The function f is thus a contraction mapping from 1 1

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The function f is thus a contraction mapping from [ - 1 , 1] into itself, hence has a unique fixed point which is the limit of every sequence obtained by successive applications of f . Key things I’d like to see when grading this exercise: f maps R into [ - 1 , 1] and after the first iteration one stays in [ - 1 , 1] (or words to this effect); | f 0 ( x ) | ≤ sin 1 < 1 for all x [ - 1 , 1]. 4
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6. Define f : R R by f ( x ) = x - arctan x + 5. (There is no significance to the number 5 other than it is > π/ 2.) Prove: f has no fixed point; that is, f ( x ) 6 = x for all x R . Prove also | f 0 ( x ) | < 1 for all x R . Explain why this does not contradict the proposition on page 172 of Rosenlicht. Or, does it? Solution. f ( x ) = x is equivalent to arctan x = 5 and since | arctan x | < π/ 2 for all x R , this is out. There is no contradiction with the Banach-Kakutani fixed point theorem, and its corollaries, because the theorem requires d ( f ( x ) , f ( y )) αd ( x, y ) where 0 α < 1; all that happens here is d ( f ( x ) , f ( y )) < d ( x, y ), and the example simply shows that this is not enough. It is not enough for the mapping to get points strictly closer than they were. It has to get them not just strictly closer, but they have to be closer by less than fixed amount. 5
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