Basic Theorems Regarding the Max/Min Functions of Two Functions

# Basic Theorems Regarding the Maximum and Minimum Functions of Two Functions

Recall from The Maximum and Minimum Functions of Two Functions page that if $f$ and $g$ are two functions defined on an interval $I$ then the maximum function of $f$ and $g$ denoted $\max (f, g)$ is defined for all $x \in I$ by:

(1)\begin{align} \quad \max (f, g)(x) = \max \{ f(x), g(x) \} \end{align}

Similarly, the minimum function of $f$ and $g$ denoted $\min (f, g)$ is defined for all $x \in I$ by:

(2)\begin{align} \quad \min (f, g)(x) = \min \{ f(x), g(x) \} \end{align}

We noted three important results regarding these functions. For $f$, $g$, and $h$ as functions defined on $I$ we saw that:

- $\min (f, g) \leq \max (f, g)$ for all $x \in I$.

- $\max (f, g) + \min (f, g) = f + g$.

- $\max (f + h, g + h) = \max (f, g) + h$ and $\min (f + h, g + h) = \min (f, g) + h$.

We will now look at a few more basic results regarding these functions.

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ be two increasing sequences of functions on $I$. Then $(\max (f_n, g_n))_{n=1}^{\infty}$ and $(\min (f_n, g_n))_{n=1}^{\infty}$ are both increasing sequences of functions on $I$. |

**Proof:**Since $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ are both increasing sequences of functions on $I$, then for all $n \in \mathbb{N}$ and for all $x \in I$ we have that both:

\begin{align} \quad f_n(x) \leq f_{n+1}(x) \quad \mathrm{and} \quad g_n(x) \leq g_{n+1}(x) \end{align}

- So $\max \{ f_n(x), g_n(x) \} \leq \max \{ f_{n+1}(x), g_{n+1}(x) \}$ for all $n \in \mathbb{N}$ and for all $x \in I$. So $(\max (f_n, g_n))_{n=1}^{\infty}$ is an increasing sequence of functions on $I$.

- Similarly, we also have that $\min \{ f_n(x), g_n(x) \} \leq \min \{ f_{n+1}(x), g_{n+1}(x) \}$ for all $n \in \mathbb{N}$ and for all $x \in I$. So $(\min (f_n, g_n))_{n=1}^{\infty}$ is an increasing sequnce of functions on $I$. $\blacksquare$

Theorem 2: Let $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ be two increasing sequences of functions on $I$ that converge to $f$ and $g$ (respectively) almost everywhere on $I$. Then $(\max (f_n, g_n))_{n=1}^{\infty}$ and $(\min (f_n, g_n))_{n=1}^{\infty}$ converge to $\max (f, g)$ and $\min (f, g)$ (respectively) almost everywhere on $I$. |

**Proof:**Suppose that $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ converge pointwise to $f$ and $g$ almost everywhere on $I$. Let $D_f$ and $D_g$ be the set of points in $I$ such that the sequences $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ do not converge to $f$ and $g$ respectively, where $m(D_f) = m(D_g) = 0$.

- Since $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$ we have that for all $x \in I \setminus D_f$ and for all $\epsilon > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:

\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \\ \quad f(x) - \epsilon < f_n(x) < f(x) + \epsilon \quad (*) \end{align}

- Similarly, since $(g_n(x))_{n=1}^{\infty}$ converges to $g$ almost everywhere on $I$ we have that for all $x \in I \setminus D_g$ and for all $\epsilon > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:

\begin{align} \quad \mid g_n(x) - g(x) \mid < \epsilon \quad \\ \quad g(x) - \epsilon < g_n(x) < g(x) + \epsilon \quad (**) \end{align}

- Take $N = \max \{ N_1, N_2 \}$. Then both $(*)$ and $(**)$ hold. Moreover, we see that for all $x \in I \setminus (D_f \cup D_g)$ and for all $n \geq N$ that:

\begin{align} \quad \max \{f(x), g(x)\} - \epsilon = \max \{ f(x) - \epsilon, g(x) - \epsilon \} \leq \max \{ f_n(x), g_n(x) \} < \max \{ f(x) + \epsilon, g(x) + \epsilon \} = \max \{f(x), g(x) \} + \epsilon \end{align}

- Hence $\mid \max \{ f_n(x), g_n(x) \} - \max \{ f(x), g(x) \} \mid < \epsilon$. This shows that $(\max (f_n, g_n))_{n=1}^{\infty}$ converges to $\max \{ f, g \}$ increasingly (by Theorem 1).

- Similarly, if we take $N = \max \{ N_1, N_2 \}$, we see that for all $x \in I \setminus D_f \cup D_g)$ and for all $n \geq N$ that:

\begin{align} \quad \min \{f(x), g(x)\} - \epsilon = \min \{ f(x) - \epsilon, g(x) - \epsilon \} \leq \min \{ f_n(x), g_n(x) \} < \min \{ f(x) + \epsilon, g(x) + \epsilon \} = \min\{f(x), g(x) \} + \epsilon \end{align}

- Hence $\mid \min \{ f_n(x), g_n(x) \} - \min \{ f(x), g(x) \} \mid < \epsilon$. This shows that $(\min (f_n, g_n))_{n=1}^{\infty}$ converges to $\min \{ f, g \}$ increasingly (by Theorem 1). $\blacksquare$