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:
 * $$\sup_{x \in S} \left({f \left({x}\right) + c}\right) = c + \sup_{x \in S} \left({f \left({x}\right)}\right)$$;
 * $$\sup_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \le \sup_{x \in S} \left({f \left({x}\right)}\right) + \sup_{x \in S} \left({g \left({x}\right)}\right)$$

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

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

Then:
 * $$\inf_{x \in S} \left({f \left({x}\right) + c}\right) = c + \inf_{x \in S} \left({f \left({x}\right)}\right)$$;
 * $$\inf_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) \ge \inf_{x \in S} \left({f \left({x}\right)}\right) + \inf_{x \in S} \left({g \left({x}\right)}\right)$$

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

Proof for Bounded Above

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

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

Then:

$$ $$ $$


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

Let:
 * $$H = \sup_{x \in S} \left({f \left({x}\right)}\right)$$;
 * $$K = \sup_{x \in S} \left({g \left({x}\right)}\right)$$.

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 $$\inf_{x \in S} \left({f \left({x}\right) + c}\right) = c + \inf_{x \in S} \left({f \left({x}\right)}\right)$$:

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

Then:

$$ $$ $$


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

Let:
 * $$H = \inf_{x \in S} \left({f \left({x}\right)}\right)$$;
 * $$K = \inf_{x \in S} \left({g \left({x}\right)}\right)$$.

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 does not apply.

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:


 * $$\sup_{x \in S} \left({f \left({x}\right)}\right) = \sup_{x \in S} \left({g \left({x}\right)}\right) = 1$$;
 * $$\inf_{x \in S} \left({f \left({x}\right)}\right) = \inf_{x \in S} \left({g \left({x}\right)}\right) = -1$$.

So:


 * $$\sup_{x \in S} \left({f \left({x}\right)}\right) + \sup_{x \in S} \left({g \left({x}\right)}\right) = 2$$;
 * $$\inf_{x \in S} \left({f \left({x}\right)}\right) + \inf_{x \in S} \left({g \left({x}\right)}\right) = - 2$$.

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

So $$\sup_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) = \inf_{x \in S} \left({f \left({x}\right) + g \left({x}\right)}\right) = 0$$ and it is immediately obvious that the equality does not hold.