可微函数
f
f
f 是凸函数 当且仅当
d
o
m
f
domf
domf 是凸集,且
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▽
f
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−
▽
f
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T
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−
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;
0
,
    
∀
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,
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∈
d
o
m
f
(\bigtriangledown f(x)-\bigtriangledown f(y))^T(x-y);0, \;\; \forall x,y \in dom f
(▽f(x)−▽f(y))T(x−y)>0,∀x,y∈domf 即
▽
f
:
R
n
→
R
n
\bigtriangledown f: \R^n \rightarrow \R^n
▽f:Rn→Rn 是单调映射(monotone mapping)。
证明:
如果
f
f
f 是可微的凸函数,则有
f
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≥
f
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+
▽
f
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T
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,
f
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≥
f
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▽
f
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T
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.
f(y) \geq f(x) + \bigtriangledown f(x)^T(y-x),\\ f(x) \geq f(y) + \bigtriangledown f(y)^T(x-y).
f(y)≥f(x)+▽f(x)T(y−x),f(x)≥f(y)+▽f(y)T(x−y).将上面两式相加得
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▽
f
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−
▽
f
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)
T
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−
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;
0
(\bigtriangledown f(x)-\bigtriangledown f(y))^T(x-y);0
(▽f(x)−▽f(y))T(x−y)>0如果
▽
f
\bigtriangledown f
▽f 是单调的,定义函数
g
g
g :
g
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=
f
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,
  
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∈
[
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]
g
′
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=
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f
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T
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g(t) = f(x+t(y-x)), \;t \in [0,1]\\ g'(t) = \bigtriangledown f(x+t(y-x))^T(y-x)
g(t)=f(x+t(y−x)),t∈[0,1]g′(t)=▽f(x+t(y−x))T(y−x)则由
g
′
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t
)
g'(t)
g′(t) 的连续性以及
g
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−
g
′
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;
0
  
且
  
g
′
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−
g
′
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=
0
g'(1)-g'(0) ;0 \;且\; g'(0)-g'(0) = 0
g′(1)−g′(0)>0且g′(0)−g′(0)=0得
g
′
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t
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−
g
′
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0
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≥
0
,
    
g'(t) -g'(0) \geq 0,\;\;
g′(t)−g′(0)≥0,因此
f
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y
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=
g
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1
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=
g
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+
∫
0
1
g
′
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d
t
≥
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g
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=
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f
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T
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f(y) = g(1) = g(0) + \int_0^1 g'(t)dt \geq g(0) + g'(0) \\= f(x) + \bigtriangledown f(x))^T(y-x)
f(y)=g(1)=g(0)+∫01g′(t)dt≥g(0)+g′(0)=f(x)+▽f(x))T(y−x) 即
f
f
f 为凸函数。
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