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I found this problem on an Italian forum and since then I struggled to solve it. The autor there claims it was proposed by Borel. At any rate the problem is as follows

(Borel?) For any $\{a_n\}_{n=0}^{\infty}\subseteq \mathbb R,\: x_0\in\mathbb R\,$ and $\,\varepsilon>0$ does there exist a $C^\infty$ function $f\colon\mathbb R\to\mathbb R$ such that $$f^{(n)}(x_0)=a_n,\; \forall n\in\mathbb N\cup\{0\}\tag{1}$$ and moreover $$\left|f(x)-a_0\right|<\varepsilon,\;\forall x\in\mathbb R?$$

Clearly even reference about problem are welcomed, but I strongly encourage anybody to think about it because really, when I met it for the first time, I thought: "this is a wonderful problem".

I do not have any clue towards the solution. I hope you will have fun with this.

Cheers.

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(please convert into a comment) Borel, Émile: Sur quelques points de la théorie des fonctions. Annales scientifiques de l'École Normale Supérieure, Sér. 3, 12 (1895), p. 9-55, bottom of page 44. numdam.org/numdam-bin/fitem?id=ASENS_1895_3_12__9_0 –  user42754 Sep 26 '12 at 15:56
    
See also: ncatlab.org/nlab/show/Borel's+theorem –  user42754 Sep 26 '12 at 15:58
    
Thanks for the references –  uforoboa Sep 27 '12 at 9:53

1 Answer 1

up vote 4 down vote accepted

Here is a counterexample.

For arbitrary interval $I\subset\mathbb{R}$ denote $M_k(f,I)=\max_{t\in I}|f^{(k)}(t)|$, then from exercise 14 Chapter 5 in Rudin's Principles of Mathematical analysis we know that $$ M_1^2(f,I)\leq 4M_0(f,I)M_2(f,I) $$ for any interval $I$ and $f\in C^3(I)$

Let's take $a_1>2\sqrt{(|a_0|+\varepsilon)(|a_2|+\varepsilon)}$ and the rest whatever you want. Assume that there exist a smooth function $f$ with predescribed properties. Since $f\in C^\infty(\mathbb{R})$ then we can find neighbourhood $U$ of $x_0$ where $$ M_2(f,U)<|f^{(2)}(x_0)|+\varepsilon=|a_2|+\varepsilon $$ By construction $|f(x)-a_0|<\varepsilon$ for all $x\in\mathbb{R}$, so $$ M_0(f,U)<|a_0|+\varepsilon $$ Finally $$ M_1^2(f,U)>|f'(x_0)|^2=a_1^2>4(|a_0|+\varepsilon)(|a_2|+\varepsilon)>4M_0(f,U)M_2(f,U) $$ Contradiction.

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Good! This show that boundedness is restrictive in this case, as we should have $|a_n|\leq 4(|a_n|+\varepsilon)(|a_{n+1}|+\varepsilon)$ for all $n$. –  Davide Giraudo Sep 26 '12 at 17:10
    
@DavideGiraudo, thanks! –  no identity Sep 26 '12 at 17:37
    
@Norbert I see your point. I was aware of that result you quoted however I never linked it to the solution of this problem. Thank you. –  uforoboa Sep 27 '12 at 9:52

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