Hi,
I am trying to prove the above argument and have this thus far:
Assume f(ax+(1-a)y)>=0. Then
max{0,f(ax+(1-a)y)}=f(ax+(1-a)y)<=af(x)+(1-a)f(y) since f is convex
I now need f(x)>=0 and f(y)>=0 so that
af(x)+(1-a)f(y)=amax{0,f(x)}+(1-a)max(0,f(y)}
but I don't know how to show that f(x)>=0 and f(y)>=0.
Can anyone please help?