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Homelocally bounded

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# locally bounded

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

###### Definition.

A set ${\mathcal{F}}$ of functions $f\colon X\to Y$ 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 $f\colon G\subset{\mathbb{C}}\to{\mathbb{C}}$ 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 $\lvert z-a\rvert<\delta$, $\lvert f(z)\rvert<M$ 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 $\lvert f(z)\rvert\leq(\lvert z_{0}\rvert+1)^{2}+1$ 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. Functions of One Complex Variable I. Springer-Verlag, New York, New York, 1978.

## Mathematics Subject Classification

30A99*no label found*54-00

*no label found*

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