Suprema and Infima of Combined Bounded Functions

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

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

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

Then:
 * $\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
 * $\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$

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

Proof for Bounded Above
First we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$:

Let $T = \set {\map f x: x \in S}$.

Then:

Next we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:

Let:
 * $\displaystyle H = \map {\sup_{x \mathop \in S} } {\map f x}$
 * $\displaystyle K = \map {\sup_{x \mathop \in S} } {\map g x}$

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.

Proof for Bounded Below
This follows exactly the same lines.

First we show that:
 * $\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$

Let $T = \set {\map f x: x \in S}$.

Then:

Next we show that:
 * $\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$

Let:
 * $\displaystyle H = \map {\inf_{x \mathop \in S} } {\map f x}$
 * $\displaystyle K = \map {\inf_{x \mathop \in S} } {\map g x}$

Then:
 * $\forall x \in S: \map f x + g \map x \ge H + K$

Hence $H + K$ is a lower bound for $\set {\map f x + \map g x: x \in S}$.

The result follows.

Note
The equality versions of the inequalities stated 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:


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

So:


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

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

So:
 * $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + \map g x} = \map {\inf_{x \mathop \in S} } {\map f x + \map g x} = 0$

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