Let $f$ be a function defined on $[0,1]$ which is continuous for each point in $[0,1]$ and differentiable for each point in $(0,1)$. Suppose that $f^\prime (x) \neq 1$ for every $x \in (0,1)$. Prove that there exists exactly one point $x \in [0,1]$ such that $f(x) = x$ [Hint: You might find both the intermediate value theorem and the mean value theorem useful.]

Could someone kindly assist me with this?

  • $\begingroup$ Do you also assume that $f:[0,1]\to [0,1]$? If $f$ is an arbitrary real-valued function, then it is of course not necessarily the case that $f$ has a fixed point (e.g., take $f:[0,1]\to \mathbb{R}$ defined by $f(x)=2x+1$). $\endgroup$ – Amitesh Datta Apr 27 '14 at 10:37
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    $\begingroup$ @user146068: You should not deface your Questions (which you've done three times in rapid succession) after getting an Answer. A goal of this Community is to provide content that is useful to future Readers as well as to you. $\endgroup$ – hardmath Apr 27 '14 at 11:26


Consider the function $g(x) = x - f(x) $. Notice

$g(0) = - f(0) < 0 $

$g(1) = 1 - f(1) > 0 $

Now, can you finish the problem?

To show uniqueness, another hint: suppose $x,y$ are both fixed points, then by Mean value theorem, we can find $\xi \in (x,y) $ such that

$$ |x-y| = |f(x) - f(y)| = |f'(\xi) |x-y| $$

Then, .....

  • $\begingroup$ Also note, in addition to Lemur's excellent answer, that the condition $f'(x)\neq 1$ for all $x\in (0,1)$ is necessary to establish uniqueness of the fixed point, in conjunction with the mean value theorem. $\endgroup$ – Amitesh Datta Apr 27 '14 at 10:36

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