参考连接: 共轭函数两个性质的证明open in new window

性质1

无论fff是否是凸函数,f∗f^*f∗恒为凸函数.

证明:

证明"无论fff是否是凸函数,f∗f^*f∗恒为凸函数.",即证:

f∗(θt1+(1−θ)t2)≤θf∗(t1)+(1−θ)f∗(t2) f^*(\theta t_1 + (1-\theta)t_2) \leq \theta f^*(t_1) + (1-\theta)f^*(t_2) f∗(θt1​+(1−θ)t2​)≤θf∗(t1​)+(1−θ)f∗(t2​)

由定义可知:

sup⁡x∈dom f{[θt1+(1−θ)t2]x−f(x)}≤θsup⁡x∈dom f{xt1−f(x)}+(1−θ)sup⁡x∈dom f{xt2−f(x)}⇔sup⁡x∈dom f{(1−θ)[xt2−f(x)]+θ[xt1−f(x)]}≤θsup⁡x∈dom f{xt1−f(x)}+(1−θ)sup⁡x∈dom f{xt2−f(x)} \begin{aligned} &\qquad \sup_{x\in \mathrm{dom}\,f}\{ [\theta t_1 + (1-\theta)t_2]x - f(x) \} \leq \theta\sup_{x\in \mathrm{dom}\,f}\{ xt_1 - f(x) \} + (1-\theta)\sup_{x\in \mathrm{dom}\,f}\{ xt_2 - f(x) \} \\ &\Leftrightarrow \sup_{x\in \mathrm{dom}\,f} \{ (1-\theta)[xt_2-f(x)] + \theta[xt_1 - f(x)] \} \leq \theta\sup_{x\in \mathrm{dom}\,f}\{ xt_1 - f(x) \} + (1-\theta)\sup_{x\in \mathrm{dom}\,f}\{ xt_2 - f(x) \} \end{aligned} ​x∈domfsup​{[θt1​+(1−θ)t2​]x−f(x)}≤θx∈domfsup​{xt1​−f(x)}+(1−θ)x∈domfsup​{xt2​−f(x)}⇔x∈domfsup​{(1−θ)[xt2​−f(x)]+θ[xt1​−f(x)]}≤θx∈domfsup​{xt1​−f(x)}+(1−θ)x∈domfsup​{xt2​−f(x)}​

设在x→x0x\to x_0x→x0​时不等式左式取得最小上界,则有:

(1−θ)[x0t2−f(x0)]+θ[x0t1−f(x0)]≤θsup⁡x∈dom f{t1x−f(x)}+(1−θ)sup⁡x∈dom f{t2x−f(x)} (1-\theta)[x_0t_2-f(x_0)] + \theta[x_0t_1 - f(x_0)] \leq \theta\sup_{x\in \mathrm{dom}\,f}\{ t_1x - f(x) \} + (1-\theta)\sup_{x\in \mathrm{dom}\,f}\{ t_2x - f(x) \} (1−θ)[x0​t2​−f(x0​)]+θ[x0​t1​−f(x0​)]≤θx∈domfsup​{t1​x−f(x)}+(1−θ)x∈domfsup​{t2​x−f(x)}

注意到下式显然成立:

x0t1−f(x0)≤sup⁡x∈dom f{t1x−f(x)}(1)x0t2−f(x0)≤sup⁡x∈dom f{t2x−f(x)}(2) \begin{aligned} &x_0t_1 - f(x_0)\leq \sup_{x\in \mathrm{dom}\,f}\{ t_1x-f(x) \} \qquad \text{(1)} \\ &x_0t_2 - f(x_0)\leq \sup_{x\in \mathrm{dom}\,f}\{ t_2x-f(x) \} \qquad \text{(2)} \end{aligned} ​x0​t1​−f(x0​)≤x∈domfsup​{t1​x−f(x)}(1)x0​t2​−f(x0​)≤x∈domfsup​{t2​x−f(x)}(2)​

则θ(1)+(1−θ)(2)\theta(1) + (1-\theta)(2)θ(1)+(1−θ)(2)这一线性组合即可得到以上不等式成立,由此性质1得证.

性质2

凸函数的共轭函数的共轭函数是它自己.

证明:

已知f(x)f(x)f(x)为凸函数,共轭函数的定义如下:

f∗(t)=sup⁡x∈dom f{tx−f(x)} f^*(t) = \sup_{x\in \mathrm{dom}\,f}\{ tx - f(x) \} f∗(t)=x∈domfsup​{tx−f(x)}

∵ddxf∗(t)=0⇒ddx(tx−f(x))=0∴t=ddxf(x) \because \frac{d}{dx} f^*(t) = 0 \Rightarrow \frac{d}{dx}(tx - f(x)) = 0 \\ \therefore t = \frac{d}{dx}f(x) ∵dxd​f∗(t)=0⇒dxd​(tx−f(x))=0∴t=dxd​f(x)

又有f∗(t)f^*(t)f∗(t)的共轭函数为

f∗∗(s)=sup⁡t∈dom f{st−f∗(t)} f^{**}(s) = \sup_{t\in \mathrm{dom}\,f}\{ st - f^*(t) \} f∗∗(s)=t∈domfsup​{st−f∗(t)}

∵ddtf∗∗(s)=0⇒ddt[st−f∗(t)]=0 \because \frac{d}{dt}f^{**}(s) = 0 \Rightarrow \frac{d}{dt}[st - f^*(t)] = 0 ∵dtd​f∗∗(s)=0⇒dtd​[st−f∗(t)]=0

∴s=ddtf∗(t)=ddt[tx−f(x)]=x+tdxdt−ddtf(x)=x+tdxdt−ddxf(x)⋅dxdt=x+tdxdt−tdxdt=x \begin{aligned} \therefore s &= \frac{d}{dt}f^*(t) \\ &= \frac{d}{dt}[tx - f(x)] \\ &= x + t\frac{dx}{dt} - \frac{d}{dt}f(x) \\ &= x + t\frac{dx}{dt} - \frac{d}{dx}f(x)\cdot \frac{dx}{dt} \\ &= x + t\frac{dx}{dt} - t\frac{dx}{dt} \\ &= x \end{aligned} ∴s​=dtd​f∗(t)=dtd​[tx−f(x)]=x+tdtdx​−dtd​f(x)=x+tdtdx​−dxd​f(x)⋅dtdx​=x+tdtdx​−tdtdx​=x​

∴f∗∗(s)=sup⁡t∈dom f{st−f∗(t)}=sup⁡t∈dom f{tx−f∗(t)}=sup⁡t∈dom f{tx−[tx−f(x)]}=sup⁡t∈dom ff(x)=f(x)=f(s) \begin{aligned} \therefore f^{**}(s) &= \sup_{t\in \mathrm{dom}\,f}\{ st - f^*(t) \} \\ &= \sup_{t\in \mathrm{dom}\,f} \{ tx - f^*(t) \} \\ &= \sup_{t\in \mathrm{dom}\,f} \{ tx - [tx - f(x)] \} \\ &= \sup_{t\in \mathrm{dom}\,f}f(x) \\ &= f(x) \\ &= f(s) \end{aligned} ∴f∗∗(s)​=t∈domfsup​{st−f∗(t)}=t∈domfsup​{tx−f∗(t)}=t∈domfsup​{tx−[tx−f(x)]}=t∈domfsup​f(x)=f(x)=f(s)​

故该性质成立.