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A multivariate function $f:{\mathbb R}^d \to {\mathbb R}$ is quasi-convex on a convex set $S \subset {\mathbb R}^d$ if $f(\lambda z + (1-\lambda)z') \leq \max\{f(z),f(z')\}$ for all $z,z' \in S$ and $0 \leq \lambda \leq 1$. Equivalently $f$ is quasi-convex on $S$ if $f^{-1}((-\infty,a]) \cap S$ is convex for all real $a$.

For a real-valued positive exponent $\beta$, if $\beta \geq 2$ then the function of two variables $f(x,y) = x^\beta / y$ is not just quasi-convex but actually convex, for positive $x,y$. But $f$ is not convex for $0 < \beta < 2$. But it is obvious that $f$ is still at least quasi-convex for positive $x,y$ when $\beta = 1$, so intuitively $f$ should be quasi-convex for any $\beta \geq 1$ as well. Is this true and how can we show it?

user2566092
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1 Answers1

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I realized I can use the fact that a function $f$ is quasi-convex if and only if $g(f)$ is quasi-convex for some increasing function $g$. So I can consider $g(w) = w^n$ for a sufficiently large positive integer $n$ and get $g(f(x,y)) = x^\alpha / y^\gamma$ where $\alpha > \gamma - 1$. Then it is easy to show that $g(f(x,y))$ is convex by computing the Hessian, hence $g(f(x,y))$ is quasi-convex, so $f(x,y)$ is quasi-convex as well.

user2566092
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