Convergence proof on Newton's method from Boyd & Vandenberghe's book on convex optimization From Boyd & Vandenberghe's book, page 491:
Applying the Lipschitz condition,we have
$$\begin{aligned}
\left\|\nabla f\left(x^{+}\right)\right\|_{2} &=\left\|\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}\right\|_{2} \\
&=\left\|\int_{0}^{1}\left(\nabla^{2} f\left(x+t \Delta x_{\mathrm{nt}}\right)-\nabla^{2} f(x)\right) \Delta x_{\mathrm{nt}} d t\right\|_{2} \\
& \leq \frac{L}{2}\left\|\Delta x_{\mathrm{nt}}\right\|_{2}^{2} \\
&=\frac{L}{2}\left\|\nabla^{2} f(x)^{-1} \nabla f(x)\right\|_{2}^{2} \\
& \leq \frac{L}{2 m^{2}}\|\nabla f(x)\|_{2}^{2}
\end{aligned}$$
to get the inequality assumption (9.33), which is:
$$\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k+1)}\right)\right\|_{2} \leq\left(\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k)}\right)\right\|_{2}\right)^{2}$$
I dont understand how he get the first two lines of the derivation? 
My understanding is that $\nabla f\left(x^{+}\right)$ equals $\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)$ already when t=1, so where do the rest of two terms come from?
please anyone help me, really appreciate.
 A: I agree with Deepak's point, but I will expand his answer.
As defined on Page 484, the Newton step is
$$
\Delta x_{\mathrm{nt}}=-\nabla^{2} f(x)^{-1} \nabla f(x)
$$
So,
$$
\begin{aligned}
-\nabla f(x)-\nabla^{2} f(x) \Delta x_{nt}
&=-\nabla f(x)-\nabla^{2} f(x)( -\nabla^{2} f(x)^{-1} \nabla f(x))\\
&=-\nabla f(x)+\nabla^{2} f(x)\nabla^{2} f(x)^{-1} \nabla f(x)\\
&=0
\end{aligned}
$$
The above answered the fist line of the derivation in the question.
For the second line, since there is no $t$ in $\nabla^{2} f(x) \Delta x_{nt}$, we have
$$
\int_0^1 \nabla^{2} f(x) \Delta x_{nt}dt = \nabla^{2} f(x) \Delta x_{nt}(1-0) = \nabla^{2} f(x) \Delta x_{nt}
$$
As we know that $\nabla_t(\nabla f(x+t\Delta x_{nt}))=\nabla^2f(x+t\Delta x_{nt})\Delta x_{nt}$, we have the following derivation
$$
\begin{aligned}
\int_0^1 \nabla^2f(x+t\Delta x_{nt})\Delta x_{nt}dt
&=\int_0^1 d(\nabla f(x+t\Delta x_{nt})+c)\\
&=\nabla f(x+t\Delta x_{nt})+c|_0^1\\
&=(\nabla f(x+\Delta x_{nt})+c)-(\nabla f(x)+c)\\
&=0
\end{aligned}
$$
where c is a constant, and the integral interval is $[0,1]$ because t is supposed to be in that range.
A: $-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}=0$ since $\Delta x _{nt}$ is the Newton search direction.
