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:

• $\displaystyle \sup_{x \in S} \left({f \left({x}\right) + c}\right) = c + \sup_{x \in S} \left({f \left({x}\right)}\right)$
• $\displaystyle \sup_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \le \sup_{x \in S} \left({f \left({x}\right)}\right) + \sup_{x \in S} \left({g \left({x}\right)}\right)$

where $\displaystyle \sup \left({f \left({x}\right)}\right)$ is the supremum of $f \left({x}\right)$.

Bounded Below

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

Then:

• $\displaystyle \inf_{x \in S} \left({f \left({x}\right) + c}\right) = c + \inf_{x \in S} \left({f \left({x}\right)}\right)$
• $\displaystyle \inf_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \ge \inf_{x \in S} \left({f \left({x}\right)}\right) + \inf_{x \in S} \left({g \left({x}\right)}\right)$

where $\displaystyle \inf \left({f \left({x}\right)}\right)$ is the infimum of $f \left({x}\right)$.

Proof

Proof for Bounded Above

• First we show that $\displaystyle \sup_{x \in S} \left({f \left({x}\right) + c}\right) = c + \sup_{x \in S} \left({f \left({x}\right)}\right)$:

Let $T = \left\{{f \left({x}\right): x \in S}\right\}$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sup_{x \in S} \left({f \left({x}\right) + c}\right)\)	\(=\)	\(\displaystyle \sup_{y \in T} \left({y + c}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle c + \sup_{y \in T} \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Supremum Plus Constant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle c + \sup_{x \in S} \left({f \left({x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• Next we show that $\displaystyle \sup_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \le \sup_{x \in S} \left({f \left({x}\right)}\right) + \sup_{x \in S} \left({g \left({x}\right)}\right)$:

Let:

• $\displaystyle H = \sup_{x \in S} \left({f \left({x}\right)}\right)$
• $\displaystyle K = \sup_{x \in S} \left({g \left({x}\right)}\right)$

Then:

$\forall x \in S: f \left({x}\right) + g \left({x}\right) \le H + K$

Hence $H + K$ is an upper bound for $\left\{{f \left({x}\right) + g \left({x}\right): x \in S}\right\}$.

The result follows.

$\blacksquare$

Proof for Bounded Below

This follows exactly the same lines.

• First we show that $\displaystyle \inf_{x \in S} \left({f \left({x}\right) + c}\right) = c + \inf_{x \in S} \left({f \left({x}\right)}\right)$:

Let $T = \left\{{f \left({x}\right): x \in S}\right\}$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \inf_{x \in S} \left({f \left({x}\right) + c}\right)\)	\(=\)	\(\displaystyle \inf_{y \in T} \left({y + c}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle c + \sup_{y \in T} \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Infimum Plus Constant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle c + \inf_{x \in S} \left({f \left({x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• Next we show that $\displaystyle \inf_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \ge \inf_{x \in S} \left({f \left({x}\right)}\right) + \inf_{x \in S} \left({g \left({x}\right)}\right)$:

Let:

• $\displaystyle H = \inf_{x \in S} \left({f \left({x}\right)}\right)$
• $\displaystyle K = \inf_{x \in S} \left({g \left({x}\right)}\right)$

Then:

$\forall x \in S: f \left({x}\right) + g \left({x}\right) \ge H + K$

Hence $H + K$ is a lower bound for $\left\{{f \left({x}\right) + g \left({x}\right): x \in S}\right\}$.

The result follows.

$\blacksquare$

Note

The equality versions of the inequalities stated do not apply in general.

Let us take as an example:

• $S = \left[{-1 .. 1}\right]$
• $f \left({x}\right) = x$
• $g \left({x}\right) = -x$

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

Then:

• $\displaystyle \sup_{x \in S} \left({f \left({x}\right)}\right) = \sup_{x \in S} \left({g \left({x}\right)}\right) = 1$
• $\displaystyle \inf_{x \in S} \left({f \left({x}\right)}\right) = \inf_{x \in S} \left({g \left({x}\right)}\right) = -1$

So:

• $\displaystyle \sup_{x \in S} \left({f \left({x}\right)}\right) + \sup_{x \in S} \left({g \left({x}\right)}\right) = 2$
• $\displaystyle \inf_{x \in S} \left({f \left({x}\right)}\right) + \inf_{x \in S} \left({g \left({x}\right)}\right) = - 2$

However:

$\forall x \in S: f \left({x}\right) + g \left({x}\right) = x + \left({-x}\right) = 0$

So:

$\displaystyle \sup_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) = \inf_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) = 0$

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

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 7.16 \ (6)$