# If $f$ satisfies a particular differential equation then its derivatives satisfies a certain condition

$f:\mathbb{R}\rightarrow\mathbb{R}$ 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 $|f^{(j)}(x)|\le M$? – Gerry Myerson Oct 18 '11 at 7:53
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, you may also find the fact that any finite family of continuous functions are automatically equicontinuous somewhat useful. – Willie Wong Oct 18 '11 at 8:00
@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. – Did Oct 18 '11 at 8:13
@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.) – Willie Wong Oct 18 '11 at 12:07
yeah, you were right, it's $M$, I edited the question – Alex M Oct 18 '11 at 22:56

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 $\bar B(0,R)$. – Did Oct 18 '11 at 8:15