# Weak topologies in normed spaces

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 $$X$$?

• 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. – 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.
• 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. – golomorfMath Dec 13 '19 at 5:17
• $V$ 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 $f_0 = 0$). – Daniel Fischer Dec 13 '19 at 12:52