Continuous functions satisfying $2f(x) = f(2x)$

Question

Can I just say that continuous solutions to the equation $$ f(x+y) = f(x) + f(y) $$ (for all $x$,$y$) are of the form $f(x) = kx$ so our solution set is of the form $kx$ as well? Then plugging in $kx$ gives $k=0,2$ but I'm not sure if this is a valid way to do it..

Answer 1

I'll address the equation in the title, since it leads to an important subtlety people often overlook.

Let $g(x):=x^{-1}f(x)$ so $g(2x)=g(x)$. Let $h(x):=g(e^x)$ so $h(x+\ln 2)=h(x)$. Any periodic function, its argument scaled so the period is $\ln 2$, gives us a solution $f$. Continuity narrows the choices of $h$ somewhat, but it needn't be constant. A less trivial solution, for example, is $$f(x)=x\cos\frac{2\pi \ln x}{\ln 2}.$$One may object this breaks for $x\le 0$, but $f(2x)=2f(x)$ requires only that $f(0)=0$, and if we define $f(x)$ for positive and negative $x$ separately there is no issue. For example, we could take $f$ to be odd and simply replace $\ln x$ in the above with $\lvert\ln x\rvert$.