Rolle bound for a divided difference

If $q=0$, which is allowed by [F4], then $f[t_0]=f(t_0)$, so the claim is $\mathrm{\mid }f(t_0)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },[a,b]}$. The norm is finite by [F5].

Suppose $q\ge 1$, and let $P$ be the Lagrange interpolant in [F1], with leading coefficient $c=f[t_0,\dots ,t_q]$. The function $g=f-P$ has a zero at each of the $q+1$ distinct ordered nodes. Order those nodes increasingly and apply [F3] on each consecutive interval to obtain $q$ distinct zeros of $g^{\mathrm{\prime }}$. Applying [F3] between consecutive zeros of $g^{\mathrm{\prime }}$ gives $q-1$ distinct zeros of $g^{\mathrm{\prime }\mathrm{\prime }}$; continuing, after $r$ rounds $g^{(r)}$ has $q+1-r$ distinct zeros. Because $f\in C^q$ by [F2] and $P$ is a polynomial, each derivative through order $q$ is continuous and its predecessor is differentiable, so every application satisfies [F3]. After $q$ rounds one zero $\xi$ of $g^{(q)}$ remains. Since $P$ has degree at most $q$ and leading coefficient $c$, its $q$th derivative is the constant $q!c$. Hence $f^{(q)}(\xi )=q!c$, and the claimed bound follows by taking absolute values. The derivative norm is finite by [F5].