# Prove that $\mathbb{R}^n \ni (x_1, \cdots, x_n) \mapsto \max\{|x_1|, \cdots, |x_n|\} \in \mathbb{R}$ is a contiuous function. Simpler Proof Please.

Prove that $\mathbb{R}^n \ni (x_1, \cdots, x_n) \mapsto \max\{|x_1|, \cdots, |x_n|\} \in \mathbb{R}$ is a contiuous function.

I proved this like the following:

Let $(a_1, \cdots, a_n)$ be any point in $\mathbb{R}^n$.
Let $I := \{i \in \{1, \cdots, n\} | |a_i| = \max\{|a_1|, \cdots, |a_n|\}\}$.

Case 1:

Assume that $I = \{1, \cdots, n\}$.
Let $\epsilon$ be any positive real number.
Let $\delta := \epsilon$.
Let $(x_1, \cdots, x_n)$ be any point in $\mathbb{R}^n$ such that $\sqrt{(x_1-a_1)^2 + \cdots + (x_n-a_n)^2} < \delta$
Let $J := \{j \in \{1, \cdots, n\} | |x_j| = \max\{|x_1|, \cdots, |x_n|\}\}$.
Let $j \in J$.

$$|\max\{|x_1|,\cdots, |x_n|\} - max\{|a_1|, \cdots, |a_n|\}| = |x_j - a_j| \leq \sqrt{(x_1 - a_1)^2 + … (x_n - a_n)^2} < δ = ε.$$

Case 2:

Assume that $I \ne \{1, \cdots, n\}$.
Let $i \in I$.
Let $j \in \{1, \cdots, n\} - I$.
Let $\epsilon$ be any positive real number.
Let $\delta := \min\{\epsilon, \frac{|a_i| - |a_j|}{3}\}$.
Let $(x_1, \cdots, x_n)$ be any point in $\mathbb{R}^n$ such that $\sqrt{(x_1-a_1)^2 + \cdots + (x_n-a_n)^2} < \delta$
Let $J := \{j \in \{1, \cdots, n\} | |x_j| = \max\{|x_1|, \cdots, |x_n|\}\}$.

$$|x_i - a_i| \leq \sqrt{(x_1 - a_1)^2 + … (x_n - a_n)^2} < δ.$$ $$|x_j - a_j| \leq \sqrt{(x_1 - a_1)^2 + … (x_n - a_n)^2} < δ.$$ $$|a_i| - \delta \leq |x_i| \leq |a_i| + \delta.$$ $$|a_j| - \delta \leq |x_j| \leq |a_j| + \delta.$$

So,

$$|x_j| \leq |a_j| + \delta < |a_i| - \delta \leq |x_i|.$$

So,

$$J \subset I.$$

Let $k$ be any element of $J$.

Then,

$$|a_k| = \max\{|a_1|, \cdots, |a_n|\}.$$ $$|x_k| = \max\{|x_1|, \cdots, |x_n|\}.$$

So,

$$|\max\{|x_1|, \cdots, |x_n|\} - \max\{|a_1|, \cdots, |a_n|\}| = ||x_k| - |a_k|| \leq |x_k - a_k| \leq \sqrt{(x_1 - a_1)^2 + \cdots + (x_n - a_n)^2} < \delta \leq \epsilon$$

I want a simpler proof.

By induction. For $n=2$, $\max\{|x_{1}|,|x_{2}|\}=\dfrac{|x_{1}|+|x_{2}|+||x_{1}|-|x_{2}||}{2}$, the function $(x_{1},x_{2})\rightarrow||x_{1}|-|x_{2}||$ is continuous, so is $(x_{1},x_{2})\rightarrow |x_{1}|+|x_{2}|$ is continuous, then the sum of these two functions is also continuous.

For higher $n$, note that $\max\{|x_{1}|,...,|x_{n-1}|,|x_{n}|\}=\max\{\max\{|x_{1}|,...,|x_{n-1}|\},|x_{n}|\}$.

More details: The map $A:u\rightarrow|u|$ is continuous. The map $B:(x_{1},x_{2})\rightarrow x_{1}$ is continuous, then so is $A\circ B$. And the continuity of $(x_{1},x_{2})\rightarrow||x_{1}|-|x_{2}||$ is deduced in similar steps.

• Thank you very much, Mr. user284331 for very quick answer. – tchappy ha Jan 25 '18 at 5:36
• Oh, $\max\{|x_{1}|,|x_{2}|\}\neq\dfrac{x_{1}+x_{2}+|x_{1}-x_{2}|}{2}$ – tchappy ha Jan 25 '18 at 5:42
• Just put all the absolute values to all those terms, it still goes through. – user284331 Jan 25 '18 at 5:43
• What I wished to indicate was that, $\max\{a,b\}=\dfrac{a+b+|a-b|}{2}$, this is the formula. – user284331 Jan 25 '18 at 5:47
• Thank you very much again. – tchappy ha Jan 25 '18 at 5:47

$x_k - \max(y_1,...,y_n) \le x_k-y_k \le \|x-y\|_2$ for all $k$. Now choose the $k$ for which $\max(x_1,...,x_n)$ is attained to get $\max(x_1,...,x_n) - \max(y_1,...,y_n) \le\|x-y\|_2$. Switching the roles of $x,y$ shows that $|\max(x_1,...,x_n) - \max(y_1,...,y_n)| \le\|x-y\|_2$ from which (Lipschitz) continuity follows.

Since $\max(|x_1|,...,|x_n|)= \max(x_1,-x_1,...,x_n,-x_n)$, it follows that $x \mapsto \max(|x_1|,...,|x_n|)$ is continuous.

Depending on how much you're willing to take for granted, the following works:

Note that $||x|| \doteq \max (|x_1|,...,|x_n|)$ is a norm on $\mathbb{R}^n$. All norms on finite dimensional vector spaces are equivalent, i.e. they induce the same topology. $x \mapsto ||x||$ is always continuous in the topology induced by $|| \cdot ||$, but this is the same topology as the standard one.