# How to prove the set $\{(x,y,z)\in \mathbb R^3|z\geq 0,x^2+y^2\leq z^2\}$ is convex?

How to prove the set $$S=\{(x,y,z)\in \mathbb R^3|z\geq 0,x^2+y^2\leq z^2\}$$ is convex?

So to prove this I took $$(x,y,z),(x_1,y_1,z_1)\in S$$ and $$\lambda\in (0,1).$$

Then $$x^2+y^2\leq z^2$$ and $$x_1^2+y_1^2\leq z_1^2$$ and $$z,z_1\geq 0.$$

Now to prove $$\{\lambda x+(1-\lambda)x_1\}^2+\{\lambda y+(1-\lambda)y_1\}^2\leq \{\lambda z+(1-\lambda)z_1\}^2$$ , it's coming like $$\lambda^2(x^2+y^2-z^2)+(1-\lambda)^2(x_1^2+y_1^2-z_1^2)+2\lambda(1-\lambda)(xx_1+yy_1-zz_1)\leq 0.$$

So from here, I was trying to prove $$(xx_1+yy_1-zz_1)\leq 0$$ because the former terms are already negative in the above inequality.

I'm not sure whether $$(xx_1+yy_1-zz_1)\leq 0$$ is true or not?

Can we prove the inequality $$xx_1+yy_1-zz_1\leq 0$$? Or there is another way to prove the convexity of the set $$S$$?

Any help is appreciated. Thank you.

## 4 Answers

Yes, you can show the inequality $$x x_1+y y_1 \leq z z_1.$$ It is sufficient to show the inequality $$(x x_1 + y y_1)^2 \leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $$z, z_1 \geq 0$$.) To show the squared form, multiply the inequalities $$x^2 + y^2 \leq z^2$$ $$x_1^2 + y_1^2 \leq z_1^2$$ together. Then use the inequality $$a^2 + b^2 \geq 2ab$$ with for $$a=x y_1$$ and $$b= x_1y$$ . This gives $$z^2 z_1^2 \geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 \geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$

Note that $$(x,y,z)\in S\implies \lambda (x,y,z)\in S$$ since $$(\lambda x)^2+(\lambda y)^2=\lambda^2(x^2+y^2)\le \lambda^2z^2=(\lambda z)^2$$

Thus $$S$$ is a cone.

Prove that a cone is convex if and only if it is closed under addition.

Then you see that your condition $$xx_1+yy_1\le zz_1$$ is the closure under addition of $$(x,y,z)$$ and $$(x_1,y_1,z_1)$$.

This one is obtained using $$ab\le \frac {a^2+b^2}2$$ since both $$\begin{cases}x^2+y^2\le z^2\\{x_1}^2+{y_1}^2\le{z_1}^2\end{cases}$$ are verified for points of $$S$$.

Here we have $$xx_1 + yy_1 \le \sqrt{(x^2+y^2)(x_1^2+y_1^2)} \le \sqrt{z^2}\sqrt{z_1^2}= zz_1\,.$$ The final equality relies on $$z,z_1 \ge 0$$.

The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $$xx_1+yy_1 < 0$$.)

$$x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 \ge 0 \\ x^2y_1^2 + y^2x_1^2 \ge 2xyx_1y_1 \\ x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 \ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \\ (x^2+y^2)(x_1^2+y_1^2) \ge (xx_1+yy_1)^2$$

Let us consider the usual norm $$\|\vec a\|=\sqrt{a_1^2+a_2^2}$$ on $$\mathbb R^2$$

Notice also that the condition describing $$S$$ is exactly $$\sqrt{x^2+y^2} \le z$$, so $$S=\{(x,y,z); z\ge0, \|(x,y)\|\le z\}.$$ So now if we have two points $$(x_1,y_1,z_1),(x_2,y_2,z_2)\in S$$, then we get for any $$\lambda\in(0,1)$$ $$\|(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2)\| \overset{(*)}\le \lambda\|(x_1,y_1)\| + (1-\lambda)\|(x_2,y_2)\| \le \lambda z_1+(1-\lambda)z_2.$$

The inequality $$(*)$$ follows from triangle inequality applied to vectors $$(\lambda x_1,\lambda y_1)$$ and $$((1-\lambda)x_2,(1-\lambda)y_2)$$. But we can view it also as convexity of the function $$(x,y)\mapsto\|(x,y)\|$$. See, for example Why is every $$p$$-norm convex? and other posts linked there.

We have shown that $$\|(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2)\| \le \lambda z_1+(1-\lambda)z_2,$$ which means that also the convex combination $$(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2,\lambda z_1+(1-\lambda)z_2)$$ belongs to $$S$$.

Notice that similar approach would work to prove convexity of $$S=\{(x_1,\dots,x_{n+1}); x_{n+1}\ge 0, \sum_{k=1}^n x_k^2 \le x_{n+1}^2\}$$ in $$\mathbb R^{n+1}$$. (This condition can be equivalently stated as $$\|(x_1,\dots,x_n)\|\le\sqrt{x_{n+1}}$$.)