Question

prove the following equality:
sup(f) = -inf(-f).

Answer 1

.

$\sup(f)=-\inf(-f).$

Task. Let $X$ be a set and $f:X\to\mathbb{R}$ be a function. Prove the following equality:

$\sup(f)=-\inf(-f).$

Solution. By definition $\sup(f)$ is a number such that $f(x)\leq\sup(f)$ for all $x\in X$, and for every $\varepsilon>0$ there exists $x\in X$ such that

$\sup(f)<f(x)+\varepsilon.$

Similarly, $\inf(f)$ is the number such that $\sup(f)\leq f(x)$ for every $x\in X$, and for every $\varepsilon>0$ there exists $x\in X$ such that

$f(x)-\varepsilon<\inf(f).$

In particular, if we know that

$A\leq f(x)\leq B$

for all $x\in A$, then

$A\leq\inf(f)\leq\sup(f)\leq B.$

Now we can deduce from latter property that

$\sup(f)=-\inf(-f).$

First we show that

$\sup(f)\leq-\inf(-f).$

We have that $f(x)\leq\sup(f)$ for all $x\in X$, which is equivalent to

$-f(x)\geq-\sup(f)$

whence

$-f(x)\geq\inf(-f)\geq-\sup(f)$

and so

$-\inf(-f)\leq\sup(f).$ (1)

Conversely, we have that $\inf(-f)\leq-f(x)$ for all $x\in X$, which is equivalent to

$-\inf(-f)\geq f(x)$

whence

$-\inf(-f)\geq\sup(f)\geq f(x)$

and so

$-\inf(-f)\geq\sup(f).$ (2)

Therefore from (1) and (2) we obtain that

$\sup(f)=-\inf(-f).$