How to go from $\frac{\|g_{k+1}\|}{\|g_{k}\|^2} \leq c$ to $\frac{\|x_{k+1} - x_*\|}{\|x_{k} - x_*\|^2} \leq c$ as $k\to \infty$ I am reading a paper where
$$\frac{\|g_{k+1}\|}{\|g_{k}\|^2} \leq c \tag{1}$$ as $k\to \infty$ where $g_k = g(x_k) \stackrel{\Delta}{=} \nabla f(x_k)$ and $\| \nabla^2 f(x_k)\| \leq L_H$. Then, it is stated that from Taylor expansion of $g_k$ and $g_{k+1}$ around $x_*$ and from $g(x_*)=0$ we get
$$\frac{\|x_{k+1} - x_*\|}{\|x_{k} - x_*\|^2} \leq c \tag{2}$$ as $k\to \infty$. I know that the Taylor expansion is
$$g_{k+1} = \nabla f (x_k + p_k) = \nabla f (x_k) + \int_{0}^{1} \nabla^2 f (x_k + t p_k) p_k dt$$ with $p_k = x_{k+1} - x_k$ but I found it confusing to apply it in $x_*$ instead of $x_k$. The most possibly useful relation I have found is
$$\nabla f_k -\nabla f_* = \int_{0}^{1} \nabla^2 f (x_k + t(x_* -x_k))(x_k - x_*)dt \tag{3}$$ in the proof of Theorem 3.5 in [1] but again I could not understand how it is derived from Taylor's theorem in Theorem 2.1 in [1]. Taking the norm of $(3)$ we get
$$\begin{aligned}\| \nabla f_k\| \leq & \underbrace{\|\nabla f_k \|}_{0}+ \|\int_{0}^{1} \nabla^2 f (x_k + t(x_* -x_k))(x_k - x_*)dt \| \\ \leq & \underbrace{\|\nabla f_k \|}_{0}+ \int_{0}^{1} \|\nabla^2 f (x_k + t(x_* -x_k))(x_k - x_*)\|dt \leq L_H \|x_k - x_* \|\end{aligned}\tag{4}$$ where $\|\nabla^2 f(x) \| \leq L_H$ is used. Similarly for step $k+1$ we get
$$\| \nabla f_{k+1}\| \leq \|x_{k+1} - x_* \|$$ which get as close to $(2)$. Could you please someone help to proceed?
[1]Jorge Nocedal, Numerical Optimization
 A: For simplicity in notation assume $x_*=0$ from now on. It looks as if you work in Euclidean space $V={\Bbb R}^n$ (but the result in fact holds in any Banach space). $|\cdot|$ denotes norm in $V$ and $\|\cdot\|$ is the induced  operator (or matrix) norm in $L(V)$.
In order for this to work you  need that
the Jacobian $J=Dg(0)=D^2f(0)$ be invertible, i.e. there exists $0<\alpha\leq \beta<+\infty$ so that
$$ \alpha |x| \leq |J x| \leq \beta |x|, \ \ \forall\ x\in V.$$
$f$ should also be $C^2$ so that for any $\epsilon>0$ there is $\delta(\epsilon)>0$ so that $\|Dg(x)-J\| < \epsilon$ for $|x|<\delta(\epsilon)$.We have
$$ g(x) = g(x)-g(0) = \int_0^1 \left( \frac{d}{dt} g(tx)\right) dt = 
\left(\int_0^1 Dg(tx) \; dt \right)x. $$ Taking $\epsilon=\alpha/2$ we get  for $|x|<\delta(\alpha/2)$
$$ |g(x)-Jx| = \left| \int_0^1 \left( Dg(tx)-Dg(0)\right) dt \ x \right|  
 \leq \left\|  \int_0^1 \left( Dg(tx)-Dg(0)\right) dt \right\|\ |x| \leq \alpha/2 |x|,$$
so we get for such $x$:
$$ \alpha/2 \; |x| \leq |g(x)| \leq (\beta+\alpha/2) \; |x| .$$
From this the equivalence of (1) and (2) follows (with distinct constants $c$) when the points are close enough to $x_*$ ($=0$).
For example, if $|x_k|,|x_{k+1}|<\delta$ and
$|g(x_{k+1})|/|g(x_k)|^2 \leq c$ then
$|x_{k+1}| / |x_k|^2 \leq c \times \frac{2}{\alpha} \times (\beta+\alpha/2)^2$.
