3

Is there a straightforward way to determine if the following set is convex?

$$Z = \left\{x\in\mathbb{R}^2:3x_1^4-x_1x_2+x_2^4\le x_2,x_1>2,x_2>2\right\}$$

I know I can try by manipulation of linear combination of two points of this set Is there any simpler way than using the definition $$f(x_1, y_1)<0 \wedge f(x_2, y_2)<0 \Rightarrow f(t x_1+(1-t)x_2, t y_1+(1-t)y_2)<0\;$$ $$\text{ for }\; 0 < t < 1?$$ enter image description here

enter image description here

Rodrigo de Azevedo
  • 1
4pie0
  • 481
  • maybe you can use something like "the intersection of convex sets is again convex" – Quickbeam2k1 Mar 05 '13 at 20:00
  • I think this is not convex – 4pie0 Mar 05 '13 at 20:01
  • did you try to differentiate twice the function that determines your first relation? – Quickbeam2k1 Mar 05 '13 at 20:05
  • 1
    I'm removing the [set-theory] tag once more as this is not related at all. It would be just as appropriate to use the tag [number-theory], since you mentioned $2$, $3$, and $4$. Please do not add it again. The tag is for technical questions in the area of set theory, to which yours does not belong. "This tag is for set theory topics typically studied at the advanced undergraduate or graduate level. These include cofinality, axioms of ZFC, axiom of choice, forcing, set-theoretic independence, large cardinals, models of set theory, ultrafilters, ultrapowers, constructible universe, inner models" – Andrés E. Caicedo Apr 06 '14 at 02:31

2 Answers2

3

The set is empty, because $$(x_1+1)x_2\le 2x_1x_2\le x_1^2+x_2^2<3x_1^4+x_2^4$$ in the range $x_1,x_2>2$. The empty set is usually considered convex.

  • yes, I have seen that this is empty, but shoudn't we take the sum of these two sets? I think this meant to be sum – 4pie0 Mar 05 '13 at 20:15
  • 1
    @cf16 No. The notation ${x: P,Q,R}$ always means the set of points $x$ that satisfy the conditions $P$ and $Q$ and $R$. If this notation is misused here to mean or, then the set is still not convex: it contains $(0,10)$ and $(10,0)$, but not $(5,5)$. –  Mar 05 '13 at 20:19
  • yes, I have also another example. I think I will show both explanations, in any case – 4pie0 Mar 05 '13 at 20:34
1

yes, so if this is empty, this is convex by definition. if this was meant to be sum of two sets then my solution is take one point that belongs to set of function solutions and one from other set, we see $f(X1)<0$ so it is solution of $f$, and $X2=(2.2,2.2)$ so it belongs to the second set, now $f(\alpha X_1+(1-\alpha)X_2)>0$ so it doesn't belong to f solutions and also not to the second set as $X=(1.105,1.35)$

    x        y       f
X1  0,01    0,5 -0,44249997
X2  2,2     2,2   86,6624

alfa   0,5  0,5 

X   1,105 1,35   4,952462402
4pie0
  • 481