First-Order Necessary Conditions for Extrema of Multivariable Functions 📂Optimization

First-Order Necessary Conditions for Extrema of Multivariable Functions

Theorem¹

Let’s assume the function $f : \mathbb{R}^{n} \to \mathbb{R}$ is given. If $x^{\ast}$ is a local optimizer and $f \in C^{1}$ in the vicinity of $x^{\ast}$ , then,

$\nabla f(x^{\ast}) = 0$

$\nabla f$ is the gradient of $f$ . Note here that $0$ is not the numeric zero, but a zero vector.

Explanation

The first-order necessary condition tells us about the property of the gradient, which is the first-order derivative of $f$ , when $x^{\ast}$ is a local minimizer of $f$ . Named and extended to multivariable functions, the concept that differentiation at a maximum or minimum yields $0$ is something we learn even in high school.

Proof

We prove by contradiction. Assume $\nabla f (x^{\ast}) \ne 0$ . And denote as follows:

$p = - \nabla f (x^{\ast}),\quad p^{t}\nabla f (x^{\ast}) = - \left\| \nabla f (x^{\ast}) \right\|^{2} \lt 0$

Here, $p^{t}$ denotes the transpose matrix. Then, because $\nabla f$ is continuous, there exists $s \gt 0$ such that the following equation holds:

$\begin{equation} p^{t}\nabla f (x^{\ast} + \xi p) \lt 0, \qquad \forall \xi \in [0, s] \end{equation}$

Furthermore, by the Taylor expansion formula for multivariable functions,

$\begin{equation} f(x^{\ast} + \xi p) = f (x^{\ast}) + \xi p^{t} \nabla f(x^{\ast} + \bar{\xi} p),\quad \text{for some } \bar{\xi} \in (0, \xi) \end{equation}$

Then, by $(1)$ and $(2)$ , we obtain the following:

$f(x^{\ast} + \xi p) \lt f (x^{\ast}), \qquad \forall \xi \in [0, s]$

This contradicts the fact that $x^{\ast}$ is a local minimizer. Therefore, the assumption is wrong, and $\nabla f (x^{\ast}) = 0$ is true.

■