### 数学代写|凸优化作业代写Convex Optimization代考|CS 531

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

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

The norm ball of a point $\mathbf{x} \in \mathbb{R}^{n}$ is defined as the following set ${ }^{1}$
$$B(\mathbf{x}, r)=\left{\mathbf{y} \in \mathbb{R}^{n} \mid|\mathbf{y}-\mathbf{x}| \leq r\right} .$$
where $r$ is the radius and $\mathbf{x}$ is the center of the norm ball. It is also called the neighborhood of the point $\mathbf{x}$. For the case of $n=2, \mathbf{x}=\mathbf{0}{2}$, and $r=1$, the 2 norm ball is $B(\mathbf{x}, r)=\left{\mathbf{y} \mid y{1}^{2}+y_{2}^{2} \leq 1\right}$ (a circular disk of radius equal to 1 ), the 1-norm ball is $B(\mathbf{x}, r)=\left{\mathbf{y}|| y_{1}|+| y_{2} \mid \leq 1\right}$ (a 2-dimensional cross-polytope of area equal to 2 ), and the $\infty$-norm ball is $B(\mathbf{x}, r)=\left{\mathbf{y}|| y_{1}|\leq 1,| y_{2} \mid \leq 1\right$,$} (a$ square of area equal to 4) (see Figure 1.1). Note that the norm ball is symmetric with respect to (w.r.t.) the origin, convex, closed, bounded and has nonempty interior. Moreover, the 1-norm ball is a subset of the 2-norm ball which is a subset of $\infty$-norm ball, due to the following inequality:
$$|\mathbf{v}|_{p} \leq|\mathbf{v}|_{q}$$
where $\mathbf{v} \in \mathbb{R}^{n}, p$ and $q$ are real and $p>q \geq 1$, and the equality holds when $\mathbf{v}=$ re $_{i}$, i.e., all the $p$-norm balls of constant radius $r$ have intersections at $r \mathbf{e}{i}, i=$ $1, \ldots, n$. For instance, in Figure 1.1, $\left|\mathbf{x}{1}\right|_{p}=1$ for all $p \geq 1$, and $\left|\mathbf{x}{2}\right|{\infty}=1<$ $\left|\mathbf{x}{2}\right|{2}=\sqrt{2}<\left|\mathbf{x}{2}\right|{1}=2$. The inequality (1.17) is proven as follows.

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

A point $\mathbf{x}$ in a set $C \subseteq \mathbb{R}^{n}$ is an interior point of the set $C$ if there exists an $\epsilon>0$ for which $B(\mathbf{x}, \epsilon) \subseteq C$ (see Figure 1.3). In other words, a point $\mathbf{x} \in C$ is said to be an interior point of the set $C$ if the set $C$ contains some neighborhood of $\mathbf{x}$, that is, if all points within some neighborhood of $\mathbf{x}$ are also in $C$.

Remark $1.7$ The set of all the interior points of $C$ is called the interior of $C$ and is represented as int $C$, which can also be expressed as
$$\text { int } C={\mathbf{x} \in C \mid B(\mathbf{x}, r) \subseteq C \text {, for some } r>0} \text {, }$$
which will be frequently used in many proofs directly or indirectly in the ensuing chapters.
Complement, scaled sets, and sum of sets
The complement of a set $C \subset \mathbb{R}^{n}$ is defined as follows (see Figure 1.3):
$$\mathbb{R}^{n} \backslash C=\left{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{x} \notin C\right},$$
where ” $\backslash$ ” denotes the set difference, i.e., $A \backslash B={\mathbf{x} \in A \mid \mathbf{x} \notin B}$. The set $C \subset$ $\mathbb{R}^{n}$ scaled by a real number $\alpha$ is a set defined as
$$\alpha \cdot C \triangleq{\alpha \mathbf{x} \mid \mathbf{x} \in C}$$

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

$$|\mathbf{A}| \mathbf{F}=\left(\sum i=1^{m} \sum_{j=1}^{n}|[\mathbf{A}] i j|^{2}\right)^{1 / 2}=\sqrt{\operatorname{Tr}\left(\mathbf{A}^{T} \mathbf{A}\right)}$$

$$\operatorname{Tr}(\mathbf{X})=\sum i=1^{n}[\mathbf{X}] i i$$

$\mid \backslash$ mathbf $A \mid-a, b=\backslash$ sup $\backslash$ left

$|\mathbf{A}| 1=\max |\mathbf{u}| 1 \leq 1 \sum_{j=1}^{n} u_{j} \mathbf{a} j|1, \quad(a=b=1) \quad \leq \max | \mathbf{u}\left|1 \leq 1 \sum_{j=1}^{n}\right| u_{j}|\cdot| \mathbf{a} j \mid 1$ (by triangle in

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

$$\langle\mathbf{x}, \mathbf{y}\rangle=\mathbf{y}^{T} \mathbf{x}=\sum_{i=1}^{n} x_{i} y_{i}$$

Cauchy-Schwartz 不等式：对于任意两个向量 $\mathbf{x}$ 和 $\mathbf{y}$ 在 $\mathbb{R}^{n}$ ， CauchySchwartz 不等式
$$|\langle\mathbf{x}, \mathbf{y}\rangle| \leq|\mathbf{x}| 2 \cdot|\mathbf{y}| 2$$

$$|\mathbf{x}+\mathbf{y}| 2^{2}=(\mathbf{x}+\mathbf{y})^{T}(\mathbf{x}+\mathbf{y})=|\mathbf{x}| 2^{2}+2(\mathbf{x}, \mathbf{y}\rangle+|\mathbf{y}| 2^{2}=|\mathbf{x}| 2^{2}+|\mathbf{y}|_{2}^{2}$$

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## MATLAB代写

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