An inequality involving the $\sup$ and $\inf$ of a bounded function and its absolute value 
Let $f:[a,b]\to\mathbb{R}$ be a bounded function and set $$M=\sup_{[a,b]}f(x)\,,\; m=\inf_{[a,b]}f(x)\,,\;M^*=\sup_{[a,b]}|f(x)|\,,\;m^*=\inf_{[a,b]}|f(x)|\,.$$ Prove that $M^*-m^*\le M-m$.

First, since $f$ is bounded, all of $M$, $M^*$, $m$, $m^*$ are finite.
Next, we have $M^*-m^*\le M-m$ if and only if $$0\le M-m+m^*-M^*=(M-M^*)+(m^*-m)\,,$$ so it suffices to show that $M-M^*\ge 0$ and $m^*-m\ge 0$.
We obviously have $f\le |f|$, so it follows that $\inf f\le \inf|f|$; that is, $m\le m^*$. So $0\le m^*-m$. 
But I can't figure out the other inequality. Thanks!
 A: *

*For every bounded function $g:S\to\mathbb R$, $\sup_S g-\inf_S g=K(g)$ where 
$$
K(g)=\sup\{g(x)-g(y)\,;\,(x,y)\in S\times S\}=\sup\{|g(x)-g(y)|\,;\,(x,y)\in S\times S\}.
$$

*By the triangular inequality $|f(x)|-|f(y)|\leqslant|f(x)-f(y)|$.

*Applying 1. to $g=f$ and to $g=|f|$, and 2., yields 
$$
K(|f|)\leqslant\sup\{|f(x)-f(y)|\,;\,(x,y)\in S\times S\}=K(f).
$$
Edit: Another, shorter, approach is to note that $t\mapsto|t|$ is 1-Lipschitz.
A: For any real $x$ and $y$, there holds $||x|-|y|| \leq |x-y|$. If $|f(x_n)| \to M^*$ and $|f(y_n)| \to m^*$, then $||f(x_n)|-|f(y_n)|| \leq |f(x_n)-f(y_n)|$. But $|f(x_n)-f(y_n)| \leq M-m$, and therefore, letting $n \to +\infty$, $M^*-m^* \leq M-m$.
A: Unless I'm missing something, this is far simpler than the other answers have made it.
Notice that $M^*$ is either $|M|$ or $|m|$. In fact, it's either $M$ or $-m$. Considering $-f$ if necessary (it's not hard to see that if the statement holds for $f$ it also holds for $-f$ and vice versa), let's say $M^*=M$. Then the inequality is $m \le m^*$, which follows from the definition of $m$.
