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!! :)
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.