# 数学代写|凸优化作业代写Convex Optimization代考|EE364a

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## 数学代写|凸优化作业代写Convex Optimization代考|Norm cones

Let $|\cdot|$ represent a norm on $\mathbb{R}^{n}$. A lot of commonly known convex sets involve norm, such as norm balls (cf. (1.16)). Norm cones (convex sets to be defined next) are particularly important in many applications.
The norm cone associated with the norm $|\cdot|$ is the convex set
$$C=\left{(\mathbf{x}, t) \in \mathbb{R}^{n+1} \mid|\mathbf{x}| \leq t\right} \subseteq \mathbb{R}^{n+1}$$
Its convexity can be easily proven by the definition of convex set. Note that it is not strictly convex because the boundary of this cone is a set of rays
$$\mathbf{b d} C=\bigcup_{|\mathbf{u}|=1}\left{t(\mathbf{u}, 1) \in \mathbb{R}^{n+1}, t \geq 0\right} \subseteq \mathbb{R}^{n+1}$$
(containing line segments).
Remark 2.5 In the special case when the norm in (2.55) is Euclidean norm (i.e., 2-norm), the norm cone is called Lorentz cone or quadratic cone, secondorder cone, or ice-cream cone (see Figure 2.11). The second-order cone has been prevalent in many signal processing and communication problems that will be presented in the subsequent chapters.

## 数学代写|凸优化作业代写Convex Optimization代考|Affine function

A function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is affine if it takes the form
$$\boldsymbol{f}(\mathbf{x})=\mathbf{A} \mathbf{x}+\mathbf{b}$$
where $\mathbf{A} \in \mathbb{R}^{m \times n}$ and $\mathbf{b} \in \mathbb{R}^{m}$. The affine function, for which $f($ dom $f$ ) is an affine set if dom $f$ is an affine set, also called the affine transformation or the affine mapping, has been implicitly used in defining the affine hull given by (2.7) in the preceding Subsection 2.1.2. It preserves points, straight lines, and planes, but not necessarily preserves angles between lines or distances between points. The affine mapping plays an important role in a variety of convex sets and convex functions, problem reformulations to be introduced in the subsequent chapters.
Suppose $S \subseteq \mathbb{R}^{n}$ is convex and $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is an affine function (see Figure 2.12). Then the image of $S$ under $f$
$$f(S)={f(\mathbf{x}) \mid \mathbf{x} \in S}$$
is convex. The converse is also true, i.e., the inverse image of the convex set $C$
$$f^{-1}(C)={\mathbf{x} \mid \boldsymbol{f}(\mathbf{x}) \in C}$$
is convex. The proof is given below.
Proof: Let $\mathbf{y}{1}$ and $\mathbf{y}{2} \in C$. Then there exist $\mathbf{x}{1}$ and $\mathbf{x}{2} \in f^{-1}(C)$ such that $\mathbf{y}{1}=\mathbf{A} \mathbf{x}{1}+\mathbf{b}$ and $\mathbf{y}{2}=\mathbf{A} \mathbf{x}{2}+\mathbf{b} .$ Our aim is to show that the set $\boldsymbol{f}^{-1}(C)$, which is the inverse image of $f$, is convex. For $\theta \in[0,1]$,
\begin{aligned} \theta \mathbf{y}{1}+(1-\theta) \mathbf{y}{2} &=\theta\left(\mathbf{A} \mathbf{x}{1}+\mathbf{b}\right)+(1-\theta)\left(\mathbf{A} \mathbf{x}{2}+\mathbf{b}\right) \ &=\mathbf{A}\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}\right)+\mathbf{b} \in C \end{aligned}
which implies that $\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2} \in f^{-1}(C)$, and that the convex combination of $\mathbf{x}{1}$ and $\mathbf{x}{2}$ is in $f^{-1}(C)$, and hence $f^{-1}(C)$ is convex.

Remark $2.9$ If $S_{1} \subset \mathbb{R}^{n}$ and $S_{2} \subset \mathbb{R}^{n}$ are convex and $\alpha_{1}, \alpha_{2} \in \mathbb{R}$, then the set $S=\left{(\mathbf{x}, \mathbf{y}) \mid \mathbf{x} \in S_{1}, \mathbf{y} \in S_{2}\right}$ is convex. Furthermore, the set
$$\alpha_{1} S_{1}+\alpha_{2} S_{2}=\left{\mathbf{z}=\alpha_{1} \mathbf{x}+\alpha_{2} \mathbf{y} \mid \mathbf{x} \in S_{1}, \mathbf{y} \in S_{2}\right} \quad(c f . ~(1.22) \text { and (1.23)) }$$
is also convex (since this set can be thought of as the image of the convex set $S$ through the affine mapping given by (2.58) from $S$ to $\alpha_{1} S_{1}+\alpha_{2} S_{2}$ with $\mathbf{A}=\left[\begin{array}{lll}\alpha_{1} & \mathbf{I}{n} & \alpha{2} \ \mathbf{I}_{n}\end{array}\right]$ and $\left.\mathbf{b}=\mathbf{0}\right)$

## 数学代写|凸优化作业代写Convex Optimization代考| Norm cones

(1.16) )。范数雉（接下来要定义的凸集）在许多应用中尤为重要。

$C=\backslash \backslash$ eft ${\backslash$ mathbf ${x}, ~ t)$ \in $\backslash$ mathbb ${R} \backslash{n+1} \backslash$ mid $\mid \backslash$ mathbf ${x} \mid \backslash$ leq tright $} \backslash$ subseteq $\backslash$ math

$\backslash$ mathbf ${b \mathrm{~d}} \mathrm{C}=\backslash$ bigcup_${\mid \backslash$ mathbf ${u} \mid=1} \backslash \operatorname{eft}{\mathrm{t}$ ( \mathbf ${\mathrm{u}}, \quad 1) \operatorname{lin} \backslash \operatorname{mathbb}{R} \wedge{n+1}, \quad \mathrm{t} \backslash g$
(包含线殷)。

## 数学代写|凸优化作业代写Convex Optimization代考| Affine function

$$\boldsymbol{f}(\mathbf{x})=\mathbf{A} \mathbf{x}+\mathbf{b}$$

$$f(S)=f(\mathbf{x}) \mid \mathbf{x} \in S$$

$$f^{-1}(C)=\mathbf{x} \mid f(\mathbf{x}) \in C$$

$\mathbf{y} 2=\mathbf{A x} 2+\mathbf{b}$.我们的目标是表明 $\boldsymbol{f}^{-1}(C)$ ，这是的逆像 $f$ ，是凸的。为 $\theta \in[0,1]$ ，
$$\theta \mathbf{y} 1+(1-\theta) \mathbf{y} 2=\theta(\mathbf{A} \mathbf{x} 1+\mathbf{b})+(1-\theta)(\mathbf{A} \mathbf{x} 2+\mathbf{b}) \quad=\mathbf{A}(\theta \mathbf{x} 1+(1-\theta) \mathbf{x} 2)+\mathbf{b} \in C$$

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