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Where $f(x,y,z):=x+z-1\text{ and }g(x,y,z):=y^2-xz$

I tried to prove the convexity of the set $X$ by showing it is the intersection of two convex sets. $f(x,y,z) = 0$ is a hyperplane and $g(x,y,z)\le 0$ is a half-space. Since both of these types of sets are convex, the intersection is convex. But I think this pseudo-proof is not specific enough. Should I try to prove that $f(x,y,z)$ and $g(x,y,z)$ are convex functions?

clathratus
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intrepid
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  • What do you mean by 'half-space'? Drawing the graph of $g$ doesn't look like it cuts $\mathbb R^3$ in half. – WishofStar Oct 28 '20 at 05:02
  • The equation $$f(x,y,z)=0$$ defines a plane, which is clearly convex. The equation $$g(x,y,z)\le 0$$ looks parabolic in planes parallel to xy and yz planes, but hyperbolic in planes parallel to xz plane. – Math Keeps Me Busy Oct 28 '20 at 05:18

1 Answers1

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Let $(x,y,z)\in X$. Then since $z=1-x$, $y^2-x(1-x)=x^2-x+y^2\le 0$. Hence $$\left(x-\dfrac{1}{2}\right)^2+y^2\le 1, z=1-x$$ implies that $X$ is actually convex.

WishofStar
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