f bounded above implies -f bounded below

**Connected** · March 25th 2011, 02:02 PM

Let $f:A\to\mathbb R.$ Show that $f$ is bounded above, then $\underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).$

Okay, so I first showed that $-f$ is bounded below, that's easy, but as for second part, I want to know whether my work is correct or not:

Since $f$ is bounded above, then it has supremum, so $f\le \sup f$ (from this and the rest, sup and inf are taken for $x\in A$ ), and on the other hand $\inf(-f)\le -f,$ since $f$ and $-f$ are bounded above and below, respectively, then it exists a $k\in\mathbb R$ so that $f\le k$ and $-k\le -f,$ namely we have $k=\sup f$ and $\inf(-f)=-k,$ hence $\underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).$

If it's not correct, what's the way to do this?

**Drexel28** · March 25th 2011, 02:25 PM

Originally Posted by Connected

Let $f:A\to\mathbb R.$ Show that $f$ is bounded above, then $\underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).$

Okay, so I first showed that $-f$ is bounded below, that's easy, but as for second part, I want to know whether my work is correct or not:

Since $f$ is bounded above, then it has supremum, so $f\le \sup f$ (from this and the rest, sup and inf are taken for $x\in A$ ), and on the other hand $\inf(-f)\le -f,$ since $f$ and $-f$ are bounded above and below, respectively, then it exists a $k\in\mathbb R$ so that $f\le k$ and $-k\le -f,$ namely we have $k=\sup f$ and $\inf(-f)=-k,$ hence $\underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).$

If it's not correct, what's the way to do this?

Looks fine to me.

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