Bounded function

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

${\displaystyle |f(x)|\leq M}${\displaystyle |f(x)|\leq M}

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

If ${\displaystyle f}${\displaystyle f} is real-valued and ${\displaystyle f(x)\leq A}${\displaystyle f(x)\leq A} for all ${\displaystyle x}${\displaystyle x} in ${\displaystyle X}${\displaystyle X}, then the function is said to be bounded (from) above by ${\displaystyle A}${\displaystyle A}. If ${\displaystyle f(x)\geq B}${\displaystyle f(x)\geq B} for all ${\displaystyle x}${\displaystyle x} in ${\displaystyle X}${\displaystyle X}, then the function is said to be bounded (from) below by ${\displaystyle B}${\displaystyle 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 ${\displaystyle X}${\displaystyle X} is taken to be the set ${\displaystyle \mathbb {N} }${\displaystyle \mathbb {N} } of natural numbers. Thus a sequence ${\displaystyle f=(a_{0},a_{1},a_{2},\ldots )}${\displaystyle f=(a_{0},a_{1},a_{2},\ldots )} is bounded if there exists a real number ${\displaystyle M}${\displaystyle M} such that

${\displaystyle |a_{n}|\leq M}${\displaystyle |a_{n}|\leq M}

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

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

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

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

• The sine function ${\displaystyle \sin :\mathbb {R} \rightarrow \mathbb {R} }${\displaystyle \sin :\mathbb {R} \rightarrow \mathbb {R} } is bounded since ${\displaystyle |\sin(x)|\leq 1}${\displaystyle |\sin(x)|\leq 1} for all ${\displaystyle x\in \mathbb {R} }${\displaystyle x\in \mathbb {R} }.^{[1] [2]}
• The function ${\displaystyle f(x)=(x^{2}-1)^{-1}}${\displaystyle f(x)=(x^{2}-1)^{-1}}, defined for all real ${\displaystyle x}${\displaystyle x} except for −1 and 1, is unbounded. As ${\displaystyle x}${\displaystyle 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, ${\displaystyle [2,\infty )}${\displaystyle [2,\infty )} or ${\displaystyle (-\infty ,-2]}${\displaystyle (-\infty ,-2]}.^{[citation needed ]}
• The function ${\textstyle f(x)=(x^{2}+1)^{-1}}${\textstyle f(x)=(x^{2}+1)^{-1}}, defined for all real ${\displaystyle x}${\displaystyle x}, is bounded, since ${\textstyle |f(x)|\leq 1}${\textstyle |f(x)|\leq 1} for all ${\displaystyle x}${\displaystyle x}.^{[citation needed ]}
• The inverse trigonometric function arctangent defined as: ${\displaystyle y=\arctan(x)}${\displaystyle y=\arctan(x)} or ${\displaystyle x=\tan(y)}${\displaystyle x=\tan(y)} is increasing for all real numbers ${\displaystyle x}${\displaystyle x} and bounded with ${\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}}${\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}} radians ^[3]
• By the boundedness theorem, every continuous function on a closed interval, such as ${\displaystyle f:[0,1]\rightarrow \mathbb {R} }${\displaystyle f:[0,1]\rightarrow \mathbb {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 ${\displaystyle f:\mathbb {C} \rightarrow \mathbb {C} }${\displaystyle f:\mathbb {C} \rightarrow \mathbb {C} } which are entire are either unbounded or constant as a consequence of Liouville's theorem.^[5] In particular, the complex ${\displaystyle \sin :\mathbb {C} \rightarrow \mathbb {C} }${\displaystyle \sin :\mathbb {C} \rightarrow \mathbb {C} } must be unbounded since it is entire.^{[citation needed ]}
• The function ${\displaystyle f}${\displaystyle f} which takes the value 0 for ${\displaystyle x}${\displaystyle x} rational number and 1 for ${\displaystyle x}${\displaystyle 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 ${\displaystyle [0,1]}${\displaystyle [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 ${\displaystyle g:\mathbb {R} ^{2}\to \mathbb {R} }${\displaystyle g:\mathbb {R} ^{2}\to \mathbb {R} } and ${\displaystyle h:(0,1)^{2}\to \mathbb {R} }${\displaystyle h:(0,1)^{2}\to \mathbb {R} } defined by ${\displaystyle g(x,y):=x+y}${\displaystyle g(x,y):=x+y} and ${\displaystyle h(x,y):={\frac {1}{x+y}}}${\displaystyle 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] )

• Bounded set
• Compact support
• Local boundedness
• Uniform boundedness

• Complex analysis
• Real analysis
• Types of functions

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