Let here be two convex functions: $f(x)$ and $g(x)$ let there be two real numbers: $a$ and $b$ so it is known that: $f(x) + g(x) = ax + b$
Prove that $f(x)$ and $g(x)$ are both affine
*meaning that there are $a_f$,$b_f$,$a_g$,$b_g$, so that:
$f(x) = a_fx+b_f$
$g(x) = a_gx+b_g$
** This has something to do with the fact that a function is both convex and concave IFF it is affine(had to prove it just now)
Consider two real numbers $x$ and $y$, and $\lambda\in[0,1]$ then $$\eqalign{ I_1&=\lambda f(x)+(1-\lambda) f(y)-f(\lambda x+(1-\lambda)y)\geq0\cr I_2&=\lambda g(x)+(1-\lambda) g(y)-g(\lambda x+(1-\lambda)y)\geq0 } $$ But $$ I_1+I_2=\lambda(ax+b)+(1-\lambda)(ay+b)-a(\lambda x+(1-\lambda)y)-b=0 $$ So, we must have $I_1=I_2=0$ that is, for all $(x,y)$ and $\lambda\in[0,1]$ we have $$\eqalign{f(\lambda x+(1-\lambda)y)&= \lambda f(x)+(1-\lambda) f(y)\cr g(\lambda x+(1-\lambda)y)&= \lambda g(x)+(1-\lambda) g(y) } $$ This means that both $f$ and $g$ are affine functions.