0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thank you very much, Mr. user284331 for very quick answer. $\endgroup$ – tchappy ha Jan 25 '18 at 5:36
  • $\begingroup$ Oh, $\max\{|x_{1}|,|x_{2}|\}\neq\dfrac{x_{1}+x_{2}+|x_{1}-x_{2}|}{2}$ $\endgroup$ – tchappy ha Jan 25 '18 at 5:42
  • 1
    $\begingroup$ Just put all the absolute values to all those terms, it still goes through. $\endgroup$ – user284331 Jan 25 '18 at 5:43
  • 1
    $\begingroup$ What I wished to indicate was that, $\max\{a,b\}=\dfrac{a+b+|a-b|}{2}$, this is the formula. $\endgroup$ – user284331 Jan 25 '18 at 5:47
  • $\begingroup$ Thank you very much again. $\endgroup$ – tchappy ha Jan 25 '18 at 5:47
0
$\begingroup$

$x_k - \max(y_1,...,y_n) \le x_k-y_k \le \|x-y\|_2$ 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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.