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Given a $C^{k}$ function $f:\mathbb{R}^d\to\mathbb{R},$ we can use Taylor's theorem to write it as

$$f(x)=\sum_{|\alpha|\le k-1} c_\alpha x^\alpha + R(x),$$ where $R$ is $C^k$ and can be expressed as $$R(x)= \sum_{|\alpha|=k} h_\alpha(x) x^\alpha$$ for some coefficients $h_\alpha$ that are not unique in general.

My question is (for $d>1$): Can we always choose $h_\alpha x^\alpha$ to be $C^k$?

Thank you.

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    $\begingroup$ Max, it may be good to mention what happens in the case $d = 1$ to give some motivation. // By the way, after we talked in the afternoon, it occurred to me that your statement looks a little bit like Malgrange preparation. A bit of digging revealed that some finite regularity versions of it are known in the literature: MR312520 and MR478212 are examples. I have not actually tried too hard to see whether indeed we can convert your question to a division problem, but that may be a direction to try. $\endgroup$
    – Willie Wong
    Commented Sep 29, 2016 at 1:22
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    $\begingroup$ The problem is at the origin: if we could do it, then we could approximate each $h_\alpha$ by its Taylor polynomial too and get for $f$ a polynomial approximation with an error of order $o(|x|^{2k})$ near the origin, but that is too much for a general $C^k$-function. $\endgroup$
    – fedja
    Commented Sep 29, 2016 at 2:12
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    $\begingroup$ Does this hold for $d = 1$? In that case, the integral form of the remainder term is $$ R(x) = \frac{x^k}{(k-1)!}\int_0^1 f^{(k)}(tx)(1-t)^{k-1}\,dt $$ As far as I can tell, the integral factor is in general no better than $C^0$. $\endgroup$
    – Deane Yang
    Commented Sep 29, 2016 at 3:57
  • $\begingroup$ For $d = 1$ and $x \ne 0$, $R(x)/x^k$ is obviously $C^k$. The only thing is that it can blow up at $x = 0$. Something similar is probably true in higher dimensions. $\endgroup$
    – Deane Yang
    Commented Sep 29, 2016 at 16:54
  • $\begingroup$ Thank you everyone for your comments. Indeed I should have motivated what happens when $d=1$ as Willie suggested (now it has been motivated by Deane) and should have instead asked what happens away from coordinate hyperplanes. I have instead changed the question to whether $h_\alpha x^\alpha$ can be chosen to be $C^k$. Willie -- I will take a look! Thank you again. $\endgroup$
    – Maxim Gilula
    Commented Sep 29, 2016 at 19:40

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