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How to prove the following set is convex? $$\{(x,y,z)|x^2+y^2-z^2 \leqslant 0,z \geqslant 0\}$$

I try to make it by prove that $$[\theta x_1 + (1-\theta)x_2]^2 + [\theta y_1 + (1-\theta)y_2]^2 - [\theta z_1 + (1-\theta)z_2]^2 \leqslant 0$$, but then stop at $$\theta^2(x_1^2+y_1^2-z_1^2)+(1-\theta)^2(x_2^2+y_2^2-z_2^2)-2\theta(1-\theta)(x_1x_2+y_1y_2-z_1z_2) \leqslant 0$$.

How can I continue or is there some other way to do this?

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$$\mathit{S} = \{(x,y,z)|x^2+y^2-z^2 \leqslant 0,\ z \geqslant 0\}$$ $$ \implies \mathit{S} = \{(x,y,z)|\sqrt{x^2+y^2}\leqslant z,\ z \geqslant 0\}$$ $$ \implies \mathit{S} = \{(a,z)| \|a\|_2\leqslant z,\ z \geqslant 0\}$$ Lest say, $(a_1,z_1) ,(a_2,z_2) \in \mathit{S}.$ So we have to show the following $$(\theta a_1 + (1-\theta)a_2,\theta z_1 + (1-\theta)z_2) \in \mathit{S} \text{ (where, } \theta \in [0,1]).$$ $$\|\theta a_1 + (1-\theta)a_2\|_2\leqslant \|\theta a_1\| + \|(1-\theta)a_2\|_2 \leqslant\theta z_1+(1-\theta) z_2.$$ $$(1-\theta),\theta, z_1,z_2 \geq 0 \implies \theta z_1+(1-\theta) z_2 \geq 0.$$

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This set is the epigraph of the $2$-norm, which is a convex function. Hence, this set is convex.

By the way, this set is called the "second-order cone" or the "ice-cream cone". (More precisely, this is the ice-cream cone in $\mathbb R^3$.)

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  • $\begingroup$ Thank u! But can this be a prove? Would u please supply some details? $\endgroup$ – Oswin Sep 22 '15 at 9:57
  • $\begingroup$ This is a proof, if you have already established that: 1) every norm is a convex function; 2) the epigraph of a convex function is a convex set. $\endgroup$ – littleO Sep 22 '15 at 10:27

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