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# 数学代写|凸优化代写Convex Optimization代考|Uniformly Convex Functions

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## 数学代写|凸优化代写Convex Optimization代考|Uniformly Convex Functions

In this section, we will often use the cubic power function
$$d_3(x)=\frac{1}{3}\left|x-x_0\right|^3, \quad \nabla d_3(x)=\left|x-x_0\right| \cdot B\left(x-x_0\right), \quad x \in \mathbb{E} .$$
This is the simplest example of the uniformly convex function. In order to understand their properties, we need to develop some theory.

Let the function $d(\cdot)$ be differentiable on a closed convex set $Q$. We call it uniformly convex on $Q$ of degree $p \geq 2$ if there exists a constant $\sigma_p=\sigma_p(d)>0$ such that $^1$
$$d(y) \geq d(x)+\langle\nabla d(x), y-x\rangle+\frac{1}{p} \sigma_p|y-x|^p, \quad \forall x, y \in Q .$$
The constant $\sigma_p$ is called the parameter of uniform convexity of this function. By adding such a function to an arbitrary convex function, we get a uniformly convex function of the same degree and with the same value of parameter. Recall that degree $p=2$ corresponds to strongly convex functions (see (2.1.20)). In our old notation, the parameter $\mu$ of strong convexity for the function $f$ corresponds to $\sigma_2(f)$.
Note that any uniformly convex function grows faster than any linear function. Therefore, its level sets are always bounded. This implies that any minimization problem with uniformly convex objective is always solvable provided that its feasible set is nonempty. Moreover, its solution is always unique.
Adding two copies of inequality (4.2.10) with $x$ and $y$ interchanged, we get
$$\langle\nabla d(x)-\nabla d(y), x-y\rangle \geq \frac{2}{p} \sigma_p|x-y|^p, \quad \forall x, y \in Q .$$
It appears that this condition is sufficient for uniform convexity (however, for $p>2$ the convexity parameter is changing).

## 数学代写|凸优化代写Convex Optimization代考|Cubic Regularization of Newton Iteration

Consider the following minimization problem:
$$\min _{x \in \mathbb{E}} f(x)$$
where $\mathbb{E}$ is a finite-dimension real vector space, and $f$ is a twice differentiable convex function with Lipschitz-continuous Hessian. As was shown in Sect. 4.1, the global rate of convergence of the Cubic Newton Method (CNM) on this problem class is of the order $O\left(\frac{1}{k^2}\right)$, where $k$ is the iteration counter (see Theorem 4.1.4). However, note that $\mathrm{CNM}$ is a local one-step second-order method. From the complexity theory of smooth Convex Optimization, it is known that the rate of convergence of the local one-step first-order method (this is just the Gradient Method, see Theorem 2.1.14) can be improved from $O\left(\frac{1}{k}\right)$ to $O\left(\frac{1}{k^2}\right)$ by applying a multi-step strategy (see, for example, Theorem 2.2.3). In this section we show that a similar trick also works with CNM. As a result, we get a new method, which converges on the specified problem class as $O\left(\frac{1}{k^3}\right)$.

Let us recall the most important properties of cubic regularization of Newton’s method, taking into account the convexity of the objective function.
As suggested in Sect. 4.1, we introduce the following mapping:
$$T_M(x) \stackrel{\text { def }}{=} \operatorname{Arg} \min {y \in \mathbb{E}}\left[\hat{f}_M(x ; y) \stackrel{\text { def }}{=} f_2(x ; y)+\frac{M}{6}|y-x|^3\right] .$$ Note that $T=T_M(x)$ is a unique solution of the following equation $$\nabla f(x)+\nabla^2 f(x)(T-x)+\frac{1}{2} M \cdot|T-x| \cdot B(T-x)=0$$ Define $r_M(x)=\left|x-T_M(x)\right|$. Then, \begin{aligned} |\nabla f(T)|* & \stackrel{(4.2 .25)}{=}\left|\nabla f(T)-\nabla f(x)-\nabla^2 f(x)(T-x)-\frac{M}{2} r_M(x) B(T-x)\right|_* \ & \stackrel{(4.2 .8)}{\leq} \frac{L_3+M}{2} r_M^2(x) \end{aligned}
Further, multiplying (4.2.25) by $T-x$, we obtain
$$\langle\nabla f(x), x-T\rangle=\left\langle\nabla^2 f(x)(T-x), T-x\right\rangle+\frac{1}{2} M r_M^3(x)$$

## 数学代写|凸优化代写Convex Optimization代考|Uniformly Convex Functions

$$d_3(x)=\frac{1}{3}\left|x-x_0\right|^3, \quad \nabla d_3(x)=\left|x-x_0\right| \cdot B\left(x-x_0\right), \quad x \in \mathbb{E} .$$

$$d(y) \geq d(x)+\langle\nabla d(x), y-x\rangle+\frac{1}{p} \sigma_p|y-x|^p, \quad \forall x, y \in Q .$$

$$\langle\nabla d(x)-\nabla d(y), x-y\rangle \geq \frac{2}{p} \sigma_p|x-y|^p, \quad \forall x, y \in Q .$$

## 数学代写|凸优化代写Convex Optimization代考|Cubic Regularization of Newton Iteration

$$\min {x \in \mathbb{E}} f(x)$$ 在哪里止是有限维实向量空间，并且 $f$ 是具有 Lipschitz 连续 Hessian 矩阵的二次可微凸函数。如节所示。 4.1、三次牛顿法 (CNM) 在该问题坣上的全局收敛速度为 $O\left(\frac{1}{k^2}\right)$ ，在哪里 $k$ 是迭代计数器（见定理 4.1.4) 。但是，请注意 $\mathrm{CNM}$ 是局部一步二阶方法。从平滑凸优化的复杂性理论可知，局部一步一阶法（这只 是梯度法，见定理2.1.14) 的收敛速度可以提高为 $O\left(\frac{1}{k}\right)$ 到 $O\left(\frac{1}{k^2}\right)$ 通过应用多步策略 (例如，参见定理 2.2.3) 。在本节中，我们将展示类似的技巧也适用于 $\mathrm{CNM}{\text {。 }}$ 结果，我们得到了一种新方法，它收敛于指定的 问题类，如下所示 $O\left(\frac{1}{k^3}\right)$.

$$T_M(x) \stackrel{\text { def }}{=} \operatorname{Arg} \min y \in \mathbb{E}\left[\hat{f}M(x ; y) \stackrel{\text { def }}{=} f_2(x ; y)+\frac{M}{6}|y-x|^3\right] .$$ 注意 $T=T_M(x)$ 是下列方程的唯一解 $$\nabla f(x)+\nabla^2 f(x)(T-x)+\frac{1}{2} M \cdot|T-x| \cdot B(T-x)=0$$ 定义 $r_M(x)=\left|x-T_M(x)\right|$ 然后， $$|\nabla f(T)| * \stackrel{(4.2 .25)}{=}\left|\nabla f(T)-\nabla f(x)-\nabla^2 f(x)(T-x)-\frac{M}{2} r_M(x) B(T-x)\right|* \stackrel{(4.2 .8)}{\leq} \frac{L_3+M}{2} r_M^2(x)$$

$$\langle\nabla f(x), x-T\rangle=\left\langle\nabla^2 f(x)(T-x), T-x\right\rangle+\frac{1}{2} M r_M^3(x)$$

## MATLAB代写

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