### 数学代写|凸优化作业代写Convex Optimization代考|Convexity preserving operations

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

If $S_{1}$ and $S_{2}$ are convex sets, then $S_{1} \cap S_{2}$ is also convex. This property extends to the intersection of an infinite number of convex sets, i.e., if $S_{a}$ is convex for every $\alpha \in \mathcal{A}$, then $\cap_{\alpha \in \mathcal{A}} S_{\alpha}$ is convex. Let us illuminate the usefulness of this convexity preserving operation with the following remarks and examples.

Remark 2.6 A polyhedron can be considered as intersection of a finite number of halfspaces and hyperplanes (which are convex) and hence the polyhedron is convex.

Remark 2.7 Subspaces are closed under arbitrary intersections; so are affine sets and convex cones. So they all are convex sets.

Remark 2.8 A closed convex set $S$ is the intersection of all (possibly an infinite number of) closed halfspaces that contain $S$. This can be proven by the separating hyperplane theory (to be introduced in Subsection 2.6.1).

Example 2.4 The PSD cone $\mathbb{S}{+}^{n}$ is known to be convex. The proof of its convexity by the intersection property is given as follows. It is easy to see that $S{+}^{n}$ can be expressed as
$$\mathbb{S}{+}^{n}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf{z} \geq 0, \forall \mathbf{z} \in \mathbb{R}^{n}\right}=\bigcap{\mathbf{z} \in \mathbb{R}^{n}} S_{\mathbf{z}}$$
where
\begin{aligned} S_{\mathbf{z}} &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf{z} \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left(\mathbf{z}^{T} \mathbf{X} \mathbf{z}\right) \geq 0\right} \ &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left(\mathbf{X} \mathbf{z z}^{T}\right) \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}(\mathbf{X} \mathbf{Z}) \geq 0\right} \end{aligned}
in which $\mathbf{Z}=\mathbf{z z}^{T}$, implying that $S_{\mathbf{z}}$ is a halfspace if $\mathbf{z} \neq \mathbf{0}{n}$. As the intersection of halfspaces is also convex, $\mathbb{S}{+}^{n}$ (intersection of infinite number of halfspaces) is a convex set. It is even easier to prove the convexity of $S_{\mathbf{z}}$ by the definition of convex sets.
Example 2.5 Consider
$$P(\mathbf{x}, \omega)=\sum_{i=1}^{n} x_{i} \cos (i \omega)$$
and a set
\begin{aligned} C &=\left{\mathbf{x} \in \mathbb{R}^{n} \mid l(\omega) \leq P(\mathbf{x}, \omega) \leq u(\omega) \quad \forall \omega \in \Omega\right} \ &=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \mid l(\omega) \leq \sum_{i=1}^{n} x_{i} \cos (i \omega) \leq u(\omega)\right} \end{aligned}
Let
$$\mathbf{a}(\omega)=[\cos (\omega), \cos (2 \omega), \ldots, \cos (n \omega)]^{T}$$

Then we have
$C=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{a}^{T}(\omega) \mathbf{x} \geq l(\omega), \mathbf{a}^{T}(\omega) \mathbf{x} \leq u(\omega)\right}$ (intersection of halfspaces), which implies that $C$ is convex. Note that the set $C$ is a polyhedron only when the set size $|\Omega|$ is finite.

## 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 $\boldsymbol{f}$ (dom $\boldsymbol{f}$ ) 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 $C$
$$\boldsymbol{f}^{-1}(C)={\mathbf{x} \mid \boldsymbol{f}(\mathbf{x}) \in C}$$
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 $f^{-1}(C)$, which is the inverse image of $\boldsymbol{f}$, 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
$$\alpha_{1} S_{1}+\alpha_{2} S_{2}=\left{\mathbf{z}=\alpha_{1} \mathbf{x}+\alpha_{2} \mathbf{y} \mid \mathbf{x} \in S_{1}, \mathbf{y} \in S_{2}\right} \quad(\text { cf. (1.22) and (1.23)) }$$
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

## 数学代写|凸优化作业代写Convex Optimization代考|Perspective function and linear-fractional function

Linear-fractional functions are functions which are more general than affine but still preserve convexity. The perspective function scales or normalizes vectors so that the last component is one, and then drops the last component.

The perspective function $\boldsymbol{p}: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n}$, with $\operatorname{dom} \boldsymbol{p}=\mathbb{R}^{n} \times \mathbb{R}{++}$, is defined as $$\boldsymbol{p}(\mathbf{z}, t)=\frac{\mathbf{z}}{t} .$$ The perspective function $\boldsymbol{p}$ preserves the convexity of the convex set. Proof: Consider two points $\left(\mathbf{z}{1}, t_{1}\right)$ and $\left(\mathbf{z}{2}, t{2}\right)$ in a convex set $C$ and so $\mathbf{z}{1} / t{1}$ and $\mathbf{z}{2} / t{2} \in \boldsymbol{p}(C)$. Then
$$\theta\left(\mathbf{z}{1}, t{1}\right)+(1-\theta)\left(\mathbf{z}{2}, t{2}\right)=\left(\theta \mathbf{z}{1}+(1-\theta) \mathbf{z}{2}, \theta t_{1}+(1-\theta) t_{2}\right) \in C,$$
for any $\theta \in[0,1]$ implying
$$\frac{\theta \mathbf{z}{1}+(1-\theta) \mathbf{z}{2}}{\theta t_{1}+(1-\theta) t_{2}} \in \boldsymbol{p}(C)$$
Now, by defining
$$\mu=\frac{\theta t_{1}}{\theta t_{1}+(1-\theta) t_{2}} \in[0,1],$$
we get
$$\frac{\theta \mathbf{z}{1}+(1-\theta) \mathbf{z}{2}}{\theta t_{1}+(1-\theta) t_{2}}=\mu \frac{\mathbf{z}{1}}{t{1}}+(1-\mu) \frac{\mathbf{z}{2}}{t{2}} \in \boldsymbol{p}(C),$$
which implies $\boldsymbol{p}(C)$ is convex.
A linear-fractional function is formed by composing the perspective function with an affine function. Suppose $\boldsymbol{g}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m+1}$ is affine, i.e.,
$$\boldsymbol{g}(\mathbf{x})=\left[\begin{array}{c} \mathbf{A} \ \mathbf{c}^{T} \end{array}\right] \mathbf{x}+\left[\begin{array}{l} \mathbf{b} \ d \end{array}\right]$$

where $\mathbf{A} \in \mathbb{R}^{m \times n}, \mathbf{b} \in \mathbb{R}^{m}, \mathbf{c} \in \mathbb{R}^{n}$, and $d \in \mathbb{R}$. The function $\boldsymbol{f}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is given by $f=\boldsymbol{p} \circ \boldsymbol{g}$, i.e.,
$$\boldsymbol{f}(\mathbf{x})=\boldsymbol{p}(\boldsymbol{g}(\mathbf{x}))=\frac{\mathbf{A} \mathbf{x}+\mathbf{b}}{\mathbf{c}^{T} \mathbf{x}+d}, \quad \operatorname{dom} \boldsymbol{f}=\left{\mathbf{x} \mid \mathbf{c}^{T} \mathbf{x}+d>0\right},$$
is called a linear-fractional (or projective) function. Hence, linear-fractional functions preserve the convexity.

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

\mathbb{S}{+}^{n}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf {z} \geq 0, \forall \mathbf{z} \in \mathbb{R}^{n}\right}=\bigcap{\mathbf{z} \in \mathbb{R}^{n}} S_ {\mathbf{z}}\mathbb{S}{+}^{n}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf {z} \geq 0, \forall \mathbf{z} \in \mathbb{R}^{n}\right}=\bigcap{\mathbf{z} \in \mathbb{R}^{n}} S_ {\mathbf{z}}

\begin{aligned} S_{\mathbf{z}} &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf{z} \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left(\mathbf{z}^{T} \mathbf{X} \mathbf{z}\right) \geq 0\right} \ &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left (\mathbf{X} \mathbf{z z}^{T}\right) \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr }(\mathbf{X} \mathbf{Z}) \geq 0\right} \end{对齐}\begin{aligned} S_{\mathbf{z}} &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \mathbf{z}^{T} \mathbf{X} \mathbf{z} \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left(\mathbf{z}^{T} \mathbf{X} \mathbf{z}\right) \geq 0\right} \ &=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr}\left (\mathbf{X} \mathbf{z z}^{T}\right) \geq 0\right}=\left{\mathbf{X} \in \mathbb{S}^{n} \mid \operatorname{Tr }(\mathbf{X} \mathbf{Z}) \geq 0\right} \end{对齐}

\begin{aligned} C &=\left{\mathbf{x} \in \mathbb{R}^{n} \mid l(\omega) \leq P(\mathbf{x}, \omega) \leq u (\omega) \quad \forall \omega \in \Omega\right} \ &=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \ mid l(\omega) \leq \sum_{i=1}^{n} x_{i} \cos (i \omega) \leq u(\omega)\right} \end{对齐}\begin{aligned} C &=\left{\mathbf{x} \in \mathbb{R}^{n} \mid l(\omega) \leq P(\mathbf{x}, \omega) \leq u (\omega) \quad \forall \omega \in \Omega\right} \ &=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \ mid l(\omega) \leq \sum_{i=1}^{n} x_{i} \cos (i \omega) \leq u(\omega)\right} \end{对齐}

C=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{a}^{T}(\omega) \mathbf{x } \geq l(\omega), \mathbf{a}^{T}(\omega) \mathbf{x} \leq u(\omega)\right}C=\bigcap_{\omega \in \Omega}\left{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{a}^{T}(\omega) \mathbf{x } \geq l(\omega), \mathbf{a}^{T}(\omega) \mathbf{x} \leq u(\omega)\right}（半空间的交集），这意味着C是凸的。请注意，集C仅当设定大小时才为多面体|Ω|是有限的。

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

F(X)=一种X+b

F(小号)=F(X)∣X∈小号

F−1(C)=X∣F(X)∈C

θ是1+(1−θ)是2=θ(一种X1+b)+(1−θ)(一种X2+b) =一种(θX1+(1−θ)X2)+b∈C

\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(\text { 参见 (1.22) 和 (1.23)) }\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(\text { 参见 (1.22) 和 (1.23)) }

## 数学代写|凸优化作业代写Convex Optimization代考|Perspective function and linear-fractional function

θ(和1,吨1)+(1−θ)(和2,吨2)=(θ和1+(1−θ)和2,θ吨1+(1−θ)吨2)∈C,

θ和1+(1−θ)和2θ吨1+(1−θ)吨2∈p(C)

μ=θ吨1θ吨1+(1−θ)吨2∈[0,1],

θ和1+(1−θ)和2θ吨1+(1−θ)吨2=μ和1吨1+(1−μ)和2吨2∈p(C),

G(X)=[一种 C吨]X+[b d]

\boldsymbol{f}(\mathbf{x})=\boldsymbol{p}(\boldsymbol{g}(\mathbf{x}))=\frac{\mathbf{A} \mathbf{x}+\mathbf{ b}}{\mathbf{c}^{T} \mathbf{x}+d}, \quad \operatorname{dom} \boldsymbol{f}=\left{\mathbf{x} \mid \mathbf{c} ^{T} \mathbf{x}+d>0\right},\boldsymbol{f}(\mathbf{x})=\boldsymbol{p}(\boldsymbol{g}(\mathbf{x}))=\frac{\mathbf{A} \mathbf{x}+\mathbf{ b}}{\mathbf{c}^{T} \mathbf{x}+d}, \quad \operatorname{dom} \boldsymbol{f}=\left{\mathbf{x} \mid \mathbf{c} ^{T} \mathbf{x}+d>0\right},

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