令S为$ \ mathbb {R} ^ n $和$ f：S \ rightarrow \ mathbb {R} ^ n $中的非空凸集。当且仅当对于每个整数$ k> 0 $时，f才是凸的

$ x_1，x_2，… x_k \ in S，\ displaystyle \ sum \ limits_ {i = 1} ^ k \ lambda_i = 1，\ lambda_i \ geq 0，\ forall i = 1,2，s，k $，我们有$ f \ left(\ displaystyle \ sum \ limits_ {i = 1} ^ k \ lambda_ix_i \ right)\ leq \ displaystyle \ sum \ limits_ {i = 1} ^ k \ lambda _if \ left(x \ right) $

证明

通过对k的归纳。

$ k = 1：x_1 \ in S $因此$ f \ left(\ lambda_1 x_1 \ right)\ leq \ lambda_i f \ left(x_1 \ right)$因为$ \ lambda_i = 1 $。

$ k = 2：\ lambda_1 + \ lambda_2 = 1 $和$ x_1，x_2 \ in S $

因此，$ \ lambda_1x_1 + \ lambda_2x_2 \ in S $

因此根据定义，$ f \ left(\ lambda_1 x_1 + \ lambda_2 x_2 \ right)\ leq \ lambda _1f \ left(x_1 \ right)+ \ lambda _2f \ left(x_2 \ right)$

令语句对于$ n

因此，

$ f \ left(\ lambda_1 x_1 + \ lambda_2 x_2 + …. + \ lambda_k x_k \ right)\ leq \ lambda_1 f \ left(x_1 \ right)+ \ lambda_2 f \ left(x_2 \ right)+ … + \ lambda_k f \ left(x_k \ right)$

$ k = n + 1：$让$ x_1，x_2，…. x_n，x_ {n + 1} \ in S $和$ \ displaystyle \ sum \ limits_ {i = 1} ^ {n + 1} = 1元

因此$ \ mu_1x_1 + \ mu_2x_2 + ……. + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ in S $

因此，$ f \ left(\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ right)$

$ = f \ left(\ left(\ mu_1 + \ mu_2 + … + \ mu_n \ right)\ frac {\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n} {\ mu_1 + \ mu_2 + \ mu_3} + \ mu_ {n + 1} x_ {n + 1} \ right)$

$ = f \ left(\ mu_y + \ mu_ {n + 1} x_ {n + 1} \ right)$其中$ \ mu = \ mu_1 + \ mu_2 + … + \ mu_n $和

$ y = \ frac {\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n} {\ mu_1 + \ mu_2 + … + \ mu_n} $以及$ \ mu_1 + \ mu_ {n + 1} = 1，y \ in S $

$ \ Rightarrow f \ left(\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ right)\ leq \ mu f \ left(y \ right)+ \ mu_ {n +1} f \ left(x_ {n + 1} \ right)$

$ \ Rightarrow f \ left(\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ right)\ leq $

$ \ left(\ mu_1 + \ mu_2 + … + \ mu_n \ right)f \ left(\ frac {\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n} {\ mu_1 + \ mu_2 + … + \ mu_n} \ right) + \ mu_ {n + 1} f \ left(x_ {n + 1} \ right)$

$ \ Rightarrow f \ left(\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ right)\ leq \ left(\ mu_1 + \ mu_2 + … + \ mu_n \右)$

$ \ left [\ frac {\ mu_1} {\ mu_1 + \ mu_2 + … + \ mu_n} f \ left(x_1 \ right)+ … + \ frac {\ mu_n} {\ mu_1 + \ mu_2 + … + \ mu_n} f \ left(x_n \ right)\ right] + \ mu_ {n + 1} f \ left(x_ {n + 1} \ right)$

$ \ Rightarrow f \ left(\ mu_1x_1 + \ mu_2x_2 + … + \ mu_nx_n + \ mu_ {n + 1} x_ {n + 1} \ right)\ leq \ mu_1f \ left(x_1 \ right)+ \ mu_2f \ left( x_2 \ right)+ …. $

因此证明。