### 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|Nonnegative weighted sum

Let $f_{1}, \ldots, f_{m}$ be convex functions and $w_{1}, \ldots, w_{m} \geq 0$. Then $\sum_{i=1}^{m} w_{i} f_{i}$ is convex.

Proof: $\operatorname{dom}\left(\sum_{i=1}^{m} w_{i} f_{i}\right)=\bigcap_{i=1}^{m}$ dom $f_{i}$ is convex because dom $f_{i}$ is convex for all $i$. For $0 \leq \theta \leq 1$, and $\mathbf{x}, \mathbf{y} \in \operatorname{dom}\left(\sum_{i=1}^{m} w_{i} f_{i}\right)$, we have
\begin{aligned} \sum_{i=1}^{m} w_{i} f_{i}(\theta \mathbf{x}+(1-\theta) \mathbf{y}) & \leq \sum_{i=1}^{m} w_{i}\left(\theta f_{i}(\mathbf{x})+(1-\theta) f_{i}(\mathbf{y})\right) \ &=\theta \sum_{i=1}^{m} w_{i} f_{i}(\mathbf{x})+(1-\theta) \sum_{i=1}^{m} w_{i} f_{i}(\mathbf{y}) \end{aligned}
Hence proved.
Remark 3.27 $f(\mathbf{x}, \mathbf{y})$ is convex in $\mathbf{x}$ for each $\mathbf{y} \in \mathcal{A}$ and $w(\mathbf{y}) \geq 0$. Then,
$$g(\mathbf{x})=\int_{\mathcal{A}} w(\mathbf{y}) f(\mathbf{x}, \mathbf{y}) d \mathbf{y}$$
is convex on $\bigcap_{y \in \mathcal{A}} \operatorname{dom} f$.
Composition with affine mapping
If $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is a convex function, then for $\mathbf{A} \in \mathbb{R}^{n \times m}$ and $\mathbf{b} \in \mathbb{R}^{n}$, the function $g: \mathbb{R}^{m} \rightarrow \mathbb{R}$, defined as
$$g(\mathbf{x})=f(\mathbf{A} \mathbf{x}+\mathbf{b}),$$
is also convex and its domain can be expressed as
\begin{aligned} \operatorname{dom} g &=\left{\mathbf{x} \in \mathbb{R}^{m} \mid \mathbf{A} \mathbf{x}+\mathbf{b} \in \operatorname{dom} f\right} \ &=\left{\mathbf{A}^{\dagger}(\mathbf{y}-\mathbf{b}) \mid \mathbf{y} \in \mathbf{d o m} f\right}+\mathcal{N}(\mathbf{A}) \quad(\text { cf. }(2.62)) \end{aligned}
which is also a convex set by Remark $2.9$.
Proof (using epigraph): Since $g(\mathbf{x})=f(\mathbf{A x}+\mathbf{b})$ and epi $f={(\mathbf{y}, t) \mid f(\mathbf{y}) \leq t}$, we have
\text { epi } \begin{aligned} g &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid f(\mathbf{A} \mathbf{x}+\mathbf{b}) \leq t\right} \ &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid(\mathbf{A} \mathbf{x}+\mathbf{b}, t) \in \mathbf{e p i} f\right} \end{aligned}

Now, define
$$\mathcal{S}=\left{(\mathbf{x}, \mathbf{y}, t) \in \mathbb{R}^{m+n+1} \mid \mathbf{y}=\mathbf{A} \mathbf{x}+\mathbf{b}, f(\mathbf{y}) \leq t\right}$$
so that
$$\text { epi } g=\left{\left[\begin{array}{lll} \mathbf{I}{m} & \mathbf{0}{m \times n} & \mathbf{0}{m} \ \mathbf{0}{m}^{T} & \mathbf{0}{\mathrm{n}}^{T} & 1 \end{array}\right](\mathbf{x}, \mathbf{y}, t) \mid(\mathbf{x}, \mathbf{y}, t) \in \mathcal{S}\right}$$ which is nothing but the image of $\mathcal{S}$ via an affine mapping. It can be easily shown, by the definition of convex sets, that $\mathcal{S}$ is convex if $f$ is convex. Therefore epi $g$ is convex (due to affine mapping from the convex set $\mathcal{S}$ ) implying that $g$ is convex (by Fact 3.2). Alternative proof: For $0 \leq \theta \leq 1$, we have \begin{aligned} g\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}\right) &=f\left(\mathbf{A}\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}\right)+\mathbf{b}\right) \ &=f\left(\theta\left(\mathbf{A} \mathbf{x}{1}+\mathbf{b}\right)+(1-\theta)\left(\mathbf{A} \mathbf{x}{2}+\mathbf{b}\right)\right) \ & \leq \theta f\left(\mathbf{A} \mathbf{x}{1}+\mathbf{b}\right)+(1-\theta) f\left(\mathbf{A} \mathbf{x}{2}+\mathbf{b}\right) \ &=\theta g\left(\mathbf{x}{1}\right)+(1-\theta) g\left(\mathbf{x}_{2}\right) \end{aligned}
Moreover, dom $g$ (cf. (3.72)) is also a convex set, and so we conclude that $f(\mathbf{A} \mathbf{x}+\mathbf{b})$ is a convex function.

## 数学代写|凸优化作业代写Convex Optimization代考|Composition (scalar)

Suppose that $h: \operatorname{dom} h \rightarrow \mathbb{R}$ is a convex (concave) function and dom $h \subset \mathbb{R}^{n}$. The extended-value extension of $h$, denoted as $\tilde{h}$, with $\operatorname{dom} \tilde{h}=\mathbb{R}^{n}$ aids in simple representation as its domain is the entire $\mathbb{R}^{n}$, which need not be explicitly mentioned. The extended-valued function $\tilde{h}$ is a function taking the same value of $h(\mathbf{x})$ for $\mathbf{x} \in$ dom $h$, otherwise taking the value of $+\infty(-\infty)$. Specifically, if $h$ is convex,
$$\tilde{h}(\mathbf{x})=\left{\begin{array}{l} h(\mathbf{x}), \mathbf{x} \in \operatorname{dom} h \ +\infty, \mathbf{x} \notin \operatorname{dom} h \end{array}\right.$$
and if $h$ is concave,
$$\tilde{h}(\mathbf{x})=\left{\begin{array}{l} h(\mathbf{x}), \mathbf{x} \in \operatorname{dom} h \ -\infty, \mathbf{x} \notin \operatorname{dom} h \end{array}\right.$$
Then the extended-valued function $\tilde{h}$ does not affect the convexity (or concavity) of the original function $h$ and Eff-dom $\tilde{h}=$ Eff-dom $h$.

Some examples for illustrating properties of an extended-value extension of a function are as follows.

• $h(x)=\log x$, $\operatorname{dom} h=\mathbb{R}_{++}$. Then $h(x)$ is concave and $\tilde{h}(x)$ is concave and nondecreasing.
• $h(x)=x^{1 / 2}$, dom $h=\mathbb{R}_{+}$. Then $h(x)$ is concave and $\tilde{h}(x)$ is concave and nondecreasing.
• In the function
$$h(x)=x^{2}, x \geq 0,$$
i.e., dom $h=\mathbb{R}_{+}, h(x)$ is convex and $\tilde{h}(x)$ is convex but neither nondecreasing nor nonincreasing.

Let $f(\mathbf{x})=h(g(\mathbf{x}))$, where $h: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{n} \rightarrow \mathbb{R}$. Then we have the following four composition rules about the convexity or concavity of $f$.
(a) $f$ is convex if $h$ is convex, $\bar{h}$ nondecreasing, and $g$ convex.
(3.76a)
(b) $f$ is convex if $h$ is convex, $\tilde{h}$ nonincreasing, and $g$ concave.
(3.76b)
(c) $f$ is concave if $h$ is concave, $\tilde{h}$ nondecreasing, and $g$ concave.
(3.76c)
(d) $f$ is concave if $h$ is concave, $\tilde{h}$ nonincreasing, and $g$ convex.
(3.76d)
Consider the case that $g$ and $h$ are twice differentiable and $\tilde{h}(x)=h(x)$. Then,
$$\nabla f(\mathbf{x})=h^{\prime}(g(\mathbf{x})) \nabla g(\mathbf{x})$$
and
\begin{aligned} \nabla^{2} f(\mathbf{x}) &=D(\nabla f(\mathbf{x}))=D\left(h^{\prime}(g(\mathbf{x})) \cdot \nabla g(\mathbf{x})\right) \quad(\text { by }(1.46)) \ &=\nabla g(\mathbf{x}) D\left(h^{\prime}(g(\mathbf{x}))\right)+h^{\prime}(g(\mathbf{x})) \cdot D(\nabla g(\mathbf{x})) \ &=h^{\prime \prime}(g(\mathbf{x})) \nabla g(\mathbf{x}) \nabla g(\mathbf{x})^{T}+h^{\prime}(g(\mathbf{x})) \nabla^{2} g(\mathbf{x}) \end{aligned}
The composition rules (a) (cf. (3.76a)) and (b) (cf. (3.76b)) can be proven for convexity of $f$ by checking if $\nabla^{2} f(\mathbf{x}) \succeq \mathbf{0}$, and the composition rules (c) (cf. $(3.76 \mathrm{c}))$ and (d) (cf. $(3.76 \mathrm{~d}))$ for concavity of $f$ by checking if $\nabla^{2} f(\mathbf{x}) \preceq \mathbf{0}$. Let us conclude this subsection with a simple example.
Example $3.3$ Let $g(\mathbf{x})=|\mathbf{x}|_{2}$ (convex) and
$$h(x)= \begin{cases}x^{2}, & x \geq 0 \ 0, & \text { otherwise }\end{cases}$$
which is convex. So $\tilde{h}(x)=h(x)$ is nondecreasing. Then, $f(\mathbf{x})=h(g(\mathbf{x}))=$ $|\mathbf{x}|_{2}^{2}=\mathbf{x}^{T} \mathbf{x}$ is convex by $(3.76 \mathrm{a})$, or by the second-order condition $\nabla^{2} f(\mathbf{x})=$ $2 \mathbf{I}_{n} \succ \mathbf{0}, f$ is indeed convex.

## 数学代写|凸优化作业代写Convex Optimization代考|Pointwise minimum and infimum

If $f(\mathbf{x}, \mathbf{y})$ is convex in $(\mathbf{x}, \mathbf{y}) \in \mathbb{R}^{m} \times \mathbb{R}^{n}$ and $C \subset \mathbb{R}^{n}$ is convex and nonempty, then
$$g(\mathbf{x})=\inf {\mathbf{y} \in C} f(\mathbf{x}, \mathbf{y})$$ is convex, provided that $g(\mathbf{x})>-\infty$ for some $\mathbf{x}$. Similarly, if $f(\mathbf{x}, \mathbf{y})$ is concave in $(\mathbf{x}, \mathbf{y}) \in \mathbb{R}^{m} \times \mathbb{R}^{n}$, then $$\tilde{g}(\mathbf{x})=\sup {\mathbf{y} \in C} f(\mathbf{x}, \mathbf{y})$$
is concave provided that $C \subset \mathbb{R}^{n}$ is convex and nonempty and $\tilde{g}(\mathbf{x})<\infty$ for some $\mathbf{x}$. Next, we present the proof for the former.

Proof of (3.87): Since $f$ is continuous over int(dom $f$ ) (cf. Remark 3.7), for any $\epsilon>0$ and $\mathbf{x}{1}, \mathbf{x}{2} \in \operatorname{dom} g$, there exist $\mathbf{y}{1}, \mathbf{y}{2} \in C$ (depending on $\epsilon$ ) such that
$$f\left(\mathbf{x}{i}, \mathbf{y}{i}\right) \leq g\left(\mathbf{x}{i}\right)+\epsilon, i=1,2 .$$ Let $\left(\mathbf{x}{1}, t_{1}\right),\left(\mathbf{x}{2}, t{2}\right) \in$ epi $g$. Then $g\left(\mathbf{x}{i}\right)=\inf {\mathbf{y} \in C} f\left(\mathbf{x}{i}, \mathbf{y}\right) \leq t{i}, i=1,2$. Then for any $\theta \in[0,1]$, we have
\begin{aligned} g\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}\right) &=\inf {\mathbf{y} \in C} f\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}, \mathbf{y}\right) \ & \leq f\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}, \theta \mathbf{y}{1}+(1-\theta) \mathbf{y}{2}\right) \ & \leq \theta f\left(\mathbf{x}{1}, \mathbf{y}{1}\right)+(1-\theta) f\left(\mathbf{x}{2}, \mathbf{y}{2}\right) \quad \text { (since } f \text { is convex) } \ & \leq \theta g\left(\mathbf{x}{1}\right)+(1-\theta) g\left(\mathbf{x}{2}\right)+\epsilon \quad \text { (by (3.89)) } \ & \leq \theta t{1}+(1-\theta) t_{2}+\epsilon . \end{aligned}
It can be seen that as $\epsilon \rightarrow 0, g\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}\right) \leq \theta t_{1}+(1-\theta) t_{2}$, implying $\left(\theta \mathbf{x}{1}+(1-\theta) \mathbf{x}{2}, \theta t_{1}+(1-\theta) t_{2}\right) \in$ epi $g$. Hence epi $g$ is a convex set, and thus $g(\mathbf{x})$ is a convex function by Fact 3.2.

Alternative proof of (3.87): Because dom $g={\mathbf{x} \mid(\mathbf{x}, \mathbf{y}) \in \operatorname{dom} f, \mathbf{y} \in C}$ is the projection of the convex set ${(\mathbf{x}, \mathbf{y}) \mid(\mathbf{x}, \mathbf{y}) \in \operatorname{dom} f, \mathbf{y} \in C}$ on the $\mathbf{x}-$ coordinate, it must be a convex set (cf. Remark 2.11).

## 数学代写|凸优化作业代写Convex Optimization代考|Nonnegative weighted sum

∑一世=1米在一世F一世(θX+(1−θ)是)≤∑一世=1米在一世(θF一世(X)+(1−θ)F一世(是)) =θ∑一世=1米在一世F一世(X)+(1−θ)∑一世=1米在一世F一世(是)

G(X)=∫一种在(是)F(X,是)d是

IfF:Rn→R是一个凸函数，那么对于一种∈Rn×米和b∈Rn， 功能G:R米→R， 定义为
G(X)=F(一种X+b),

\begin{aligned} \operatorname{dom} g &=\left{\mathbf{x} \in \mathbb{R}^{m} \mid \mathbf{A} \mathbf{x}+\mathbf{b} \in \operatorname{dom} f\right} \ &=\left{\mathbf{A}^{\dagger}(\mathbf{y}-\mathbf{b}) \mid \mathbf{y} \in \ mathbf{d o m} f\right}+\mathcal{N}(\mathbf{A}) \quad(\text { cf. }(2.62)) \end{aligned}\begin{aligned} \operatorname{dom} g &=\left{\mathbf{x} \in \mathbb{R}^{m} \mid \mathbf{A} \mathbf{x}+\mathbf{b} \in \operatorname{dom} f\right} \ &=\left{\mathbf{A}^{\dagger}(\mathbf{y}-\mathbf{b}) \mid \mathbf{y} \in \ mathbf{d o m} f\right}+\mathcal{N}(\mathbf{A}) \quad(\text { cf. }(2.62)) \end{aligned}

\text { epi } \begin{aligned} g &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid f(\mathbf{A} \mathbf{ x}+\mathbf{b}) \leq t\right} \ &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid(\mathbf{A } \mathbf{x}+\mathbf{b}, t) \in \mathbf{e p i} f\right} \end{aligned}\text { epi } \begin{aligned} g &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid f(\mathbf{A} \mathbf{ x}+\mathbf{b}) \leq t\right} \ &=\left{(\mathbf{x}, t) \in \mathbb{R}^{m+1} \mid(\mathbf{A } \mathbf{x}+\mathbf{b}, t) \in \mathbf{e p i} f\right} \end{aligned}

\mathcal{S}=\left{(\mathbf{x}, \mathbf{y}, t) \in \mathbb{R}^{m+n+1} \mid \mathbf{y}=\mathbf{ A} \mathbf{x}+\mathbf{b}, f(\mathbf{y}) \leq t\right}\mathcal{S}=\left{(\mathbf{x}, \mathbf{y}, t) \in \mathbb{R}^{m+n+1} \mid \mathbf{y}=\mathbf{ A} \mathbf{x}+\mathbf{b}, f(\mathbf{y}) \leq t\right}

\text { epi } g=\left{\left[\begin{array}{lll} \mathbf{I}{m} & \mathbf{0}{m \times n} & \mathbf{0}{m} \ \mathbf{0}{m}^{T} & \mathbf{0}{\mathrm{n}}^{T} & 1 \end{array}\right](\mathbf{x}, \mathbf{ y}, t) \mid(\mathbf{x}, \mathbf{y}, t) \in \mathcal{S}\right}\text { epi } g=\left{\left[\begin{array}{lll} \mathbf{I}{m} & \mathbf{0}{m \times n} & \mathbf{0}{m} \ \mathbf{0}{m}^{T} & \mathbf{0}{\mathrm{n}}^{T} & 1 \end{array}\right](\mathbf{x}, \mathbf{ y}, t) \mid(\mathbf{x}, \mathbf{y}, t) \in \mathcal{S}\right}这不过是小号通过仿射映射。通过凸集的定义可以很容易地证明，小号是凸的，如果F是凸的。因此epiG是凸的（由于来自凸集的仿射映射小号) 暗示G是凸的（根据事实 3.2）。替代证明：对于0≤θ≤1， 我们有G(θX1+(1−θ)X2)=F(一种(θX1+(1−θ)X2)+b) =F(θ(一种X1+b)+(1−θ)(一种X2+b)) ≤θF(一种X1+b)+(1−θ)F(一种X2+b) =θG(X1)+(1−θ)G(X2)

## 数学代写|凸优化作业代写Convex Optimization代考|Composition (scalar)

$$\tilde{h}(\mathbf{x})=\left{H(X),X∈dom⁡H +∞,X∉dom⁡H\对。 一种nd一世FH一世sC这nC一种在和, \波浪号{h}(\mathbf{x})=\left{H(X),X∈dom⁡H −∞,X∉dom⁡H\对。$$

• H(X)=日志⁡X, dom⁡H=R++. 然后H(X)是凹的并且H~(X)是凹的且不减的。
• H(X)=X1/2, 域H=R+. 然后H(X)是凹的并且H~(X)是凹的且不减的。
• 在函数中
H(X)=X2,X≥0,
即，domH=R+,H(X)是凸的并且H~(X)是凸的，但既非非减也非非增。

（一种）F是凸的，如果H是凸的，H¯非递减，并且G凸的。
(3.76a)
(b)F是凸的，如果H是凸的，H~不增加，和G凹。
(3.76b)
(c)F如果是凹的H是凹的，H~非递减，并且G凹。
(3.76c)
(d)F如果是凹的H是凹的，H~不增加，和G凸的。
(3.76d)

∇F(X)=H′(G(X))∇G(X)

∇2F(X)=D(∇F(X))=D(H′(G(X))⋅∇G(X))( 经过 (1.46)) =∇G(X)D(H′(G(X)))+H′(G(X))⋅D(∇G(X)) =H′′(G(X))∇G(X)∇G(X)吨+H′(G(X))∇2G(X)

H(X)={X2,X≥0 0, 除此以外

## 数学代写|凸优化作业代写Convex Optimization代考|Pointwise minimum and infimum

G(X)=信息是∈CF(X,是)是凸的，前提是G(X)>−∞对于一些X. 同样，如果F(X,是)是凹进去的(X,是)∈R米×Rn， 然后G~(X)=支持是∈CF(X,是)

(3.87) 的证明：因为F在 int(dom 上是连续的F)（参见备注 3.7），对于任何ε>0和X1,X2∈dom⁡G， 存在是1,是2∈C（根据ε) 使得
F(X一世,是一世)≤G(X一世)+ε,一世=1,2.让(X1,吨1),(X2,吨2)∈和G. 然后G(X一世)=信息是∈CF(X一世,是)≤吨一世,一世=1,2. 那么对于任何θ∈[0,1]， 我们有
G(θX1+(1−θ)X2)=信息是∈CF(θX1+(1−θ)X2,是) ≤F(θX1+(1−θ)X2,θ是1+(1−θ)是2) ≤θF(X1,是1)+(1−θ)F(X2,是2) （自从 F 是凸的）  ≤θG(X1)+(1−θ)G(X2)+ε （由（3.89））  ≤θ吨1+(1−θ)吨2+ε.

(3.87) 的替代证明：因为 domG=X∣(X,是)∈dom⁡F,是∈C是凸集的投影(X,是)∣(X,是)∈dom⁡F,是∈C在X−坐标，它必须是一个凸集（参见备注 2.11）。

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