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.
On
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 $$ g(x)=\lim_{n\to\infty}2^{-n}g(2^{-n}x)=0 $$
On
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.
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 $f(x)=x/3$.