Let $\alpha\in(0,1)$ and $\eta\in\Lambda\_0^\alpha(\mathbb{R})$ be a compactly supported Hölder continuous function of order $\alpha$. I would like to show that, for any $n\in\mathbb{N}$, it is possible to decompose $$\eta=f+g$$ 
in such a way that $f\in C^n(\mathbb{R})$ and $||f||\_{C^n}=O(R^C)$, and $g\in L^\infty(\mathbb{R})$ with $\|g\|\_{L^\infty}=O(R^{-1})$. 


Here $C$ is a universal constant. On the other hand, the real parameter $R$ can be chosen as large as we want (at the expense of increasing $\|f\|\_{C^n}$).


Thank you!



Decomposition of Hölder continuous functions
I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose $\phi\in C^\infty(\mathbb{R})$, $\phi\ge0$ and $\int\_{\mathbb{R}}\phi(x)dx=1$, and let $\phi\_R(x)=R\phi(Rx)$. Define


$$ f=\phi\_R\star\eta,\quad g=\eta-f.$$


Then it is easy to see that


$$ \|f\|\\_{C^n}=O(R^{n-\alpha}),\quad \|g\|\\_\infty=O(R^{-\alpha}),$$


but this is not what you are asking for.


My feeling is that the constant $C$ must show some dependence on $n$.


In response to your last comment, let me prove the estimate on $\|f\|\_{C^n}$. We have


$$f^{(n)}=(\phi\_R)^{(n)}\star\eta=R^n(\phi^{(n)})\_R\star\eta.$$


Since $(\phi^{(n)})\_R$ has mean zero, for any $x\in\mathbb{R}$:


$$ |f^{(n)}(x)|\le R^n\int\_{\mathbb{R}}|\phi^{(n)}(y)||\eta(x-\frac{y}{R})-\eta(x)|dy\le HR^{n-\alpha}\int\_{\mathbb{R}}|\phi^{(n)}(y)||y|^\alpha dy,$$


where $H$ is $\eta$'s Hölder constant.


