Question

(1) Consider the polynomial map C → C defined by z-Plz) 22 + c, c E C. In class, we proved the following two facts: Suppose lel < 2. If an orbit [z0, 21,22,..} (where zn p"(20)) contains an iterate zn such that 2n2, then the orbit diverges to oo. (Thus if the orbit of zo ever strays outside the disk of radius 2 about the origin, 20 does not belong to the filled Julia set for any p(z) with | (P -2 + ε, e> 0. Using |zHc2 lZn12-1c| |from the roof sketch: Assume 1.2n| triangle inequality] and lcl 2, deduce that n+ 2+2. Apply this inequality repeatedly to deduce that the orbit diverges to oo.) . Suppose 2. Then the orbit of 0 diverges to oo. (Thus c is not an element of the Mandelbrot set for polynomials of the form 22 c.) (Proof sketch: Observe that the orbit of 0 is 0, c, c2 +c, (c2 + c)2 +c,... J. As above, 2lnfrom the triangle inequality. Assume l2n| 2 lc| (and observe that for the orbit of 0, this is in fact true of p(0-c). Then (applying zn c, factoring, and applying 2nl 2 Icl again), we obtain |22 + cn (e). Apply this inequality repeatedly to deduce that further iterates of 2n will diverge to oo) (Taken together, the two facts imply that for any cE C, if the orbit of 0 under p(z)2 not belong to the Mandelbrot set for p(z).) c ever strays outside the disk of radius 2 centered at the origin, then c does Observe that the methods used in proving these two facts are very similar, though not identical. By remixing the arguments used, prove that for any cE C, if an orbit z0, 21,22,... J (where 2np"() contains an iterate 2n such that |zn| > max (2, lc), then the orbit diverges to oo. [Note that rather than coming up with a new argument, this problem is about critically understanding exactly what the proofs sketched above logically imply

Answer 1

Suppose that
c < 2
. Then
max(2, c
. Thus, there is an iterate $z_n$ such that
2
. By fact 1 given above (in question) this implies that the orbit ${z_0,z_1,cdots}$ diverges to infinity.

Now suppose that
2
, so that
max2. C
. Since
해> 7t
, we have

Equivalently,
m+1
. Notice that this implies $|z_{n+1}|>|c|$ because
C-1 1
and
해> 7t
. Thus, the above calculation holds for $z_{n+1}$, and we get

By induction, we get $|z_{n+r}|>|z_n|(|c|-1)^r$ for all integers . Since
2
, and
해> 7t
, we have $|z_{n+r}|>2(|c|-1)^r$ and

Therefore,

which shows that the orbit ${z_0,z_1,cdots}$ diverges to infinity.