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

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## 数学代写|凸优化作业代写Convex Optimization代考|Basic properties and examples of convex functions

Prior to introducing the definition, properties and various conditions of convex functions together with illustrative examples, we need to clarify the role of $+\infty$ and $-\infty$ for a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$. In spite of $+\infty,-\infty \notin \mathbb{R}, f(\mathbf{x})$ is allowed to take a value of $+\infty$ or $-\infty$ for some $\mathrm{x} \in \operatorname{dom} f$, hereafter. For instance, the following functions
$f_{1}(\mathbf{x})=\left{\begin{array}{l}|\mathbf{x}|_{2}^{2},|\mathbf{x}|_{2} \leq 1 \ +\infty, \quad 1<|\mathbf{x}|_{2} \leq 2\end{array}, \quad\right.$ dom $f_{1}=\left{\mathbf{x} \in \mathbb{R}^{n} \mid|\mathbf{x}|_{2} \leq 2\right}$ $f_{2}(x)=\left{\begin{array}{l}-\infty, x=0 \ \log x, x>0\end{array}, \quad \operatorname{dom} f_{2}=\mathbb{R}{+}\right.$ are well-defined functions, and $f{1}$ is a convex function and $f_{2}$ is a concave function. The convexity of functions will be presented next in detail.
Definition and fundamental properties
A function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is said to be convex if the following conditions are satisfied

• dom $f$ is convex.
• For all $\mathbf{x}, \mathbf{y} \in \operatorname{dom} f, \theta \in[0,1]$.
$$f(\theta \mathbf{x}+(1-\theta) \mathbf{y}) \leq \theta f(\mathbf{x})+(1-\theta) f(\mathbf{y})$$

A convex function basically looks like a faceup bowl as illustrated in Figure 3.1, and it may be differentiable, or continuous but nonsmooth or a nondifferentiable function (e.g., with some discontinuities or with $f(\mathbf{x})=+\infty$ for some $\mathbf{x}$ ). Note that for a given $\theta \in[0,1], \mathbf{z} \triangleq \theta \mathbf{x}+(1-\theta) \mathbf{y}$ is a point on the line segment from $\mathbf{x}$ to $\mathbf{y}$ with
$$\frac{|\mathbf{z}-\mathbf{y}|_{2}}{|\mathbf{y}-\mathbf{x}|_{2}}=\theta \text {, and } \frac{|\mathbf{z}-\mathbf{x}|_{2}}{|\mathbf{y}-\mathbf{x}|_{2}}=1-\theta \text {, }$$
and $f(\mathbf{z})$ is upper bounded by the sum of $100 \times \theta \%$ of $f(\mathbf{x})$ and $100 \times(1-\theta) \%$ of $f(\mathbf{y})$ (i.e., the closer (further) the $\mathbf{z}$ to $\mathbf{x}$, the larger (smaller) the contribution of $f(\mathbf{x})$ to the upper bound of $f(\mathbf{z})$, and this also applies to the contribution of $f(\mathbf{y})$ as shown in Figure 3.1). Note that when $\mathbf{z}$ is given instead of $\theta$, the value of $\theta$ in the upper bound of $f(\mathbf{z})$ can also be determined by (3.4). Various convex function examples will be provided in Subsection 3.1.4.

## 数学代写|凸优化作业代写Convex Optimization代考|First-order condition

Suppose that $f$ is differentiable. Then $f$ is convex if and only if $\operatorname{dom} f$ is convex and
$$f(\mathbf{y}) \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x}) \quad \forall \mathbf{x}, \mathbf{y} \in \operatorname{dom} f$$
This is called the first-order condition, which means that the first-order Taylor series approximation of $f(\mathbf{y})$ w.r.t. $\mathbf{y}=\mathbf{x}$ is always below the original function (see Figure $3.3$ for the one-dimensional case), i.e., the first-order condition (3.16) provides a tight lower bound (which is an affine function in $\mathbf{y}$ ) over the entire domain for a differentiable convex function. Moreover, it can be seen from (3.16) that
$$f(\mathbf{y})=\max _{\mathbf{x} \in \operatorname{dom} f} f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x}) \quad \forall \mathbf{y} \in \operatorname{dom} f$$
For instance, as illustrated in Figure $3.3, f(b) \geq f(a)+f^{\prime}(a)(b-a)$ for any $a$ and the equality holds only when $a=b$. Next, let us prove the first-order condition.

Proof of (3.16): Let us prove the sufficiency followed by necessity.

• Sufficiency: (i.e., if (3.16) holds, then $f$ is convex) From (3.16), we have, for all $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \operatorname{dom} f$ which is convex and $0 \leq \lambda \leq 1$,
\begin{aligned} f(\mathbf{y}) & \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x}), \ f(\mathbf{z}) & \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{z}-\mathbf{x}), \ \Rightarrow \lambda f(\mathbf{y})+(1-\lambda) f(\mathbf{z}) & \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\lambda \mathbf{y}+(1-\lambda) \mathbf{z}-\mathbf{x}) . \end{aligned}

By setting $\mathbf{x}=\lambda \mathbf{y}+(1-\lambda) \mathbf{z} \in \operatorname{dom} f$ in the above inequality, we obtain $\lambda f(\mathbf{y})+(1-\lambda) f(\mathbf{z}) \geq f(\lambda \mathbf{y}+(1-\lambda) \mathbf{z})$. So $f$ is convex.

• Necessity: (i.e., if $f$ is convex, then (3.16) holds) For $\mathbf{x}, \mathbf{y} \in \mathbf{d o m} f$ and $0 \leq$ $\lambda \leq 1$,
\begin{aligned} f((1-\lambda) \mathbf{x}+\lambda \mathbf{y}) &=f(\mathbf{x}+\lambda(\mathbf{y}-\mathbf{x})) \ &=f(\mathbf{x})+\lambda \nabla f(\mathbf{x}+\theta \lambda(\mathbf{y}-\mathbf{x}))^{T}(\mathbf{y}-\mathbf{x}) \end{aligned}
for some $\theta \in[0,1]$ (from the first-order expansion of Taylor series (1.53)). Since $f$ is convex, we have
$$f((1-\lambda) \mathbf{x}+\lambda \mathbf{y}) \leq(1-\lambda) f(\mathbf{x})+\lambda f(\mathbf{y})$$
Substituting (3.18) on the left-hand side of this inequality yields
$$\lambda f(\mathbf{y}) \geq \lambda f(\mathbf{x})+\lambda \nabla f(\mathbf{x}+\theta \lambda(\mathbf{y}-\mathbf{x}))^{T}(\mathbf{y}-\mathbf{x})$$
For $\lambda>0$, we get (after dividing by $\lambda$ ),
\begin{aligned} f(\mathbf{y}) & \geq f(\mathbf{x})+\nabla f(\mathbf{x}+\theta \lambda(\mathbf{y}-\mathbf{x}))^{T}(\mathbf{y}-\mathbf{x}) \ &=f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x}) \quad\left(\text { as } \lambda \rightarrow 0^{+}\right) \end{aligned}
because $\nabla f$ is continuous due to the fact that $f$ is differentiable and convex (cf. Remark $3.13$ below). Hence (3.16) has been proved.

## 数学代写|凸优化作业代写Convex Optimization代考|Second-order condition

Suppose that $f$ is twice differentiable. Then $f$ is convex if and only if dom $f$ is convex and the Hessian of $f$ is PSD for all $\mathbf{x} \in \operatorname{dom} f$, that is,
$$\nabla^{2} f(\mathbf{x}) \succeq \mathbf{0}, \forall \mathbf{x} \in \operatorname{dom} f$$
Proof: Let us prove the sufficiency followed by necessity.

• Sufficiency: (i.e., if $\nabla^{2} f(\mathbf{x}) \succeq \mathbf{0}, \forall \mathbf{x} \in \operatorname{dom} f$, then $f$ is convex) From the second-order expansion of Taylor series of $f(\mathrm{x})$ (cf. (1.54)), we have
\begin{aligned} f(\mathbf{x}+\mathbf{v}) &=f(\mathbf{x})+\nabla f(\mathbf{x})^{T} \mathbf{v}+\frac{1}{2} \mathbf{v}^{T} \nabla^{2} f(\mathbf{x}+\theta \mathbf{v}) \mathbf{v} \ & \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T} \mathbf{v} \quad(\text { by }(3.27)) \end{aligned}
for some $\theta \in[0,1]$. Let $\mathbf{y}=\mathbf{x}+\mathbf{v}$, i.e., $\mathbf{v}=\mathbf{y}-\mathbf{x}$. Then we have
$$f(\mathbf{y}) \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x})$$
which is the exactly first-order condition for the convexity of $f(\mathbf{x})$, implying that $f$ is convex.
• Necessity: Since $f(\mathbf{x})$ is convex, from the first-order condition we have
$$f(\mathbf{x}+\mathbf{v}) \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T} \mathbf{v}$$
which together with the second-order expansion of Taylor series of $f(\mathbf{x})$ given by (3.28) implies
$$\mathbf{v}^{T} \nabla^{2} f(\mathbf{x}+\theta \mathbf{v}) \mathbf{v} \geq 0$$
By letting $|\mathbf{v}|_{2} \rightarrow 0$, it can be inferred that $\nabla^{2} f(\mathbf{x}) \succeq \mathbf{0}$ because $\nabla^{2} f(\mathbf{x})$ is continuous for a convex twice differentiable function $f(\mathbf{x})$.

Remark 3.16 If the second-order condition given by (3.27) holds true with the strict inequality for all $\mathbf{x} \in \operatorname{dom} f$, the function $f$ is strictly convex; moreover, under the second-order condition given by (3.27) for the case that $f: \mathbb{R} \rightarrow \mathbb{R}$,

the first derivative $f^{\prime}$ must be continuous and nondecreasing if $f$ is convex, and continuous and strictly increasing if $f$ is strictly convex.

Remark $3.17$ (Strong convexity) A convex function $f$ is strongly convex on a set $C$ if there exists an $m>0$ such that either $\nabla^{2} f(\mathbf{x}) \succeq m \mathbf{I}$ for all $\mathbf{x} \in C$, or equivalently the following second-order condition holds true:
$$f(\mathbf{y}) \geq f(\mathbf{x})+\nabla f(\mathbf{x})^{T}(\mathbf{y}-\mathbf{x})+\frac{m}{2}|\mathbf{y}-\mathbf{x}|_{2}^{2} \quad \forall \mathbf{x}, \mathbf{y} \in C$$
which is directly implied from (3.28). So if $f$ is strongly convex, it must be strictly convex, but the reverse is not necessarily true.

## 数学代写|凸优化作业代写Convex Optimization代考|Basic properties and examples of convex functions

$f_{1}(\mathbf{x})=\left{|X|22,|X|2≤1 +∞,1<|X|2≤2, \四\右。d这米f_{1}=\left{\mathbf{x} \in \mathbb{R}^{n} \mid|\mathbf{x}|_{2} \leq 2\right}f_{2}(x)=\左{−∞,X=0 日志⁡X,X>0, \quad \operatorname{dom} f_{2}=\mathbb{R}{+}\right.一种r和在和ll−d和F一世n和dF在nC吨一世这ns,一种ndf{1}一世s一种C这n在和XF在nC吨一世这n一种ndf_{2}一世s一种C这nC一种在和F在nC吨一世这n.吨H和C这n在和X一世吨是这FF在nC吨一世这ns在一世llb和pr和s和n吨和dn和X吨一世nd和吨一种一世l.D和F一世n一世吨一世这n一种ndF在nd一种米和n吨一种lpr这p和r吨一世和s一种F在nC吨一世这nf: \mathbb{R}^{n} \rightarrow \mathbb{R}$ 如果满足以下条件，则称其为凸的

• domF是凸的。
• 对全部X,是∈dom⁡F,θ∈[0,1].
F(θX+(1−θ)是)≤θF(X)+(1−θ)F(是)

|和−是|2|是−X|2=θ， 和 |和−X|2|是−X|2=1−θ,

## 数学代写|凸优化作业代写Convex Optimization代考|First-order condition

F(是)≥F(X)+∇F(X)吨(是−X)∀X,是∈dom⁡F

F(是)=最大限度X∈dom⁡FF(X)+∇F(X)吨(是−X)∀是∈dom⁡F

(3.16) 的证明：让我们证明充分性和必然性。

• 充分性：（即，如果 (3.16) 成立，则F是凸的）从（3.16），我们有，对于所有X,是,和∈dom⁡F这是凸的并且0≤λ≤1,
F(是)≥F(X)+∇F(X)吨(是−X), F(和)≥F(X)+∇F(X)吨(和−X), ⇒λF(是)+(1−λ)F(和)≥F(X)+∇F(X)吨(λ是+(1−λ)和−X).

• 必要性：（即，如果F是凸的，那么 (3.16) 成立）对于X,是∈d这米F和0≤ λ≤1,
F((1−λ)X+λ是)=F(X+λ(是−X)) =F(X)+λ∇F(X+θλ(是−X))吨(是−X)
对于一些θ∈[0,1]（来自泰勒级数（1.53）的一阶展开式）。自从F是凸的，我们有
F((1−λ)X+λ是)≤(1−λ)F(X)+λF(是)
将 (3.18) 代入该不等式的左侧，得到
λF(是)≥λF(X)+λ∇F(X+θλ(是−X))吨(是−X)
为了λ>0，我们得到（除以λ ),
F(是)≥F(X)+∇F(X+θλ(是−X))吨(是−X) =F(X)+∇F(X)吨(是−X)( 作为 λ→0+)
因为∇F是连续的，因为F是可微的和凸的（参见备注3.13以下）。因此 (3.16) 已被证明。

## 数学代写|凸优化作业代写Convex Optimization代考|Second-order condition

∇2F(X)⪰0,∀X∈dom⁡F

• 充分性：（即，如果∇2F(X)⪰0,∀X∈dom⁡F， 然后F是凸的）从泰勒级数的二阶展开F(X)（参见（1.54）），我们有
F(X+在)=F(X)+∇F(X)吨在+12在吨∇2F(X+θ在)在 ≥F(X)+∇F(X)吨在( 经过 (3.27))
对于一些θ∈[0,1]. 让是=X+在， IE，在=是−X. 然后我们有
F(是)≥F(X)+∇F(X)吨(是−X)
这正是凸性的一阶条件F(X), 意味着F是凸的。
• 必要性：因为F(X)是凸的，从我们有的一阶条件
F(X+在)≥F(X)+∇F(X)吨在
连同泰勒级数的二阶展开F(X)(3.28) 给出的暗示
在吨∇2F(X+θ在)在≥0
通过让|在|2→0, 可以推断出∇2F(X)⪰0因为∇2F(X)对于凸两次可微函数是连续的F(X).

F(是)≥F(X)+∇F(X)吨(是−X)+米2|是−X|22∀X,是∈C

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