Konvexe Optimierung - Jensens Ungleichung

Sei S eine nicht leere konvexe Menge in $ \ mathbb {R} ^ n $ und $ f: S \ rightarrow \ mathbb {R} ^ n $. Dann ist f genau dann konvex, wenn für jede ganze Zahl $ k> 0 $

$ x_1, x_2, ... x_k \ in S, \ Anzeigestil \ Summe \ Grenzen_ {i = 1} ^ k \ lambda_i = 1, \ lambda_i \ geq 0, \ forall i = 1,2, s, k $, wir haben $ f \ left (\ displaystyle \ sum \ limit_ {i = 1} ^ k \ lambda_ix_i \ right) \ leq \ displaystyle \ sum \ limit_ {i = 1} ^ k \ lambda _if \ left (x \ right) $

Beweis

Durch Induktion auf k.

$ k = 1: x_1 \ in S $ Daher $ f \ left (\ lambda_1 x_1 \ right) \ leq \ lambda_i f \ left (x_1 \ right) $, weil $ \ lambda_i = 1 $.

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

Daher $ \ lambda_1x_1 + \ lambda_2x_2 \ in S $

Daher ist per Definition $ f \ left (\ lambda_1 x_1 + \ lambda_2 x_2 \ right) \ leq \ lambda _1f \ left (x_1 \ right) + \ lambda _2f \ left (x_2 \ right) $

Die Aussage sei wahr für $ n <k $

Deshalb,

$ 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: $ Sei $ x_1, x_2, .... x_n, x_ {n + 1} \ in S $ und $ \ displaystyle \ sum \ limit_ {i = 1} ^ {n + 1} = 1 $

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

also $ 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) $ wobei $ \ mu = \ mu_1 + \ mu_2 + ... + \ mu_n $ und

$ y = \ frac {\ mu_1x_1 + \ mu_2x_2 + ... + \ mu_nx_n} {\ mu_1 + \ mu_2 + ... + \ mu_n} $ sowie $ \ 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 \ rechts) $

$ \ 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) + .... $

Daher bewiesen.