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Let $X$ be a vector space. For two linear topologies $\tau_1$ and $\tau_2$ on $X$ we have the following implications: $\tau_1$ and $\tau_2$ have the

same closed convex sets $\Rightarrow$ same closed linear subspaces $\Rightarrow$ same dual (same closed hyperplanes)

If both $\tau_1$ and $\tau_2$ are locally convex then the above implications are equivalences (by the Mackey-Arens theorem). In this case, the topologies $\tau_1$ and $\tau_2$ are called compatible. Moreover, given a locally convex topology on $X$, there exists a finest locally convex topology with the same dual $X'$, namely the Mackey topology $\tau(X,X')$.

In contrast, if at least one of the topologies $\tau_1$ or $\tau_2$ is linear and not locally convex then in general the above implications are not equivalences. As an example, on $X = l^{\frac{1}{2}}$ consider its usual linear non-locally convex topology $\tau_{\frac{1}{2}}$ induced by the $F$-norm $\lVert \cdot \rVert_{\frac{1}{2}}$ and the norm topology $\tau_1$ induced from the Banach space $l^1$ with its usual norm $\lVert \cdot \rVert_1$. Then $\tau_1 \subsetneq \tau_{\frac{1}{2}}$ and both topologies have the same dual $l^\infty$ but $\tau_{\frac{1}{2}}$ has strictly more closed linear subspaces than $\tau_1$.

Moreover, given a locally convex (or just linear) topology on $X$, there need not exist a finest linear topology with the same dual [Kakol, "Note on compatible vector topologies" (1987)].

Questions:

  1. Does there exist a finest linear topology with the same closed linear subspaces?
  2. Does there exist a finest linear topology with the same closed convex sets?
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