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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!

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1 Answer 1

up vote 1 down vote accepted

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 $Q$.

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@Pietro Majer: thank you very much. So I see, so we can (albeit non-constructively)define a kind of limit even though my functions do not converge on borders. It is actually worse than that, since the kind of functions I consider in my application usually have a limit. What other conditions on the function space could I add to forbid such a counter-example, if you have an idea? –  Aldric Degorre Jun 23 '11 at 12:28
    
For instance, in $L^p(\Omega)$ with $1\le p < \infty$ your property is true: $u(t)>0$ a.e. if and only if $\int_\Omega uv dt > 0 $ for all $0\neq v\ge 0$ in $L^q(\Omega) $. –  Pietro Majer Jun 23 '11 at 14:22

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