I have a problem and the problem is:
Determine whether or not there exists an infinitely differentiable function $f(x)$ such that for every real $x$ and for every positive integer $n$
$$1\le f^{(n)}(x)+f^{(n+1)}(x)+\cdots+f^{(3n)}(x) \le 3.$$
where $f^{(n)}(x)$ denotes the $n$-th derivative of $f$.
I want some hints to solve the problem or where this problem comes from.
Thanks!
The following is the pirture my friends sends me.
I have find the problem. This problem comes from First HKUST Undergraduate Math Competition-Junior Level.
I paste the solution:
Lemma 1: If there exixts $M\in \mathbb{R}$ and $\varepsilon>0$ such that $f^{n}(x)>\varepsilon$,the $f$ is unbounded above.
Proof: Let $c_n=f^{n-1}(M)$. Since $f^{(n-1)}(x)>c_{n-1}+\varepsilon x$ for all $x> M$ by the mean value theorem, $f^{(n-1)}(x)$ is unbounded above.Then there exists $M'\in \mathbb{R}$ such that $f^{(n-1)}(x)>\varepsilon$ for all $x\ge M'$.Repeating this $n-1$ times more , we get $f$ is unbounded above. This proved the lemma.
Now assume the such a function $f$ exists, Consider
$A(x)=f^{(1)}(x)+f^{(2)}(x)+f^{(3)}(x)$, $~~~~~~~~~~~~~~B(x)=f^{(4)}(x)+f^{(5)}(x)+\cdots+f^{(12)}(x)$
$C(x)=f^{(13)}(x)+f^{(14)}(x)+\cdots+f^{(39)}(x)$,$D(x)=f^{(19)}(x)+f^{(20)}(x)+\cdots+f^{(57)}(x)$.
Let $E(x)=A(x)+B(x)+C(x)$.We are given $1\le A(x),B(x),C(x)\le 3$(hence $3\le E(x)\le 9$) and $D(x)\ge 1$ for all real $x$.Now
$$D(x)=A^{(18)}(x)+B^{(18)}(x)+D^{(18)}(x)=E^{(18)}(x).$$
By the lemma, $E$ is unbounded above,a contradiction to $E(x) \le 9$ for all real $x$.
The link is https://www.math.hkust.edu.hk/ug/competition/award.php where you can find the solutions to other problems.
