Does there always exist an injective map in the algebraic dual space $X^*$ of an infinite-dimensional t.v.s $X$?

Question

Let $X$ be a topological vector space with $\mathrm{dim}\ X=\infty$ and let $X^*$ be its algebraic dual space, that is the set of all linear maps from $X \to \mathbb C$.

Does there always exist a injective map in $X^*$? If so, why? Or can you give me an exmaple?

Thanks!

Answer 1

If an element $f$ of $X^{*}$ is injective then $X$ is necessarily one dimensional since any non-zero linear functinal $f$ is also surjective.

Answer 2

Let $(b_i)_{i \in I}$ be a (linear/Hamel) basis for $X$.

For each $i$ define a linear map $f_i: X \to \Bbb C$ by sending $b_i$ to $1 \in \Bbb C$ (and all $b_j, j \neq i$ to $0$) and extending linearly.

Then define $i: X \to X^\ast$ by linearity: Write $x=\sum_{i \in F} c_i b_i$ for some finite subset $F \subseteq I$ and complex coefficients $c_i$, in a unique way. Then $i(x)=\sum_{i \in F}c_i f_i \in X^\ast$ is a well-defined member of $X^\ast$ and $x \neq y$ implies $i(x) \neq i(y)$, so this defines a linear injection from $X$ into $X^\ast$.

The size of the basis doesn't matter (we need AC to get a basis of course) and I don't see any topological connection: the $f_i$ will not be continuous in general, and neither will $i$ be. But there was no question of that.

Answer 3

Disclaimer: This answer assumes that $V^*$ consists of continuous linear functionals.

The other answers explain why there is never an injective map. The situation can be worse. It may happen that there are no surjective maps, i.e. no non-zero functionals.

Consider $V:= L^p([0,1])$ with $0 < p < 1$, which becomes a topological vector space for the metric

$$d(f,g) = \int_{[0,1]} |f(x)-g(x)|^p dx$$

Then one can show that $V^*= \{0\}$ (by showing for example that the only convex neighborhood of $0$ is $V$ itself). In particular, there are no injective/surjective functionals.