# Thread: f bounded above implies -f bounded below

1. ## f bounded above implies -f bounded below

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

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

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

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

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

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

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

If it's not correct, what's the way to do this?
Looks fine to me.