$f(ct)=cf(t)$ for all rational numbers $c$ and $f$ is continuous Given that $f(ct)=cf(t)$ for all rational numbers $c$ and that $f$ is continuous, it must be that $f$ is a straight line. Can the same conclusion can be drawn if $f(2t)=2f(t)$ and $f$ is continuous, i.e. the statement holds only for $c=2$? Clearly it cannot if $f$ is not continuous, as you could set $f(0)=0, f(1)=1, f(2)=2, f(4)=4, f(8)=8$, and then for another "sequence" of numbers, set $f(0)=0, f(3)=1, f(6)=2$, and you would end up with many dots that lie on lines. But I was wondering if the assertion that $f$ is continuous forces $f$ to be a straight line even when the condition is reduced from all rational numbers $c$ to just one number.
 A: If $f(ct)=cf(t)$ is only required to be true for $c=2$, then there are many possibilities for what $f$ can be.
Take any continuous function $f_1$ on $[1,2]$ such that $f_1(2)=2f_1(1)$. Then we can extend this uniquely to any positive real number input. For instance, $$f(3)=2f_1(1.5)\\
f(10)=8f_1(1.25)\\
f(1/3)=\frac14f_1(4/3)$$ And then you can do something similar for the negative real numbers separately: start with a continuous $f_2$ on $[-2,-1]$ with $f_2(-2)=2f_2(-1)$, and expand as above.
To ensure continuity of $f$ at $0$, it is quite obvious from the above expansion construction that $\lim_{t\to0}f(t)=0$, so we need $f(0)=0$.
A: No. Define $f$ by
$$
f(x) = (x-1)(2 -x) \text{ on } [1,2)
$$
and for every $k\neq0$,
$$
f(x) = 2^k f\left(\frac{x}{2^k}\right) \text{ on } [2^k,2^{k+1})
$$
That $f$ isn't linear.

Now suppose that $c_1$ and $c_2$ are co-prime, as Arthur suggested, for example $2$ and $3$. The set $\{2^l/3^k; l,k\in \mathbb Z\}$ is dense in $\mathbb R^+$, as it is pointed out here so then for a dense set of points in $R^+$, $f(x)=xf(1)$ therefore the conclusion follows.
The only requirement for this argument to hold is
$\ln c_1 /\ln c_2$ irrational, which is true when they are co-prime.
A: $|2x|=2|x|$ but it's not a straight line.
