Is it true that $\max(-x_{(1)},x_{(n)})=\max_{1\le i\le n}|x_i|$?

Question

Let $x_1,x_2,\cdots,x_n$ be a set of $n$ observations where $x_i\in(-a,a)\,,i=1,2,\cdots,n$ for some $a>0$. Suppose $x_{(1)}<x_{(2)}<\cdots<x_{(n)}$ are the ordered observations. Is it true that $$\max(-x_{(1)},x_{(n)})=\max_{1\le i\le n}|x_i|\quad?$$

I suspect this is true, and it might be obvious. But I could not put together a rigorous proof.

The reason I felt it is true was that if $-a<x_1,x_2,\cdots,x_n<a$, then $x_{(1)}>-a$ and $x_{(n)}<a$, together implying $a>\max(-x_{(1)},x_{(n)})$.

Again, $-a<x_1,\cdots,x_n<a\implies 0<|x_1|,\cdots,|x_n|<a\implies \max|x_i|<a$.

So we have $a>\max(-x_{(1)},x_{(n)})$ and $a>\max|x_i|$.

But I cannot conclude that both lower bounds of $a$ are equal from here.

We certainly have $$\max(-x_{(1)},x_{(n)})=\frac{1}{2}\left(-x_{(1)}+x_{(n)}+|-x_{(1)}-x_{(n)}|\right)$$

I would also like to know what happens if I include the terminal points, i.e. what if $x_i\in[-a,a]$ for all $i$ ? Will the result still hold?

Answer 1

As $-x_1\le |x_1|$ and $x_n\le |x_n|$, certainly $$ \max\{-x_1,x_n\}\le\max_i|x_i|.$$ There exists $k$ such that $\max_i|x_i|=|x_k|=\max\{-x_k,x_k\}$. From the given ordering, $x_1\le x_k\le x_n$ and $-x_1\ge -x_k\ge -x_n$, hence $$\max_i|x_i|=|x_k|=\max\{-x_k,x_k\}\le \max\{x_n,-x_1\}.$$

Answer 2

Perhaps something to get you started?

From the fact that your sequence is monotone increasing, we have $\max{\{{x_n}\}} = x_n$. As a consequence, we have $-x_n < -x_{n-1} < ... < -x_1 = \{{-x_n}\}$ implying $\max{\{{-x_n}\}} = -x_1$.

Answer 3

The largest $x_k$ is obviously $x_n$, and the largest of the $-x_k$ is minus the smallest, $-x_1$.

Then the largest among the $x_k$ and $-x_k$ is the largest of $x_n,-x_1$.

Answer 4

$$x_1<x_2\implies \max(x_1,x_2)=x_2.$$

$$-x_1>-x_2\implies \max(-x_1,-x_2)=-x_1.$$

$$|x_k|=\max(x_k,-x_k)\implies\max(|x_1|,|x_2|)=\max(\max(x_1,-x_1),\max(x_2,-x_2)) \\=\max(\max(x_1,x_2),\max(-x_1,-x_2))=\max(x_2,-x_1).$$

These deductions generalize to arbitrary $n$, as the intermediate elements play no role.