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Prove that $C = \{(x,y) \in \mathbb{R}^2:\max \{ |x|,|y|\}\leq 1 \}$ is a convex set.

I am using the following definition for a convex set

Let $D \subseteq \mathbb{R}^n$. The set $D$ is said to be convex if, given $\bar{x},\bar{y}\in D$, we have that $$(1-t)\bar{x} \ +t\bar{y} \in D $$ for all $t\in [0,1].$

Let $\bar{x}=(x_1,y_1)$ and $\bar{y}=(x_2,y_2)$ be arbitrary points such that $\bar{x},\bar{y} \in C $, and let $t\in [0,1].$ Now, we want to see that $(1-t)\bar{x} \ +t\bar{y} \in C.$ So then \begin{align} (1-t)\bar{x} \ +t\bar{y} &=(1-t)(x_1,y_1)+t(x_2,y_2)\\ &= ((1-t)x_1,(1-t)y_1)+ (tx_2,ty_2)\\ &= ((1-t)x_1+tx_2,(1-t)y_1+ty_2) \end{align}

And here is where I get stuck. I want to prove that $ ((1-t)x_1+tx_2,(1-t)y_1+ty_2)\in C$, which I believe is the same as proving that $\max\{|(1-t)x_1+tx_2|,|(1-t)y_1+ty_2| \}\leq 1$, but I don't know how to proceed.

Any help would be appreciated!

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    $\begingroup$ Use triangle inequality. $\endgroup$ – Youem Apr 13 '18 at 3:57
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So $|(1-t)x_{1}+tx_{2}|\leq(1-t)|x_{1}|+t|x_{2}|\leq(1-t)+t=1$ and $|(1-t)y_{1}+ty_{2}|\leq(1-t)|y_{1}|+t|y_{2}|\leq(1-t)+t=1$ because $|x_{1}|,|x_{2},|y_{1}|,|y_{2}|\leq 1$, this shows that $(1-t)\overline{x}+t\overline{y}\in C$.

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$|(1-t)x_1 + tx_2| \le (1-t)|x_1| + t |x_2| \le (1-t) + t = 1$ the same for $y$

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Let $C_x = \{(x,y) \in \mathbb{R}^2:|x|\leq 1 \}$ and $C_y = \{(x,y) \in \mathbb{R}^2:|y|\leq 1 \}$. Then,

  • $C_x$ and $C_y$ are obviously convex
  • $C = C_x \cap C_y$ is convex, too
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