### 数学代写|凸优化作业代写Convex Optimization代考|Separating and supporting hyperplanes

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## 数学代写|凸优化作业代写Convex Optimization代考|Separating hyperplane theorem

Suppose that $C$ and $D$ are convex sets in $\mathbb{R}^{n}$ and $C \cap D=\emptyset$. Then there exists a hyperplane $H(\mathbf{a}, b)=\left{\mathbf{x} \mid \mathbf{a}^{T} \mathbf{x}=b\right}$ where $\mathbf{a} \in \mathbb{R}^{n}$ is a nonzero vector and $b \in \mathbb{R}$ such that
$$C \subseteq H_{-}(\mathbf{a}, b) \text {, i.e., } \mathbf{a}^{T} \mathbf{x} \leq b, \forall \mathbf{x} \in C$$
and
$$D \subseteq H_{+}(\mathbf{a}, b), \text { i.e., } \mathbf{a}^{T} \mathbf{x} \geq b, \forall \mathbf{x} \in D$$
Only the proof of (2.123) will be given below since the proof of (2.122) can be proven similarly. In the proof, we implicitly assume that the two convex sets $C$ and $D$ are closed without loss of generality. The reasons are that $\mathbf{c l} C$, int $C$, cl $D$, and int $D$ are also convex by Property $2.5$ in Subsection 2.1.4, and thus the same hyperplane that separates $\mathbf{c l} C$ (or int $C$ ) and $\mathbf{c l} D$ (or int $D$ ) can also separate $C$ and $D$.
Proof: Let
$$\operatorname{dist}(C, D)=\inf \left{|\mathbf{u}-\mathbf{v}|_{2} \mid \mathbf{v} \in C, \mathbf{u} \in D\right} .$$
Assume that $\operatorname{dist}(C, D)>0$, and that there exists a point $\mathbf{c} \in C$ and a point $\mathbf{d} \in D$ such that
$$|\mathbf{c}-\mathbf{d}|_{2}=\operatorname{dist}(C, D)$$
(as illustrated in Figure $2.20$ ). These assumptions will be satisfied if $C$ and $D$ are closed, and one of $C$ and $D$ is bounded. Note that it is possible that if both $C$ and $D$ are not bounded, such $\mathbf{c} \in C$ and $\mathbf{d} \in D$ may not exist. For instance, $C=\left{(x, y) \in \mathbb{R}^{2} \mid y \geq e^{-x}+1, x \geq 0\right}$ and $D=\left{(x, y) \in \mathbb{R}^{2} \mid y \leq-e^{-x}, x \geq 0\right}$ are convex, closed, and unbounded with $\operatorname{dist}(C, D)=1$, but $\mathbf{c} \in C$ and $\mathbf{d} \in D$ satisfying (2.125) do not exist.

## 数学代写|凸优化作业代写Convex Optimization代考|Supporting hyperplanes

For any nonempty convex set $C$ and for any $\mathbf{x}{0} \in \mathbf{b d} C$, there exists an $\mathbf{a} \neq \mathbf{0}$, such that $\mathbf{a}^{T} \mathbf{x} \leq \mathbf{a}^{T} \mathbf{x}{0}$, for all $\mathbf{x} \in C$; namely, the convex set $C$ is supported by

Proof: Assume that $C$ is a convex set, $A=$ int $C$ (which is open and convex), and $\mathbf{x}{0} \in \mathbf{b d} C$. Let $B=\left{\mathbf{x}{0}\right}$ (which is convex). Then $A \cap B=\emptyset$. By the separating hyperplane theorem, there exists a separating hyperplane $H={\mathbf{x} \mid$ $\left.\mathbf{a}^{T} \mathbf{x}=\mathbf{a}^{T} \mathbf{x}{0}\right}$ (since the distance between the set $A$ and the set $B$ is equal to zero), where $\mathbf{a} \neq \mathbf{0}$, between $A$ and $B$, such that $\mathbf{a}^{T}\left(\mathbf{x}-\mathbf{x}{0}\right) \leq 0$ for all $\mathbf{x} \in C$ (i.e., $C \subseteq H_{-}$by Remark $2.23$ ). Therefore, the hyperplane $H$ is a supporting hyperplane of the convex set $C$ which passes $\mathbf{x}_{0} \in$ bd $C$.

It is now easy to prove, by the supporting hyperplane theorem, that a closed convex set $S$ with int $S \neq \emptyset$ is the intersection of all (possibly an infinite number of) closed halfspaces that contain it (cf. Remark 2.8). Let
$$\mathcal{H}\left(\mathbf{x}{0}\right) \triangleq\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf{x}-\mathbf{x}{0}\right)=0\right}$$
be a supporting hyperplane of $S$ passing $\mathbf{x}{0} \in \mathbf{b d} S$. This implies, by the hyperplane supporting theorem, that the associated closed halfspace $\mathcal{H}{-}\left(\mathbf{x}{0}\right)$, which contains the closed convex set $S$, is given by $$\mathcal{H}{-}\left(\mathbf{x}{0}\right) \triangleq\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf{x}-\mathbf{x}{0}\right) \leq 0\right}, \quad \mathbf{x}{0} \in \mathbf{b d} S$$ Thus it must be true that $$S=\bigcap{\mathbf{x}{0} \in \mathbf{b d} S} \mathcal{H}{-}\left(\mathbf{x}{0}\right)$$ implying that a closed convex set $S$ can be defined by all of its supporting hyperplanes $\mathcal{H}\left(\mathrm{x}{0}\right)$, though the expression (2.136) may not be unique, thereby justifying Remark $2.8$. When the number of supporting halfspaces containing the closed convex set $S$ is finite, $S$ is a polyhedron. When $S$ is compact and convex,

the supporting hyperplane representation (2.136) can also be expressed as
$$\left.S=\bigcap_{\mathbf{x}{0} \in S{\text {extr }}} \mathcal{H}{-}\left(\mathbf{x}{0}\right) \text { (cf. }(2.24)\right)$$
where the intersection also contains those halfspaces whose boundaries may contain multiple extreme points of $S$. Let us conclude this section with the following three remarks.

## 数学代写|凸优化作业代写Convex Optimization代考|Summary and discussion

In this chapter, we have introduced convex sets and their properties (mostly geometric properties). Various convexity preserving operations were introduced together with many examples. In addition, the concepts of proper cones on which the generalized equality is defined, dual norms, and dual cones were introduced in detail. Finally, we presented the separating hyperplane theorem, which corroborates the existence of a hyperplane separating two disjoint convex sets, and the existence of the supporting hyperplane of any nonempty convex set. These fundamentals on convex sets along with convex functions to be introduced in the next chapter will be highly instrumental in understanding the concepts of convex optimization. The convex geometry properties introduced in this chapter have been applied to blind hyperspectral unmixing for material identification in remote sensing. Some will be introduced in Chapter 6 .

## 数学代写|凸优化作业代写Convex Optimization代考|Separating hyperplane theorem

C⊆H−(一种,b)， IE， 一种吨X≤b,∀X∈C

D⊆H+(一种,b), IE， 一种吨X≥b,∀X∈D

\operatorname{dist}(C, D)=\inf \left{|\mathbf{u}-\mathbf{v}|_{2} \mid \mathbf{v} \in C, \mathbf{u} \在 D\right} 中。\operatorname{dist}(C, D)=\inf \left{|\mathbf{u}-\mathbf{v}|_{2} \mid \mathbf{v} \in C, \mathbf{u} \在 D\right} 中。

|C−d|2=距离⁡(C,D)
（如图所示2.20）。如果满足这些假设C和D关闭，并且其中之一C和D是有界的。请注意，如果两者都C和D没有界限，例如C∈C和d∈D可能不存在。例如，C=\left{(x, y) \in \mathbb{R}^{2} \mid y \geq e^{-x}+1, x \geq 0\right}C=\left{(x, y) \in \mathbb{R}^{2} \mid y \geq e^{-x}+1, x \geq 0\right}和D=\left{(x, y) \in \mathbb{R}^{2} \mid y \leq-e^{-x}, x \geq 0\right}D=\left{(x, y) \in \mathbb{R}^{2} \mid y \leq-e^{-x}, x \geq 0\right}是凸的、封闭的和无界的距离⁡(C,D)=1， 但C∈C和d∈D满足 (2.125) 不存在。

## 数学代写|凸优化作业代写Convex Optimization代考|Supporting hyperplanes

\mathcal{H}\left(\mathbf{x}{0}\right) \triangleq\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf{x}-\数学bf{x}{0}\right)=0\right}\mathcal{H}\left(\mathbf{x}{0}\right) \triangleq\left{\mathbf{x} \mid \mathbf{a}^{T}\left(\mathbf{x}-\数学bf{x}{0}\right)=0\right}

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