For a rational map , denotes the Julia set of *R* , the set and the set of that are not in the closure of a connected component of . A point is said to be *eventually periodic* if there exists a such that is periodic. By *sink* we mean a connected component of an attracting immediate basin.

*Let f be a rational map with two distinct sinks* *and* *(not necessarily in the same cycle) such that* *. Assume that* *and* *are simply connected, and* *and* *are locally connected.*

As a particular case of part 2 of Theorem 1, if then the set of eventually periodic points in is non-empty and dense in . Nevertheless, the theorem does not require to be finite.

Here is an example of a non-empty intersection between two sink boundaries with no periodic point in the intersection. Let us consider , where and ( denotes the unit circle in ) is such that has rotation number *θ* . The map has been studied in [[3]]. The map has two attracting fixed points 0 and ∞. The intersection between the boundaries of the corresponding sinks is non-empty and included in . This intersection contains no periodic point since is topologically conjugate to . One notes that in this example the intersection contains the point 1, which is a critical point with an infinite orbit.

To prove the theorem, we assume that and are fixed, for otherwise we work with an iterate of *f* . Since there will not be confusion, we will note , and .

We assume that does not contain a critical point with finite orbit nor a parabolic point, for otherwise would contain a periodic point.

A point is said to be *multiple* if it belongs to the impression of at least two prime ends in . Using the expansion of *f* on , it is easy to show that a multiple point of in is eventually periodic. Thus we assume that contains no multiple point of nor .

In this context we show, using Theorem 3, that is *distance-expanding* with respect to the spherical metric, that is there exist , and such that for any , if then . Then we find a periodic point in using the Theorem 4 dealing with periodic points for distance-expanding maps.

*The restriction* *is distance-expanding with respect to the spherical metric.*

By Theorem 3 below, there exists an integer such that . By continuity of the map , there exist and a neighborhood *U* of such that . By compactness of , there exists such that if , then the geodesic Γ between and lifts to a path *γ* from *x* to *y* with . Thus we get . □

**(**[[2]]**)** *Let g be a rational map of degree at least* 2*, and* *be a compact forward invariant set containing no critical point nor parabolic point. If* Λ *is disjoint from the ω-limit set of every recurrent critical point, then there exists* *such that* *for every* *.*

**(**[[5]]**, chapter 4)** *Let* *be a compact metric space. If* *is continuous, open and distance-expanding, then there exists* *such that the following holds: if there exist* *and* *such that* *, then X contains a periodic point.*

*The restriction* *is open.*

Let and assume is not open.

There exists a sequence converging to some . Let be such that . Since contains no critical point, there exist a neighborhood *U* of *x* and a neighborhood *V* of *y* such that is a homeomorphism. Thus for *n* large enough , the point is well defined and .

We show now that so that *O* is not open. It is clear that since . For any *n* , there exists a Fatou component such that and . The following assertion finishes the proof of the lemma.

*For n large enough* *,* *.*

Otherwise, for some there exists a Fatou component *B* such that , and . The boundary ∂*B* has finitely many connected components, thus each one of them is locally connected. Let be either *B* or . There exists a connected component of such that . Since is simple in , there exists a unique connected component of such that . Hence . Since , we have and , which contradicts the injectivity of . □

Now we apply Theorem 4 to finish the proof of part 1. Let *w* be an accumulation point of the orbit of some . There exist such that , where *α* is the constant in Theorem 4. Hence , and we get a periodic point in .

The proof of part 2 uses ideas and techniques developed by K. Pilgrim in his thesis ([[4]], chapter 5). In case where *f* is hyperbolic and , part 2 is a corollary of his work.

We assume that , for otherwise *f* is conjugate to for some , and the conclusion follows. Up to make a quasi-conformal deformation, we also assume that all the critical points in have a finite orbit (see [[1]], theorem VI 5.1; this is why we assume that each component of that is eventually mapped to or to is simply connected).

Let be the degree of and let be an isomorphism conjugating *f* with . For , set . Since is locally connected, extends continuously to .

Denote *χ* the set of *chords* , that is the set of such that . If is periodic, then the point is periodic. For any chord *α* and any set , will denote the isotopy class of *α* rel *X* . For any distinct , the complement has at least two connected components and at most three, with points of in each of them. For any , if and only if one connected component of contains all but two points of (these two points being the extremities of the chords).

Set . From the hypothesis of part 2, the set is finite. Thus if is such that , then the point is eventually periodic. We denote the set . For any we denote the set .

The proof of part 2 is as follows. We equip *χ* with the Hausdorff distance , so that it is a compact metric space. Pick . If the sequence is eventually cyclical then *α* is eventually periodic (Lemma 10). Otherwise, noting that contains twice the same element (Lemma 11), we build a sequence by a series of adjustments (Lemma 7) such that converges (Lemma 8) to a chord *β* with the following property: either , or is eventually cyclical. This proves the existence of a periodic point in . The density part will follow from the fact that we can build *β* as close as we want to *α* .

Let (resp. ). A *lift of α* is the closure of a connected component of . If a lift of *α* is a chord, then it belongs to (resp. ).

*Let* *,* *and* *be such that* *. There exists a unique chord* *isotopic to α rel* *, such that* *and* *for every* *. Furthermore,* *.*

Since is a lift of , there exists a unique lift of such that . In particular, . For each , we construct inductively a unique such that and for any . Note that . □

*For any* *, there exists* *such that:* *, if* *then* *. As a consequence, if* *for every* *then* *.*

It follows from the following assertion:

*For every* *there exists* *such that : for any* *, if* *then in at least two connected components of* *lie an open ball centered at a point of* *and with radius η.*

By contradiction, assume that there exists , a sequence tending to 0, and a sequence such that, for any : and there does not exist two connected components of in which lies an open ball centered at a point of and with radius . By compactness of , we choose an accumulation point of and up to extraction . We have . If has three connected components, then we note one of the two connected components that are Jordan domains and we note the connected component that is not a Jordan domain. If has two connected components, then we note them and . In any case, there exist and such that , . For *n* large enough, and are included in two distinct connected components of . This is a contradiction as soon as . □

Let and *η* as in the assertion. Since , there exists such that each one of the two balls of the assertion contains a point of . Hence there is a point of in at least two connected components of , thus . □

*For any* *, if the sequence* *is cyclical, then α is periodic.*

Assume that there exists such that for any . In particular, . By Lemma 7, there exists a unique chord such that and for all , . This chord is . Thanks to Lemma 7, we also have . Since this is true for any , we conclude by Lemma 8 that . □

*For any* *, there exist* *distinct such that* *.*

Assume that for any distinct, we have . Let us show that the set or the set accumulate on , which contradicts the hypothesis of part 2 of Theorem 1.

Since is compact, up to extraction the sequence accumulates on a chord *β* . Since for any distinct at least two connected components of contain a point of , one can construct a non-stationary sequence , which accumulates on a point .

Assume that is infinite and up to extraction that . There exist finitely many distinct connected components of , which are distinct from and and such that . Each is included in , but by construction there is an infinite subset of whose elements are pairwise separated by chords, thus there is an infinite subset of included in . Since we assume that the extremities of the chords are the only points of in , we conclude that there is an infinite subset of included in .

Hence, up to extraction, we have or . In particular, , and . □

Let *α* be a chord. We have three cases.

*Case 1:* *.* Thus is eventually periodic, as explained before.

*Case 2:* *and* *is eventually cyclical.* Then *α* is eventually periodic by Lemma 10, and the point is eventually periodic.

*Case 3:* *and* *is not eventually cyclical.* Let us build from *α* a chord *β* fitting case 1 or 2.

By Lemma 11 there exist and such that . Set . By Lemma 7, there exists a chord such that for any , and . Thus for any , and . We build inductively a sequence of chords such that:

(i) | for any , and |

(ii) | |

*The sequence* *converges to a chord β whose point* *is eventually periodic.*

The convergence follows from (i) and Lemma 8. The limit *β* is a chord since *χ* is compact. If then *β* fits case 1. If , then we get for the limit for every and . Hence is cyclical, and *β* fits case 2. □

Thus there exists an eventually periodic point in .

To finish the proof, let us explain now the density. The chord *β* is in . Applying Lemma 7 to , we obtain a chord such that is eventually periodic. Using Lemma 11, we can have *N* as large as we want. Thus we can build a sequence converging to *α* , such that every is eventually periodic. Since and are locally connected, the sequence converge to . □