A function is bounded on a set if its output values are confined between two horizontal lines. This concept captures the idea that the function cannot grow without limit on that set.
Quick Reference
| Concept | Meaning |
|---|---|
| $f$ bounded on $E$ | $\exists M > 0$ such that $ |
| $f$ unbounded on $E$ | No single $M$ can bound $\lvert f(x) \rvert$ for all $x \in E$ |
| Geometric interpretation | Graph of $f$ above $E$ lies between the lines $y = M$ and $y = -M$ |
| "Bounded" without qualifier | Refers to boundedness on the entire domain |
Definition
If $E$ is a subset of the domain of $f$, we say $f$ is bounded on $E$ (by $M$) if there exists a positive number $M$ such that
$|f(x)| < M \quad \text{for all } x \in E.$Otherwise, $f$ is said to be unbounded on $E$.
Geometrically, $f$ is bounded on $E$ when the portion of the graph above $E$ lies entirely between the horizontal lines $y = M$ and $y = -M$ for some $M > 0$.
Examples
The floor function $f(x) = \lfloor x \rfloor$ (which outputs the greatest integer less than or equal to $x$) is bounded on the interval $E = (-1, 1)$.
Solution
For all $x \in (-1, 1)$, the value $\lfloor x \rfloor$ is either $0$ (when $0 \leq x < 1$) or $-1$ (when $-1 < x < 0$). In either case: $|f(x)| \leq 1.$ So $f$ is bounded on $(-1, 1)$ with $M = 1$ (or any $M > 1$). However, $f(x) = \lfloor x \rfloor$ is unbounded on all of $\mathbb{R}$: for any candidate bound $M$, there exist integers $n > M$, so $|f(n)| = n > M$.
Let $g(x) = \dfrac{1}{x}$. Determine whether $g$ is bounded on $[1, \infty)$ and on its full domain $\mathbb{R} - \{0\}$.
Solution
On $[1, \infty)$: Since $x \geq 1$, we have $0 < g(x) = 1/x \leq 1$. Therefore: $|g(x)| \leq 1 \quad \text{for all } x \in [1, \infty).$ So $g$ is bounded on $[1, \infty)$ with $M = 1$.
Let $h(x) = \dfrac{1}{x^2 + 1}$. Show that $h$ is bounded on all of $\mathbb{R}$.
Solution
For all $x \in \mathbb{R}$: $x^2 + 1 \geq 1 \quad \text{(with equality at } x = 0\text{)},$ so $0 < h(x) = \frac{1}{x^2+1} \leq 1.$ Therefore $|h(x)| \leq 1$ for all $x \in \mathbb{R}$, and $h$ is bounded on its entire domain with $M = 1$. As $|x| \to \infty$, the denominator grows without bound, causing $h(x) \to 0$. The maximum value of $h$ is $h(0) = 1$.
Remark: When we say a function is "bounded" or "unbounded" without specifying a set, we refer to its behavior on its entire domain. For instance, $g(x) = 1/x$ is unbounded and $h(x) = 1/(x^2+1)$ is bounded.
