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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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    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