For each
the escape set
and the keep set
of the basic quadratic function
have the following properties
-
and
-
is open and
is close -
and
-
and
are both invariant under
-
and
are both symmetric under rotation by
about 0. -
is connected and
has no holes in it
Proof
a) By the backwards form of the triangle inequality
(1)
is the positive solution of the quadratic equation
We claim that if
then
(2) for![]()
Indeed if
then![]()
as required by (2). (1) and (2) now give
for![]()
If
then we can apply this inequality successively to
to deduce that
is increasing and
as![]()
b) Suppose
then
as
so for some
Let
Since
and
is a polynomial
is continuous at
so there exists
such that
implies
hence
It follows that
implies
as
so that
and
is open.
c) The set
is not the whole of
because it does not include the fixed points of![]()
d)
as![]()
as![]()
as![]()
![]()
e) Since the set
is connected and
it is sufficient to show that each point
can be joined to some point of
by a path in
The proof is by contradiction.
Suppose that
cannot be joined to
Define
then
since![]()
is open because if
can be joined to
then so can points of any open disc in
with centre
and
is connected because pairs of points in
can be joined in
via
R is thus a subset of
Since
cannot be joined in
to
and
is open we deduce![]()
Now use the maximum principle. If
then
else we increase the size of
so
for![]()
Applying the maximum principle to each polynomial function
on
we obtain
for
and
which contradicts that
hence
is connected.