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UA SIE545 优化理论基础 函数凸性的一些有趣的判断方法

發布時間:2025/4/14 编程问答 44 豆豆
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UA SIE545 優化理論基礎 函數凸性的一些有趣的判斷方法

Convex function f:S→Rf:S \to \mathbb{R}f:SR where SSS is a nonempty convex set. Call fff a convex function on SSS if ?x1,x2∈S\forall x_1,x_2 \in S?x1?,x2?S, λ∈(0,1)\lambda \in (0,1)λ(0,1)
f(λx1+(1?λ)x2)≤λf(x1)+(1?λ)f(x2)f(\lambda x_1+(1-\lambda)x_2) \le \lambda f(x_1)+(1-\lambda)f(x_2)f(λx1?+(1?λ)x2?)λf(x1?)+(1?λ)f(x2?)

Call fff a strictly convex function on SSS if ?x1,x2∈S\forall x_1,x_2 \in S?x1?,x2?S, λ∈(0,1)\lambda \in (0,1)λ(0,1)
f(λx1+(1?λ)x2)<λf(x1)+(1?λ)f(x2)f(\lambda x_1+(1-\lambda)x_2) < \lambda f(x_1)+(1-\lambda)f(x_2)f(λx1?+(1?λ)x2?)<λf(x1?)+(1?λ)f(x2?)

Call fff a (strictly) concave function on SSS if ?f-f?f is (strictly) convex.


hypograph f:S→Rf:S \to \mathbb{R}f:SR where SSS is a nonempty convex set.
hypf={(x,y):x∈S,y∈R,y≤f(x)}hyp \ f = \{(x,y):x \in S, y \in \mathbb{R}, y \le f(x)\}hyp?f={(x,y):xS,yR,yf(x)}

方法1 f:S→Rf:S \to \mathbb{R}f:SRSSS是非空凸集,則fff concave的充要條件是hypfhyp \ fhyp?f是凸集。
證明
?\Rightarrow?: 假設fff是凹函數,?x1,x2∈S\forall x_1,x_2 \in S?x1?,x2?S, λ∈(0,1)\lambda \in (0,1)λ(0,1)
f(λx1+(1?λ)x2)≥λf(x1)+(1?λ)f(x2)f(\lambda x_1+(1-\lambda)x_2) \ge \lambda f(x_1)+(1-\lambda)f(x_2)f(λx1?+(1?λ)x2?)λf(x1?)+(1?λ)f(x2?)考慮?y1,y2∈R\forall y_1,y_2 \in \mathbb{R}?y1?,y2?R,滿足(x1,y1)∈hypf,(x2,y2)∈hypf(x_1,y_1) \in hyp\ f,(x_2,y_2) \in hyp\ f(x1?,y1?)hyp?f,(x2?,y2?)hyp?f,即
y1≤f(x1),y2≤f(x2)y_1 \le f(x_1),y_2 \le f(x_2)y1?f(x1?),y2?f(x2?)

于是
λy1+(1?λ)y2≤λf(x1)+(1?λ)f(x2)≤f(λx1+(1?λ)x2)\lambda y_1 + (1-\lambda)y_2 \le \lambda f(x_1)+(1-\lambda)f(x_2) \\ \le f(\lambda x_1+(1-\lambda)x_2)λy1?+(1?λ)y2?λf(x1?)+(1?λ)f(x2?)f(λx1?+(1?λ)x2?)

所以λ(x1,y1)+(1?λ)(x2,y2)∈hypf\lambda (x_1,y_1)+(1-\lambda)(x_2,y_2) \in hyp\ fλ(x1?,y1?)+(1?λ)(x2?,y2?)hyp?f,即hypfhyp \ fhyp?f是凸集。

?\Leftarrow?: 假設hypfhyp \ fhyp?f是凸集,(x1,f(x1))∈hypf,(x2,f(x2))∈hypf(x_1,f(x_1)) \in hyp\ f,(x_2,f(x_2)) \in hyp\ f(x1?,f(x1?))hyp?f,(x2?,f(x2?))hyp?f?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1), λ(x1,f(x1))+(1?λ)(x2,f(x2))∈hypf\lambda (x_1,f(x_1))+(1-\lambda)(x_2,f(x_2)) \in hyp\ fλ(x1?,f(x1?))+(1?λ)(x2?,f(x2?))hyp?f,也就是
λf(x1)+(1?λ)f(x2)≤f(λx1+(1?λ)x2)\lambda f(x_1) + (1-\lambda)f(x_2) \le f(\lambda x_1+(1-\lambda)x_2)λf(x1?)+(1?λ)f(x2?)f(λx1?+(1?λ)x2?)于是fff是凹函數。


方法2 凸函數的線性組合 假設f1,?,fkf_1,\cdots,f_kf1?,?,fk?是一列凸函數,f=∑j=1kαjfj,αj>0,?jf = \sum_{j=1}^k \alpha_j f_j,\alpha_j >0,\forall jf=j=1k?αj?fj?,αj?>0,?j,則fff是凸函數。
證明
因為fj:S→R,?jf_j:S \to \mathbb{R},\forall jfj?:SR,?j是凸函數,所以?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1), ?x1,x2∈S\forall x_1,x_2 \in S?x1?,x2?S
fj(λx1+(1?λ)x2)≤λfj(x1)+(1?λ)fj(x2)αjfj(λx1+(1?λ)x2)≤λαjfj(x1)+(1?λ)αjfj(x2)∑j=1kαjfj(λx1+(1?λ)x2)≤λ∑j=1kαjfj(x1)+(1?λ)∑j=1kαjfj(x2)f_j(\lambda x_1 + (1-\lambda)x_2) \le \lambda f_j(x_1)+(1-\lambda)f_j(x_2) \\ \alpha_jf_j(\lambda x_1 + (1-\lambda)x_2) \le \lambda \alpha_j f_j(x_1)+(1-\lambda) \alpha_jf_j(x_2) \\ \sum_{j=1}^k \alpha_jf_j(\lambda x_1 + (1-\lambda)x_2) \le \lambda \sum_{j=1}^k\alpha_j f_j(x_1)+(1-\lambda)\sum_{j=1}^k\alpha_j f_j(x_2)fj?(λx1?+(1?λ)x2?)λfj?(x1?)+(1?λ)fj?(x2?)αj?fj?(λx1?+(1?λ)x2?)λαj?fj?(x1?)+(1?λ)αj?fj?(x2?)j=1k?αj?fj?(λx1?+(1?λ)x2?)λj=1k?αj?fj?(x1?)+(1?λ)j=1k?αj?fj?(x2?)

因此fff也是凸函數。


方法3 凸函數的最值 假設f1,?,fkf_1,\cdots,f_kf1?,?,fk?是一列凸函數,f=max?(f1,?,fk)f = \max(f_1,\cdots,f_k)f=max(f1?,?,fk?),則fff是凸函數。
證明
因為fj:S→R,?jf_j:S \to \mathbb{R},\forall jfj?:SR,?j是凸函數,所以?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1), ?x1,x2∈S\forall x_1,x_2 \in S?x1?,x2?S
fj(λx1+(1?λ)x2)≤λfj(x1)+(1?λ)fj(x2)f_j(\lambda x_1 + (1-\lambda)x_2) \le \lambda f_j(x_1)+(1-\lambda)f_j(x_2) fj?(λx1?+(1?λ)x2?)λfj?(x1?)+(1?λ)fj?(x2?)

因為f=max?(f1,?,fk)f = \max(f_1,\cdots,f_k)f=max(f1?,?,fk?)
f(λx1+(1?λ)x2)=max?jfj(λx1+(1?λ)x2)≤max?j[λfj(x1)+(1?λ)fj(x2)]≤max?jλfj(x1)+max?j(1?λ)fj(x2)f(\lambda x_1 + (1-\lambda)x_2) = \max_j f_j(\lambda x_1 + (1-\lambda)x_2) \\ \le \max_j[ \lambda f_j(x_1)+(1-\lambda)f_j(x_2)] \le \max_j \lambda f_j(x_1) +\max_j(1-\lambda)f_j(x_2) f(λx1?+(1?λ)x2?)=jmax?fj?(λx1?+(1?λ)x2?)jmax?[λfj?(x1?)+(1?λ)fj?(x2?)]jmax?λfj?(x1?)+jmax?(1?λ)fj?(x2?)

所以fff是凸函數。


distance function f:Rn→Rf:\mathbb{R}^n \to \mathbb{R}f:RnRSSS是非空凸集,distance function的定義是
f(y)=inf?{∥y?x∥:x∈S}f(y) = \inf\{ \left\| y-x \right\|:x \in S\}f(y)=inf{y?x:xS}

方法4 distance function是凸函數
證明
?y1,y2∈Rn\forall y_1,y_2 \in \mathbb{R}^n?y1?,y2?Rn?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1)
f(λy1+(1?λ)y2)=inf?{∥λy1+(1?λ)y2?x∥:x∈S}≤inf?{λ∥y1?x∥+(1?λ)∥y2?x∥:x∈S}≤λinf?{∥y1?x∥:x∈S}+(1?λ)inf?{∥y2?x∥:x∈S}≤λf(y1)+(1?λ)f(y2)f(\lambda y_1+(1-\lambda)y_2) = \inf\{ \left\| \lambda y_1+(1-\lambda)y_2-x \right\|:x \in S\} \\ \le \inf\{\lambda\left\| y_1- x \right\|+(1-\lambda) \left\|y_2-x \right\|:x \in S\} \\ \le \lambda\inf\{\left\| y_1- x \right\|:x \in S\} + (1-\lambda)\inf\{\left\| y_2- x \right\|:x \in S\} \\ \le \lambda f(y_1)+(1-\lambda)f(y_2) f(λy1?+(1?λ)y2?)=inf{λy1?+(1?λ)y2??x:xS}inf{λy1??x+(1?λ)y2??x:xS}λinf{y1??x:xS}+(1?λ)inf{y2??x:xS}λf(y1?)+(1?λ)f(y2?)

所以fff是凸函數。


Support function f:Rn→Rf:\mathbb{R}^n \to \mathbb{R}f:RnRSSS是非空有界凸集,support function的定義是
f(y)=sup?{yTx:x∈S}f(y) = \sup \{y^Tx:x \in S\}f(y)=sup{yTx:xS}

方法5 Support function是凸函數
證明
?y1,y2∈Rn\forall y_1,y_2 \in \mathbb{R}^n?y1?,y2?Rn?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1)
f(λy1+(1?λ)y2)=sup?{[λy1+(1?λ)y2]Tx:x∈S}=sup?{λy1Tx+(1?λ)y2Tx:x∈S}≤λsup?{y1Tx:x∈S}+(1?λ)sup?{y2Tx:x∈S}≤λf(y1)+(1?λ)f(y2)f(\lambda y_1 + (1-\lambda)y_2)=\sup \{[\lambda y_1 + (1-\lambda)y_2]^Tx:x \in S\} \\ =\sup \{\lambda y_1^Tx + (1-\lambda)y_2^Tx:x \in S\} \\ \le \lambda \sup \{ y_1^Tx:x \in S\}+(1-\lambda)\sup \{ y_2^Tx:x \in S\}\le \lambda f(y_1)+(1-\lambda)f(y_2)f(λy1?+(1?λ)y2?)=sup{[λy1?+(1?λ)y2?]Tx:xS}=sup{λy1T?x+(1?λ)y2T?x:xS}λsup{y1T?x:xS}+(1?λ)sup{y2T?x:xS}λf(y1?)+(1?λ)f(y2?)所以fff是凸函數。


perturbation function f:Rn→Rf:\mathbb{R}^n \to \mathbb{R}f:RnR, g:Rn→Rmg: \mathbb{R}^n \to \mathbb{R}^mg:RnRm是凸函數,SSS是非空凸集,perturbation function的定義是
?(y)=inf?{f(x):g(x)≤y,x∈S}\phi(y) = \inf \{f(x):g(x) \le y,x \in S\}?(y)=inf{f(x):g(x)y,xS}

方法6 perturbation function是凸函數
證明
?y1,y2∈Rn\forall y_1,y_2 \in \mathbb{R}^n?y1?,y2?Rn?λ∈(0,1)\forall \lambda \in (0,1)?λ(0,1)
?(λy1+(1?λ)y2)=inf?{f(x):g(x)≤λy1+(1?λ)y2,x∈S}\phi(\lambda y_1+(1-\lambda)y_2) = \inf\{f(x):g(x) \le \lambda y_1+(1-\lambda)y_2,x \in S \}?(λy1?+(1?λ)y2?)=inf{f(x):g(x)λy1?+(1?λ)y2?,xS}

考慮集合{x∈S:g(x)≤λy1+(1?λ)y2}={x∈S:λg(x)+(1?λ)g(x)≤λy1+(1?λ)y2}?{x∈S:g(λx+(1?λ)x)≤λy1+(1?λ)y2}\{x \in S:g(x) \le\lambda y_1+(1-\lambda)y_2 \} \\ = \{x \in S:\lambda g(x)+(1-\lambda)g(x) \le\lambda y_1+(1-\lambda)y_2 \} \\ \subset \{x \in S:g(\lambda x+(1-\lambda)x) \le\lambda y_1+(1-\lambda)y_2 \}{xS:g(x)λy1?+(1?λ)y2?}={xS:λg(x)+(1?λ)g(x)λy1?+(1?λ)y2?}?{xS:g(λx+(1?λ)x)λy1?+(1?λ)y2?}

因此
inf?{f(x):g(x)≤λy1+(1?λ)y2,x∈S}≤inf?{f(λx+(1?λ)x):g(λx+(1?λ)x)≤λy1+(1?λ)y2,x∈S}≤λinf?{f(x):g(λx+(1?λ)x)≤λy1+(1?λ)y2,x∈S}+(1?λ)inf?{f(x):g(λx+(1?λ)x)≤λy1+(1?λ)y2,x∈S}=λinf?{f(x):g(x)≤λy1,x∈S}+(1?λ)inf?{f(x):g(x)≤λy2,x∈S}\inf\{f(x):g(x) \le \lambda y_1+(1-\lambda)y_2,x \in S \} \\ \le \inf\{f(\lambda x+(1-\lambda)x):g(\lambda x+(1-\lambda)x) \le \lambda y_1+(1-\lambda)y_2,x \in S \} \\ \le \lambda \inf\{f(x):g(\lambda x+(1-\lambda)x) \le \lambda y_1+(1-\lambda)y_2,x \in S \} \\ + (1-\lambda)\inf\{f(x):g(\lambda x+(1-\lambda)x) \le \lambda y_1+(1-\lambda)y_2,x \in S \} \\ = \lambda \inf\{f(x):g(x) \le \lambda y_1,x \in S \} + (1-\lambda)\inf\{f(x):g(x) \le \lambda y_2,x \in S \}inf{f(x):g(x)λy1?+(1?λ)y2?,xS}inf{f(λx+(1?λ)x):g(λx+(1?λ)x)λy1?+(1?λ)y2?,xS}λinf{f(x):g(λx+(1?λ)x)λy1?+(1?λ)y2?,xS}+(1?λ)inf{f(x):g(λx+(1?λ)x)λy1?+(1?λ)y2?,xS}=λinf{f(x):g(x)λy1?,xS}+(1?λ)inf{f(x):g(x)λy2?,xS}

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