Justify: since $\nabla f(\cdot)$ is continuous, therefore $\int\limits_0^1 (\nabla^2 f(x+ tp) - \nabla^2 f(x))pdt = o(\|p\|)$

Question

This is a claim from Nocedal and Wright's text on optimization 2nd ed p.23.

Suppose $f$ is $C^2$, then $\exists t \in (0,1)$, such that

$\nabla f(x+p) = \nabla f(x) + \int\limits_0^1 \nabla^2 f(x+ p)pdt$

Equivalently,

$\nabla f(x+p) = \nabla f(x) + \nabla^2f(x)p + \int\limits_0^1 (\nabla^2 f(x+ p) - \nabla^2 f(x))pdt$

However, next comes the confounding claim:

Claim: Since $\nabla f(\cdot)$ is continuous, therefore $$\int\limits_0^1 (\nabla^2 f(x+ tp) - \nabla^2 f(x))pdt = o(\|p\|)$$

Note: in the text, it was defined: if $\eta(\cdot):\mathbb{R} \to \mathbb{R}, \eta(\nu) = o(\nu)$ if $\eta(\nu)/\nu \to 0$ as $\nu \to 0$ or $\nu \to \infty$

Questions:

• How was continuity used? For me, a vector valued function is continuous if all its components are continuity. I can't see how to go from $\dfrac{\partial f(\cdot)}{\partial x_i}, \forall i \in \{1, \ldots, n\}$ is continuous to that integral with the Hessian being continuous.

• How do we justify the $o(\|p\|)$ bound?

Answer 1

$$| \int\limits_0^1 (\nabla^2 f(x+ tp) - \nabla^2 f(x))~p ~dt | \leq \int\limits_0^1 |(\nabla^2 f(x+ tp) - \nabla^2 f(x))p |~dt$$

By Cauchy Swartz inequality we get;

$$| \int\limits_0^1 (\nabla^2 f(x+ tp) - \nabla^2 f(x))~p ~dt | \leq \int\limits_0^1 \| \nabla^2 f(x+ tp) - \nabla^2 f(x) \|~\|p \| ~ dt $$

By mean value theorem for integrals; for all $p$ there is a scalar $s_p \in [0,1]$ such that $$\int\limits_0^1 \| \nabla^2 f(x+ tp) - \nabla^2 f(x) \|~\|p \| ~ dt = \| \nabla^2 f(x+ s_pp) - \nabla^2 f(x) \|~\|p \| $$ Therefore we finally arrive

$$ \frac{| \int\limits_0^1 (\nabla^2 f(x+ tp) - \nabla^2 f(x))~p ~dt |}{\|p\|} \leq \| \nabla^2 f(x+ s_pp) - \nabla^2 f(x) \| $$

Now by letting $p \to 0$ we get the desired result.