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If $f(x) \in C^\infty(\Bbb{R})$,and $f(a)=0$, do we have $$f(x)=(x-a)g(x)$$? where $g(x) \in C^\infty(\Bbb{R})$ and $g(a)=f'(a)$

Laura
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2 Answers2

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For $a=0$ (to simplify things): $f(x)=\int_0^1\frac{d}{dt}f(tx)dt=x\int_0^1f'(tx)dt$, i.e. $g(x)=\int_0^1 f'(tx)dt$.

user8268
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The answer is yes. For simplicity and without loss of generality I will assume $a=0$. Define $$ g(x)=\frac{f(x)}{x}\quad\text{if}\quad x\ne0,\quad g(0)=f'(0). $$ $g$ is $C^\infty$ in $\mathbf{R}\setminus\{0\}$, and since $\lim_{x\to0}f(x)/x=f´(0)$, it is continuous at $x=0$. Let's see that $g$ is also $C^\infty$ at $x=0$. Fot any $k\in\mathbb{N}$ we have $$ f(x)=f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(k+1)}(0)}{(k+1)!}x^{k+1}+O(x^{k+2}). $$ Then $$ g(x)=f'(0)+\frac{f''(0)}{2!}x+\dots+\frac{f^{(k+1)}(0)}{(k+1)!}x^k+O(x^{k+1}). $$ This shows that $g$ has derivatives of order $k$ at $x=0$ and that $$ g^{(k)}(0)=\frac{f^{(k+1)}(0)}{k+1}. $$ All is left is to show that each $g^{(k)}$ is continuous at $x=0$. But this is easy: $g^{(k)}$ has a derivative at $x=0$, which implies that it is continuous at $x=0$.

Ingeneral, if $f$ is $C^n$, $g$ is $c^{n-1}$.

Julián Aguirre
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  • Thanks very much. I think for any $k\in \Bbb{N}$, we get $g^{(k)}(0)=f^{(k+1)}(0)/(k+1)$ have alrealy prove that $g(x)$ is also $C^\infty$ at $x=0$. Why we need to show that each $g^{(k)}$ is continuous at $x=0$? What's wrong with me? – Laura Oct 14 '13 at 02:18
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    There is nothing wrong. I did it just to show hat no examples like $x^2\sin(1/x)$, which is differentiable but not $C^1$ can occur. – Julián Aguirre Oct 14 '13 at 09:46
  • Thanks very much ,I have understood. If $f(x)=\begin{cases}x^2\sin(1/x) & x \not= 0\0 & x = 0\end{cases}$, then $f'(x)=\begin{cases}2x\sin(1/x)-\cos(1/x) & x \not= 0\0 & x = 0\end{cases}$ is not continuous. – Laura Oct 16 '13 at 02:21
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    "This shows that $g$ has derivatives of order $k$ at $x=0$" is correct only when $k=1$. The existence of polynomial approximation $g(x)=p(x)+O(x^{k+1})$ does not imply the existence of $g^{(k)}(0)$ when $k\ge 2$. For example, $g(x)=x^3 \sin (1/x^3)$ is $O(x^3)$ but $g''(0)$ does not exist. – user103402 Oct 27 '13 at 04:22
  • @user103402 Thank you for your comment. I was too hasty in writing my answer. – Julián Aguirre Oct 28 '13 at 10:06
  • @user103402.thanks very much. If $g(x)=\begin{cases}x^3\sin(1/x^3) & x \not= 0\1 & x = 0\end{cases}$. then $g'(0)=\lim \frac{x^3\sin(1/x^3) - 1}{x}$ does not exist. How to calculate $g^{''}(0)=\lim \frac{g'(x)-g'(0)}{x}$ and conclude that it doesn't exist ? – Laura Nov 04 '13 at 04:07