Bounded function

Examples

• The sine function ${\displaystyle \sin :\mathbb{R}\to \mathbb{R}}$ is bounded since ${\displaystyle |\sin (x)|\le 1}$ for all ${\displaystyle x\in \mathbb{R}}$.
• The function ${\displaystyle f(x)=(x^2-1{)}^{-1}}$, defined for all real ${\displaystyle x}$ except for −1 and 1, is unbounded. As ${\displaystyle x}$ 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, ${\displaystyle [2,\mathrm{\infty })}$ or ${\displaystyle (-\mathrm{\infty },-2]}$.
• The function $f(x)=(x^2+1{)}^{-1}$, defined for all real ${\displaystyle x}$, is bounded, since $|f(x)|\le 1$ for all ${\displaystyle x}$.
• The inverse trigonometric function arctangent defined as: ${\displaystyle y=\arctan (x)}$ or ${\displaystyle x=\tan (y)}$ is increasing for all real numbers ${\displaystyle x}$ and bounded with ${\displaystyle -\frac{\pi }{2}<y<\frac{\pi }{2}}$ radians
• By the boundedness theorem, every continuous function on a closed interval, such as ${\displaystyle f:[0,1]\to \mathbb{R}}$, is bounded. More generally, any continuous function from a compact space into a metric space is bounded.
• All complex-valued functions ${\displaystyle f:\mathbb{C}\to \mathbb{C}}$ which are entire are either unbounded or constant as a consequence of Liouville's theorem. In particular, the complex ${\displaystyle \sin :\mathbb{C}\to \mathbb{C}}$ must be unbounded since it is entire.
• The function ${\displaystyle f}$ which takes the value 0 for ${\displaystyle x}$ rational number and 1 for ${\displaystyle x}$ 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 ${\displaystyle [0,1]}$ is much larger than the set of continuous functions on that interval. Moreover, continuous functions need not be bounded; for example, the functions ${\displaystyle g:{\mathbb{R}}^2\to \mathbb{R}}$ and ${\displaystyle h:(0,1{)}^2\to \mathbb{R}}$ defined by ${\displaystyle g(x,y):=x+y}$ and ${\displaystyle h(x,y):=\frac{1}{x+y}}$ are both continuous, but neither is bounded. (However, a continuous function must be bounded if its domain is both closed and bounded.)