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Boundedness of Functions

Definition

Boundedness of a Function

Bounded function: If there exists a constant $M$ such that for every $x \in D$ , $|f(x)| \leq M$ , then $f(x)$ is said to be bounded on $D$ .

Classification

Bounded above: if there exists a constant $M$ such that for every $x \in D$ , $f(x) \leq M$
Bounded below: if there exists a constant $m$ such that for every $x \in D$ , $f(x) \geq m$
Bounded: both bounded above and bounded below

Geometric Meaning

The graph of a bounded function lies between two horizontal lines—there is a horizontal strip that contains the entire graph.

Examples

$f(x) = \sin x$ is bounded on $\mathbb{R}$ because $|\sin x| \leq 1$
$f(x) = x^2$ is bounded below (by 0) on $\mathbb{R}$ but not bounded above
$f(x) = \frac{1}{x}$ is unbounded on $(0, 1]$

$\mathbb{R}$ (double-struck R): This is the standard math symbol for the set of real numbers. The blackboard-bold font is used specifically to distinguish sets from ordinary variables.

How to Test Boundedness

Direct method: inspect the expression and estimate the range of values
Derivative method: for continuous functions, locate extreme points using the derivative
Inequality method: use known inequalities to bound the expression

The derivative method may be unfamiliar—that is fine; you will meet it when you study calculus.

Important Properties

Operations on Bounded Functions

Operations on bounded functions

\text{The sum, difference, and product of two bounded functions are still bounded}

证明

Let $f(x)$ and $g(x)$ be bounded, so there exist constants $M_1, M_2$ with $|f(x)| \leq M_1$ and $|g(x)| \leq M_2$ .

Sum: $|f(x) + g(x)| \leq |f(x)| + |g(x)| \leq M_1 + M_2$
Difference: $|f(x) - g(x)| \leq |f(x)| + |g(x)| \leq M_1 + M_2$
Product: $|f(x) \cdot g(x)| \leq M_1 \cdot M_2$

Hence addition, subtraction, and multiplication all preserve boundedness.

Bounded Function Times a Constant

A bounded function times a constant

\text{A bounded function times a constant is still bounded}

证明

Let $f(x)$ be bounded, so there exists a constant $M$ with $|f(x)| \leq M$ , and let $k$ be any constant.

Proof: $|k \cdot f(x)| = |k| \cdot |f(x)| \leq |k| \cdot M$

Therefore $k \cdot f(x)$ is bounded, with a new bound of $|k| \cdot M$ .

The Boundedness Theorem

Boundedness theorem

\text{A continuous function on a closed interval is necessarily bounded}

If $f(x)$ is continuous on the closed interval $[a, b]$ , then $f(x)$ is bounded on $[a, b]$ ; that is, there exists a constant $M > 0$ such that for all $x \in [a, b]$ , $|f(x)| \leq M$ .

证明

Proof: By contradiction.

Assume $f(x)$ is continuous on $[a, b]$ but unbounded on $[a, b]$ .

Then for every arbitrarily large positive number $M$ , there is some $x_M \in [a, b]$ with $|f(x_M)| > M$ .

Take $M = 1, 2, 3, \dots$ , producing a sequence $\{x_n\}$ with $x_n \in [a, b]$ and $|f(x_n)| > n$ .

Since $[a, b]$ is closed and bounded, $\{x_n\}$ has a convergent subsequence; say $\{x_{n_k}\}$ converges to $x_0 \in [a, b]$ .

Because $f$ is continuous at $x_0$ , $\lim_{k \to \infty} f(x_{n_k}) = f(x_0)$ .

But on the other hand $|f(x_{n_k})| > n_k \to +\infty$ , contradicting the existence of the limit.

Therefore the assumption is false, so $f(x)$ is bounded on $[a, b]$ .

Alternative proof (using the extreme value theorem):

A continuous function on a closed interval attains a maximum and a minimum, so its values have an upper bound and a lower bound—hence the function is bounded.

Exercises

Exercise 1

Determine whether $f(x) = \frac{x}{x^2 + 1}$ is bounded on $\mathbb{R}$ .

Answer and Explanation(3 个标签)

boundednessrangeinequality

Approach: Analyze the range of the function to see whether upper and lower bounds exist.

Detailed steps:

Analyze the expression: $f(x) = \frac{x}{x^2 + 1}$
When $x = 0$ , $f(0) = 0$
When $x \neq 0$ , $f(x) = \frac{1}{x + \frac{1}{x}}$
Use the inequality $x + \frac{1}{x} \geq 2$ (for $x > 0$ ) or $x + \frac{1}{x} \leq -2$ (for $x < 0$ )
Therefore $|f(x)| \leq \frac{1}{2}$

Answer: The function is bounded on $\mathbb{R}$ , with upper bound $\frac{1}{2}$ and lower bound $-\frac{1}{2}$ .

Exercise 2

Determine whether $f(x) = \ln x$ is bounded on $(0, 1]$ .

Answer and Explanation(3 个标签)

boundednesslogarithmmonotonicity

Approach: Analyze the range of the logarithm on the given interval.

Detailed steps:

$\ln x$ is strictly increasing on $(0, +\infty)$
As $x \to 0^+$ , $\ln x \to -\infty$
At $x = 1$ , $\ln 1 = 0$
So on $(0, 1]$ the range is $(-\infty, 0]$

Answer: The function is bounded above (by 0) on $(0, 1]$ but not bounded below, so it is not a bounded function.

Exercise 3

Determine whether $f(x) = \frac{\sin x}{x}$ is bounded on $(0, +\infty)$ .

Answer and Explanation(3 个标签)

boundednesstrigonometric functionlimit

Approach: Analyze the behavior of the ratio of a sine function to a linear function.

Detailed steps:

For $0<x\leq1$ , the standard inequality $|\sin x|\leq x$ holds; for $x\geq1$ , $|\sin x|\leq1\leq x$ .
Hence for all $x>0$ : $\left|\frac{\sin x}{x}\right|=\frac{|\sin x|}{x}\leq1$
Moreover, $\frac{\sin x}{x}\to1$ as $x\to0^+$ and $\frac{\sin x}{x}\to0$ as $x\to+\infty$ , consistent with the global estimate above.

Answer: The function is bounded on $(0, +\infty)$ , with $|f(x)| \leq 1$ .

Exercise 4

Let $f(x) = x^2 - 2x + 3$ . Determine its boundedness on $[0, 3]$ and find its supremum and infimum.

Answer and Explanation(4 个标签)

boundednessquadratic functioncompleting the squareextreme values

Approach: This is a boundedness problem for a quadratic on a closed interval. Complete the square or use the derivative to find the extreme values.

Detailed steps:

Complete the square: $f(x) = x^2 - 2x + 3 = (x-1)^2 + 2$
From the completed square:
- When $x = 1$ , $(x-1)^2 = 0$ , so the minimum is $f(1) = 2$
- Since $(x-1)^2 \geq 0$ , we have $f(x) \geq 2$
Evaluate the endpoints on $[0, 3]$ $[0, 3]$ :
- $f(0) = 0^2 - 2 \cdot 0 + 3 = 3$
- $f(3) = 3^2 - 2 \cdot 3 + 3 = 9 - 6 + 3 = 6$
Compare the stationary point and endpoints:
- Minimum: $f(1) = 2$
- Maximum: $f(3) = 6$
Therefore on $[0, 3]$ $[0, 3]$ :
- Infimum (minimum) is 2
- Supremum (maximum) is 6

Answer: The function is bounded on $[0, 3]$ , with infimum 2 and supremum 6.

Summary

Symbols Used in This Article

Symbol	Type	Reading/Explanation	Meaning in This Article
$f(x)$	Math symbol	f of x	A function of variable $x$
$D$	Math symbol	D	The domain of the function
$M$	Math symbol	M	Upper bound constant
$m$	Math symbol	m	Lower bound constant
$\mathbb{R}$	Math symbol	Real numbers	The set of all real numbers
$[a, b]$	Math symbol	closed interval	Interval including endpoints
$(a, b)$	Math symbol	open interval	Interval excluding endpoints