# Is it true that $\dim X\leq \dim X^*$ for locally convex topological vector spaces?

A couple of years back, remilt asked whether the dimension of the topological dual of a normed space is always greater or equal than the dimension of the original space. The answer turned out to be positive. It feels natural to ask the same question for the wider class of topological vector spaces.

Since there exist topological vector spaces $X$ for which $X^*$ is trivial (for example the $\ell_p$ spaces for $0<p<1$), lets concentrate on locally convex topological vector spaces, where it is a standard fact that $X^*$ is always nontrivial. Also in such spaces we have Hahn-Banach type theorems, which are always useful for constructing distinct elements of $X^*$.

Definition: For a topological vector space $(X, \tau)$, we denote by $X^*$ the vector space of linear and continuous functionals from $X$ to $\mathbb{R}$.

Question: Let $(X,\tau)$ be a locally convex topological vector space. Is it true that $\dim X\leq \dim X^*$?

In the case of normed spaces the proof consisted of three simple steps: The dimension of the space is equal to its cardinality, the cardinality is determined by the density character and finally density character increases when we take the dual. The first step is still true in the case of t.v.s., but the second and third need to be modified a lot, or changed altogether.

Partial positive answers, ex. Frechet spaces, are also welcomed.

Consider the Fréchet space

$$F := \mathbb{C}^{\mathbb{N}} = \prod_{n\in \mathbb{N}} \mathbb{C}.$$

Its dual space is

$$F^{\ast} = \bigoplus_{n \in \mathbb{N}} \mathbb{C},$$

which has dimension $\aleph_0 < 2^{\aleph_0} = \dim F$.

So the answer is negative even for Fréchet spaces.

• I didn't see that one coming! – tree detective Mar 7 '17 at 14:07