Suprema and Infima of Combined Bounded Functions

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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 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:


 * 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.

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:


 * 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.

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.