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!
Best Answer
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 $n \to +\infty$, $M^*-m^* \leq M-m$.