Showing that $ \sup_{x \in A}(|f(x)|-|g(x)|)=\sup_{x\in A} |f(x)|-\inf_{x\in A}|g(x)|$

Question

Let $f,g$ be real, bounded functions. I'm not sure that this equation was true or if there exists a counter-example.

For proving that $$\sup_{x \in A}\big(|f(x)|-|g(x)|\big) \,=\, \sup_{x\in A} |f(x)|-\inf_{x\in A}|g(x)|,$$ I use the definition of $\sup$ and $\inf$, but i don't have the result. I prove only that $$\sup_{x \in A}\big(|f(x)|-|g(x)|\big)\,\leq\,\sup_{x\in A} |f(x)|-\inf_{x\in A}|g(x)|.$$

Answer 1

\begin{align} \sup_{x \in A}(|f(x)|-|g(x)|)&=\sup_{x\in A} |f(x)|-\inf_{x\in A}|g(x)|\end{align}

Let, $h= |f|=|g|$ and $A=I\subset \Bbb{R} \text { an interval }$

Then, \begin{align} \sup_{x \in I}(h(x)-h(x))&=\sup_{x\in I} h(x)-\inf_{x\in I} h(x)\end{align}

$0=\sup_{x\in I} h(x)-\inf_{x\in I} h(x)$ $0=\omega_h(I)$

Now, choose a function $h:I\to \Bbb{R}$ such that $\omega_h(I) \neq 0$

Does this provides a counter example?

$h: [0, 1]\to \Bbb{R}$ defined by

$$h(x)=\begin{cases} 1 &\text{ if } x\in [0,\frac{1}{2}] \\ 2 &\text{ if } x\in (\frac{1}{2},1]\end{cases}$$

Then, \begin{align} \omega_h(I) &=\sup_{x\in I} h(x)-\inf_{x\in I} h(x)\\&=2-1\\&=1\end{align}