A Lipschitz Fréchet derivative gives a quadratic linearization remainder

Convexity gives $z+th\in K\subseteq U$ for $0\le t\le 1$. If $h\mathrm{\ne }0$ and $0<t<1$, put $u=z+th$. For real $s\to 0$ with $t+s\in [0,1]$, Fréchet differentiability at $u$ gives $\parallel \frac{F(u+sh)-F(u)}{s}-F^{\mathrm{\prime }}(u)h\parallel =\frac{\mathrm{\parallel }F(u+sh)-F(u)-F^{\mathrm{\prime }}(u)(sh)\mathrm{\parallel }}{\mathrm{\mid }s\mathrm{\mid }}⟶0,$ because $F^{\mathrm{\prime }}(u)$ is linear over the common field and hence over real scalars. Thus ${\gamma }^{\mathrm{\prime }}(t)=F^{\mathrm{\prime }}(z+th)h$. If $h=0$, $\gamma$ is constant and the same formula holds. Differentiability at every point of $U$ also makes $F$ and hence $\gamma$ continuous on the segment, directly from the Fréchet remainder definition. In either case this derivative is continuous in $t$ and extends to the endpoints, since $F^{\mathrm{\prime }}$ is Lipschitz on $K$. The Banach-valued fundamental theorem [F1] therefore gives $F(x)-F(z)={\int }_0^1{F}^{\mathrm{\prime }}(z+th)hdt$. The constant vector function $t\mapsto F^{\mathrm{\prime }}(z)h$ has integral $F^{\mathrm{\prime }}(z)h$ by [F6]. Subtracting that integral and using linearity [F2] gives $F(x)-F(z)-F^{\mathrm{\prime }}(z)h={\int }_0^1(F^{\mathrm{\prime }}(z+th)-F^{\mathrm{\prime }}(z))hdt$.

For each $t\in [0,1]$, the Lipschitz hypothesis and bounded linearity give $\mathrm{\parallel }(F^{\mathrm{\prime }}(z+th)-F^{\mathrm{\prime }}(z))h\mathrm{\parallel }\le Lt\mathrm{\parallel }h{\mathrm{\parallel }}^2$. Therefore [F3] and scalar integration give $\mathrm{\parallel }F(x)-F(z)-F^{\mathrm{\prime }}(z)h\mathrm{\parallel }\le {\int }_0^{1}Lt\mathrm{\parallel }h{\mathrm{\parallel }}^{2}dt=(L\mathrm{/}2)\mathrm{\parallel }h{\mathrm{\parallel }}^2$.