Suprema and Infima of Combined Bounded Functions/Bounded Below

Theorem

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

Let $c$ be a constant.

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 $\map f x$.

Proof

First we show that:

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

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

Then:

\(\ds \map {\inf_{x \mathop \in S} } {\map f x + c}\)

\(=\)

\(\ds \map {\inf_{y \mathop \in T} } {y + c}\)

\(\ds \)

\(=\)

\(\ds c + \map {\sup_{y \mathop \in T} } y\)

Infimum Plus Constant

\(\ds \)

\(=\)

\(\ds c + \map {\inf_{x \mathop \in S} } {\map f x}\)

Next we show that:

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

Let:

$\ds H = \map {\inf_{x \mathop \in S} } {\map f x}$

$\ds K = \map {\inf_{x \mathop \in S} } {\map g x}$

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.

$\blacksquare$

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.