Prove that $\omega(|f|,[a,b]) \le \omega(f,[a,b])$

Question

Definition :

$\omega(f,[a,b]):=\sup\{f(x):x \in [a,b]\} - \inf \{f(x):x \in [a,b] \}$

Question :

Assume that $f$ is a bounded function on $[a,b]$.

(i) Prove that $\omega(|f|,[a,b]) \le \omega(f,[a,b])$

(ii) Assume that $M$ is a bound of $f$. ( Meaning $\forall x \space |f(x)| \le M$ ) Prove that :

$\omega(f^2,[a,b]) \le 2M\omega(f,[a,b])$

(iii) Assume that $m \gt 0$ is a lower bound of $f$. Prove that :

$\omega(\frac{1}{f},[a,b]) \le \frac{\omega(f,[a,b])}{m^2}$

Note 1 : About part (i), I know that the difference between absolute values is less than the previous values, i mean, i have the image in my mind, but i don't know how to write in a formal way.

Note 2 : About part (ii) & (iii), I'm completely out of ideas. What's the relation between $f^2$ and $2M\omega(f,[a,b])$?! Or between $\frac{1}{f}$ and $\frac{\omega(f,[a,b])}{m^2}$ !

Answer 1

(i) You have $||f(x)| - |f(y)| | \le |f(x)-f(y)|$, hence $|f(x)| - |f(y)| \le ||f(x)| - |f(y)| | \le \omega(f,[a,b])$ for all $x,y \in [a,b]$, from which the desired result follows.

(ii) Note that $f^2(x)-f^2(y) = (f(x)-f(y))(f(x)+f(y)) \le 2M (f(x)-f(y))$.

(iii) Note that ${1 \over f(x)} - {1 \over f(y)} = {f(y)-f(x) \over f(x)f(y)} \le {f(y)-f(x) \over m^2}$.