How can I find the solutions of $$f(x) + f(qx) = 0,$$ where $q \in \mathbb{Q}, q\neq1, x \in \mathbb{R}$, with $f$ being a continuous function?


1 Answer 1


Let's let $x_0$ be an arbitrary real number.

For $|q| < 1$, we find that by substituting $x=x_0$ that

$$f(x_0) = -f(qx_0)$$ and, substituting $x=qx_0$, $$f(q^2x_0) = -f(qx_0)$$ Thus, $f(x_0) = f(x_0q^2)$, or more generally $$f(x_0) = f(x_0q^{2n})$$

Now, $\lim_{n\to\infty}x_0q^{2n} = 0$, so by continuity, $$\lim_{n\to\infty}f(x_0q^{2n}) = f(\lim_{n\to\infty}xq^{2n}) = f(0)$$ But, on the other hand $f(x_0q^{2n} = f(x_0)$ for all $n$, so $$f(x_0) = \lim_{n\to\infty}f(x_0q^{2n}) = f(0)$$ By substituting $x = 0$ into the function equation, we find $2f(0) = 0$, so $$f(x_0) = f(0) = 0$$ and $f$ is the zero function. Now, let's suppose $|q| > 1$. Then, if we substitute $x = \frac{y}{q}$, we find $$f(\frac{1}{q}y) + f(y) = 0$$ and since $\left|\frac{1}{q}\right| < 1$, we can use the previous argument to show that $f \equiv 0$.

The 'interesting' case is $q = -1$. Then, we can rearrange to find $$f(x) = -f(-x)$$ so any continuous odd function (for example, $f(x) = x$ or $f(x) = \sin x$) is a solution.

  • $\begingroup$ By continuity, $\lim_{n\to\infty} f(xq^{2n}) = f(\lim_{n\to\infty}xq^{2n}) = f(0)$. Where do you need more details? What have you tried so far? $\endgroup$
    – user88319
    Jan 20, 2015 at 0:51
  • $\begingroup$ I actually would appreciate a complete explanation: this is my first functional equation and I don't really know what to do. $\endgroup$
    – user62029
    Jan 20, 2015 at 1:00
  • $\begingroup$ @user62029 OK, I've given a complete answer. $\endgroup$
    – user88319
    Jan 20, 2015 at 1:11
  • $\begingroup$ Thank you very much. As I said in another comment I now understand this solution, but I still don't see how I should have thought of this myself. Could you recommend some material (for example, lecture notes available online) that teaches how to go about solving a functional equation? $\endgroup$
    – user62029
    Jan 20, 2015 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.