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# 数学代写|凸优化代写Convex Optimization代考|Definition of self-concordant barriers

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## 数学代写|凸优化代写Convex Optimization代考|Definition of self-concordant barriers

DeFINITION 4.2.2 Let $F(x)$ be a standard self-concordant function. We call it a $\nu$-self-concordant barrier for set $\operatorname{Dom} F$, if
$$\sup _{u \in R^n}\left[2\left\langle F^{\prime}(x), u\right\rangle-\left\langle F^{\prime \prime}(x) u, u\right\rangle\right] \leq \nu$$

for all $x \in \operatorname{dom} F$. The value $\nu$ is called the parameter of the barrier.
Note that we do not assume $F^{\prime \prime}(x)$ to be nondegenerate. However, if this is the case, then the inequality (4.2.3) is equivalent to
$$\left\langle\left[F^{\prime \prime}(x)\right]^{-1} F^{\prime}(x), F^{\prime}(x)\right\rangle \leq \nu .$$
We will use also another equivalent form of inequality (4.2.3):
$$\left\langle F^{\prime}(x), u\right\rangle^2 \leq \nu\left\langle F^{\prime \prime}(x) u, u\right\rangle \quad \forall u \in R^n \text {. }$$
(To see that for $u$ with $\left\langle F^{\prime \prime}(x) u, u\right\rangle>0$, replace $u$ in (4.2.3) by $\lambda u$ and find the maximum of the left-hand side in $\lambda$.) Note that the condition (4.2.5) can be written in a matrix notation:
$$F^{\prime \prime}(x) \succeq \frac{1}{\nu} F^{\prime}(x) F^{\prime}(x)^T$$
Let us check now which self-concordant functions given by Example 4.1.1 are also self-concordant barriers.

## 数学代写|凸优化代写Convex Optimization代考|Main inequalities

Let us show that the local characteristics of a self-concordant barrier (the gradient and the Hessian) provide us with global information about the structure of the domain.

THEOREM 4.2.4 1 . Let $F(x)$ be a $\nu$-self-concordant barrier. For any $x$ and $y$ from $\operatorname{dom} F$, we have
$$\left\langle F^{\prime}(x), y-x\right\rangle<\nu .$$

Moreover, if $\left\langle F^{\prime}(x), y-x\right\rangle \geq 0$, then
$$\left\langle F^{\prime}(y)-F^{\prime}(x), y-x\right\rangle \geq \frac{\left\langle F^{\prime}(x), y-x\right\rangle^2}{\nu-\left\langle F^{\prime}(x), y-x\right\rangle} .$$

A standard self-concordant function $F(x)$ is a $\nu$-self-concordant barrier if and only if
$$F(y) \geq F(x)-\nu \ln \left(1-\frac{1}{\nu}\left\langle F^{\prime}(x), y-x\right\rangle\right) \quad \forall x, y \in \operatorname{dom} F .$$

up to a logarithmic transformation of both sides of the inequality.
THEOREM 4.2.5 Let $F(x)$ be a $\nu$-self-concordant barrier. Then for any $x \in \operatorname{dom} F$ and $y \in \operatorname{Dom} F$ such that
$$\left\langle F^{\prime}(x), y-x\right\rangle \geq 0,$$
we have
$$|y-x|_x \leq \nu+2 \sqrt{\nu} \text {. }$$

## 数学代写|凸优化代写Convex Optimization代考|Definition of self-concordant barriers

4.2.2设$F(x)$为标准自调和函数。我们称它为$\nu$ -自和谐势垒对于集合$\operatorname{Dom} F$，如果
$$\sup _{u \in R^n}\left[2\left\langle F^{\prime}(x), u\right\rangle-\left\langle F^{\prime \prime}(x) u, u\right\rangle\right] \leq \nu$$

$$\left\langle\left[F^{\prime \prime}(x)\right]^{-1} F^{\prime}(x), F^{\prime}(x)\right\rangle \leq \nu .$$

$$\left\langle F^{\prime}(x), u\right\rangle^2 \leq \nu\left\langle F^{\prime \prime}(x) u, u\right\rangle \quad \forall u \in R^n \text {. }$$
(要用$\left\langle F^{\prime \prime}(x) u, u\right\rangle>0$查看$u$，请将(4.2.3)中的$u$替换为$\lambda u$，并在$\lambda$中找到左侧的最大值。)注意，条件(4.2.5)可以写成矩阵表示法:
$$F^{\prime \prime}(x) \succeq \frac{1}{\nu} F^{\prime}(x) F^{\prime}(x)^T$$

## 数学代写|凸优化代写Convex Optimization代考|Main inequalities

$$\left\langle F^{\prime}(x), y-x\right\rangle<\nu .$$

$$\left\langle F^{\prime}(y)-F^{\prime}(x), y-x\right\rangle \geq \frac{\left\langle F^{\prime}(x), y-x\right\rangle^2}{\nu-\left\langle F^{\prime}(x), y-x\right\rangle} .$$

$$F(y) \geq F(x)-\nu \ln \left(1-\frac{1}{\nu}\left\langle F^{\prime}(x), y-x\right\rangle\right) \quad \forall x, y \in \operatorname{dom} F .$$

$$\left\langle F^{\prime}(x), y-x\right\rangle \geq 0,$$

$$|y-x|_x \leq \nu+2 \sqrt{\nu} \text {. }$$

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