linear functional-equation $ \,2f\left(x+1\right)=f\left(x\right)+f\left(2x\right) $ I'm looking for all functions : $ \ R\rightarrow R\ $ satisfying: $ \,2f\left(x+1\right)=f\left(x\right)+f\left(2x\right) $
 A: Assume we have an interval $[a,b]\subset\mathbb R$ with $a<0$ and $b>2$ and a (if you like: continuous) function $f_{a,b}\colon[a,b]\to\mathbb R$ that satisfies the functional equation for all $x$ with $\{x,2x,x+1\}\subset[a,b]$.
Let $a'=\max\{2a,a-1\}<a$, $b'=2(b-1)>b$ and define $f_{a',b'}\colon[a',b']\to\mathbb R$ by
$$f_{a',b'}(x)=\begin{cases}f_{a,b}(x)&\text{if }a\le x\le b,\\
{2f_{a,b}(\tfrac x2+1)-f_{a,b}(\tfrac x2)}&\text{if }a'\le x\le a\text{ or }b\le x\le b'.\end{cases}$$
Note that this is possible because $a'\le x\le a$ implies $\frac x2\ge \frac{a'}2\ge a$ and $\frac x2+1\le 1\le b$, and $b\le x\le b'$ implies $\frac x2\ge 0>a$ and $\frac x2+1\le \frac{b'}2+1=b$.
Then if $x$ is a real number with $\{x,2x,x+1\}\subset[a',b']$, the following cases are possible:


*

*$\{x,2x,x+1\}\subset[a,b]$. Then the functional equatin holds for $f_{a',b'}$ because it hilds for $f_{a,b}$.

*$b<\max\{x,2x,x+1\}\le b'$. Then $b>2$ implies that $2x$ is this maximal element and $x,x+1\in[a,b]$. The definition of $f_{a',b'}$ for $b\le x\le b'$ is specifically taylored to make the functional equation hold

*$a'\le\lim\{x,2x,x+1\}\le a$. Then this minimal element must again be $2x$ and the definition fo $f_{a',b'}$ for $a'\le x<a$  ensures again that the functional equation holds.


We conclude that the $f_{a',b'}$ satisfies the functional equation for all $x$ with $\{x,2x,x+1\}\subset[a',b']$.
If one repeats the process above, one obtains a seuence of intervals $[a_n,b_n]$ and corresponding functions. If $a_1<0$ and $b_1>2$, then $a_n\to-\infty$ and $b_n\to+\infty$. 
Since all functions agree with the previous ones on the common domain, we can define $f\colon\mathbb R\to\mathbb R$ by
$$ f(x)=f_{[a_n,b_n]}(x)\qquad\text{for some $n$ with $a_n\le x\le b_n$}.$$
Then $f$ satisfis the functional equation for all $x\in\mathbb R$ (and is continuous if we started with a continuous function).
For some  $0<r<\frac23$ select a continuos function $g\colon[-r,r]\to\mathbb R$ and let $h_0\colon [-r,r]\cup[1-\frac r2,1+\frac r2]\cup [2-r,2+r]$ be given by
$$h_0(x)=\begin{cases}
g(x)&\text{if }-r\le x\le r,\\
g(x-2)&\text{if }2-r\le x\le 2+r,\\
\frac{g(x-1)+g(2x-2)}2&\text{if }1-\frac r2\le x\le 1+\frac r2.\end{cases} $$
Then $h_0$ satisfies the functional equation for all cases where $x,2x,x+1$ are in its domain.
Let $h_1\colon[-r,1+\frac r2]\cup[2-r,2+r]$ be a continuous extension of $h_0$.
Then $h_1$ still satisfies the functional equation for all $x$ with $x,2x,x+1$ in its domain (there are no new such $x$).
Next define $h_2\colon[-r,2+r]$ by
$$h_2(x)=\begin{cases}
h_1(x)&\text{if }-r\le x\le 1+\frac r2\text { or }2-r\le x\le 2+r,\\
\frac{h_1(x-1)+h_1(2x-2)}2&\text{if }1+\frac r2\le x\le 2-r.\end{cases} $$
Then we still have the functional equation for $h_2$, i.e. $a=-r$, $ b=2+r$, $f_{a,b}=h_2$ is a valid starting point for the procedure described above and gives us a solution.
A: Here is an uncountable class of discontinuous solutions.  The irrationals can be partitioned into equivalence classes
of the form $C(\alpha) = \{2^m \alpha + r: m \in {\mathbb Z}, r \in {\mathbb Q}\}$
for irrational numbers $\alpha$.  On each $C(\alpha)$ define
$f(2^m \alpha + r) = k(\alpha) 2^{-m} (r - 4)$ where $k(\alpha)$ is arbitrary.  On the rationals take $f(r) = 0$.  
