Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 $C_0(R^n)$ be the space of continuous functions from $R^n$ to $R$ which vanish at infinity. Let $D$ be a subset of $C_0(R^n)$, I'd like to prove that if D is not dense in $C_0(R^n)$, then there exists a bounded measure $\mu$ such that $\int f d\mu=0, \forall f \in D$ and $\int g d\mu \neq 0 $ for some $g \in C_0(R^n)$. It this true?

All I can say is $\exists \epsilon >0$ and $g \in C_0(R^n)$ such that $||g - f||>\epsilon, \forall f \in D$, but then I don't how what to do.

share|cite|improve this question
If $D$ is a dense subset of the unit ball of $C_0(R^n)$ endowed with the uniform norm, then such a measure doesn't exist. – Davide Giraudo Mar 30 '14 at 9:31
@DavideGiraudo Thanks for this comment. So if I add the condition that $D$ is a linear subspace of $C_0(R^n)$, your previous answer is valid, right?(yes, I have a quick eye to catch it) – Petite Etincelle Mar 30 '14 at 9:45
up vote 2 down vote accepted

Assume that $D$ is a subspace of $C_0(\mathbf R^n)$. Denote $F\subset C_0(\mathbf R^n)$ the closure of the vector subspace generated by $D$. Then $F$ is strictly contained in $C_0(\mathbf R^n)$: pick some $g\in C_0(\mathbf R^n)\setminus F$ and with Hahn-Banach theorem, define a linear continuous functional $L\colon C_0(\mathbf R^n)\to\mathbf R$ such that $L(g)\neq 0$ and $L(f)=0$ for each $f\in F$.

Then use Riesz theorem to represent this linear functional as a bounded measure.

If $D$ is not a subspace, the result doesn't necessarily hold: take $D$ the unit ball.

share|cite|improve this answer
hmm -- why is the vector subspace generated by $D$ also not dense? What if $D$ is the unit ball? – Thomas Mar 30 '14 at 9:27
I misred the question. – Davide Giraudo Mar 30 '14 at 9:29
I've edited. ${}{}$ – Davide Giraudo Mar 30 '14 at 9:54

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.