Difference between sup and inf of $|f(x)|$ less than difference between sup and inf of $f(x)$

Question

If $A$ is a non empty subset of the reals and $f$ is a bounded function from $A$ to the reals, how can we show that:

$\sup|f(x)| - \inf|f(x)| \le \sup(f(x)) - \inf(f(x))$?

I started by stating that since $f$ is bounded, $\inf(f(x)) \le f(x) \le \sup(f(x))$. And then $|f(x)| \le \max\{|\sup(f(x))|,|\inf(f(x))|\}$.

So $\sup|f(x)| = \max\{|\sup(f(x))|,|\inf(f(x))|\}$ and $\inf|f(x)| = \max\{\min\{|\sup(f(x))|,|\inf(f(x))|\},0\}$.

I feel like my $\inf|f(x)|$ is wrong though, and I don't know where to get it.

Any help would be really appreciated!

Answer 1

Hint:

LHS= $\sup \{|f(x)|-|f(y)|\}$

RHS=$\sup\{f(x)- f(y)\}= \sup\{|f(x) - f(y)|\}$

Answer 2

1. If $\inf f(x)\geq 0$ then $f(x)$ is never negative so $|f(x)|=f(x)$ for all $x,$ so $\sup f(x) =\sup |f(x)|$ and $\inf f(x)=\inf |f(x)|.$

2. If $\sup f(x)\leq 0$ then $f(x)$ is never positive so $|f(x)|=-f(x)$ for all $x,$ so $$\sup f(x)=\sup |f(x)|=\sup (-f(x))=-\inf f(x)$$ $$\text { and also }\quad \inf |f(x)|=\inf (-f(x))=-\sup f(x).$$

3. If $\sup f(x)>0>\inf f(x)$ then let $A=\{x:f(x)\geq 0\}$ and $B=\{x:f(x)<0\}.$

Now if $\sup_{x\in A}f(x)\geq \sup_{x\in B} (-f(x))$ then $$\sup f(x)=\sup_{x\in A}f(x)=\sup |f(x)|\geq \inf |f(x)|\geq 0>\inf f(x).$$

Or if $\sup_{x\in A}<\sup _{x\in B}(-f(x))$ then $-\inf f(x)=\sup_{x\in B}(-f(x)) =\sup |f(x)|$. So we have $$\sup |f(x)|-\inf |f(x)|=-\inf f(x)-\inf |f(x)|\leq -\inf f(x)<-\inf f(x)+\sup f(x).$$ The last inequality holds because $\sup f(x)>0$ by hypothesis.