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Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Assume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Assume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Assume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I tried to find a counterexample but all my attempts have ended up with $f$ convex.

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Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Furthermore, assumeAssume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Furthermore, assume that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Assume also that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

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Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Furthermore, assume that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n\geq 3$$n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Furthermore, assume that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n\geq 3$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

Consider a function $f:\mathbb{R}^n_{\geq 0}\rightarrow \mathbb{R}$ that is separately convex, i.e. such that $\frac{d^2f}{dx_i^2}\geq 0$ for all $i\in \{1,\dots n\}$. Furthermore, assume that $f$ is decreasing in each of its arguments and is positively homogenous of degree 1 so that $f(tx)=tf(x)$ for all $t\geq 0$. Is $f$ quasiconvex?

According to this article, if $n=2$ then $f$ is convex (so also quasiconvex) but this result does not generalize to $n>2$. I have tried to find a counterexample but all my attempts have ended up with $f$ convex.

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