# 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.

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 $$X$$.