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Let $U\subset \mathbb{R}^n$ be open and $f:U\to \mathbb{R}^n$ with $x\in U$ and $\xi$ sufficiently small. Suppose that the following hold:

$f(x+\xi)=\sum_{\alpha=0}^k c_\alpha\xi^\alpha+o(\|\xi\|^k)$

$f(x+\xi)=\sum_{\alpha=0}^k \hat{c}_\alpha\xi^\alpha+o(\|\xi\|^k)$. Prove that $c_\alpha=\hat{c}_\alpha$.


I was told we should use Taylor expansion here but I don't see how Taylor's theorem is used since the function is not assumed to be differentiable. I tried induction on $\alpha$ but it wasn't so useful. Your help is appreciated.

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I would try subtracting the two quantities, and assume $\xi \neq 0$ –  Nameless Mar 16 at 19:23

1 Answer 1

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You are right: the assumptions are not sufficient for Taylor's theorem to apply. As Nameless suggested, you should subtract the two expressions: $$P(\xi):=\sum_{|\alpha|=0}^k (c_\alpha -\hat c_\alpha) \xi^\alpha = o(\|\xi \|^k) \tag{1}$$ (By the way, we are dealing with multiindices here: $\alpha = (\alpha_1,\dots,\alpha_n)$ and $|\alpha| = \alpha_1+\dots+\alpha_n$). It helps to separate the polynomial $P$ into homogeneous polynomials $P_d$ where $$P_d(\xi) = \sum_{|\alpha|=d} (c_\alpha -\hat c_\alpha) \xi^\alpha$$ Suppose $P$ is not identically zero. Let $d$ be the smallest degree for which $P_d$ is not identically zero. The number $M=\max_{|\xi|=1}|P_d(\xi)|$ must be positive then; otherwise $P_d$ would vanish everywhere by virtue of $P_d(t\xi)=t^d P_d(\xi)$.

The maximum of $|P_d|$ on the sphere $|\xi|=r$ is $Mr^d$. All higher degree polynomials present in (1) are $O(r^{d+1})$ as $r\to 0$. Thus, $$\lim_{r\to 0} \, r^{-d}\max_{|\xi|=r}|P(\xi)| =M>0$$ contradicting (1).

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