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MathJax: \langle, \rangle
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edit approved Jun 19, 2017 at 9:30
Martin Sleziak
edit approved Jun 19, 2017 at 9:30
Martin Sleziak
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For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}<\xi,x>.$$$$h_K(\xi):=\sup_{x\in K}\langle\xi,x\rangle.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$.

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}<\xi,x>.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$.

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}\langle\xi,x\rangle.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$.

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

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asv
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When minimum of two supporting functionals of convex bodies is convex?

deleted 1 character in body
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asv
asv
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For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}<\xi,x>.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex (inconvex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$).

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}<\xi,x>.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex (in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$).

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}<\xi,x>.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function.

Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$.

Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex?

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asv
asv
  • 21.8k
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  • 121