I want to believe that this has an easy answer, but I’ve never considered it before and can’t seem to answer it now either.



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> Does every infinite-dimensional Banach space admit a locally convex vector topology that is strictly coarser than the norm topology and strictly finer than the weak topology?
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If this is non-trivial and constructive, I’d be much obliged if an explicit example (or reference) is provided.



Strict topology between weak and norm topologies
In a normed space, every weak neighborhood of zero contains a finite codimensional subspace. Consequently, if you have a weak to norm continuous linear operator between normed spaces, then the operator has finite rank.


Suppose that $X$ is an infinite dimensional Banach space. Take a closed subspace $Y$ that has infinite dimension and infinite codimension, and let $Q:X\to X/Y$ be the quotient map. Let $\tau$ be the topology on $X$ generated by $X^\*$ and the seminorm $|x| = \|Qx\|$. Obviously $\tau$ is a locally convex topology between the norm and weak topologies on $X$. On $Y$, the topology $\tau$ is the weak topology, so $\tau$ is not the norm topology on $X$. OTOH, $Q$ is $\tau $ to norm continuous, so by my opening remark, $\tau$ cannot be the weak topology on $X$.


