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# 数学代写|凸优化代写Convex Optimization代考|Gradient-Dominated Functions

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## 数学代写|凸优化代写Convex Optimization代考|Gradient-Dominated Functions

Let us now look at another interesting class of nonconvex functions.
Definition 4.1.3 A function $f(\cdot)$ is called gradient dominated of degree $p \in[1,2]$ if it attains a global minimum at some point $x^$ and for any $x \in \mathscr{F}$ we have $$f(x)-f\left(x^\right) \leq \tau_f|\nabla f(x)|^p$$
where $\tau_f$ is a positive constant. The parameter $p$ is called the degree of domination.
We do not assume here that the global minimum of function $f$ is unique. Let us give several examples of gradient dominated functions.

Example 4.1.1 (Convex Functions) Let $f$ be convex on $\mathbb{R}^n$. Assume it achieves its minimum at point $x^$. Then, for any $x \in \mathbb{R}^n$ with $\left|x-x^\right|<R$, we have
$$f(x)-f\left(x^\right) \stackrel{(2.1 .2)}{\leq}\left\langle\nabla f(x), x-x^\right\rangle \leq|\nabla f(x)| \cdot R$$
Thus, the function $f$ is a gradient dominated function of degree one on the set $\mathscr{F}=\left{x:\left|x-x^*\right|<R\right}$ with $\tau_f=R$.

Example 4.1.2 (Strongly Convex Functions) Let $f$ be differentiable and strongly convex on $\mathbb{R}^n$. This means that there exists a constant $\mu>0$ such that
$$f(y) \stackrel{(2.1 .20)}{\geq} f(x)+\langle\nabla f(x), y-x\rangle+\frac{1}{2} \mu|y-x|^2$$
for all $x, y \in \mathbb{R}^n$. Then, minimizing both sides of this inequality in $y$, we obtain,
$$f(x)-f\left(x^*\right) \leq \frac{1}{2 \mu}|\nabla f(x)|^2 \quad \forall x \in \mathbb{R}^n$$

## 数学代写|凸优化代写Convex Optimization代考|Nonlinear Transformations of Convex Functions

Let $u(x): \quad \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a non-degenerate vector function. Denote by $v(u)$ its inverse:
$$v(u): \mathbb{R}^n \rightarrow \mathbb{R}^n, \quad v(u(x)) \equiv x$$
Consider the following function:
$$f(x)=\phi(u(x))$$
where $\phi(u)$ is a convex function with bounded level sets. Denote by $x^* \equiv v\left(u^\right)$ its minimum. Let us fix some $x_0 \in \mathbb{R}^n$. Define \begin{aligned} & \sigma=\max _u\left{\left|v^{\prime}(u)\right|: \phi(u) \leq f\left(x_0\right)\right}, \ & D=\max _u\left{\left|u-u^\right|: \phi(u) \leq f\left(x_0\right)\right} . \end{aligned}
The following result is straightforward.

Lemma 4.1.9 For any $x, y \in \mathscr{L}\left(f\left(x_0\right)\right)$ we have
$$|x-y| \leq \sigma|u(x)-u(y)| .$$
Proof Indeed, for $x, y \in \mathscr{L}\left(f\left(x_0\right)\right)$, we have $\phi(u(x)) \leq f\left(x_0\right)$ and $\phi(u(y)) \leq$ $f\left(x_0\right)$. Consider the trajectory $x(t)=v(t u(y)+(1-t) u(x)), t \in[0,1]$. Then
$$y-x=\int_0^1 x^{\prime}(t) d t=\left(\int_0^1 v^{\prime}(t u(y)+(1-t) u(x)) d t\right) \cdot(u(y)-u(x)),$$
and (4.1.42) follows.
The following result is very similar to Theorem 4.1.4.

## 数学代写|凸优化代写Convex Optimization代考|Gradient-Dominated Functions

$$f(x)-f\left(x^\right) \stackrel{(2.1 .2)}{\leq}\left\langle\nabla f(x), x-x^\right\rangle \leq|\nabla f(x)| \cdot R$$

$$f(y) \stackrel{(2.1 .20)}{\geq} f(x)+\langle\nabla f(x), y-x\rangle+\frac{1}{2} \mu|y-x|^2$$

$$f(x)-f\left(x^*\right) \leq \frac{1}{2 \mu}|\nabla f(x)|^2 \quad \forall x \in \mathbb{R}^n$$

## 数学代写|凸优化代写Convex Optimization代考|Nonlinear Transformations of Convex Functions

$$v(u): \mathbb{R}^n \rightarrow \mathbb{R}^n, \quad v(u(x)) \equiv x$$

$$f(x)=\phi(u(x))$$

$$|x-y| \leq \sigma|u(x)-u(y)| .$$

(4.1.42)

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