The Supremum and Infimum of a Function

# The Supremum and Infimum of a Function

We will now begin to look at some applications of the definition of a supremum and infimum with regards to functions.

 Definition: Let $f : A \to \mathbb{R}$ be a function. Then define the supremum of $f$ to be $\underset{A} \sup f := \sup \{ f(x) : x \in A \} = \sup R(f)$, and define the infimum of $f$ to be $\underset{A} \inf f := \inf \{ f(x) : x \in A \} = \inf R(f)$ where $R(f)$ is the range of $f$.

From the definition above, we acknowledge that the supremum and infimum of a function $f$ pertain to the set $R(f)$ that is the range of $f$. The diagram below illustrates the supremum and infimum of a function:

We will now look at some important theorems.

 Theorem 1: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be functions such that $g$ is bounded above. If $f(x) ≤ g(x)$ for all $x \in A$, then $\underset{A} \sup f ≤ \underset{A} \sup g$.
• Proof: Let $f : A \to \mathbb{R}$. Let $g : A \to \mathbb{R}$ be a function that is bounded above. Since the range of $g$ is nonempty we have that for all $x \in A$, $g(x) ≤ \underset{A} \sup g$. Now since $f(x) ≤ g(x)$ for all $x \in A$, we have that $f(x) ≤ g(x) ≤ \underset{A} \sup G$.
• Furthermore we note that since $f(x) ≤ g(x)$ for all $x \in A$, then $g$ is bounded below by $f$, and so $\underset{A} \sup f ≤ g(x) ≤ \underset{A} \sup G$ and thus $\underset{A} \sup f ≤ \underset{A} \sup g$. $\blacksquare$
 Theorem 2: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be functions such that $f$ is bounded below. If $f(x) ≤ g(x)$ for all $x \in A$, then $\underset{A} \inf f ≤ \underset{A} \inf g$.
• Proof: Let $f : A \to \mathbb{R}$ be a function that is bounded below. Let $g : A \to \mathbb{R}$. Since the range of $f$ is nonempty we have that for all $x \in A$, $\underset{A} \inf f ≤ f(x)$. Now since $f(x) ≤ g(x)$ for all $x \in A$ we have that $\underset{A} \inf f ≤ f(x) ≤ g(x)$.
• Furthermore we note that since $f(x) ≤ g(x)$ for all $x \in A$, then $f$ is bounded above by $g$, and so $\underset{A} \inf f ≤ f(x) ≤ \underset{A} \inf g$ and thus $\underset{A} \inf f ≤ \underset{A} \inf g$. $\blacksquare$
 Theorem 3: Let $f : A \to \mathbb{R}$ be a bounded function and let $a \in \mathbb{R}$. If $a > 0$ then $\underset{A} \sup (af) = a \cdot \underset{A} \sup f$, and $\underset{A} \inf (af) = a \cdot \underset{A} \inf f$. If $a < 0$ then $\underset{A} \sup (af) = a \cdot \underset{A} \inf f$, and $\underset{A} \inf (af) = a \cdot \underset{A} \sup f$.

Theorem 3 immediately follows from the theorems we've already proven on The Supremum and Infimum of The Bounded Set (aS) page where $S = R(f)$.