LECTURE 3. THE WEINSTEIN CONJECTURE IN THE OVERTWISTED CASE 85along the leaves vanish. On the other hand, by Stokes,
(30)
where Df c 1) is the small piece of surface between the two close by leaves. Hence
the integral (28) is also small, as claimed. We have proved that ml :;::: 1 - 8 for
every singular point Zi· Since uk(8D) is transversal to the leaves, we have
n
(31) 'Lm7::;1,
i=l
so that there exists at most one singular point on 8D. Assume there is precisely
one, say zo E 8D. Then the associated mk is at least 1-&, if k is large. Arguing as
before, zo can belong to at most two of the closed segments [1, i], [i, 1] and [-1, 1]
on 8D. On the segment S not containing z 0 , the sequence Uk converges in C^00
to u, and the measure of parameter values belonging to uk(S) is , in view of our
normalization of the parametrisation at least ~. Hence
(32) 1 > - mk + ~ 4 -> 1 - 8 + ~. 4
Choosing 8 sufficiently small we have the desired contradiction. This proves that
there is no singular point on 8D. The proof is complete.
To sum up: if bubbling off occurs in the Bishop family of embeddings u: D ---+
~ x M, it necessarily occurs at points in the interior of D. D
Now the Weinstein conjecture in the overtwisted case follows immediately. Take
the maximal Bishop family. Then we obtain a sequence of disk maps uk satisfying
(33) uk : D ---+ ~ x M
8suk + f(uk)8tuk = 0
E(uk):::; C < oo
Z k ---+ zo E int(D)
l(\7uk)(zk)I---+ oo.From the section about the bubbling-off analysis we obtain a nontrivial finite energy
plane.