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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.

Vons
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    Next question is now if we fix for two values $c_1$ and $c_2$. I guess it will be similar to the question of "subgroups of $\mathbb R$ are either $a\mathbb Z$ or dense, and we get either some leeway on a fixed interval or $f$ is linear. – zwim Mar 26 '21 at 20:45

3 Answers3

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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 $f(0)=0$.

Arthur
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  • What is the exension to the real number $0$? – GEdgar Mar 26 '21 at 20:45
  • @GEdgar I forgot entirely. Thanks for reminding me. – Arthur Mar 26 '21 at 20:50
  • Thank you. It is clear. – Vons Mar 26 '21 at 21:38
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    @Stacker I strongly suspect that requiring that two values, for instance $c = 2$ and $c = 3$, both work (together with continuity) forces us to have a straight line. But not all pairs work. For instance requiring that $c = 2$ and $c = 4$ work is the same as just $c = 2$. – Arthur Mar 26 '21 at 21:41
  • @Arthur good point, it is true, I added a line in my answer. –  Mar 27 '21 at 11:43
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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.

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$|2x|=2|x|$ but it's not a straight line.

Vasili
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