Vector Space with Trivial Dual

Question

How to construct a Vector Space $E$ (non trivial) such that, the only continuous linear functional in $E$ is the function $f=0$?

Answer 1

In addition to what Asaf has written: There are non-trivial topological vector spaces with non-trivial topologies which have a trivial dual. I think in Rudin's "Functional Analysis" it is shown that the $L^p$-spaces with $0<p<1$ are an example of this.

Answer 2

Such vector space is not going to be a normed vector space, and for this you really just need any vector space whose topology is trivial.

However note that assuming the axiom of choice every vector space is isomorphic to a normed vector space (e.g. by embedding into an appropriate subspace of $L^2(X)$ for some large enough $X$), so there will be a topology which is richer in continuous functionals.

If the axiom of choice fails badly then it is consistent that there is a vector space that is not only without continuous functionals, but without any functionals. This means that it cannot be topologized in such way that it has any nontrivial functionals.

Answer 3

Continuing Asaf's answer ... You cannot exhibit a nonzero continuous linear functional (nor, indeed, any nonzero linear functional) for the Banach space $l^\infty / c_0$. Such things exist by the Axiom of Choice.

Answer 4

Here is an explicit example doesn't seem too artificial: Let $$E = \mathcal{L}^0[0,1] = \{f:[0,1] \to \mathbb{C} \mid f\text{ measurable}\}$$ with the topology of convergence in measure (Lebesgue measure on $[0,1]$). This topology is (pseudo-)metrizable with $$d(f,g) = \int_{[0,1]} \min\{|f(t)-g(t)|,1\}\,dt\,.$$ Then $E' = \{0\}$. Indeed, if $\phi \in E'$ such that there is some $f \in E$ with $\phi(f) \neq 0$, let $g_n$ be the classical typewriter sequence: $$g_{2^k+m}(t) = \begin{cases} 1 & \text{if } t \in \left[\tfrac{m}{2^k},\tfrac{m+1}{2^k}\right] \\ 0 & \text{else} \end{cases} $$ for $k \in \mathbb{N}$ and $m \in \{0,\ldots,2^k-1\}$. Then $$0 < |\phi(f)| = \left|\phi\left(\sum_{n=2^k}^{2^{k+1}-1} g_nf\right)\right| \leq \sum_{n=2^k}^{2^{k+1}-1} |\phi(g_n f)|$$ for each $k \in \mathbb{N}$, i.e. for each $k \in \mathbb{N}$ there exits some $n_k \in \{2^k,\ldots,2^{k+1}-1\}$ such that $$|\phi(g_{n_k} f)| \geq \frac{1}{2^k} |\phi(f)|\,.$$ Let $h_k = 2^k g_{n_k} f$. Then $h_k \to 0$ in measure, as $\mu(|h_k| > 0) \leq \frac{1}{2^k}$, but $|\phi(h_k)| \geq |\phi(f)| > 0$, a contradiction.