We create sequence $\{a_n\}_{n=0}^{+\infty}$ satisfying: $\begin{cases} a_0 = 2 \\ a_n = 2a_{n-1} - 1\end{cases}$. We will prove by induction that $\begin{cases} f\left(a_nx - (a_n-1)f(x)\right) = (a_{n} - 1)x - (a_{n}-2)f(x),\forall x\in\mathbb R \\ a_nx - (a_n-1)f(x) \text{ is an increasing function}\end{cases}(*)$, for all $n\in\mathbb N$. If $n=0$, the two conditions is obvious. Suppose (*) is true up to $n,n\in\mathbb N$. Since $f(a_nx - (a_n-1)f(x)) = (a_n-1)x - (a_n-2)f(x),\forall x\in\mathbb R$, replace $x$ with $a_nx - (a_n-1)f(x)$ we have: $f((2a_n - 1)x - (2a_n - 2)f(x)) = 2(a_n-1)x - (2a_n-3)f(x),\forall x\in\mathbb R$ $\Leftrightarrow f\left(a_{n+1} x - (a_{n+1}-1)f(x)\right) = (a_{n+1} - 1)x - (a_{n+1} - 2)f(x)$. Since $a_{n+1}x - (a_{n+1} -1)f(x) = a_n(a_n - (a_n-1)f(x)) - (a_n - 1)f(a_n - (a_n-1)f(x))$, and $a_nx - (a_n-1)f(x)$ is increasing, $a_{n+1}x - (a_{n+1} - 1)f(x)$ is also increasing $\Rightarrow a_{n+1}x - (a_{n+1} - 1)f(x) > a_{n+1}y - (a_{n+1} - 1)f(y),\forall x > y,n\in\mathbb N^*$ $\Rightarrow \frac{f(x) - f(y)}{x-y}\leq \frac{a_{n+1} - 1}{a_{n+1}},\forall x > y,n\in\mathbb N^*$. For $n\to+\infty$ we have $\frac{f(x) - f(y)}{x-y}\leq 1\Rightarrow f(x) - x \leq f(y) - y,\forall x >y$. Let $g(x) = f(x) - x,\forall x\in\mathbb R$. Then $g$ is a non-increasing function. We have $g(x) = g(a_nx - (a_n - 1)f(x)),\forall x\in\mathbb R,n\in\mathbb N^*$ $\Rightarrow g(c) = g(x),\forall c\in (a_nx - (a_n-1)f(x), x)$. Moreover, $a_nx - (a_n-1)f(x) = x + (a_n - 1)(x - f(x))\to -\infty$ when $n\to +\infty$. Therefore $g(y) = g(x),\forall y < x$. Hence, $g$ is const and $f(x) = x + f(0),\forall x\in\mathbb R$.