30A99-LocallyBounded.tex

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\begin{document}
Suppose that $X$ is a topological space and $Y$ a metric space.  

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

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

\begin{defn}
A set ${\mathcal{F}}$ of functions $f\colon G \subset {\mathbb{C}} \to {\mathbb{C}}$ is said to be {\em 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 $f \in {\mathcal{F}}$.
\end{defn}

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.

\begin{thebibliography}{9}
\bibitem{Conway:complexI}
John~B. Conway.
{\em \PMlinkescapetext{Functions of One Complex Variable I}}.
Springer-Verlag, New York, New York, 1978.
\end{thebibliography}
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