The Maximum and Minimum Functions of Two Functions

# The Maximum and Minimum Functions of Two Functions

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 $x \in I$ by $\max (f, g)(x) = \max \{ f(x), g(x) \}$, and similarly, the Minimum Function of $f$ and $g$ is defined for all $x \in I$ by $\min (f, g)(x) = \min \{ f(x), g(x) \}$. 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) $\min (f + h, g + h) = \min (f, g) + h$.
• Proof of a) Let $f$, $g$, and $h$ be defined on $I$.
• 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. $\blacksquare$