 Concave Newton iteration from below

Suppose $f\in C^2([a,r])$, $f(r)=0$, $f^{\prime }\ge m>0$ and $-K\le f^{\prime \prime }<0$. Starting at $x_0\in [a,r)$, the Newton root-finding iteration remains below $r$ and increases strictly to it. To prove this, assume $x_n<r$. Strict increase of $f$ gives $f(x_n)<0$, so the Newton step is positive. Taylor's theorem between $x_n$ and $r$ gives, with $e_n=x_n-r$,Its coefficient is negative and at least $-K/(2m)$, proving both the stated error bound and $x_n<x_{n+1}<r$. Mathematical induction establishes the invariant. The increasing bounded iterates have a limit $L\le r$; continuity in $x_{n+1}-x_n=-f(x_n)/f^{\prime }(x_n)$ forces $f(L)=0$, hence $L=r$. The error bound supplies local quadratic convergence.