Local Hessian comparison, damped Newton decrement, and center distance

Fix $h\mathrm{\ne }0$ and put $q(t)=\mathrm{\parallel }h{\mathrm{\parallel }}_{x+th}$. Differentiation gives $q^{\mathrm{\prime }}(t)=D^{3}F(x+th)[h,h,h]\mathrm{/}(2q(t))$, so [F1] implies $\mathrm{\mid }{q}^{\mathrm{\prime }}(t)\mathrm{\mid }\le q(t{)}^2$. Therefore $\mathrm{\mid }(1\mathrm{/}q{)}^{\mathrm{\prime }}\mathrm{\mid }\le 1$. Integrating from $0$ to $t$ yields $1\mathrm{/}r-t\le 1\mathrm{/}q(t)\le 1\mathrm{/}r+t$ whenever $tr<1$, and the lower inequality $1\mathrm{/}q(t)\le 1\mathrm{/}r+t$ holds along every segment in $D$. Taking reciprocals gives the asserted two-sided metric bound for $tr<1$ and the lower bound for all $t\ge 0$.

For a fixed $v\mathrm{\ne }0$, put $p(t)=\sqrt{D^{2}F(x+th)[v,v]}$. By [F1], $\mathrm{\mid }(\log p{)}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{D}^{3}F[h,v,v]\mathrm{\mid }\mathrm{/}(2p^2)\le q(t)$. When $tr<1$, step 1.1 gives $q(t)\le r\mathrm{/}(1-tr)$; integration gives $\mathrm{\mid }\log (p(t)\mathrm{/}p(0))\mathrm{\mid }\le -\log (1-tr)$. Squaring this inequality for every $v$ proves the stated Hessian comparison.

For any $y\in D$, put $h=y-x$ and $r=\mathrm{\parallel }h{\mathrm{\parallel }}_x$. Taylor's formula with integral remainder and step 1.1 give $F(y)\ge F(x)+DF(x)[h]+r^2{\int }_0^1\frac{1-t}{(1+tr{)}^2}dt=F(x)+DF(x)[h]+r-\log (1+r)$. The dual-norm inequality $DF(x)[h]\ge -{\lambda }_F(x)r$ makes the right side at least $F(x)+(1-{\lambda }_F(x))r-\log (1+r)$, which tends to $+\mathrm{\infty }$ as $r\to \mathrm{\infty }$ when ${\lambda }_F(x)<1$. Under this condition the sublevel set $\{y\in D:F(y)\le F(x)\}$ is bounded.

Since $\mathrm{\parallel }d{\mathrm{\parallel }}_x=\lambda$, the damped displacement $h=d\mathrm{/}(1+\lambda )$ has local length $r=\lambda \mathrm{/}(1+\lambda )<1$, so Dikin inclusion places $y$ in $D$. The gradient expansion is $\mathrm{\nabla }F(y)=\mathrm{\nabla }F(x)+H_{x}h+R$, where $R={\int }_0^1(H_{x+th}-H_x)hdt$. The first two terms have $x$-dual norm ${\lambda }^2\mathrm{/}(1+\lambda )$. By step 2.1, $\mathrm{\parallel }R{\mathrm{\parallel }}_{x,\ast }\le r{\int }_0^1((1-tr{)}^{-2}-1)dt=r^2\mathrm{/}(1-r)={\lambda }^2\mathrm{/}(1+\lambda )$. Thus $\mathrm{\parallel }\mathrm{\nabla }F(y){\mathrm{\parallel }}_{x,\ast }\le 2{\lambda }^2\mathrm{/}(1+\lambda )$. Since $H_y⪰(1-r{)}^2{H}_x$, the dual norm at $y$ is at most $(1-r{)}^{-1}=1+\lambda$ times the $x$-dual norm, giving ${\lambda }_F(y)\le 2{\lambda }^2$.

Assume now ${\lambda }_F(x)<1$. The sublevel set in step 2.2 cannot accumulate at a finite boundary point of $D$, because $F$ blows up there while its values on the sublevel are at most $F(x)$. It is closed in $E$ and bounded, hence compact. Continuity gives a minimizer on it, which is a global minimizer on $D$; positive-definite Hessian makes it unique.

Let $h=x_{\ast }-x$ and $r=\mathrm{\parallel }h{\mathrm{\parallel }}_x$. If $r=0$ the claim holds. Otherwise, stationarity at the minimizer from step 3.2 and the fundamental theorem of calculus give $-\mathrm{\nabla }F(x{)}^{T}h={\int }_0^1\mathrm{\parallel }h{\mathrm{\parallel }}_{x+th}^{2}dt$. By the global lower bound from step 1.1, this is at least $r^2{\int }_0^1(1+tr{)}^{-2}dt=r^2\mathrm{/}(1+r)$. Cauchy–Schwarz in the local metric bounds it above by ${\lambda }_F(x)r$. Dividing by $r$ and rearranging gives $r\le {\lambda }_F(x)\mathrm{/}(1-{\lambda }_F(x))$.