Solutions of the functional equation $f(x) + f(qx) = 0$

Question

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?

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 $f \equiv 0$.

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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.