locally bounded

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

Definition.

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

In the special case of functions on the complex plane where itis often used, the definition can be given as follows.

Definition.

A set $ℱ$ of functions $f:G\subset ℂ\to ℂ$ 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 , for all $f\in ℱ$.

As an example we can look at the set $ℱ$ of entire functions where$f(z)=z^2+t$ for any $t\in [0,1]$. Obviously each such $f$ is unboundeditself, however if we take a small neighbourhood around any point we canbound all $f\in ℱ$. Say on an open ball $B(z_0,1)$ we can showby 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 fromsome 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 $f_n(z)=n$ forall natural numbers $n$. While each of these functions is itself bounded,we can never find a uniform bound for all such functions.