Past exam of the numerical linear algebra and continuous optimisation course of the University of Oxford / 2007 / B / 5 / iv / Solution

The true quadratic has $g(x)=c+Hx$ and the exact expansionSubtracting the model with Hessian $I$ yields $f(x+s)-m(s)=\frac{1}{2}{s}^T(H-I)s$. Since $H$ is symmetric positive definite and $\Vert H{\Vert }_2=1$, its eigenvalues lie in $(0,1]$, so $\Vert H-I{\Vert }_2\le 1$. ThereforeIn fact the stated assumptions give a factor one-half better than the requested bound.