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I am having trouble understanding a proof to establish a specific version of Taylor's formula.

I'll first give the statement and then below cite the part where I am stuck, so here is what I'd like to prove:

Let m be a non-negative integer, $f \in S(\mathbb{R})$ (Schwartz Space) and $y \in \mathbb{R}$ be a fixed point. If $f$ and all its derivatives up to the order $m$ vanish at $y$ then there exist functions $h_\beta \in S(\mathbb{R})$ such that \begin{equation} f(x) = \sum_{\beta \colon\\, |\beta| = m + 1} (x-y)^\beta h_\beta(x), \qquad \forall x \in \mathbb{R} \end{equation}

And this is how the proof starts:

Let $\zeta \in C^\infty_0(\mathbb{R}^n)$ and $\zeta \equiv 1$ in a neighborhood of the point $y$. Denote $f_1 = (1 - \zeta)f$ and $f_2 = \zeta f$. Obviously, the function \begin{equation} h(x) := |x-y|^{-2m-2}f_1(x) \end{equation} belongs to $S(\mathbb{R})$.

The last sentence is what troubles me - how is it obvious that the function above is Schwartz in $\mathbb{R}^n$? In particular, what is the value of $h(x)$ at $x = y$ ? I would say it's not defined, and even if the limit exists because $f_1(y) = 0$ I cannot have $h$ to be smooth then ...

What am I missing ? Thks alot for helping me!

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Since $x$ and $y$ are real numbers, what is $n$? –  Davide Giraudo Feb 13 '12 at 15:21

1 Answer 1

up vote 0 down vote accepted

Let $r_0$ such that $\zeta(x)=0$ if $x\in (y-r_0,y+r_0)$ and $A$ such that $\operatorname{supp}\zeta\subset [-A,A]$. We have $$h(x)=\begin{cases}0&\mbox{ if }x\in (y-r_0,y+r_0)\\\ \frac{(1-\zeta(x))f(x)}{(x-y)^{2(m+1)}}&\mbox{ otherwise}.\end{cases}$$

This shows that $h$ is smooth. Let $d,p\in\mathbb N$. The map $x\mapsto x^ph^{(d)}(x)$ is bounded on $[-A,A]$ as a continuous map on a compact set, and if $x\notin [-A,A]$ then $h(x)=\frac{f(x)}{(x-y)^{2(m+1)}}$, so \begin{align*}x^ph^{(d)}(x)&=x^p\sum_{k=0}^d\binom dkf^{(k)}(x)\frac{\partial^{d-k}}{\partial x}\frac 1{(x-y)^{2(m+1)}}\\ &=x^p\sum_{k=0}^d\binom dkf^{(k)}(x)(x-y)^{-2(m+1)-(d-k)}C_{m,d,k}, \end{align*} so $$|x^ph^{(d)}(x)|\leq \sum_{k=0}^d\binom dkN_{p,d}(f)C_{m,d,k}(|A|-|y|)^{-2(m+1)-(d-k)},$$ where $N_{p,d}(f)=\sup_{x\in\mathbb R}|x^pf^{(d)}(x)|$. This show that $h\in\mathcal S(\mathbb R)$.

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aah, great thks alot ! –  harlekin Feb 13 '12 at 14:49

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