Prove that $M_{|f|}-m_{|f|} \le M_f-m_f$

Question

I want to prove this statment:

Let $f:[a,b] \to \mathbb{R}$ be a bounded function.

Prove that if $f$ is integrable in $[a,b]$ then $|f|$ is also integrable in $[a,b]$ - HINT: first prove that if $M_f=\sup(f(x):x \in [a,b])$ and $m_f=\inf(f(x):x \in [a,b])$, then $M_{|f|}-m_{|f|} \le M_f-m_f$

Unfortunately I have been trying to prove the hint for over than 3 hours now with no luck.

Any help will be amazing!!

Thanks!

Answer 1

The triangle inequality gives $$|f(x)|\le|f(x)-f(x')|+|f(x')|$$ with $x,x' \in [a,b]$.

Case 1: $\;f(x) \ge f(x')$

Then $\;|f(x)|\le f(x)-f(x')+|f(x')|\le M_f-f(x')+|f(x')|\;$ for every $x\;\; $ ($x'$ fixed)

So $\;M_{|f|}\le M_f-f(x')+|f(x')|$

and then $\;|f(x')|\ge M_{|f|}-M_f+f(x')\ge M_{|f|}-M_f+m_f\;$ for every $x'$

It follows that $\;m_{|f|}\ge M_{|f|}-M_f+m_f$

Case 2: $\;f(x) < f(x')$

Then $\;|f(x)|\le f(x')-f(x)+|f(x')|\le f(x')-m_f+|f(x')|\;$ for every $x\;\;$ ($x'$ fixed)

So $\;M_{|f|}\le f(x')-m_f+|f(x')|$

and then $\;|f(x')|\ge M_{|f|}-f(x')+m_f\ge M_{|f|}-M_f+m_f\;$ for every $x'$

It follows that $\;m_{|f|}\ge M_{|f|}-M_f+m_f$

Answer 2

$(I)$ If $M_f \geq 0$ and $m_f \geq 0$ then $M_{|f|}=M_f$ and $m_{|f|}=m_f$. We have a equality.

$(II)$ If $M_f \geq 0$ but $m_f < 0$ then $M_{|f|}=M_f$ and $m_{|f|}\geq m_f$. In this case $M_{|f|}-m_{|f|} \leq M_f-m_f$.

$(III)$ If $M_f,~m_f<0$ note that $|m_f|=M_{|f|}$ and $|M_f|=m_{|f|}$. We get $M_{|f|}-m_{|f|}=|m_f|-|M_f|=-m_f-(-M_f)=M_f-m_f$.