1. THE LOCAL THEORY OF THE CAUCHY PROBLEM 7 For the last term, we use older’s inequality and the chain rule, to get, since |F (u)| |u|4 , |F (u)| |u|3 [F (u + w) F (u)] Lq Ij Lx q C u 4 S(Ij) + w 4 S(Ij) Dβw Lq I j Lxq + u 3 S(Ij) + w 3 S(Ij) Dβu L q I j Lx q + Dβw L q I j Lx q · w S(Ij) C(η) w S(Ij) + Dβw Lq Ij Lq x + C w S(Ij) + Dβw Lq Ij Lq x 5 , C(η) η→0 0. Then, if η is so small that C(η) 1 3 , and letting γj = S (t aj) ((w(aj),∂tw(aj))) S(R) + DβS(t aj) (w(aj),∂tw(aj),∂tw(aj)) LqLxq R + Cε. We see that w S(Ij) + Dβw Lq I j Lx q 3 2 γj + C w S(Ij) + Dβw Lq I j Lxq 5 . Note that the choice of η fixed. But then, a standard continuity argument shows that ∃C0, depending only on q such that if γj C0, we have w S(Ij) + Dβw Lq I j Lxq 3γj and C w S(Ij) + Dβw Lq I j Lxq 3γj. Thus, if γj C0, w S(Ij) + Dβw Lq Ij Lq x 3 S(t aj) (w(aj),∂tw(aj)) S(R) + DβS(t aj) ((w(aj),∂tw(aj))) L q R Lx q + 3Cε. To continue the iteration, we put t = aj+1 in the integral formula, apply S (t aj+1) and use trigonometric addition formulas. We then have S (t aj+1) (w(aj+1),∂tw(aj+1)) = S(t aj) ((w(aj),∂tw(aj))) aj+1 aj sin ( (t t ) −Δ ) −Δ e(t )dt aj+1 aj sin ( (t t ) −Δ ) −Δ [F (u + w) F (u)] dt . Applying the same argument as before, we see that S(t aj+1) (w(aj),∂tw(aj)) S(R) + DβS(t aj) (w(aj+1),∂tw(aj+1)) L q R Lx q S(t aj) (w(aj),∂tw(aj)) S(R) + DβS(t aj) (w(aj),∂tw(aj)) LqLxq R + +C(η) w S(Ij) + Dβw Lq Ij Lx q + C w S(Ij) + Dβw Lq Ij Lxq 5 . Using our previous bounds and taking again η small, we find that if γj C0,γj+1 10γj. Recall that, by assumption γ0 ε + ε , so that iterating γj
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