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I have to show whether the following set is convex subset of $\mathbb{R^5}$ or not.

$$S(x_1,x_2,x_3,x_4)=\{ (x_1^2+x_2^2+x_3^2+x_4^2+u_1^2+u_2^2,x_3,x_4,u_1,u_2) \mid -1\leq u_1 \leq 1, -1\leq u_2\leq 1 \}$$

for fixed $x_1$, $x_2$, $x_3$ and $x_4$.

Up to $\mathbb{R}^3$, one way to show the convexity of any set is by plotting its graph, but how to check for sets like above which is in $\mathbb{R}^5$?

Rodrigo de Azevedo
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Phoenix8128
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  • How do you define that for $\mathbb{R}^3$? Could you also put it in set builder notation? – gist076923 Sep 29 '22 at 14:07
  • $@$gst076923 I mean for any general set in $\mathbb{R^3}$, we can plot it's graph and check the convexity. I am not defining the above set in $\mathbb{R^3}$. – Phoenix8128 Sep 29 '22 at 14:39
  • Are you familiar with the definition of convex sets? – gist076923 Sep 29 '22 at 14:58
  • $@$ gst076923 Yes I am. Do you want me to write the definition? – Phoenix8128 Sep 29 '22 at 15:05
  • I would suggest attempting to prove it via the definition. Also, I would like to throw some doubt at the claim, as in my attempt to prove it, I think that you could come up with a counterexample. I will leave it to you to prove or disprove – gist076923 Sep 29 '22 at 15:12
  • $@$gst076923 Here's how I tried. Since $x_1$, $x_2$, $x_3$ and $x_4$ are fixed, to make things simpler, I can just replace the term $x_1^2+x_2^2+x_3^2+x_4^2$ by some scalar say $a$ and $x_3$, $x_4$ by $b$ and $c$ respectively.Now to show the above set is convex, I take two elements in above set say $s_1$ =$(a_1+u_1^2+u_2^2,b_1,c_1,u_1,u_2)$ and $s_2$=$(a_2+v_1^2+v_2^2,b_2,c_2,v_1,v_2)$ where $|u_1| \leq 1, |u_2| \leq 1, |v_1| \leq 1$ and $|v_2| \leq 1$, and then I need to show $ps_1 +(1-p)s_2 \in S$ where $0 \leq p \leq 1$. Is this how should I proceed? – Phoenix8128 Oct 01 '22 at 06:34
  • I just realized that for fixed $x_1, x_2,x_3$ and $x_4$, I don't need to vary the replaced scalars $a, b$ and $c$. So can I just take $s_1 = (u_1,u_2)$ and $s_2=(v_1,v_2)$, in which case its easy to show that the given set is convex. – Phoenix8128 Oct 01 '22 at 06:56

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