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:

\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. $\blacksquare$