Splitting Function Across Sum With Inequality Given Convex Function

Theorem

If f:X→R is a convex function where X as a convex subset of R and f(0)≤0 then for all a,b∈X

f(a+b)≥f(a)+f(b).
Proof

Let f:X→R be a convex function as above. Then consider for any a,b∈X that because f is convex,

aa+bf(0)+ba+bf(a+b)≥f(aa+b(0)+ba+b(a+b))=f(b)

given that

aa+b+ba+b=a+ba+b=1.

Because f(0)≤0, we then have the inequality

ba+bf(a+b)≥f(b).

Interchanging the role of a and b because of symmetry we get

aa+bf(a+b)≥f(a)

and then adding these two inequalities we acquire

f(a+b)=aa+bf(a+b)+ba+bf(a+b)≥f(a)+f(b).
Theorem

If f:X→R is a concave function where X is a convex subset of R and f(0)≥0 then for all a,b∈X

f(a+b)≤f(a)+f(b).