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Let $g^1\colon\mathbb{R}\to\mathbb{R}$ and $g^2\colon\mathbb{R}\to\mathbb{R}$ be concave functions, and let $f\colon\mathbb{R}\to\mathbb{R}$ be a non-decreasing function (i.e., $f(x)≥f(y)$ whenever $x≥y$).

Let $h\colon\mathbb{R}^2\to\mathbb{R}$ be defined by: $h(x_1,x_2)=f(g^1(x_1)+g^2(x_2)).$

How do I prove that $h$ is quasi-concave?

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    Prove that for all $x_1,x_2,x_1',x_2'$ and $0\leq\alpha\leq 1$: $h(\alpha(x_1,x_2)+(1-\alpha)(x_1',x_2'))\geq\min \{f(x_1,x_2),f(x_1',x_2')\}$.2011-10-09

2 Answers 2

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Note that if $\{x:g(x)>\alpha\}$ is convex for all $\alpha\in\mathbb{R}$, then $\{x:g(x)\ge\alpha\}=\bigcap\limits_{\beta<\alpha}\{x:g(x)>\beta\}$ is also convex.

Since $f$ is non-decreasing, $h$ is quasi-concave if $g^1(x_1)+g^2(x_2)$ is quasi-concave; i.e. either $ \small\{(x_1,x_2):f(g^1(x_1)+g^2(x_2))>\alpha\}=\{(x_1,x_2):g^1(x_1)+g^2(x_2)>\inf\{x:f(x)>\alpha\}\}\tag{1} $ or $ \small\{(x_1,x_2):f(g^1(x_1)+g^2(x_2))>\alpha\}=\{(x_1,x_2):g^1(x_1)+g^2(x_2)\ge\inf\{x:f(x)>\alpha\}\}\tag{2} $ where $(1)$ holds when $f(\inf\{x:f(x)>\alpha\})\le\alpha$ and $(2)$ holds otherwise.

Since both $g_1(x_1,x_2)=g^1(x_1)$ and $g_2(x_1,x_2)=g^2(x_2)$ are concave from $\mathbb{R}^2\to\mathbb{R}$, $g^1(x_1)+g^2(x_2)$ is also concave, hence quasi-concave.

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There is an alternative. A function is quasi-concave if all super-level sets of it are convex. A super-level set for a function $z(x)$ is defined as $S_\alpha(z) = \{ x\, | \,z(x) > \alpha\}.$

Therefore we consider $S_\alpha(h) = \{ (x_1,x_2)\, | \,h(x_1,x_2) > \alpha\} = \{(x_1,x_2)\, |\, f(g^1(x_1)+g^2(x_2)) > \alpha\}$

and since $f$ is a non-decreasing function so we have $S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$ Since according to problem statement, $g^1$ and $g^2$ are concave functions, so sum of them is also concave which have the pleasant indication that $S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$ is a convex set and therefore $h$ is a quasi-concave function.