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3.10 Let $f\in\mathcal{C}^1[-1,1]$ such that $|f(t)|\leq 1$ and $|f'(t)|\leq\frac{1}{2}$ for all $t\in[-1,1]$. Let $$A=\{t\in[-1,1]\colon f(t)=t\}.$$ Is $A$ nonempty? If the answer is 'yes', what is its cardinality?

Well, I was trying like suppose $\exists t_1 \ni f(t_1)=t_1$ then if I apply mean value theorem on $[t_1,x]$ then $$f(x)-f(t_1)=f'(c)(x-t_1)$$ $$|f(x)-t_1|\le \frac{1}{2}|x-t_1|$$ then I can not proceed, where I can use the fact $|f(t)|\le 1$?

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The first half of your question can be answered easily enough: consider the Intermediate Value Theorem applied to $g(t) = f(t)-t$. What can you say about the values of $g(-1)$ and $g(1)$? Similarly, the fact that $|f'(t)| \leq 1/2$ should let you produce bounds on $g'(t)$ on your interval; can you see what that matters? – Steven Stadnicki Jul 20 '12 at 18:34
yes sir I got it thank you. – Un Chien Andalou Jul 20 '12 at 18:39
up vote 3 down vote accepted

First let $g(x)=f(x)-x$. Then $g(1) \geq 0$ and $g(-1) \leq 0$ imply that $g$ has a root.

Now, assume by contradiction that $f$ has two fixed points $t_1,t_2$, and apply the MVT on the interval between these two points.

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ah! feeling bad :( why I did not take 2 fixed points :(. :) thank you for the answer :) – Un Chien Andalou Jul 20 '12 at 18:41

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