real analysis - 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
  $$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?

Answer

A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-\frac13$ the equality simplifies to
$$g(2x)=\frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$

$$g\left(\frac{x}{2^n}\right)=2^ng(x).\tag{1}$$
If $g$ is not identically zero, say $g(x_0)\neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(\frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.

Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=\frac13 x$.