定理

令S为$ \ mathbb {R} ^ n $中的一个非空凸集，并且$ f：S \ rightarrow \ mathbb {R} $在S上是可微的，那么当且仅当对于任何$ x_1，x_2，f是拟凸的\ in S $和$ f \ left(x_1 \ right)\ leq f \ left(x_2 \ right)$，我们有$ \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_2-x_1 \ right) \ leq 0 $

证明

设f为拟凸函数。

令$ x_1，x_2 \ in S $为$ f \ left(x_1 \ right)\ leq f \ left(x_2 \ right)$

通过f在$ x_2处的可分性，\ lambda \ in \ left(0，1 \ right)$

$ f \ left(\ lambda x_1 + \ left(1- \ lambda \ right)x_2 \ right)= f \ left(x_2 + \ lambda \ left(x_1-x_2 \ right)\ right)= f \ left(x_2 \ right )+ \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_1-x_2 \ right)$

$ + \ lambda \ left \ | x_1-x_2 \ right \ | \ alpha \ left(x_2，\ lambda \ left(x_1-x_2 \ right)\ right)$

$ \ Rightarrow f \ left(\ lambda x_1 + \ left(1- \ lambda \ right)x_2 \ right)-f \ left(x_2 \ right)-f \ left(x_2 \ right)= \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_1-x_2 \ right)$

$ + \ lambda \ left \ | x_1-x_2 \ right \ | \ alpha \ left(x2，\ lambda \ left(x_1-x_2 \ right)\ right)$

但是由于f是准凸的，$ f \ left(\ lambda x_1 + \ left(1- \ lambda \ right)x_2 \ right)\ leq f \ left(x_2 \ right)$

$ \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_1-x_2 \ right)+ \ lambda \ left \ | x_1-x_2 \ right \ | \ alpha \ left(x_2，\ lambda \ left(x_1，x_2 \ right)\ right)\ leq 0 $

但是$ \ alpha \ left(x_2，\ lambda \ left(x_1，x_2 \ right)\ right)\ rightarrow 0 $作为$ \ lambda \ rightarrow 0 $

因此，$ \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_1-x_2 \ right)\ leq 0 $

交谈

让$ x_1，x_2 \ in S $和$ f \ left(x_1 \ right)\ leq f \ left(x_2 \ right)$，$ \ bigtriangledown f \ left(x_2 \ right)^ T \ left(x_1， x_2 \ right)\ leq 0 $

为了证明f是准凸的，即$ f \ left(\ lambda x_1 + \ left(1- \ lambda \ right)x_2 \ right)\ leq f \ left(x_2 \ right)$

矛盾证明

假设存在$ x_3 = \ lambda x_1 + \ left(1- \ lambda \ right)x_2 $，使得对于某些$ \ lambda \ in，$ f \ left(x_2 \ right)

对于$ x_2 $和$ x_3，\ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_2-x_3 \ right)\ leq 0 $

$ \ Rightarrow-\ lambda \ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_2-x_3 \ right)\ leq 0 $

$ \ Rightarrow \ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_1-x_2 \ right)\ geq 0 $

对于$ x_1 $和$ x_3，\ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_1-x_3 \ right)\ leq 0 $

$ \ Rightarrow \ left(1- \ lambda \ right)\ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_1-x_2 \ right)\ leq 0 $

$ \ Rightarrow \ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_1-x_2 \ right)\ leq 0 $

因此，根据上述等式，$ \ bigtriangledown f \ left(x_3 \ right)^ T \ left(x_1-x_2 \ right)= 0 $

定义$ U = \ left \ {x：f \ left(x \ right)\ leq f \ left(x_2 \ right)，x = \ mu x_2 + \ left(1- \ mu \ right)x_3，\ mu \ in \ left(0,1 \ right)\ right \} $

因此我们可以在$中找到$ x_0 \，使得$ x_0 = \ mu_0 x_2 = \ mu x_2 + \ left(1- \ mu \ right)x_3 $ for $ \ mu _0 \ in \ left(0,1 \ right } $，它最接近$ x_3 $和$ \ hat {x} \ in \ left(x_0，x_1 \ right)$，这样根据均值定理，

$$ \ frac {f \ left(x_3 \ right)-f \ left(x_0 \ right)} {x_3-x_0} = \ bigtriangledown f \ left(\ hat {x} \ right)$$

$$ \ Rightarrow f \ left(x_3 \ right)= f \ left(x_0 \ right)+ \ bigtriangledown f \ left(\ hat {x} \ right)^ T \ left(x_3-x_0 \ right)$$

$$ \ Rightarrow f \ left(x_3 \ right)= f \ left(x_0 \ right)+ \ mu_0 \ lambda f \ left(\ hat {x} \ right)^ T \ left(x_1-x_2 \ right)$ $

由于$ x_0 $是$ x_1 $和$ x_2 $以及$ f \ left(x_2 \ right)

通过重复启动过程，$ \ bigtriangledown f \ left(\ hat {x} \ right)^ T \ left(x_1-x_2 \ right)= 0 $

因此，结合以上等式，我们得到：

$$ f \ left(x_3 \ right)= f \ left(x_0 \ right)\ leq f \ left(x_2 \ right)$$

$$ \ Rightarrow f \ left(x_3 \ right)\ leq f \ left(x_2 \ right)$$

因此，这是矛盾的。

例子

步骤1- $ f \ left(x \ right)= X ^ 3 $

$ let f \ left(x_1 \ right)\ leq f \ left(x_2 \ right)$

$ \ Rightarrow x_ {1} ^ {3} \ leq x_ {2} ^ {3} \ Rightarrow x_1 \ leq x_2 $

$ \ bigtriangledown f \左(x_2 \ right)\左(x_1-x_2 \ right)= 3x_ {2} ^ {2} \ left(x_1-x_2 \ right)\ leq 0 $

因此，$ f \ left(x \ right)$是拟凸的。

步骤2- $ f \ left(x \ right)= x_ {1} ^ {3} + x_ {2} ^ {3} $

设$ \ hat {x_1} = \ left(2，-2 \ right)$和$ \ hat {x_2} = \ left(1，0 \ right)$

因此，$ f \ left(\ hat {x_1} \ right)= 0，f \ left(\ hat {x_2} \ right)= 1 \ Rightarrow f \ left(\ hat {x_1} \ right)\ setminus

因此，$ \ bigtriangledown f \ left(\ hat {x_2} \ right)^ T \ left(\ hat {x_1}-\ hat {x_2} \ right)= \ left(3，0 \ right)^ T \ left (1，-2 \ right)= 3> 0 $

因此$ f \ left(x \ right)$不是准凸面。