3.1 Local optimization

We also must define the notion of an extremum in an arbitrary normed space.

Definition 1 Let $f$ be a real-valued functional defined on $\Omega \subseteq X$ where $X$ is a normed space. A point $x_0\in \Omega$ is a local/relative minimum of $f$ on $\Omega$ if $f\left(x_0\right)\le f\left(x\right)$ for all $x\in \Omega$ such that $\parallel x-x_0\parallel <ϵ$ for some $ϵ>0$ .

Definition 2 Let $f$ be a real-valued functional defined on $\Omega \subseteq X$ where $X$ is a normed space. A point $x_0\in \Omega$ is a local maximum of $f$ on $\Omega$ if $f\left(x_0\right)\ge f\left(x\right)$ for all $x\in \Omega$ such that $\parallel x-x_0\parallel <ϵ$ for some $ϵ>0$ .

Definition 3 Let $f$ be a real-valued functional defined on $\Omega \subseteq X$ where $X$ is a normed space. A point $x_0\in \Omega$ is a local strict minimum of $f$ on $\Omega$ if $f\left(x_0\right)<f\left(x\right)$ for all $x\in \Omega$ such that $\parallel x-x_0\parallel <ϵ$ for some $ϵ>0$ .

Definition 4 Let $f$ be a real-valued functional defined on $\Omega \subseteq X$ where $X$ is a normed space. A point $x_0\in \Omega$ is a local strict maximum of $f$ on $\Omega$ if $f\left(x_0\right)>f\left(x\right)$ for all $x\in \Omega$ such that $\parallel x-x_0\parallel <ϵ$ for some $ϵ>0$ .

It turns out the notion of a gradient is intrinsically linked to the directional derivatives we have introduced.

Definition 5 Let $X$ be a Hilbert space and $f:X\to R$ . If $f$ is a Fréchet differentiable functional, then for each $x\in X$ there exists a vector in $X$ such that $\delta f(x;h)=⟨h,\nabla f\left(x\right)⟩$ for all $h\in X$ ; the vector $\nabla f\left(x\right)$ is called the gradient of $f$ at $x$ , and can be written as a functional $\nabla f:X\to X$ .

This definition can be seen to correspond to an application of the Riesz representation theorem to the Fréchet derivative $\delta f(x;h)$ , which is a linear bounded functional on $h$ .

Example 1 We know now that:

By the Cauchy-Schwarz Inequality, we have:

If $h=\nabla f\left(x\right)$ then $\delta f(x;h)$ is maximized.

Example 2 Recall that if $f:{\mathbb{R}}^n\to \mathbb{R}$ , then

Theorem 1 Let $f:X\to \mathbb{R}$ have a Gâteaux differential on $X$ . A necessary condition for $f$ to have an extremum at $x_0\in X$ is that $\delta f(x_0;h)=0$ for all $h\in X$ . Alternatively, if $X$ is a Hilbert space, we can write $⟨h,\nabla f\left(x_0\right)⟩=0$ for all $h\in X$ , which implies $\nabla f\left(x_0\right)=0$ .

Suppose $x_0$ is a local minimum. Then there exists $ϵ>0$ such that if $\parallel x-x_0\parallel <ϵ$ then $f\left(x_0\right)\le f\left(x\right)$ . Fix $h\ne 0$ and let $\theta =\frac{ϵ}{\parallel h\parallel }$ . Next, consider $x=x_0+\alpha h$ . For $\alpha \in (-\theta ,\theta )$ :

• If $\alpha >0$ then $\frac{f(x_0+\alpha h)-f\left(x_0\right)}{\alpha }\ge 0$ , and therefore $\delta f(x_0;h)\ge 0$ .
• If $\alpha <0$ then $\frac{f(x_0+\alpha h)-f\left(x_0\right)}{\alpha }\le 0$ , and therefore $\delta f(x_0;h)\le 0$ .

Therefore, $\delta f(x_0;h)=0$ for arbitrary nonzero $h$ . Now since $\delta f(x;h)$ is linear on $h$ we must have $\delta f(x;h)=0$ for $h=0$ . Therefore, the equality is true for all $h\in X$ .

Definition 6 A point at which $\delta f(x;h)=0$ for all $h\in X$ is called a stationary point of $f$ .