Linear functionals and continuous functions on open intervals

Question

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!

Answer 1

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 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 $Q$.