Unformatted text preview: ! = !" and ! = ! !. Then !, ! and ! are all !!" positive (! is defined above [2]). Our requirement ! ! ! ! ! ! ≤ 1, from below [11], implies ! ≤ ! and ≤ ! < 1. Then it is easy to see that ! < ! , satisfying the hypotheses of Lemma 5.4. Then the !
lemma implies !" ! (!; !) ≤ 2! ! ! ! ! and ! !; ! [13] → 0 as ! → ∞. □ Proof of theorem 5.3 (Part 3). Show that the unstable components of the solution are Lipschitz functions of !. Since ! (!; !) is a fixed point of [10] it satisfies ! !; ! = ! !" ! + ! ! !! !
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! ! !; ! !" − ! ! !! !
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! !! ! ! !; ! !", [14] where ! = !! ! (0). Consider this integral equation for two different values of ! and subtract ! !, !! − ! !; !! = ! !" !! − !! + ! ! !! !
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! !; !! − ! ! !; !! !" − ! ! !! !
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! !; !! − ! ! !; !! !" Take the absolute value of both sides, use the triangle inequality and estimates [2] and [3] ! !, !! − ! !; !! ≤ ! ! !!" !! − !! + ! +! ! ! ! !! !
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! ! ! (!; !! − ! ! (!; !! !" ! ! (!; !! ) − ! ! (!; !!  !" The inequalities are retained if the magnitudes of the difference between projections of points are replaced by the magnitudes of the differences between the points themselves. Remove the projection symbols and use the Lipschitz property of !. ! !, !! − ! !; !! ≤ ! ! !!" !! − !! + !"...
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 Fall '14
 MarcEvans
 Sets, rod, the00

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