Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$.
We have $f(0) = 0$ and $f(x) = 2f(2x) - x$, but I am not sure how to convert this functional equation into something that is easier to solve. Maybe using induction may work, but I don't see an easy way to induct since we only have one variable.
Set $g(x)=f(x)-ax$; then $f(x)=g(x)+ax$ and $$ g(x)+ax+x=2(g(2x)+2ax)=2g(2x)+4ax $$ and we can choose $a=1/3$, so the equation becomes $$ g(x)=2g(2x) $$ For $x/2$, we get $$ g(x)=\frac{1}{2}g(x/2)=\frac{1}{4}g(x/4)=\dots=2^{-n}g(2^{-n}x) $$ Since $g$ is continuous, we have $$ g(x)=\lim_{n\to\infty}2^{-n}g(2^{-n}x)=0 $$
We prove that there is only one function $f:\mathbb R\to\mathbb R$ that satisfies $2f(2x)=f(x)+x$ and \begin{align*} \lim_{t\to0}tf(xt)=0 \end{align*} for all $x\in\mathbb R$, namely $f(x)=x/3$. Note that if $f$ is continuous, it satisfies this limit condition (why?)!
By iterating $f(x)=\frac{1}{2}f(x/2)+x/4$ one finds the formula \begin{align*} f(x)=\frac{1}{2^n}f\left(\frac{x}{2^n}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k,\quad n\geq 1,x\in\mathbb R.\qquad (\star) \end{align*} This can also be shown by induction: For $n=1$, this is just the definition of $f$. Assume that $(\star)$ has been proven for some $n\geq 1$. Then, again by definition, \begin{align*} f\left(\frac{x}{2^n}\right)=\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}}, \end{align*} and if we put this into our formula (induction!), we obtain \begin{align*} f(x)= \frac{1}{2^n}\left(\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k= \frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k= \frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}\sum_{k=0}^{n}4^k, \end{align*} which proves $(\star)$.
Now, if we fix $x\in\mathbb R$ and take the limit $n\to\infty$ in $(\star)$ we obtain due to the limit condition that $f(x)=x/3$.
Try with a polynomial of degree $n$ $$\sum_{k=0}^{k=n}a_kx^k$$ We have $$\sum_{k=0}^{k=n}a_kx^k+x=\sum_{k=0}^{k=n}2^{k+1}a_kx^k$$ It follows $$\begin {cases}a_k=2^{k+1}a_k;\space k=2,3,.....,n\\a_1x+x=4a_1x\\a_0=2a_0\end{cases}$$
Hence the only polynomial satisfying the property is $$f(x)=\frac x3$$
Similarly with an homography $$\frac{ax+b}{cx+d}$$ one gets at once $c=0$ and $d=3a$ which comes to $f(x)=\frac x3$
It is doubtful that the property is worth for rational or transcendental function. I stop here.
Exists an elegant and general proof to show that $f(x)=\frac{x}{3}$ is the only one solution.
Using complex variable, rewrite the functional equation as $2f(2z)-f(z)=z$. Since $2z$ and $z$ are complex-differentiable at $z\in\mathbb{C}$ it follows that $f(z)$ is necessarily complex-differentiable for all $z\in\mathbb{C}$ and hence necessarily continuous. Moreover, since $2$, $2z$, $-1$ and $z$ are finite-grade polynomials then $f(z)$ too and
$$ \deg(f) = \max( \frac{\deg(z) - \deg(2)}{\deg(2z)} , \frac{\deg(z) - \deg(-1)}{\deg(z)} ) = \max(1,1) = 1.$$
As a consequence, the general solution is $f(z) = c_0+zc_1$ where $c_0 = 0$ and $c_1 = \frac{1}{3}$ by replacing in the functional equation. Therefore, $f(z) = \frac{z}{3}$ is the only one solution.
Of course, $f(z)$ is necessarily continuous because it is necessarily complex-differentiable. Note that the fact that being a complex-differentiable function is an even stronger property than being a continuous function. Hence if you have a functional equation
$$A(z) f(U(z)) + B(z) f(V(z)) = P(z),$$
where $A(z)$, $B(z)$, $U(z)$, $V(z)$ and $P(z)$ are complex-differentiable in an open $\Omega\subset\mathbb{C}$, then $f(z)$ is also complex-differentiable for all $z\in\Omega$ (HINT: you can use the Cauchy-Riemann equations to prove this). Therefore, in order to get a general solution you can always compare the Taylor's coefficients of both members of the functional equation.
In particular, if you have that $A(z)$, $B(z)$, $U(z)$, $V(z)$ and $P(z)$ are finite order polynomials, as in your case, then $f(z)$ is complex-differentiable for all $z\in\mathbb{C}$ and its Taylor series exists, is unique and converges for all $z\in\mathbb{C}$. Moreover, for these cases it can easily be shown that $f(z)$ will always be a finite polynomial. In fact,
$$ \deg f = \max( \frac{\deg P - \deg A}{\deg U} , \frac{\deg P - \deg B}{\deg V} ),$$
that you can use to avoid comparing infinite coefficients.
