Strict topology between weak and norm topologies 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.

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?

If this is non-trivial and constructive, I’d be much obliged if an explicit example (or reference) is provided.
 A: 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 $X$.
A: Another example is the topology of uniform convergence on the norm-compact subsets of the dual. It is coarser than the norm topology (which is the topology of uniform convergence on the dual unit ball) because compact sets are bounded and strictly coarser since the dual unit ball is not compact. It is finer than the weak topology since finite sets are compact and it is strictly finer because every linear independent null sequence in the dual together with its limit is compact and not finite dimensional.
