f ( y ) ≥ f ( x ) + ▽ f ( x ) T ( y − x ) , f ( x ) ≥ f ( y ) + ▽ f ( y ) T ( x − y ) . 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).
將上面兩式相加得
( ▽ f ( x ) − ▽ f ( y ) ) T ( x − y ) > 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 是單調的，定義函式 g g g :
g ( t ) = f ( x + t ( y − x ) ) , t ∈ [ 0 , 1 ] g ′ ( t ) = ▽ f ( x + t ( y − x ) ) T ( y − x ) 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)
則由 g ′ ( t ) g'(t) g
′
(t) 的連續性以及
g ′ ( 1 ) − g ′ ( 0 ) > 0 且 g ′ ( 0 ) − g ′ ( 0 ) = 0 g'(1)-g'(0) >0 \;且\; g'(0)-g'(0) = 0
證明：
如果 f f f 是可微的凸函式，則有
f ( y ) ≥ f ( x ) + ▽ f ( x ) T ( y − x ) , f ( x ) ≥ f ( y ) + ▽ f ( y ) T ( x − y ) . 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).
將上面兩式相加得
( ▽ f ( x ) − ▽ f ( y ) ) T ( x − y ) > 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 是單調的，定義函式 g g g :
g ( t ) = f ( x + t ( y − x ) ) , t ∈ [ 0 , 1 ] g ′ ( t ) = ▽ f ( x + t ( y − x ) ) T ( y − x ) 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)
則由 g ′ ( t ) g'(t) g
′
(t) 的連續性以及
g ′ ( 1 ) − g ′ ( 0 ) > 0 且 g ′ ( 0 ) − g ′ ( 0 ) = 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 ′ ( t ) − g ′ ( 0 ) ≥ 0 , g'(t) -g