I am self studying Percolation theory from Bollobas book of the same name. To give the context I have attached the screenshots from the book: We are trying to compute critical probability for bond percolation for k-branching trees
My doubt is the following: From equation 2, how do we get to know that if $p > 1/k$, $\pi_{n-1} \geq x_0 $ ? If $p < 1/k$, why does $\pi_n$ converge to 0 ?
A general remark: It is crucial to note that any bounded real sequence with a unique accumulation point is convergent. Indeed, if $c$ is an accumulation point of $(x_n)_{n\in \mathbb{N}}$ which is not its limit, then there exist a sequence $(n_k)_{k\in \mathbb{N}}$ and an $\varepsilon>0$ such that $|x_{n_k}-c|>\varepsilon$ for all $k$. Since $(x_{n_k})_{k\in \mathbb{N}}$ is bounded, it must have an accumulation point. This accumulation point $c'$ obviously cannot be equal to $c$. Now, any accumulation point of $(x_{n_k})_{k\in\mathbb{N}}$ is also an accumulation point of the supersequence $(x_n)_{n\in\mathbb{N}}$ and thus, the latter sequence has at least two accumulation points, $c$ and $c'$.
2. Now, furthermore, a note on recursive sequences. If $x_n:=g(x_{n-1})$ for some continuous function $g$ and $x_n$ is convergent, then $$ \lim_{n\to\infty} x_n=\lim_{n\to\infty}g(x_n)=g(\lim_{n\to\infty} x_n), $$ so the only possible limits are the fixed points of $g$.
So let me answer your question in reverse order:
If $p<1/k$, then $f_{p,k}$ maps $[0,1]$ to itself and only has one fixed point ( by the remarks in the text). Using point 2. above, $\pi_n$ only has one possible accumulation point (its unique fixed point). Now, $\pi_n$ is bounded, so, appealing to point 1., we conclude that $\pi_n$ is convergent to the unique fixed point of $f_{p,k}$, which is $0$.
If $p>1/k,$ we note that, using the fact $x_0$ is a fixed point of $f_{p,k}$, that $\pi_{n-1}\geq x_0$ and that $f_{p,k}$ is increasing and concave, \begin{align} \pi_n=f_{k,p}(\pi_{n-1}) &=f_{k,p}\left(\frac{\pi_{n-1}-x_0}{1-x_0}\cdot 1+\frac{1-\pi_{n-1}}{1-x_0}\cdot x_0\right)\\ &\geq \left(\frac{\pi_{n-1}-x_0}{1-x_0}\right)f_{k,p}(1)+\left(\frac{1-\pi_{n-1}}{1-x_0}\right)f(x_0)\\ &\geq f(x_0)\\ &= x_0 \end{align}
Thus, once again, $f_{p,k}$ maps $[x_0,1]$ into itself and has a unique fixed point there, so we may refer to points $1.$ and $2.$ to conclude convergence.