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$$\sup |f(x)| -\inf |f(x)| \ge \sup f(x) -\inf f(x).$$

How can you show that it is indeed true?Sup is the lowest upper bound and inf is the greatest. lower bound

user91500
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user128576
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2 Answers2

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Take $f(x)=\left\{ \begin{array}{ll} 1, & \hbox{if $x\geq 0$;} \\ -1, & \hbox{if $x<0$.} \end{array} \right.$ Then $\sup |f(x)|=\inf |f(x)|=1$ and $\sup f(x)=1$ and $\inf f(x)=-1$. This shows that your inequality is not true in general.

Paul
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We have $\sup |f(x)| \geq \sup f(x)$ since $|f(x)| \geq f(x)$ for all $x$.

But then the overall inequality does not hold since $\inf|f(x)|$ is larger than (or equal to) $\inf f(x)$ since for example $|f(x)|$ is always $\geq 0$ whereas $\inf f(x)$ might be negative.

user50948
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  • what if we know that $f(x)$ is integrable on $[a,b]$ and bounded?is the inequality correct then?Just asking becouse i need to use it in a proof – user128576 Jun 14 '14 at 11:02