A function can fail to be useful long before it becomes discontinuous or non-measurable. The simple formula $f(x)=1/x$ on $(0,1)$ is smooth, measurable, and easy to integrate on compact subintervals away from $0$, yet it has no global ceiling. If a later argument wants to estimate $|f(x)g(x)|$ by a fixed multiple of $|g(x)|$, or wants a uniform error estimate that holds for every $x$ in the domain, this function refuses to cooperate.

The idea of a bounded function records the difference between pointwise control and uniform control. Pointwise, every finite value is harmless. Uniformly, the whole graph must live inside one horizontal strip, or more generally inside one ball of the codomain. This small distinction is one of the first places where analysis stops asking what happens at a point and starts asking what can be controlled at once.

[example: A Smooth Function with No Global Bound] Let $f:(0,1)\to\mathbb R$ be given by $f(x)=1/x$. For each fixed $x\in(0,1)$, we have $x>0$, so $f(x)=1/x$ is a finite real number. The function is differentiable at every $x\in(0,1)$ because the reciprocal function is differentiable away from $0$, with derivative $f'(x)=-1/x^2$. We show that no number $M>0$ satisfies $|f(x)|\le M$ for every $x\in(0,1)$. Let $M>0$. Since $M>0$, the number \begin{align*} x=\frac{1}{M+1} \end{align*} satisfies $x>0$. Also $M+1>1$, so \begin{align*} \frac{1}{M+1}<1 \end{align*} and therefore $x\in(0,1)$. For this choice of $x$, \begin{align*} |f(x)|=\left|\frac{1}{x}\right|=\frac{1}{x} \end{align*} because $x>0$. Substituting $x=1/(M+1)$ gives \begin{align*} \frac{1}{x}=\frac{1}{1/(M+1)}=M+1. \end{align*} Since $M+1>M$, we have $|f(x)|>M$. Thus every proposed global bound $M$ is exceeded by some point of $(0,1)$, so $f$ is not bounded on $(0,1)$. [/example]

This example is not pathological in its formula; the failure comes from the domain. On every closed interval $[a,1]$ with $0<a<1$, the same function is bounded by $1/a$. Boundedness is therefore a global assertion about a chosen domain, not just a local feature of a formula.

## Definition

The basic real-valued definition asks for one number that controls every output. The point is not that the function avoids infinity as a value, but that its finite values are uniformly trapped.

[definition: Bounded Real-Valued Function] Let $X$ be a set. A function $f:X\to\mathbb R$ is a bounded real-valued function if there exists $M\ge 0$ such that \begin{align*} |f(x)|\le M \end{align*} for every $x\in X$. [/definition]

The definition uses absolute value because the codomain is ordered and centred at $0$. In order-based arguments, however, one often receives separate estimates $m\le f(x)$ and $f(x)\le M$ rather than a symmetric estimate around zero. The obstruction is to turn those two one-sided bounds into a single absolute-value bound, and conversely to recover two-sided bounds from one number.

[quotetheorem:9148]

This theorem explains why boundedness is sometimes phrased as being bounded above and below. The absolute-value formulation is symmetric around $0$; the order formulation permits an interval whose centre is chosen by the function.

[example: Finding the Sharp Bound] Let $f:[0,2]\to\mathbb R$ be given by $f(x)=x(2-x)$. For $x\in[0,2]$, expanding around $1$ gives \begin{align*} x(2-x)=2x-x^2=1-(x^2-2x+1)=1-(x-1)^2. \end{align*} Also $0\le x$ and $0\le 2-x$, so \begin{align*} f(x)=x(2-x)\ge 0. \end{align*} Since $x\in[0,2]$, subtracting $1$ gives $-1\le x-1\le 1$, hence \begin{align*} 0\le (x-1)^2\le 1. \end{align*} Therefore \begin{align*} f(x)=1-(x-1)^2\le 1. \end{align*} Thus $0\le f(x)\le 1$ for every $x\in[0,2]$, so $|f(x)|=f(x)\le 1$ for every $x\in[0,2]$. At $x=1$, \begin{align*} f(1)=1(2-1)=1, \end{align*} so the upper bound $1$ is actually attained. Hence the least possible uniform bound is \begin{align*} \sup_{x\in[0,2]}|f(x)|=1. \end{align*} The notation for this supremal size will be introduced below as the supremum norm. The example shows that the best uniform bound is not merely an estimate here; it is reached at the midpoint of the interval. [/example]

## Metric and Normed Viewpoints

The real-valued definition is the model case, but many functions in analysis are not scalar-valued: curves take values in $\mathbb R^n$, solutions may be vector-valued, and abstract maps may land in a [metric space](/page/Metric%20Space). To keep boundedness from depending on an absolute value that may not exist, the metric version asks whether the entire image fits inside one ball. As usual for this page, metric spaces are assumed to be nonempty.

[definition: Bounded Map into a Metric Space] Let $X$ be a set and let $(Y,d)$ be a metric space. A function $f:X\to Y$ is bounded if there exist $y_0\in Y$ and $R\ge 0$ such that \begin{align*} f(X)\subset \overline{B}(y_0,R):=\{y\in Y:d(y,y_0)\le R\}. \end{align*} [/definition]

For real-valued functions, taking $Y=\mathbb R$ with its usual metric and choosing $y_0=0$ gives the absolute-value bound $|f(x)|\le R$. Conversely, if the image of $f:X\to\mathbb R$ lies in some closed ball $\overline{B}(y_0,R)$, then the triangle inequality gives $|f(x)|\le R+|y_0|$ for every $x\in X$. Thus the metric definition and the scalar definition describe the same real-valued functions.

Once a function is known to be bounded, the next question is quantitative: among all possible bounds, which number records the true uniform size of the function? For scalar functions, that number is obtained by taking the supremum of the absolute values.

[definition: Supremum Norm] Let $X$ be a set, and use the convention that the supremum of the empty set of nonnegative [real numbers](/page/Real%20Numbers) is $0$. The supremum norm is the function from the set of bounded functions $f:X\to\mathbb R$ to $[0,\infty)$ that assigns to $f$ the number \begin{align*} \sup_{x\in X}|f(x)|. \end{align*} For a bounded function $f:X\to\mathbb R$, write \begin{align*} \|f\|_\infty:=\sup_{x\in X}|f(x)|. \end{align*} [/definition]

This norm is finite exactly for bounded real-valued functions. It turns pointwise inequalities into norm estimates, and those estimates are the language used throughout the rest of the chapter.

## Uniform Control and Pointwise Control

Boundedness is a uniform condition. It is stronger than saying that every value is finite, and it is different from saying that every small part of the domain behaves well. The distinction matters whenever an estimate must survive a supremum over all inputs.