<s>[INST] Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$ 
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.
 [/INST] 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 $f(x)=x/3$.
</s>