Min and max of a product.

Let $x_i\in X_i\subset \mathbb{R}$ so that each $X_i$ is a compact set with no isolated points for all $1\leq i\leq n$.

Let:

$a_i\in X_i$ so that $|x_i|\leq|a_i|$ for all $x_i\in X_i$.

$b_i = \max\lbrace x_i \in X_i\rbrace$.

$c_i = \min\lbrace x_i \in X_i\rbrace$.

$f:\mathbb{R}^n\mapsto\mathbb{R}$ so that:

$$f(x_1,x_2,\cdots,x_n) = \prod_{i=1}^nx_i$$

Also let $$A = \prod_{i=1}^na_i$$

Then obviously:

If $A > 0\ $ (i.e. there's an even amount of negative $a_i$) then $A = \max f$

If $A < 0\ $ (i.e. there's an odd amount of negative $a_i$) then $A = \min f$.

How to find $\max f$ if $A <0$ and $\min f$ if $A > 0$ ?

1 Answer

The maximum of $|f|$ is realized by taking $x_i = |a_i| = \max\{|b_i|,|c_i|\}$ and, as you noticed, is given by $A$.

Now it is clear that to get the maximum of $f$ you must choose $x_i \in \{b_i,c_i\}$ (an extremal point) because the function $f$ is multilinear and hence intermediate points give intermediate values. So you can actually assume that $X_i = \{c_i, b_i\}$

First case. Suppose that $s(b_i) = s(c_i)$ for all $i$, where $s(x)$ is the sign of $x$. In this case the problem is easily solved... If the number of negative signs is even, the maximum is reached by taking $x_i = b_i$ if $|b_i|>|c_i|$ and $x_i=c_i$ otherwise. If you want the minimum take the point with minimum absolute value. If the number of negative signs is odd you get maximum and minimum exchanged.

Second case. if at least for one $i$ you have $s(b_i)=-s(c_i)$ you consider the point $x_i\in \{b_i,c_i\}$ which realize the maximum of $|f|$ (i.e. take the maximum between $|b_i|$ and $|c_i|$). If in this point $|f|=f$ you have found the maximum of $f$. Otherwise $|f|=-f$ but you have the possibility to change an odd number of $x_i$ so that the sign of $f$ changes. If you change the value of $x_i$ from, say, $b_i$ to $c_i$ the value of $f$ diminishes by a factor $|c_i|/|b_i|$ which is always $\le 1$. So the best choice is to change a single point and precisely choose $i$ so that $s(b_i)=-s(c_i)$ and $|c_i|/|b_i|$ is as close as possible to 1. A similar reasoning applies to the minimum.

alternative construction (presumably more clear)

As noted above you can suppose that $X_i = \{ b_i, c_i\}$. Moreover you can possibly change the sign of some $X_i$ so to obtain $c_i = |c_i| > |b_i|$ for all $i$. If you change the sign to an even number of $X_i$ then $\max f$ and $\min f$ are preserved, otherwise $\max f$ and $\min f$ are exchanged.

In this situation one obviously has $\max f = \max |f| = \prod c_i = A$. The point is to find $\min f$.

First case. If $b_i\ge 0$ for all $i$ then, obviously, $\min f = \prod b_i$.

Second case. If $b_i<0$ for some $i$, then take the index $i$ such that $b_i<0$ and $|b_i|/c_i$ (which is $\le 1$) is maximum. Then $\min f = \frac{b_i}{c_i}\prod c_i < 0$. This is justified by the fact that in this case the minimum is negative so we must choose $x_i = b_i<0$ for an odd number of $i$s. Since any choice of $b_i$ instead of $c_i$ decreases the absolute value of the product by a factor $|b_i|/c_i$ (which is less than 1) it is convenient to make only one change of sign and to choose the index so as to maximize such a factor.