understanding a statement in Gill, Murray and Wright "Practical Optimization" Hi: I'm reading the book "Practical Optimization" and there's a part in Chapter 3 that
I can't prove to myself but I'm sure it's true. On page 64, they define the Taylor expansion of $F$ about $x^{*}$:
(3.3) $F(x^{*} + \epsilon p) = F(x^{*}) + \epsilon p^{T}g(x^{*}) + \frac{1}{2} \epsilon^2 p^{T} G(x^{*} + \epsilon \theta p)p$ where $g$ is the gradient of $F$ and $G$ is the Hessian of $F$. 
They want to prove that a necessary condition for $x^{*}$ to be a local minimum of 
$F$ is that $g(x^{*}) = 0 $. i.e: $x^{*}$ is a stationary point.
They proceed using a contradiction argument as follows. Assume that $x^{*}$ is a local minimum of $F$ and that it is not a stationary point. If $g(x^{*})$ is non-zero, then there must exist a vector $p$ for which (3.4) $p^{T}g(x^{*}) < 0$. Any vector that satisfies(3.4) is called a descent direction at $x^{*}$. This is fine so far.
It is the next statement that I don't see. The statement is "Given any descent direction $p$, there exists a positive scalar $\bar\epsilon$ such that for any positive $\epsilon$ satisfying $\epsilon \le \bar\epsilon$, it holds that
$\epsilon p^{T}g(x^{*}) + \frac{1}{2} \epsilon^2 p^{T} G(x^{*} + \epsilon \theta p)p < 0$.
I'm wondering how this statement can be proven. It's obvious to the authors and probably others but not to me. Thanks.

Divide the inequality by $\epsilon$. Then for $\epsilon\to0$ the left-hand side tends to $p^Tg(x^*)<0$. (Assuming that $G$ is continuous).
Thus, there is $\bar\epsilon$ such that the expression is negative for all $\epsilon\in(0,\bar\epsilon)$.
