Let $f:\mathbb{R}\to\mathbb{R}$ be a function. Suppose:

$$\left|\sum_{k=1}^{n}3^k(f(x+ky)-f(x-ky))\right|\leqslant 1\quad\forall n\in\mathbb{N}\quad\forall x,y\in\mathbb{R}$$

Show that $f$ is a constant function.

I don't even know where to start and what is the possible approach. Any hints?

  • $\begingroup$ Can you share your thoughts on the problem, and explain what you've tried? $\endgroup$
    – user61527
    Nov 30, 2013 at 6:31
  • 1
    $\begingroup$ Well, he has mentioned that he doesn’t even know where to start. $\endgroup$ Nov 30, 2013 at 7:13

1 Answer 1


Note that $\left|a_{n}\right| \leq \left|\sum_{k=1}^{n-1}a_k\right| + \left|\sum_{k=1}^{n}a_k\right|$. It follows that, $\forall n\in \mathbb N$ and $\forall x,y\in \mathbb R$,

$$ \left| 3^n \bigl( f(x+ny) - f(x-ny) \bigr) \right| \leq 2 $$

Dividing by $3^n$ and setting $y=x/n$ gives

$$ \forall n\in \mathbb N, \quad \bigl| f(2x) - f(0) \bigr| \leq \frac2{3^n} $$

In the limit as $n\to \infty$, we conclude that for every $x\in \mathbb R$, $f(2x) = f(0)$.

  • $\begingroup$ Very clever solution. Thanks! $\endgroup$
    – Kato yu
    Dec 1, 2013 at 23:29

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