Let $Q$ be an open interval of ${\mathtt R}$ and $E$ be the space of continuous and bounded functions in $Q\to \mathtt{R}$.


I call $E^\*$ the set of linear functionals over $E$ and $E\_+^\*$ the subset of positive linear functionals.


My question is whether, for $x\in E$, the condition $\forall s\in Q,\; x(s) > 0$ is equivalent to $\forall f\in E\_+^\* \left[f\neq 0 \implies f(x)>0\right]$.


From right to left it is easy as it suffice to choose $f: x\mapsto x(s)$ and conclude.


But what about the other direction?


If $Q$ was compact, then I would just state that $x$ is greater than the constant $m = \min\_{s\in Q} x(s)$ and thus $f(x) > m\mu\_f(Q)$.


But I am interested in the case where $Q$ is open (or compact, but allowing $x$ to become zero on the border).  

Does this result still hold?  

What book/article can I reference for such a theorem or counter-example?  



(Note that in the actual problem I am trying to solve, Q is not a real interval but a finite and disjoint union of bounded and connected open sets in $\mathtt{R}^n$. I don't think this would change much of any consequence, but it is maybe better to state it now.)


Thank you in advance for your help!



Linear functionals and continuous functions on open intervals
No. Let w.l.o.g $Q:= (0,1)$. There is a bounded linear functional $f$ on $E$ such that for any $x\in E$ one has: $\liminf \_ {s\to 0} x(s)\le f(x) \le \limsup \_ {s\to 0} x(s) $. This functional is positive, still vanishes on some functions which are strictly positive on $Q$.


**rmk.** For the construction of $f$, you may directly refer to the [Banach limit](http://en.wikipedia.org/wiki/Banach_limit) functional $\phi:\ell^\infty\to\mathbb{R}$ and define $f$ by composing it with the linear bounded map $E\to\ell^\infty$ taking $x$ to the sequence $\{ x(1/n): n\in\mathbb{N}\_ + \}$. You can adapt this construction to more general non-compact $Q$.


