Suprema and Infima of Combined Bounded Functions

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ and $g$ be real functions.

Let $c$ be a constant.


Bounded Above

Let both $f$ and $g$ be bounded above on $S \subseteq \R$.


Then:

$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$

where $\ds \map \sup {\map f x}$ is the supremum of $\map f x$.


Bounded Below

Let both $f$ and $g$ be bounded below on $S \subseteq \R$.


Then:

$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$

where $\ds \map \inf {\map f x}$ is the infimum of $\map f x$.


Warning

The equalities do not apply in general.

Let us take as an example:

$S = \closedint {-1} 1$
$\map f x = x$
$\map g x = -x$

where $f$ and $g$ are real functions defined on $\R$.

Then:

$\ds \map {\sup_{x \mathop \in S} } {\map f x} = \map {\sup_{x \mathop \in S} } {\map g x} = 1$
$\ds \map {\inf_{x \mathop \in S} } {\map f x} = \map {\inf_{x \mathop \in S} } {\map g x} = -1$

So:

$\ds \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x} = 2$
$\ds \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x} = -2$

However:

$\forall x \in S: \map f x + \map g x = x + \paren {-x} = 0$

So:

$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} = \map {\inf_{x \mathop \in S} } {\map f x + \map g x} = 0$

and it is immediately clear that the equality does not hold.


Sources