Let $(X,\|\cdot\|)$ be a normed space over complex or real field and $\tau_X$ is the topology generated by the norm $\|\cdot\|$, i.e. just norm topology. 1) Is there a locally convex topology $\tau$ such that $\tau$ is weaker than $\tau_X$ and stronger than $\sigma(X,X^*)$, where $X^*$ is Banach adjoint? 2) If $\tau$ is any locally convex topology on $X$, then is it true that $\tau=\sigma(X,Y)$ for some subset $Y$ of $X^+$, where $X^+$ means the algebraic adjoint of $X$, i.e. the set of all linear functionals on $X$?

  • $\begingroup$ For the second question, have you tried to check whether the norm topology itself can be of the form $\sigma(X,Y)$? My intuition is that it cannot. $\endgroup$ – Nate Eldredge Dec 11 '19 at 13:27

Let's first consider the case that $X$ is finite-dimensional. Since then there is only one Hausdorff vector space topology on $X$, and $\sigma(X,X^{\ast})$ is Hausdorff, the answer to the first question is "no" if we understand "weaker" and "stronger" in the strict sense [in the non-strict sense the answer is trivially "yes"]. The answer to the second question is "yes" for finite-dimensional $X$, even if we consider also non-Hausdorff topologies (let $N$ be the $\tau$-closure of $\{0\}$, then $\tau = \sigma(X, N^{\perp})$).

If $X$ is infinite-dimensional, the answers are different. Every $\sigma(X,Y)$-neighbourhood of $0$ contains a linear subspace of finite codimension, thus no topology having a neighbourhood of $0$ that only contains subspaces of infinite codimension — like for example norm-topologies, that have neighbourhoods of $0$ containing no nontrivial subspace at all — can be a $\sigma(X,Y)$ for any subspace $Y \subset X^{+}$.

This property allows an easy construction of a topology strictly between the norm topology and the weak topology if $X$ contains infinite-dimensional closed subspaces with infinite codimension (all common spaces do): Let $Z$ be such a subspace. Put $U = B + Z$, where $B$ is the unit ball of $X$. Then the locally convex topology generated by $\sigma(X,X^{\ast}) \cup \{U\}$ is strictly finer than $\sigma(X,X^{\ast})$$U$ doesn't contain a subspace of finite codimension — and strictly weaker than the norm topology — every neighbourhood of $0$ contains an infinite-dimensional subspace.

If $X$ is such that every closed subspace has either finite dimension or finite codimension (I don't know whether that's possible), this construction doesn't work.

  • $\begingroup$ If $f_0\in Y$ then the set $V=\{x\in X: |f_0(x)|\leq 1\}$ is $\sigma(X,Y)$ neighbourhood of 0. How the set $V$ could contain a linear subspace? I don't understand. $\endgroup$ – golomorfMath Dec 13 '19 at 5:17
  • $\begingroup$ $V$ contains the set $\{x \in X : f_0(x) = 0\} = \ker f_0$. This is a linear subspace of $X$, since $f_0$ is linear. As $f_0 \colon X \to \mathbb{C}$ (or $\mathbb{R}$), the codimension of $\ker f_0$ is $1$ (if $f_0 \neq 0$) or $0$ (if $f_0 = 0$). $\endgroup$ – Daniel Fischer Dec 13 '19 at 12:52
  • $\begingroup$ Thank you. Understood now. $\endgroup$ – golomorfMath Dec 14 '19 at 16:06

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