Proving $\{(x,y)\in{\mathbb{R}^2}:|x|^2+|y|<1\}$ is convex

Solution 1:

\begin{align} A=\{(x,y)\in{\mathbb{R}^2}:|x|^2+|y|<1\} \\ \\ \\= \{(x,y)\in{\mathbb{R}^2}:y<1-x^2\} \cap \{(x,y)\in{\mathbb{R}^2}:y>-(1-x^2)\}\\ \\ \end{align}

and both of these sets are convex. Since the intersection of convex sets is convex, the result follows.

Solution 2:

First, the obvious simplification: $|x|^2 = x^2$.

So we want to prove that if $x^2+|y|<1$ and $a^2+|b|<1$ then $(t x+(1-t)a)^2 + |ty+(1-t) b| <1$

Consider the two inequalities $$(tx+(1-t)a)^2 \le t x^2 + (1-t)a^2 \tag{1}$$ and $$|ty+(1-t)b|\le t|y|+(1-t)|b| \tag{2}$$

To prove the inequality $(1)$ observe that $$tx^2+(1-t)a^2 - (tx+(1-t)a)^2 = t(1-t)(a-x)^2\ge 0$$

The inequality $(2)$ is just the triangle inequality.

Adding the two inequalities we get $(t x+(1-t)a)^2 + |ty+(1-t) b| \le t(x^2+|y|)+(1-t)(a^2+|b|)< t + (1-t)=1$.