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Suppose $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function.Then which of the following statements are necessarily true?

$1$. If $f'(x) \le r <1 $ $\forall x \in \mathbb{R}$ , then $f$ has at least one fixed point.
$2$. If $f$ has a unique fixed ,then $f'(x) \le r <1 $ $\forall x \in \mathbb{R}$.
$3$.If $f$ has a unique fixed ,then $f'(x) \ge r >-1 $ $\forall x \in \mathbb{R}$.
$4$. If $f'(x) \le r <1 $ $\forall x \in \mathbb{R}$ , then $f$ has unique fixed point.


My thoughts:-

$\mathbb{R}$ is a complete metric space. so we can use Banach contraction principle.so it gives $1$ and $4$ are correct.

for $2$. $f(x)=x^2$ is the counter example.
for $3$. $f(x)=-x^2$ is the counter example.

does my thoughts are correct?please somebody suggest me the right way.thanks for your time.

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  • $\begingroup$ thank you very much for your valuable suggestions.thanks a lot. $\endgroup$ – poton Apr 23 '13 at 4:53
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You cannot apply the contraction principle, as you only know an upper bound on $f'(x)$. To use the contraction principle, you would need to have $|f'(x)|\le r < 1$.

However, you can prove 1. & 4. directly as follows:

Suppose $f(x_0) > x_0$, let $\phi(x)=f(x)-x$, and choose $x> x_0$. Then $\phi(x)= \phi(x_0) + \phi'(\xi)(x-x_0) \le \phi(x_0) + (1-r) (x-x_0)$. Hence for sufficiently large $x$, $\phi(x) <0$, and by the intermediate value theorem we have $\phi(x') = 0$ for some $x'$. By a similar line of reasoning, if $x<x_0$, we have $\phi(x) \ge \phi(x_0) + (1-r) (x-x_0)$. Setting $x_0 = x'$ shows that $x'$ is the unique zero of $\phi$ and hence the unique fixed point of $f$.

If $f(x_0) < x_0$, a similar line of reasoning to the above (choose $x < x_0$ to show existence, and uniqueness follows in the same manner).

For 2., let $f(x) = 2x$. Then $f$ has a unique fixed point (zero), but $f'(x) = 2 >1$.

Similarly for 3., let $f(x) = -2x$. Then $f$ has a unique fixed point (zero), but $f'(x) = -2 < -1$.

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