### 数学代写|凸优化作业代写Convex Optimization代考|ELEN90026

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• 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|Pointwise maximum and supremum

If $f_1$ and $f_2$ are convex functions then their pointwise maximum $f$, defined by
$$f(x)=\max \left{f_1(x), f_2(x)\right},$$
with $\operatorname{dom} f=\operatorname{dom} f_1 \cap \operatorname{dom} f_2$, is also convex. This property is easily verified: if $0 \leq \theta \leq 1$ and $x, y \in \operatorname{dom} f$, then
\begin{aligned} f(\theta x+(1-\theta) y) & =\max \left{f_1(\theta x+(1-\theta) y), f_2(\theta x+(1-\theta) y)\right} \ & \leq \max \left{\theta f_1(x)+(1-\theta) f_1(y), \theta f_2(x)+(1-\theta) f_2(y)\right} \ & \leq \theta \max \left{f_1(x), f_2(x)\right}+(1-\theta) \max \left{f_1(y), f_2(y)\right} \ & =\theta f(x)+(1-\theta) f(y) \end{aligned}
which establishes convexity of $f$. It is easily shown that if $f_1, \ldots, f_m$ are convex, then their pointwise maximum
$$f(x)=\max \left{f_1(x), \ldots, f_m(x)\right}$$
is also convex.

The pointwise maximum property extends to the pointwise supremum over an infinite set of convex functions. If for each $y \in \mathcal{A}, f(x, y)$ is convex in $x$, then the function $g$, defined as
$$g(x)=\sup {y \in \mathcal{A}} f(x, y)$$ is convex in $x$. Here the domain of $g$ is $$\operatorname{dom} g=\left{x \mid(x, y) \in \operatorname{dom} f \text { for all } y \in \mathcal{A}, \sup {y \in \mathcal{A}} f(x, y)<\infty\right} .$$
Similarly, the pointwise infimum of a set of concave functions is a concave function.
In terms of epigraphs, the pointwise supremum of functions corresponds to the intersection of epigraphs: with $f, g$, and $\mathcal{A}$ as defined in (3.7), we have
$$\text { epi } g=\bigcap_{y \in \mathcal{A}} \operatorname{epi} f(\cdot, y) \text {. }$$
Thus, the result follows from the fact that the intersection of a family of convex sets is convex.

## 数学代写|凸优化作业代写Convex Optimization代考|Representation as pointwise supremum of affine functions

The examples above illustrate a good method for establishing convexity of a function: by expressing it as the pointwise supremum of a family of affine functions. Except for a technical condition, a converse holds: almost every convex function can be expressed as the pointwise supremum of a family of affine functions. For example, if $f: \mathbf{R}^n \rightarrow \mathbf{R}$ is convex, with $\operatorname{dom} f=\mathbf{R}^n$, then we have
$$f(x)=\sup {g(x) \mid g \text { affine, } g(z) \leq f(z) \text { for all } z} .$$
In other words, $f$ is the pointwise supremum of the set of all affine global underestimators of it. We give the proof of this result below, and leave the case where $\operatorname{dom} f \neq \mathbf{R}^n$ as an exercise (exercise $3.28$ ).
Suppose $f$ is convex with $\operatorname{dom} f=\mathbf{R}^n$. The inequality
$$f(x) \geq \sup {g(x) \mid g \text { affine, } g(z) \leq f(z) \text { for all } z}$$
is clear, since if $g$ is any affine underestimator of $f$, we have $g(x) \leq f(x)$. To establish equality, we will show that for each $x \in \mathbf{R}^n$, there is an affine function $g$, which is a global underestimator of $f$, and satisfies $g(x)=f(x)$.

The epigraph of $f$ is, of course, a convex set. Hence we can find a supporting hyperplane to it at $(x, f(x))$, i.e., $a \in \mathbf{R}^n$ and $b \in \mathbf{R}$ with $(a, b) \neq 0$ and
$$\left[\begin{array}{l} a \ b \end{array}\right]^T\left[\begin{array}{c} x-z \ f(x)-t \end{array}\right] \leq 0$$
for all $(z, t) \in$ epi $f$. This means that
$$a^T(x-z)+b(f(x)-f(z)-s) \leq 0$$
for all $z \in \operatorname{dom} f=\mathbf{R}^n$ and all $s \geq 0$ (since $(z, t) \in$ epi $f$ means $t=f(z)+s$ for some $s \geq 0$ ). For the inequality (3.8) to hold for all $s \geq 0$, we must have $b \geq 0$. If $b=0$, then the inequality (3.8) reduces to $a^T(x-z) \leq 0$ for all $z \in \mathbf{R}^n$, which implies $a=0$ and contradicts $(a, b) \neq 0$. We conclude that $b>0$, i.e., that the supporting hyperplane is not vertical.
Using the fact that $b>0$ we rewrite (3.8) for $s=0$ as
$$g(z)=f(x)+(a / b)^T(x-z) \leq f(z)$$
for all $z$. The function $g$ is an affine underestimator of $f$, and satisfies $g(x)=f(x)$.

# 凸优化代写

## 数学代写|凸优化作业代写Convex Optimization代考|Pointwise maximum and supremum

$f(x)=\backslash \max \backslash$ left $\left{f_{-} 1(x), f_{-} 2(x) \backslash r i g h t\right}$,

$\mathrm{f}(\mathrm{x})=\backslash \max \backslash$ left $\left{\mathrm{f} _1(\mathrm{x})\right.$, \Idots, $\mathrm{f} _\mathrm{m}(\mathrm{x}) \backslash$ right $}$

pointwise maximum 属性扩展到无限组凸函数上的 pointwise supremum。如果对于每个 $y \in \mathcal{A}, f(x, y)$ 是凸的 $x$ ，那么函数 $g$ ，定义为
$$g(x)=\sup y \in \mathcal{A} f(x, y)$$

loperatorname{dom $} g=\backslash \operatorname{geft}{x \backslash m i d(x, y) \backslash i n$ loperatorname{dom $f \backslash t e x t{$ for all $}$ y $\backslash$ in $\backslash m a t h c a l{A}$,

$$\text { epi } g=\bigcap_{y \in \mathcal{A}} \operatorname{epi} f(\cdot, y) \text {. }$$

## 数学代写|凸优化作业代写Convex Optimization代考|Representation as pointwise supremum of affine functions

$$f(x)=\sup g(x) \mid g \text { affine, } g(z) \leq f(z) \text { for all } z .$$

$$f(x) \geq \sup g(x) \mid g \text { affine, } g(z) \leq f(z) \text { for all } z$$

$$[a b]^T[x-z f(x)-t] \leq 0$$

$$a^T(x-z)+b(f(x)-f(z)-s) \leq 0$$

$a^T(x-z) \leq 0$ 对全部 $z \in \mathbf{R}^n$ ，这意味着 $a=0$ 和矛盾 $(a, b) \neq 0$. 我们的结论是 $b>0$ ，即支持超 平面不是垂直的。

$$g(z)=f(x)+(a / b)^T(x-z) \leq f(z)$$

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## MATLAB代写

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