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?