How can we prove the inequality for the supremum and infimum of f*g and f*g?

[Math Amateur]

I am reading J. J. Duistermaat and J. A. C. Kolk: Multidimensional Analysis Vol.II Chapter 6: Integration ...

I need help with the proof of Theorem 6.2.8 Part (iii) ...The Definition of Riemann integrable functions with compact support and Theorem 6.2.8 and a brief indication of its proof reads as follows:

The definition of supremum and infimum are given in the following text from D&K Vol. I ...

I cannot locate D&K's definition of sup and inf for functions so I am taking the definition from Joseph L. Taylor's book, "Foundations of Analysis".

Taylor's definition reads as follows:

If \(\displaystyle f: X \to \mathbb{R}\) is a real-valued function and \(\displaystyle A \subset X \) ... ... ...

... then we define ..

\(\displaystyle \text{ sup}_B = \text{sup} \{ f(x) \ | \ x \in B \} \)

and

\(\displaystyle \text{ inf}_B = \text{inf} \{ f(x) \ | \ x \in B \} \)
I need help to formulate a detailed, formal and rigorous proof that \(\displaystyle \text{ sup}_B \ fg - \text{ inf}_B \ fg \leq \text{ sup}_B \ f \text{ sup}_B \ g \ - \ \text{ inf}_B \ f \text{ inf}_B \ g \)I have been unable to make a meaningful start on this proof ...Help will be much appreciated ...

Peter

 

[Opalg]

For every $x\in B$, $f(x) \leqslant \sup_B f$ and $g(x) \leqslant \sup_B g$. Therefore $(fg)(x) = f(x)g(x) \leqslant \sup_B f \sup_B g$. Now take the sup over $B$ to get $\sup_B fg \leqslant \sup_B f \sup_B g$. A similar argument shows that $\inf_B fg \geqslant \inf_B f \inf_B g$ and so $-\inf_B fg \leqslant -\inf_B f \inf_B g$.

 

[Math Amateur]

Thanks Opalg …

… very much appreciate your help …

Just reflecting on what you have written …

Peter