3
$\begingroup$

Are there theorems or results to show that if for every $\varphi\in \mathcal{C}_0^\infty(\mathbb{R})$ we have, $$\int_{\mathbb{R}} \varphi^k(x)\mu(dx) \leq C$$

Then $\mu(dx) = f(x)dx$ and $f\in \mathcal{C}^{\tilde{k}}(\mathbb{R})$ ?? where $\tilde{k}$ and $k$ might be related somehow.

I mean, is it for example true that if, $$\int_{\mathbb{R}} \varphi'(x)f(x)dx \leq C$$ for all $\varphi\in \mathcal{C}_0^\infty(\mathbb{R})$ then $f\in \mathcal{C}(\mathbb{R})$ ??

Here $\mathcal{C}_0^\infty(\mathbb{R})$ means infinitely many times diff. with compact support.

Thanks a lot for your help!! :)

  • 0
    Hi, I've converted one of your posts into a comment where it properly belongs. You were not able to comment on your own post here because your account "fragmented". Please consider registering your account. This will help prevent account fragmentation and allow you to always comment on answers to your own questions, as well as to edit your own posts to correct errors, provide more details, etc.2012-12-14

1 Answers 1

2

The answer is yes, since the condition implies that $\mu\equiv0$.

Proof. Suppose that there exists $\varphi\in C^\infty_0(\mathbb{R})$ such that $$ \int_{\mathbb{R}} \varphi^k(x)\mu(dx) \ne 0. $$ Without loss of generality we may assume that $C>0$ and $\int_{\mathbb{R}} \varphi^k(x)\mu(dx) >0$. For any $\lambda>0$, $\lambda\,\varphi\in C^\infty_0(\mathbb{R})$. Then $$ \int_{\mathbb{R}} \lambda\,\varphi^k(x)\mu(dx)\le C\implies\int_{\mathbb{R}} \varphi^k(x)\mu(dx)\le\frac{C}{\lambda}\quad\forall\lambda>0. $$ Letting $\lambda\to\infty$ we get $$ \int_{\mathbb{R}} \varphi^k(x)\mu(dx)\le0, $$ which is a contradiction.

  • 0
    Thanks for your answer, but I didn't quite understand. The conjecture is: I assume that $\int_{\mathbb{R}} \varphi'(x)f(x)dx < C$ for all $\varphi \in \mathcal{C}_0^\infty(\mathbb{R})$ then $f\in\mathcal{C}(\mathbb{R})$. You say it is true because $f\equiv 0$? by supposing that the integral condition is not zero? but the integral condition is not assumed to be zero, but just bounded.2012-12-13
  • 1
    But the bound is independent of $\varphi$ (or at least that's what you wrote: same $C$ for all $\varphi$). Chanching $\varphi$ by $\lambda\,\varphi$ with $\lambda\in\mathbb{R}$, you can see that the only possibility is that the measure $\mu$ is the null measure.2012-12-13
  • 0
    You are completely right. I wrote the bound independent of $\varphi$, my mistake! I meant: $$\int_{\mathbb{R}} \varphi'(x)f(x)dx \leq C\|\varphi\|_{\infty}.$$ Soryy :(2012-12-14