Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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 $h\in\mathcal S(\mathbb R)$.

share|improve this answer
aah, great thks alot ! –  harlekin Feb 13 '12 at 14:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.