Suprema and Infima of Combined Bounded Functions/Bounded Above
Theorem
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
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 $\map f x$.
Proof
First we show that:
- $\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
\(\ds \map {\sup_{x \mathop \in S} } {\map f x + c}\) | \(=\) | \(\ds \map {\sup_{y \mathop \in T} } {y + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c + \map {\sup_{y \mathop \in T} } y\) | Supremum Plus Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds c + \map {\sup_{x \mathop \in S} } {\map f x}\) |
Next we show that $\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:
Let:
- $\ds H = \map {\sup_{x \mathop \in S} } {\map f x}$
- $\ds K = \map {\sup_{x \mathop \in S} } {\map g x}$
Then:
- $\forall x \in S: \map f x + \map g x \le H + K$
Hence $H + K$ is an upper bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
$\blacksquare$
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): $\S 7.16 \ (6)$