# locally bounded

Suppose that $X$ is a topological space^{} and $Y$ a metric space.

###### Definition.

A set $\mathcal{F}$ of functions $f:X\to Y$ is said to be locally bounded if for every $x\in X$, there exists a neighbourhood $N$ of $x$ such that $\mathcal{F}$ is uniformly bounded on $N$.

In the special case of functions on the complex plane^{} where it
is often used, the definition can be given as follows.

###### Definition.

A set $\mathcal{F}$ of functions $f:G\subset \u2102\to \u2102$ is said to be locally bounded if for every $a\in G$ there exist constants $\delta >0$ and $M>0$ such that for all $z\in G$ such that $$, $$ for all $f\in \mathcal{F}$.

As an example we can look at the set $\mathcal{F}$ of entire functions^{} where
$f(z)={z}^{2}+t$ for any $t\in [0,1]$. Obviously each such $f$ is unbounded^{}
itself, however if we take a small neighbourhood around any point we can
bound all $f\in \mathcal{F}$. Say on an open ball $B({z}_{0},1)$ we can show
by triangle inequality^{} that $|f(z)|\le {(|{z}_{0}|+1)}^{2}+1$
for all $z\in B({z}_{0},1)$. So this set of functions is locally bounded.

Another example would be say the set of all analytic functions^{} from
some region $G$ to the unit disc^{}. All those functions are bounded^{} by 1,
and so we have a uniform bound even over all of $G$.

As a counterexample suppose the we take the constant functions ${f}_{n}(z)=n$ for
all natural numbers^{} $n$. While each of these functions is itself bounded,
we can never find a uniform bound for all such functions.

## References

- 1 John B. Conway. . Springer-Verlag, New York, New York, 1978.

Title | locally bounded |
---|---|

Canonical name | LocallyBounded |

Date of creation | 2013-03-22 14:17:47 |

Last modified on | 2013-03-22 14:17:47 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 9 |

Author | jirka (4157) |

Entry type | Definition |

Classification | msc 30A99 |

Classification | msc 54-00 |