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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!! :)

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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.

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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. – Daniel Dec 13 '12 at 18:07
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. – Julián Aguirre Dec 13 '12 at 21:42
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 :( – Daniel Dec 14 '12 at 8:04

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