# 数学代写|凸优化作业代写Convex Optimization代考|CPD131

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## 数学代写|凸优化作业代写Convex Optimization代考|Minimum and minimal elements via dual inequalities

We can use dual generalized inequalities to characterize minimum and minimal elements of a (possibly nonconvex) set $S \subseteq \mathbf{R}^m$ with respect to the generalized inequality induced by a proper cone $K$.
Dual characterization of minimum element
We first consider a characterization of the minimum element: $x$ is the minimum element of $S$, with respect to the generalized inequality $\preceq_K$, if and only if for all $\lambda \succ_{K^*} 0, x$ is the unique minimizer of $\lambda^T z$ over $z \in S$. Geometrically, this means that for any $\lambda \succ_{K *} 0$, the hyperplane
$$\left{z \mid \lambda^T(z-x)=0\right}$$
is a strict supporting hyperplane to $S$ at $x$. (By strict supporting hyperplane, we mean that the hyperplane intersects $S$ only at the point $x$.) Note that convexity of the set $S$ is not required. This is illustrated in figure $2.23$.

To show this result, suppose $x$ is the minimum element of $S$, i.e., $x \preceq_K z$ for all $z \in S$, and let $\lambda \succ_{K^} 0$. Let $z \in S, z \neq x$. Since $x$ is the minimum element of $S$, we have $z-x \succeq_K 0$. From $\lambda \succ_{K^} 0$ and $z-x \succeq_K 0, z-x \neq 0$, we conclude $\lambda^T(z-x)>0$. Since $z$ is an arbitrary element of $S$, not equal to $x$, this shows that $x$ is the unique minimizer of $\lambda^T z$ over $z \in S$. Conversely, suppose that for all $\lambda \succ_{K^*} 0, x$ is the unique minimizer of $\lambda^T z$ over $z \in S$, but $x$ is not the minimum element of $S$. Then there exists $z \in S$ with $z \nsucceq_K x$. Since $z-x \nsucceq_K 0$, there exists $\tilde{\lambda} \succeq_{K^} 0$ with $\tilde{\lambda}^T(z-x)<0$. Hence $\lambda^T(z-x)<0$ for $\lambda \succ_{K^} 0$ in the neighborhood of $\tilde{\lambda}$. This contradicts the assumption that $x$ is the unique minimizer of $\lambda^T z$ over $S$.

## 数学代写|凸优化作业代写Convex Optimization代考|Dual characterization of minimal elements

We now turn to a similar characterization of minimal elements. Here there is a gap between the necessary and sufficient conditions. If $\lambda \succ_{K^*} 0$ and $x$ minimizes $\lambda^T z$ over $z \in S$, then $x$ is minimal. This is illustrated in figure $2.24$.

To show this, suppose that $\lambda \succ_{K^*} 0$, and $x$ minimizes $\lambda^T z$ over $S$, but $x$ is not minimal, i.e., there exists a $z \in S, z \neq x$, and $z \preceq_K x$. Then $\lambda^T(x-z)>0$, which contradicts our assumption that $x$ is the minimizer of $\lambda^T z$ over $S$.

The converse is in general false: a point $x$ can be minimal in $S$, but not a minimizer of $\lambda^T z$ over $z \in S$, for any $\lambda$, as shown in figure 2.25. This figure suggests that convexity plays an important role in the converse, which is correct. Provided the set $S$ is convex, we can say that for any minimal element $x$ there exists a nonzero $\lambda \succeq_{K^*} 0$ such that $x$ minimizes $\lambda^T z$ over $z \in S$.

To show this, suppose $x$ is minimal, which means that $((x-K) \backslash{x}) \cap S=\emptyset$. Applying the separating hyperplane theorem to the convex sets $(x-K) \backslash{x}$ and $S$, we conclude that there is a $\lambda \neq 0$ and $\mu$ such that $\lambda^T(x-y) \leq \mu$ for all $y \in K$, and $\lambda^T z \geq \mu$ for all $z \in S$. From the first inequality we conclude $\lambda \succeq_K \cdot 0$. Since $x \in S$ and $x \in x-K$, we have $\lambda^T x=\mu$, so the second inequality implies that $\mu$ is the minimum value of $\lambda^T z$ over $S$. Therefore, $x$ is a minimizer of $\lambda^T z$ over $S$, where $\lambda \neq 0, \lambda \succeq_{K * 0}$.

This converse theorem cannot he strengthened to $\lambda \succ_{K^*} 0$. Fixamples show that a point $x$ can be a minimal point of a convex set $S$, but not a minimizer of $\lambda^T z$ over $z \in S$ for any $\lambda \succ_{K^} 0$. (See figure $2.26$, left.) Nor is it true that any minimizer of $\lambda^T z$ over $z \in S$, with $\lambda \succeq_{K^} 0$, is minimal (see figure $2.26$, right.)

## 数学代写|凸优化作业代写Convex Optimization代考|Minimum and minimal elements via dual inequalities

$\backslash$ left ${z \backslash m i d \backslash l a m b d a \wedge T(z x)=0 \backslash$ \right } }

$z-x \succeq_K 0, z-x \neq 0$, 我们得出结论
$\lambda^T(z-x)>0$. 自从 $z$ 是任意元素 $S$ ，不等于 $x$ ，这表 明 $x$ 是的唯一极小值 $\lambda^T z$ 超过 $z \in S$. 相反，假设对于所 有 $\lambda \succ_{K^*} 0, x$ 是的唯一极小值 $\lambda^T z$ 超过 $z \in S$ ，但 $x$ 不 是的最小元素 $S$. 那么存在 $z \in S$ 和 $z \nsucceq \nsucceq_K x$. 自从 $z-x \nsucceq K 0$ ，那里存在
\代字号 $\left{\right.$ Vlambda} $\backslash$ succeq_ $\left{\mathrm{K}^{\wedge}\right} 0$ 和 $\tilde{\lambda}^T(z-x)<0$. 因 此 $\lambda^T(z-x)<0$ 为了 Ilambda Isucc $\left{\mathrm{K}^{\wedge}\right} 0$ 在附近 $\tilde{\lambda}$. 这与以下假设相矛盾 $x$ 是的唯一极小值 $\lambda^T z_{\text {超过 }} S$.

## 数学代写|凸优化作业代写Convex Optimization代考|Dual characterization of minimal elements

$z \preceq_K x$. 然后 $\lambda^T(x-z)>0$, 这与我们的假设相矛盾 $x$ 是的最小值 $\lambda^T z$ 超过 $S$.

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