 Solution

By continuity of $g^{\prime \prime }$ at $\alpha$, for any $ϵ>0$ choose $\delta (ϵ)>0$ such that $|g^{\prime \prime }(s)|\le |g^{\prime \prime }(\alpha )|+ϵ=:M$ whenever $|s-\alpha |<\delta (ϵ)$. Twice using the fundamental theorem of calculus gives the remainder asThe inner difference is ${\int }_{\alpha }^s{g}^{\prime \prime }(r)dr$ and has modulus at most $M|s-\alpha |$. Integrating this bound over the interval between $\alpha$ and $t$ proves, for either sign of $t-\alpha$,This is the local quadratic Taylor remainder estimate. The given $ϵ$ may enter the bound on the second derivative; no stronger differentiability assumption is needed.