A question about to infimum and Supermum on a bounded set

Question

show that for any bounded set $E \subset R$ we have that $$\sup_{x \in E} x - \inf_{x \in E} x \geq \sup_{x \in E} |x| - \inf_{x \in E} |x|$$

How to start this problem

Is using this can we say that $|f|$ is Riemann integrable when $f$ is Riemann integrable

Answer 1

For simplicity of notation, let $$ \sup E := \sup\{ x : x \in E\} \qquad\text{and}\qquad \sup|E| := \sup\{ |x| : x\in E\}, $$ with similar definition for the infimum.

There is probably a really slick way of proving this, but if you don't know where to start, go naive. A naive approach is to think about where $0$ fits in relation to $\sup E$ and $\inf E$, the try to make hay with that. There are only three places it can go:

1. Suppose that $\sup E \ge \inf E \ge 0$. Then $$ \sup|E| = \sup E \quad\text{and}\quad \inf|E| = \inf E.$$ From this, it immediately follows that $$\sup|E| - \inf|E| = \sup E - \inf E,$$ which is the desired result.
2. Suppose that $0 \ge \sup E \ge \inf E$. Then $$ \sup|E| = |\inf E| \quad\text{and}\quad \inf|E| = |\sup E|,$$ as the absolute value reverses the ordering of the set (do you see why?). Since $\inf E \le 0$, it follows from the definition of the absolute value that $|\inf E| = -\inf E$. Similarly, $|\sup E| = -\sup E$. But then \begin{align*} \sup|E| - \inf|E| &= |\inf E| - |\sup E| \\ &= (-\inf E) - (-\sup E) \\ &= \sup E - \inf E, \end{align*} which is the desired result.
3. Suppose that $\sup E \ge 0 \ge \inf E$. Then \begin{align} \sup|E| &= \max\{ \sup E, |\inf E| \} \\ &\le \sup E + |\inf E| && \text{(sum of two positive numbers exceeds their max)} \\ &= \sup E - \inf E. && \text{(since $\inf E < 0$)} \end{align} The left-hand side is only made smaller by subtracting a nonnegative quantity, from which it follows that $$ \sup|E| - \inf|E| \le \sup E - \inf E, $$ which is the desired result.

As these three cases are exhaustive, we are done!

Answer 2

It is a good idea to observe that $\sup_{x \in E} x -\inf_{y \in E} y$ is the diameter D of the set $E$ which is defined as $\sup_{x,y \in E} |x-y|$. To see this start with $x-y \leq |x-y|$. Take sup over x and inf over y to see that $\sup_{x \in E} x -\inf_{y \in E} y \leq D$. To prove the reverse inequality just consider that cases when $|x-y|=x-y$ and $|x-y| =y-x$. Once you have this formula you only have to show that diameter of $\{|x|:x\in E\}$ does not exceed D and this follows from the inequality $|x-y| \geq ||x|-|y||$. Yes, the inequality does tell you that if f is Riemann integrable then so is $|f|$.