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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

Escualo
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bobdylan
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1 Answers1

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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}

Ayman Hourieh
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  • is there a corresponding result for g is nonincreasing and convex? presumably not – Lost1 May 18 '13 at 00:09
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    @Lost1 Probably for nonincreasing and concave, $f$ still being convex. – AlexR Dec 09 '14 at 15:13
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    @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 09 '14 at 17:46
  • How does one prove the first rule, namely that for $f$ convex-nondecreasing and $g$ convex, then $g\circ f$ is convex? –  Jul 15 '16 at 07:09
  • @MichaelGrant you mean outer and inner function, the first one is outer and the second one is inner? – Pew Oct 07 '21 at 23:49
  • @Pew, the other way around, e.g. $\log x^{-1}$ is a concave function of a convex, nonincreasing function (example 2). It is convex. – Jake Stevens-Haas Oct 28 '22 at 20:21