MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 $Q$.

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.