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Example 3.31 Boyd's Book:

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My question is, how can we determine that the function $f$ is quasiconcave/quasiconvex?

Similarly, for $f(x_1, x_2) = x_1/x_2 $ on $R^2_{++}$ is given as both quasiconvex and quasiconcave. However, I cant see why? Can anyone please explain, how to find whether a convex function is quasiconcave or not? Thanks.

jhon_wick
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  • Your question states (correctly) that a function $f$ is quasiconcave if its upper level sets are convex. Perhaps you are asking how to check this? Or are you interested in learning why this is a characterisation? – mlc Mar 03 '17 at 21:48
  • I want to learn how to check whether a function is quasiconcave or not. – jhon_wick Mar 04 '17 at 13:48

1 Answers1

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Here is one way to prove that the proposed function is quasiconcave. (There are alternative approaches.) Using your cue, we need to check that the upper level sets $\{ \mathbf{x}: f(\mathbf{x} \ge \alpha \}$ are convex for all $\alpha$. Given $f(x_1,x_2) = x_1x_2$, this is equivalent to checking that the upper level sets $$\{ \mathbf{x} \in \mathbf{R}^2_+: x_1x_2 \ge \alpha \} = \{ \mathbf{x} \in \mathbf{R}^2_+: x_2 \ge \alpha / x_1\}$$ are convex. This follows because each set $\{ \mathbf{x} \in \mathbf{R}^2_+: x_2 \ge \alpha x_1\}$ is the intersection of two convex sets (the region above the hyperbole $x_2 = \alpha / x_1$ and the positive orthant) in the $(x_1,x_2)$-plane.

Consider now the function $f(x_1,x_2) = x_1/x_2$. Each upper level sets $\{ \mathbf{x} \in \mathbf{R}^2_+: x_1 \ge \alpha x_2\}$ is the intersection of two convex sets (namely, the half-space above the line $x_2 = \alpha x_1$ and the positive orthant) in the $(x_1,x_2)$-plane and thus is a convex set. Then $f$ is quasiconcave. Likewise, the lower level sets $\{ \mathbf{x} \in \mathbf{R}^2_+: x_1 \le \alpha x_2\}$ are also convex for all $\alpha$ and thus $f$ is also quasi convex.

mlc
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  • Considering the function $f(x_1,x_2) = \frac{x_1}{x_2}$ did you mean the half-space above the line $x_1 = \alpha x_2$ instead of $x_2 = \alpha x_1$? Also can you explain to me if while considering the set ${x \in \mathbf{R}_{+}^2 : x_1 \ge \alpha x_2}$ are we supposed to view $x_2$ as some constant? (because if no then shouldn't it be the intersection of infinitely many half-space's i.e. one for every $x_2$ and the positive orthant. In which case it is not guaranteed to be convex.) – user13 Nov 23 '20 at 12:13