TITLE: Vector Space with Trivial Dual
QUESTION [5 upvotes]: How to construct a Vector Space $E$ (non trivial) such that, the only continuous linear functional in $E$ is the function $f=0$?

REPLY [2 votes]: 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.