Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
The problem with your idea is that $m^*\geqslant m$ is true in general but not $M^*\leqslant M$, in fact $M^*\geqslant M$. – Did Aug 15 '12 at 16:06
up vote 2 down vote accepted

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

share|cite|improve this answer
  1. For every bounded function $g:S\to\mathbb R$, $\sup_S g-\inf_S g=K(g)$ where $$ K(g)=\sup\{g(x)-g(y)\,;\,(x,y)\in S\times S\}=\sup\{|g(x)-g(y)|\,;\,(x,y)\in S\times S\}. $$
  2. By the triangular inequality $|f(x)|-|f(y)|\leqslant|f(x)-f(y)|$.

  3. 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.

share|cite|improve this answer

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 $m$.

share|cite|improve this answer
Since $|z|=\max\{z,-z\}$, $\sup |f| = \max\{\sup f,-\inf f\}$. – Siminore Aug 15 '12 at 17:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.