Find all continuous functions in $0$ that $2f(2x) = f(x) + x $ I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing   in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then 
$$f(y) = \frac{1}{2}f\left(\frac{1}{2}y\right) + \frac{1}{4}$$
after induction I get:
$$f(y) = \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}}  \right)$$
I take $\lim_{n\rightarrow \infty} $
$$ \lim_{n\rightarrow \infty}f(y) = f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\cdot \lim_{n\rightarrow \infty} \left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}}  \right)$$
$$f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n} \cdot f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right) + \frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but 
function is continous in $0$ when
$$ \lim_{y\rightarrow 0^+}f(y)=f(0)=\lim_{y\rightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit

*

*Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?


This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = \frac Cx
$$
and then for the paticular we assume 
$$
f_p(x) = \frac{C(x)}{x}
$$
then
$$
a\frac{C(a x)}{a x}-\frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = \frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = \frac{C}{x}+\frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = \frac Cx+\frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = \frac x3
$$
