# The composition of two convex functions is convex

Let $f$ be a convex function on a convex domain $\Omega$ and $g$ a convex non-decreasing function on $\mathbb{R}$. prove that the composition of $g(f)$ is convex on $\Omega$. Under what conditions is $g(f)$ strictly convex.

My attempt, since $f$ is convex, $$f([1-t]x_0 +ty_0)\le [1-t]f(x_0) + tf(y_0)\:,\quad t \in [0,1] \,\text{and} \: x_0,y_0\in \Omega$$ Since $g$ is convex $$g([1-s]x_1 +sy_1) \le [1-s]g(x_1) + sg(y_1)\:,\quad s \in [0,1]\:and \: x_1,y_1 \in \mathbb{R}$$ So $$g([1-s]f([1-t]x_2 +ty_2) +sf([1-t]x_2 +ty_2)) \\\le [1-s]g([1-t]f(x_2) + tf(y_2)) + sg([1-t]f(x_3) + tf(y_3))\: for\:x_2,y_2,x_3,y_3 \in \Omega.$$ Im not sure if this is always true.

Any help would be appreciated. Thanks

We want to prove that for $x, y \in \Omega$, $(g \circ f)\left(\lambda x + (1 - \lambda) y\right) \le \lambda (g \circ f)(x) + (1 - \lambda)(g \circ f)(y)$.
• @Lost1 Probably for nonincreasing and concave, $f$ still being convex. – AlexR Dec 9 '14 at 15:13
• How does one prove the first rule, namely that for $f$ convex-nondecreasing and $g$ convex, then $g\circ f$ is convex? – user17982 Jul 15 '16 at 7:09