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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

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related: math.stackexchange.com/q/108393/21047 –  Phillip Cloud Aug 22 '13 at 23:55

1 Answer 1

up vote 9 down vote accepted

We want to prove that for $x, y \in \Omega$, $(g \circ f)\left(\lambda x + (1 - \lambda) y\right) \le \lambda (g \circ f)(x) + (1 - \lambda)(g \circ f)(y)$.

We have: \begin{align} (g \circ f)\left(\lambda x + (1 - \lambda) y\right) &= g\left(f\left(\lambda x + (1 - \lambda) y\right)\right) \\ &\le g\left(\lambda f(x) + (1 - \lambda) f(y)\right) & \text{(} f \text{ convex and } g \text{ nondecreasing)} \\ &\le \lambda g(f(x)) + (1 - \lambda)g(f(y)) & \text{(} g \text{ convex)} \\ &= \lambda (g \circ f)(x) + (1 - \lambda)(g \circ f)(y) \end{align}

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is there a corresponding result for g is nonincreasing and convex? presumably not –  Lost1 May 18 '13 at 0:09
    
@Lost1 Probably for nonincreasing and concave, $f$ still being convex. –  AlexR Dec 9 at 15:13
    
@Lost1, there are actually four such rules, for each combination of convex/concave inner and outer functions: convex-nondecreasing & convex -> convex, convex-nonincreasing & concave -> convex, concave-nondecreasing & concave -> concave, concave-nonincreasing & convex -> concave. –  Michael Grant Dec 9 at 17:46

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