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2 Included proof of the estimate on f.; added 4 characters in body; added 2 characters in body

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

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

1

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 $n$.
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