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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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    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