Bounded function

In mathematics, a function f defined on some set X with real or complex values is called bounded if the set of its values is bounded. In other words, there exists a real number M such that

|f(x)|\leq M

for all $x$ in $X$ .^[1] A function that is not bounded is said to be unbounded.^{[citation needed]}

If $f$ is real-valued and $f(x)\leq A$ for all $x$ in $X$ , then the function is said to be bounded (from) above by $A$ . If $f(x)\geq B$ for all $x$ in $X$ , then the function is said to be bounded (from) below by $B$ . A real-valued function is bounded if and only if it is bounded from above and below.^[1]^{[additional citation(s) needed]}

An important special case is a bounded sequence, where $X$ is taken to be the set $\mathbb {N}$ of natural numbers. Thus a sequence $f=(a_{0},a_{1},a_{2},\ldots )$ is bounded if there exists a real number $M$ such that

|a_{n}|\leq M

for every natural number $n$ . The set of all bounded sequences forms the sequence space $l^{\infty }$ .^{[citation needed]}

The definition of boundedness can be generalized to functions $f:X\rightarrow Y$ taking values in a more general space $Y$ by requiring that the image $f(X)$ is a bounded set in $Y$ .^{[citation needed]}

Related notions

Weaker than boundedness is local boundedness. A family of bounded functions may be uniformly bounded.

A bounded operator $T:X\rightarrow Y$ is not a bounded function in the sense of this page's definition (unless $T=0$ ), but has the weaker property of preserving boundedness; bounded sets $M\subseteq X$ are mapped to bounded sets $T(M)\subseteq Y$ . This definition can be extended to any function $f:X\rightarrow Y$ if $X$ and $Y$ allow for the concept of a bounded set. Boundedness can also be determined by looking at a graph.^{[citation needed]}

Examples

The sine function $\sin :R\rightarrow R$ is bounded since $|\sin(x)|\leq 1$ for all $x\in \mathbf {R}$ .^[1]^[2]
The function $f(x)=(x^{2}-1)^{-1}$ , defined for all real $x$ except for −1 and 1, is unbounded. As $x$ approaches −1 or 1, the values of this function get larger in magnitude. This function can be made bounded if one restricts its domain to be, for example, $[2,\infty )$ or $(-\infty ,-2]$ .^{[citation needed]}
The function ${\textstyle f(x)=(x^{2}+1)^{-1}}$ , defined for all real $x$ , is bounded, since ${\textstyle |f(x)|\leq 1}$ for all $x$ .^{[citation needed]}
The inverse trigonometric function arctangent defined as: $y=\arctan(x)$ or $x=\tan(y)$ is increasing for all real numbers $x$ and bounded with $-{\frac {\pi }{2}}<y<{\frac {\pi }{2}}$ radians^[3]
By the boundedness theorem, every continuous function on a closed interval, such as f : [0, 1] → R, is bounded.^[4] More generally, any continuous function from a compact space into a metric space is bounded.^{[citation needed]}
All complex-valued functions f : C → C which are entire are either unbounded or constant as a consequence of Liouville's theorem.^[5] In particular, the complex sin : C → C must be unbounded since it is entire.^{[citation needed]}
The function f which takes the value 0 for x rational number and 1 for x irrational number (cf. Dirichlet function) is bounded. Thus, a function does not need to be "nice" in order to be bounded. The set of all bounded functions defined on [0, 1] is much larger than the set of continuous functions on that interval.^{[citation needed]} Moreover, continuous functions need not be bounded; for example, the functions $g:\mathbb {R} ^{2}\to \mathbb {R}$ and $h:(0,1)^{2}\to \mathbb {R}$ defined by $g(x,y):=x+y$ and $h(x,y):={\frac {1}{x+y}}$ are both continuous, but neither is bounded.^[6] (However, a continuous function must be bounded if its domain is both closed and bounded.^[6])