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Next: Feasible directions, global Up: 1.2.1 Unconstrained optimization Previous: First-order necessary condition   Contents   Index Second-order conditions for optimality

We now derive another necessary condition and also a sufficient condition for optimality, under the stronger hypothesis that $ f$ is a $ \mathcal C^2$ function (twice continuously differentiable).

First, we assume as before that $ x^*$ is a local minimum and derive a necessary condition. For an arbitrary fixed $ d\in\mathbb{R}^n$ , let us consider a Taylor expansion of $ g(\alpha)=f(x^*+\alpha d)$ again, but this time include second-order terms:

$\displaystyle g(\alpha)=g(0)+g'(0)\alpha+\frac12g''(0)\alpha^2+o(\alpha^2)$ (1.12)


$\displaystyle \lim_{\alpha\to 0}\frac{o(\alpha^2)}{\alpha^2}=0.$ (1.13)

By the first-order necessary condition, $ g'(0)$ must be 0 since it is given by (1.10). We claim that

$\displaystyle g''(0)\ge 0.$ (1.14)

Indeed, suppose that $ g''(0)<0$ . By (1.13), there exists an $ \varepsilon >0$ such that

$\displaystyle \vert\alpha\vert<\varepsilon \quad\Rightarrow\quad \vert o(\alpha^2)\vert<
\frac12\vert g''(0)\vert\alpha^2.

For these values of $ \alpha$ , (1.12) reduces to $ g(\alpha)-g(0)<0$ , contradicting that fact that 0 is a minimum of $ g$ . Therefore, (1.14) must hold.

What does this result imply about the original function $ f$ ? To see what $ g''(0)$ is in terms of $ f$ , we need to differentiate the formula (1.9). The reader may find it helpful to first rewrite (1.9) more explicitly as

$\displaystyle g'(\alpha)=\sum_{i=1}^n {f}_{x_i}(x^*+\alpha d) d_i.

Differentiating both sides with respect to $ \alpha$ , we have

$\displaystyle g''(\alpha)=\sum_{i,j=1}^n {f}_{{x_i}{x_j}}(x^*+\alpha d) d_i d_j

where double subscripts are used to denote second-order partial derivatives. For $ \alpha=0$ this gives

$\displaystyle g''(0)=\sum_{i,j=1}^n {f}_{{x_i}{x_j}}(x^*) d_i d_j

or, in matrix notation,

$\displaystyle g''(0)=d^T\nabla^2 f(x^*)d$ (1.15)


$\displaystyle \nabla^2 f:=
{f}_{{x_1}{x_1}} & \dots&
...n}} \\ &\ddots& \\
{f}_{{x_n}{x_1}} & \dots &{f}_{{x_n}{x_n}}
\end{pmatrix} $

is the Hessian matrix of $ f$ . In view of (1.14), (1.15), and the fact that $ d$ was arbitrary, we conclude that the matrix $ \nabla^2 f(x^*)$ must be positive semidefinite:

$\displaystyle \fbox{$\nabla^2 f(x^*)\ge 0$} \ $    (positive semidefinite)

This is the second-order necessary condition for optimality.

Like the previous first-order necessary condition, this second-order condition only applies to the unconstrained case. But, unlike the first-order condition, it requires $ f$ to be $ \mathcal C^2$ and not just $ \mathcal C^1$ . Another difference with the first-order condition is that the second-order condition distinguishes minima from maxima: at a local maximum, the Hessian must be negative semidefinite, while the first-order condition applies to any extremum (a minimum or a maximum).

Strengthening the second-order necessary condition and combining it with the first-order necessary condition, we can obtain the following second-order sufficient condition for optimality: If a $ \mathcal C^2$ function $ f$ satisfies

$\displaystyle \nabla f(x^*)=0$   and$\displaystyle \qquad \nabla^2 f(x^*)>0 \ $    (positive definite) (1.16)

on an interior point $ x^*$ of its domain, then $ x^*$ is a strict local minimum of $ f$ . To see why this is true, take an arbitrary $ d\in\mathbb{R}^n$ and consider again the second-order expansion (1.12) for $ g(\alpha)=f(x^*+\alpha d)$ . We know that $ g'(0)$ is given by (1.10), thus it is 0 because $ \nabla f(x^*)=0$ . Next, $ g''(0)$ is given by (1.15), and so we have

$\displaystyle f(x^*+\alpha d)=f(x^*)+\frac12 d^T\nabla^2 f(x^*)d\alpha^2 +o(\alpha^2).$ (1.17)

The intuition is that since the Hessian $ \nabla^2 f(x^*)$ is a positive definite matrix, the second-order term dominates the higher-order term $ o(\alpha^2)$ . To establish this fact rigorously, note that by the definition of $ o(\alpha^2)$ we can pick an $ \varepsilon >0$ small enough so that

$\displaystyle \vert\alpha\vert<\varepsilon ,\ \alpha\ne 0 \quad\Rightarrow\quad \vert o(\alpha^2)\vert<\frac12
d^T\nabla^2 f(x^*)d\alpha^2$

and for these values of $ \alpha$ we deduce from (1.17) that $ f(x^*+\alpha d)>f(x^*).

To conclude that $ x^*$ is a (strict) local minimum, one more technical detail is needed. According to the definition of a local minimum (see page [*]), we must show that $ f(x^*)$ is the lowest value of $ f$ in some ball around $ x^*$ . But the term $ o(\alpha^2)$ and hence the value of $ \varepsilon $ in the above construction depend on the choice of the direction $ d$ . It is clear that this dependence is continuous, since all the other terms in (1.17) are continuous in $ d$ .1.2Also, without loss of generality we can restrict $ d$ to be of unit length, and then we can take the minimum of $ \varepsilon $ over all such $ d$ . Since the unit sphere in $ \mathbb{R}^n$ is compact, the minimum is well defined (thanks to the Weierstrass Theorem which is discussed below). This minimal value of $ \varepsilon $ provides the radius of the desired ball around $ x^*$ in which the lowest value of $ f$ is achieved at $ x^*$ .

next up previous contents index
Next: Feasible directions, global Up: 1.2.1 Unconstrained optimization Previous: First-order necessary condition   Contents   Index
Daniel 2010-12-20