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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?

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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.

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  • $\begingroup$ 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

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