$2f(x)=f(y) \Rightarrow 2f(tx)=f(ty)$ Find all continuous and strictly monotonic function $f:[0,\infty)\to \Bbb R$ such that:


*

*If there is a pair $(x,y)\neq (0,0)$ such that $2f(x)=f(y)$ then $2f(tx)=f(ty)$ for all $t>0$;

*There is at least one pair $(x,y)$ where the above condition hold.
What I got so far:


*

*Check that $f(x)=ax^b$ is a solution. I want to discover if there is any other; 

*$t\to 0$ then $2f(0)=f(0) \Rightarrow f(0)=0$;

*We can prove that $y> x$;

*$t:=t/x$ and then $2f(t)=f(tk)$, for all $t>0$, with $k=y/x >1$;

*$f(tk^n)=2^nf(t)$ and setting $t=1$ we get $f(k^n)=2^nf(1)$. It give us that, if $f(1)>0$ then $f$ is increasing, otherwise, $f$ is decreasing.

I found another solution and I think it is insteresting to post here. It was inspired in change of variable sugested by @Yuri Negometyanov.
Let's set $tx=2^z$ and $ty=2^w$, where $z,w \in \Bbb R$ and $t>0$. We also call $g:\Bbb R \to \Bbb R$ such that $g(s)=f(2^s)$, and then we get
$$2f(tx)=f(ty)\to 2f(2^z)=f(2^w)\to 2g(z)=g(w)$$
but $\frac{ty}{tx} =\frac{y}{x}=2^{w-z}$ and then $w=z+\log_2 (y/x)=z+d$. So,
$$2g(z)=g(z+d)$$
wich has solution
$$g(z)=2^{z/d}\phi(z)$$ 
where $\phi(z)=\phi (z+d)$, it means that $\phi$ is a periodic function with period $d$. . But $\phi$ has to follow some constrains. For example, $\phi(z)=\sin^2\left(z\frac{\pi}{d}\right)+10$ works and $\phi(z)=\sin^2\left(z\frac{\pi}{d}\right)$ don't. So, 
suppose that $f$ is increasing (if it is decreasing we work with $-f$), then $g$ is increasing, which means that for $p>q$ we have $g(p)>g(q)$, and
$$2^{p/d}\phi(p)>2^{q/d}\phi(q)\to 2^{(p-q)/d}>\frac{\phi(q)}{\phi(p)}.$$ 
