Find all $f\in C^\infty(\mathbb R)$ such that : $\forall x\in \mathbb R, f(ax)+f(bx)+f(cx)=0$ with $a,b,c>0$ and distinct Let $a,b,c\in\mathbb R^*_+$ with $a\neq b$, $a\neq c$ and $b\neq c$,
Find all $f\in C^\infty(\mathbb R)$ such that : $\forall x\in \mathbb R, f(ax)+f(bx)+f(cx)=0$ with $a,b,c>0$ and distinct.
We can make $a<b<c=1$.
Hint given : We need to show that $\forall A>0, \forall p\in\mathbb N,\exists C>0,\forall x\in[-A,A], |f(x)|\le C|x|^p$ $(\star)$
I noticed that : $a^nf^{(n)}(ax)+b^nf^{(n)}(bx)+c^nf^{(n)}(cx)=0$ so $\forall n\in\mathbb N, f^{(n)}(0)=0.$
Also : $f\in C^\infty(\mathbb R)$ so $\forall A>0,\forall n\in \mathbb N,\forall x\in [-A,A]$ :
$$|f(x)|= \left|\sum_{k=0}^{n}\dfrac{x^kf^{(k)}(0)}{k!} + R_n\right| = |R_n| \le \sup_{y\in[-A,A]}|f^{(n+1)}(y)|\dfrac{|x|^{n+1}}{(n+1)!}=C|x|^{n+1}$$
The case : $\forall A>0, \forall x\in [-A,A], \exists C>0, |f(x)| \le C$ because $f\in C^0([-A,A],\mathbb R)$.
So we did end up to the property $(\star)$.
After that I don't know what to do. Do you have some ideas ?
 A: Ok, so let us note that we can by symmetry take $0<a<b<c$ and by scaling $x\mapsto \frac{x}{c}$ we can assume that $c=1$.
Then the relation $f$ must satisfy for all $x\in\mathbb{R}$ becomes
\begin{equation}
f(ax)+f(bx)+f(x)=0 (**)
\end{equation}
We will show that $f\equiv 0$. Assume this is not the case. Then there exists some $x_0\neq 0$ such that $f(x_0)\neq 0$ ($x_0$ cannot be $0$ since clearly $f(0)=0$). Now we iteratively define a sequence $x_n$ with the starting point $x_0$.
Apply $(**)$ with $x=x_0$. We have that $|f(ax_0)+f(bx_0)|=|f(x_0)|$, so either $f(ax_0)\neq 0$ or $f(bx_0)\neq 0$. Moreover for one of the values $y\in\{ax_0,bx_0\}$ we must have that $|f(y)|\geq \frac{1}{2} |f(x_0)|$. We define $x_1$ to be this value. Now we play the same game. We plug into $(**)$ the value $x=x_1$ and choose $x_2$ and so on.
In general, we will obtain a sequence $x_n=a^j b^{n-j} x_0$ for some particular value of $j\leq n$. This sequence is strictly decreasing in absolute value (since $a,b<1$) and converges to $0$ and it will satisfy the inequality $|f(x_n)|\geq \frac{1}{2^n} |f(x_0)|$ for all $n$.
Now let us apply $(*)$ to the interval $[-|x_0|,|x_0|]$. For any natural $p$ we will get that there exists some constant $C_p$ such that $|f(x)|\leq C_p |x|^p$ holds uniformly on the interval. Setting $x=x_n$, this implies that
\begin{equation}
C_p\geq \frac{|f(x_n)|}{|x_n|^p} \geq \frac{|f(x_0)|}{|x_0|^p} \cdot \frac{1}{2^n}\cdot\frac{1}{b^{np}}=\frac{|f(x_0)|}{|x_0|^p}\cdot \frac{1}{(2b^p)^{n}}
\end{equation}
and this must hold for any $n$. We used the fact that $a^j b^{n-j}\leq b^n$ since $a<b$.
If $b<1$, then we can find some big $p$ such that $2b^p<1$. But then $\frac{1}{(2b^p)^n}$ becomes arbitrarily large as $n$ goes to $\infty$, contradiction to this being bounded by $C_p$. Therefore $b=1$, which is a contradiction to the fact that $c=1$ and $b$ and $c$ were supposed to be distinct.
