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Consider the following problem:

Is it true, that any odd continuous function satisfying $f(2x) = 2f(x)$ is linear?

Does the following function (which I found here) can serve as a counterexample?

$$ f(x) = \begin{cases} x\cos(2\pi \log_2(x)), & \mbox{if } x>0 \\ 0, & \mbox{if } x=0 \\ x\cos(2\pi \log_2(-x)) & \mbox{if } x<0 \end{cases} $$

If it can, are there any "simpler" ones? I mean, that task was given on exam, and it would be rather difficult to come up with such a function on your own. By what strategy can one construct a similar counterexample from scratch? It feels like some piecewise linear function could do the trick, but I can't figure out, how to build it.

Wunsch Punsch
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    Yes, the above function is an counterexample. In certain sense, it is one of the simplest example. Essentially, you want to rewrite $f(x)$ as $xg(x)$ such that $g(2x) = g(x)$. The cheapest way to do it is $g(x) = h(\log_2|x|)$ for some periodic function of period $1$. Among such periodic functions, $\sin(2\pi x)$ and $\cos(2 \pi x)$ are usually the first functions come to mind. – achille hui Jun 14 '18 at 16:23

2 Answers2

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$f$ DOES NOT have to be linear!

Let $g:[1,2]\to\mathbb R$ continuous and arbitrary, so that $g(1)=g(2)$, and extend $g$ to $(0,\infty)$, so that $g|_{[2^{k},2^{k+1}]}$ is defined as $$ g(x)=g(2^{-k}x),\quad x\in [2^{k},2^{k+1}]. $$ Then $$ f(x)=xg(x), \quad (0,\infty), $$ satisfies $f(2x)=2f(x)$, it is continuous, extends continuously to $x=0$, with $f(0)=0$, and extends, as an odd function continuously in $\mathbb R$, as $f(-x)=f(x)$, for $x<0$.

Yiorgos S. Smyrlis
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Let $\, g(x) := f(x)/x. \,$ Then the equation $\, f(2x) = 2f(x) \,$ leads to $\, g(2x) = g(x), \,$ a recurrence relation. We can define $\, g(x) \,$ on $\, 1 \le x \le 2 \,$ to be continuos with $\, g(1) = g(2). \,$ Now extend to all $\, x>0 \,$ using the recurrence relation. Similarly for $\, x<0. \,$

Somos
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