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We know that if $f:\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $|f^{'}(x)|\leq A$ where $0 \leq A <1$. Then $f$ has atleast one fixed point. But the above result is not true if $f^{'}(x)<1$ for example if $f(x)=x+(1+e^{x})^{-1}$ then $f^{'}(x)<1$ but $f $ has no fixed point. Now i am searching an example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f^{'}(x)\leq A$ where $0 \leq A <1$ but $f$ has no fixed point.Please help me to find such a counterexample if it is possible. I tried it but not found. Thanks for precise time in advance.

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  • $\begingroup$ Try to prove that such a function always has a fixed point in $\mathbb{R}$. If $f(0) = 0$, we have found a fixed point. Otherwise, distinguish the cases $f(0) > 0$ and $f(0) < 0$. $\endgroup$ – Daniel Fischer Jun 10 '15 at 9:18
  • $\begingroup$ Please solve it i was trying to find counterexample... $\endgroup$ – Surya Deep Mishra Jun 10 '15 at 9:20
  • $\begingroup$ You can't find a counterexample. Therefore you should direct your efforts towards proving that such an $f$ has a fixed point. I gave a hint above, try following that. If you have problems with that, tell us where you get stuck. $\endgroup$ – Daniel Fischer Jun 10 '15 at 9:23
  • $\begingroup$ i am trying to show that f-x is positive at zero and negative at some other point on right side of origin...in first case... $\endgroup$ – Surya Deep Mishra Jun 10 '15 at 9:32
  • $\begingroup$ Good. And if $f(0) < 0$, then a similar reasoning finds you $x < 0$ with $f(x) > x$. $\endgroup$ – Daniel Fischer Jun 10 '15 at 9:34
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Yes function will have unique limit point. Uniqueness is obvious as condition on derivative. If $f(0)=0$ then proved. If $f(0)>0$ then consider the function $g(x)=f(x)-x$ . Clearly $g(0)>0.$ we can show that there is a positive real number $a$ such that $g(a)<0$. For if possible let $g(x)>0 $ for all x $\in$ [0,$\infty)$ . Then $g$ is strictly decreasing positive function on $[0,\infty)$ and so $\lim_{x\to\infty} g^{'}(x)=0$ which is not possible as $g^{'}(x)=f^{'}(x)-1 \leq A-1 <0$.Hence $g(a) \leq 0 $for some positive real $a$ and so the result. Similarly we can prove the result if $f(0)<0$.

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No such counterexample exists. Use the Mean Value Theorem to conclude that $f(x)\le f(0)+Ax$ for $x\ge 0$ and $f(x)\ge f(0)+Ax$ for $x\le 0$. The line $y=f(x_0)+Ax$ intersects the diagonal $y=x$ at some $x_0$. Observe that the $\le$ and $\ge$ are just right to conclude (now using the Intermediate Value Theorem) that the graph of $f$ also intersects the diagonal somewhere between $0$ and $x_0$.

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  • $\begingroup$ Can you explane it mathematically... $\endgroup$ – neelkanth Jun 10 '15 at 9:56
  • $\begingroup$ i mean in proper algebra... $\endgroup$ – neelkanth Jun 10 '15 at 9:56

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