The Maximum and Minimum Functions of Two Functions

Table of Contents

Later on we will use two very important functions in moving forward with Lebesgue integration theory. We define these functions below.

Definition: Let

$f$

and

$g$

be two functions defined on the interval

$I$

. Then the Maximum Function of

$f$

and

$g$

is defined for all

x \in I

\max (f, g)(x) = \max \{ f(x), g(x) \}

, and similarly, the Minimum Function of

$f$

and

$g$

is defined for all

x \in I

\min (f, g)(x) = \min \{ f(x), g(x) \}

The following theorems state some elementary properties of these functions.

Theorem 1: Let

$f$

and

$g$

be two functions defined on the interval

$I$

. Then for all

x \in I

\min (f, g) \leq \max (f, g)

Proof: Suppose not, i.e., suppose that there exists an $x \in I$ such that $\min (f, g)(x) \geq \max (f, g)(x)$ . Then:

(1)

\begin{align} \quad \min \{ f(x), g(x) \} \geq \max \{ f(x), g(x) \} \end{align}

But this is a contradiction. So $\min (f, g) \leq \max (f, g)$ for all $x \in I$ . $\blacksquare$

Theorem 2: Let

$f$

and

$g$

be two functions defined on the interval

$I$

. Then

\max (f, g) + \min (f, g) = f + g

Proof: Let $x \in I$ . Then $\max (f, g)(x) = \max \{ f(x), g(x) \}$ and $\min (f, g)(x) = \min \{ f(x), g(x) \}$ .

Case 1: Suppose that $$f(x) > g(x)$$ . Then $\max \{ f(x), g(x) \} = f(x)$ and $\min \{ f(x), g(x) \} = g(x)$ , so $\max (f, g) + \min (f, g) = f + g$ .

Case 2: Suppose that $$f(x) = g(x)$$ . Then $\max \{ f(x), g(x) \} = f(x) = g(x)$ and $\min \{ f(x), g(x) \} = f(x) = g(x)$ , so $\max (f, g) + \min (f, g) = f + g$ .

Case 3: Suppose that $$f(x) < g(x)$$ . Then $\max \{ f(x), g(x) \} = g(x)$ and $\min \{ f(x), g(x) \} = f(x)$ , so $\max (f, g) + \min (f, g) = f + g$ .

In all of the three cases we conclude that $\max (f, g) + \min (f, g) = f + g$ . $\blacksquare$

Theorem 3: Let

$f$

$g$

, and

$h$

be functions defined on the interval

$I$

. Then:
a)

\max (f + h, g + h) = \max (f, g) + h

.
b)

\min (f + h, g + h) = \min (f, g) + h

Case 1: Suppose that $$f(x) > g(x)$$ . Then $$f(x) + h(x) > g(x) + h(x)$$ . So $\max \{ f(x) + h(x), g(x) + h(x) \} = f(x) + h(x)$ . But then $\max (f + h, g + h) = \max (f, g) + h$ .

Case 2: Suppose that $$f(x) = g(x)$$ . Then $$f(x) + h(x) = g(x) + h(x)$$ . So $\max \{ f(x) + h(x), g(x) + h(x) \} = f(x) + h(x) = g(x) + h(x)$ . But then $\max (f + h, g + h) = \max (f, g) + h$ .

Case 3: Suppose that $$f(x) < g(x)$$ . Then $$f(x) + h(x) < g(x) + h(x)$$ . So $\max \{ f(x) + h(x), g(x) + h(x) \} = g(x) + h(x)$ . But then $\max (f + h, g+ h) = \max (f, g) + h$ .

In all three cases we conclude that $\max (f + h, g + h) = \max (f, g) + h$ . An analogous proof can be used to show that b) is also true. $\blacksquare$

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