中科大-凸优化 笔记（lec15）-保持函数凸性的操作

全部笔记的汇总贴（视频也有传送门）：中科大-凸优化

一、函数的组合

h : R k → R          g : R n → R k f : h ⋅ g            R n → R f ( x ) = h ( g ( x ) )            d o m    f = { x ∈ d o m    g ∣ g ( x ) ∈ d o m    f } h:\R^k\rightarrow\R\;\;\;\;g:\R^n\rightarrow\R^k\\f:h\cdot g\;\;\;\;\;\R^n\rightarrow\R\\f(x)=h(g(x))\;\;\;\;\;dom\;f=\{x\in dom\; g|g(x)\in dom\; f\} h:Rk→Rg:Rn→Rkf:h⋅gRn→Rf(x)=h(g(x))domf={ x∈domg∣g(x)∈domf}

二、一维

$$k=n=1,domg=domh=domf=\mathbb{R}h,g为二阶可微$$
$f$为凸$\iff f^{\prime \prime }(x)\ge 0$
$f^{\prime }(x)=h^{\prime }(g(x))\cdot g^{\prime }(x)$
$f^{\prime \prime }(x)=\underset{\ge 0}{\underbrace{h^{\prime \prime }(g(x))}}{g}^{\prime }(x{)}^2+\underset{\ge 0}{\underbrace{h^{\prime }(g(x))}}\cdot \underset{g^{\prime \prime }(x)\ge 0}{\underbrace{g^{\prime \prime }(x)}}\ge 0$
$\Rightarrow$
$$\begin{gathered} 1.h为凸、不降、g为凸，则f为凸 \\ 2.h为凸、不增、g为凹，则f为凸 \\ 3.h为凹、不降、g为凹，则f为凹 \\ 1.h为凹、不增、g为凸，则f为凹 \end{gathered}$$

三、高维

$n,k\ge 1\underset{扩展}{\underbrace{domg,domh,domf\ne {\mathbb{R}}^n,{\mathbb{R}}^k,{\mathbb{R}}^n}},h,g均不二阶可微$
$$\begin{gathered} h(x)=-\log xdomh={\mathbb{R}}_{++} \\ \tilde{h}(x)=\{\begin{array}{l}-\log xx>0\\ \\ +\infty x\le 0\end{array} \end{gathered}$$
$\Rightarrow$
$$\begin{gathered} 1.h为凸，\tilde{h}不降，g为凸，则f为凸 \\ 2.h为凸，\tilde{h}不增，g为凹，则f为凸 \\ 3.h为凹，\tilde{h}不降，g为凹，则f为凹 \\ 4.h为凹，\tilde{h}不增，g为凸，则f为凹 \end{gathered}$$

证明 ：$\forall x,y\in domf0\le \theta \le 1$
$g$为凸，故$domg$为凸，$\theta x+(1-\theta )y\in domg$
$g(\theta x+(1-\theta )y)\le \theta g(x)+(1-\theta )g(y)$
$h$为凸，故$domh$为凸，$\theta g(x)+(1-\theta )g(y)\in domh$
$$\begin{gathered} h(\theta g(x)+(1-\theta )g(y))\le \theta \underset{f(x)}{\underbrace{h(g(x))}}+(1-\theta )\underset{f(y)}{\underbrace{h(g(x))}} \\ f(\theta x+(1-\theta )y)=h(\underset{不一定在domf中}{\underbrace{g(\theta x+(1-\theta )y)}}) \end{gathered}$$
证明：$g(\theta x+(1-\theta )y)\in domh$
反证：$g(\theta x+(1-\theta )y)\notin domh$
$$\begin{gathered} \tilde{h}不降，\tilde{h}\underset{+\infty }{\underbrace{(g(\theta x+(1-\theta )y))}}\le \tilde{h}(\theta g(x)+(1-\theta )g(y))，则\tilde{h}=+\infty \\ \Rightarrow f(\theta x+(1-\theta )y)\le h(\theta g(x)+(1-\theta )g(y))\le \theta f(y)+(1-\theta )f(y) \end{gathered}$$

例

若$g$为凸， exp ⁡ { g ( x ) } \exp\{g(x)\} exp{ g(x)}为凸
$h(x)=\exp (z)$，凸，单增
若$g$为凹， g > 0 , log ⁡ { g ( x ) } g>0,\log\{g(x)\} g>0,log{ g(x)}为凹
$$h(z)=\log (z),\tilde{h}(z)=\{\begin{array}{l}\log (z)z>0\\ \\ -\infty z\le 0\end{array}$$
若$g$为凹，$g>0,\frac{1}{g(x)}$为凸
$$h(z)=\frac{1}{z}\tilde{h}(z)=\{\begin{array}{l}\frac{1}{z}z>0\\ \\ +\infty z\le 0\end{array}$$
若$g$为凸，$g\ge 0,P\ge 0,g^P(x)为凸$
$$h(z)=z^P$$

(有问题)

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