Suprema and Infima of Combined Bounded Functions
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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 7.16 \ (6)$