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$f:\mathbb{R}\rightarrow\mathbb{R}$ infinitely differentiable and such that $f^{(3)}(x)=f(x)$. Then $\exists M=M(R)$ such that $\forall x$ with $|x|\leq R$ and all $j\geq0$ we have $|f^{(j)}(x)|\leq M$.

I really don't know where to start, could you help me please?

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    Perhaps you mean $|f^{(j)}(x)|\le M$?2011-10-18
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    with Gerry's comment, use the **hint**: continuous functions are uniformly continuous, and hence bounded, on compact sets. **hint 2**: if you know the word [equicontinuous](http://en.wikipedia.org/wiki/Equicontinuity), you may also find the fact that any finite family of continuous functions are automatically equicontinuous somewhat useful.2011-10-18
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    @Willie, come! Using equicontinuity to show that THREE bounded functions can be bounded by a same constant is a bit like... well, is a bit too much.2011-10-18
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    @Didier: yes, that second part was written rather tongue-in-cheek; it was intended to draw the OP's attention to the fact that there _are_ only a finite number of functions that are involved in this problem. (To abuse magician parlance: it is a redirection or a sleight of hand.)2011-10-18
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    yeah, you were right, it's $M$, I edited the question2011-10-18

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I suppose that should be for all $ | f^{(j)}(x)| \leq M $, not $R$.

Differentiability implies continuity, so since the function is infinitely differentiable, every derivative is continuous. Continuous functions map compact sets to compact sets, and in $\mathbb{R}$, by Heine-Borel, a set is compact iff it is closed and bounded.

Thus, the continuous functions $f, f' $ and $f''$ will map $ \overline{ B(0,R)} = \{ x\in \mathbb{R} : |x| \leq R \}$ to closed and bounded sets. Hence there exists positive constants $A,B,C$ such that $$f\left( \overline{ B(0,R)} \right) \subseteq \overline{ B(0,A)} ,$$

$$ f'\left( \overline{ B(0,R)} \right) \subseteq \overline{ B(0,B)} ,$$

$$ f''\left( \overline{ B(0,R)} \right) \subseteq \overline{ B(0,C)}. $$

Thus for all $x\in \overline{ B(0,R)}$, $$ f^{(j)} (x) = f^{ (j\mod 3) } (x) \in \overline{ B(0,\max\{A,B,C\})}$$

as required.

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    What you write, or show that $|f^{(j)}|\leqslant g$ for every $j$, with $g=|f|+|f'|+|f''|$ continuous on the compact set $\bar B(0,R)$.2011-10-18