Detrmine all solutions$\displaystyle f:R\to R$ of the functional equation:
$\displaystyle
f(2x+1)=f(x)
$
which are continuos at -1.
Define a sequence $\displaystyle \left\{x_n\right\}_{n\in\mathbb{N}}$ by $\displaystyle x_1=x,\text{ }2x_{n+1}+1=x_1$. It is relatively easy to prove that $\displaystyle \lim\text{ }x_n=-1$. Also, consider that $\displaystyle f(x_n)=f(2x_{n+1}+1)=f(x_{n+1})$ from where by induction we may conclude that $\displaystyle f(x)=f(x_n)$. Also, since $\displaystyle x_n\to -1$ and $\displaystyle f$ is continuous at $\displaystyle -1$ we see that $\displaystyle f(-1)=\lim\text{ }f(x_n)=f(x)$. Thus, $\displaystyle f$ is constant.