# Decomposition of Hölder continuous functions

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!

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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.

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I think you need to escape some underscores with a backslash in order to protect them from markdown. – Harald Hanche-Olsen Dec 3 '09 at 17:41
Oh, and when I look at your answer more closely, don't you have the thing he wanted, with C=n/α-1 (just rename R)? – Harald Hanche-Olsen Dec 3 '09 at 17:54
But then C is not a universal constant, it depends on n, unless the original poster meant independent of f. – Julián Aguirre Dec 3 '09 at 20:09
@Harald If I hit the Preview math button, everything is OK. I do not see any difference between the code in the first part of the message and the second. – Julián Aguirre Dec 3 '09 at 20:11
@Julián: Thank you for your answer. I understand the bound on the remainder since $\eta\ast\phi_R\rightarrow\eta$ uniformly as $R\rightarrow\infty$, but how do you get the bound on '$\|f\|_{C^n}$'? – user17240 Dec 4 '09 at 8:30

This is only a partial suggestion of how to get an answer, rather than a complete one; but it was getting too long for the comment box.

While I don't remember the details off-hand, it seems like your question could be answered by known results on the rate of approximation in $\Lambda_0^\alpha({\mathbb R})$ by trigonometric polynomials. This is classical stuff from Fourier series and I think one can find the details in something like Katznelson's "Introduction to Harmonic Analysis".

Roughly speaking, the idea is to move your problem onto the circle, by e.g. extending $\eta$ to a periodic function on the real line, and then take $f$ to be a trigonometric polynomial which approximates $\eta$ to within $R^{-1}$ in the Lipschitz norm. If I recall correctly - but I'm not sure - we can take $f$ to be of degree $N$ where $N^\alpha\sim R$, and then something like Bernstein's inequality would tell us that the $C^n$ norm of $f$ is at most of order $N^n = R^{n/\alpha}$.

If I've remembered these results correctly, then there should be a better and more direct way to solve our question, by convolving $\eta$ with some kind of smoothing kernel like a Jackson kernel. However, since it's been a while since I looked at this, I'm not sure: Katznelson's book ought to have enough for you to work from.

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