Citation Context

...e case K = 1 is evident and the case K =2is Lemma 4. Assume that the theorem holds for K = L where L is some natural number larger than 2. We can write min x [1,L+1]: |x [1,L+1]|=θ where F (y) =min x =-=[1,L]-=-: |x [1,L]|=y Lemma4toobtain min x [1,L+1]: |x [1,L+1]|=θ L+1 X k=1 fk(xk) = min xL+1,y: xL+1+y=θ k=1 (fL+1(xL+1)+F (y)) P L k=1 fk(xk). By Lemma 1, F is convex and we can therefore apply L+1 X k=1 fk...