Real Curve for Complex Function

Question

I was tasked with solving the functional equation $$f(x)+f(2x)=2x$$ and I found that one possible function $f$ could be $$f(x)=\frac{2x+(-1)^{\log_2 x}}{3}$$ This function is real whenever $x=2^k$, $k \in \mathbb Z$, and it is mostly complex otherwise. I was wondering if, in the real plane, there is a function $g:\mathbb R \to \mathbb R $ that connects the points of $f(x)$ that lie in the real plane with a smooth curve. Does anyone have any idea how to do this? Is it even possible? I don't even know where to start.

Answer 1

There is no smooth function defined on all of $\mathbb R$ such that $$ f(2^k) = \frac{2\cdot 2^k + (-1)^k}3 $$ for all $k\in\mathbb Z$.

By considering large negative $k$ we can see that there are $x$ arbitrarily close to $0$ where $f(x)$ must be either close to $\frac13$ or close to $-\frac13$ -- so your given values won't even extend to a continuous function defined at $0$.

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If you restrict your attention to the positive reals, you could do, for example: $$ f(x) = \frac23x + \frac13\cos\Bigl(\frac{\pi}{\ln 2} \ln x\Bigr) $$ which extends to any complex domain you can extend the natural logarithm to (and solves your original functional equation too).