whether a topology that contains a compactly generated Hausdorff topology must be c. g. – Tom Goodwillie Dec 28 '20 at 21:05
• @TomGoodwillie yeah, you are right, and this doesn't seem like it's true, so I'll refocus the question around the specific thing that i need – erz Dec 28 '20 at 22:50
• Given any space $(X,\tau)$ it's possible to refine $\tau$ to a compactly generated topology with the same compact sets as $\tau$, so every topology is contained in a very close but compactly generated one without having to go all the way to the discrete one. Going from $\tau_1$ cg to $\tau_2$ cg seems very unlikely to me even though I don't have a counterexample – Alessandro Codenotti Dec 28 '20 at 22:57

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).