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Let $f$ be a continuous function from [a, b] to [a, b], and is differentiable on (a, b). We will say that point y $\in$ [a, b] is a fixed point of f if $y = f(y)$. If the derivative $f'(x) \neq 1$ for all x $\in$ (a, b), then f has exactly one fixed point in [a,b].

I don't understand why. All I get is that if we take two points inside [a,b], there can be no fixed point since the derivative cannot be equal to one. But if we take one point inside the interval and another outside it, can there not be infinite such fixed points?

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    $\begingroup$ HINT: a fixed point for $f$ is a zero of the function $g(x)=f(x)-x$. Now apply theorems about zeroes of functions to $g$. $\endgroup$ – Crostul Jun 14 '15 at 9:56
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    $\begingroup$ f is not defined outside of [a,b], so speaking of a fixed point outside of the interval makes no sense. $\endgroup$ – Martin R Jun 14 '15 at 9:56
  • $\begingroup$ Aaah, okay so if I take f(a) negative and f(b) positive, or vice versa- I'd get a fixed point in the interval by the intermediate value theorem. And there can't be another fixed point because f'(x) can't be 1 in the interval- am i on the right track? $\endgroup$ – dexter Jun 14 '15 at 10:13
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Put $g(x):=f(x)-x$, which is still continuous on $[a,b]$ and differentiable on $(a,b)$. Observe that $x\in [a,b]$ is a fixed point for $f$ iff $g(x)=0$.

Now if $u<v$ are two distinct fixed points we can apply Rolle's theorem (since $g$ is differentiable on $(u,v)\subseteq (a,b)$) and find some $\xi\in (u,v)$ such that $g'(\xi)=0$, i.e. $f'(\xi)=1$, contradiction.

So there is at most one fixed point. The fact that at least one exists is very well-known: if $g(x)\neq 0$ for any $x\in [a,b]$ then $g(x)$ (by continuity) has always the same sign, but $f(a)\ge a$ and $f(b)\le b$, so $g(a)\ge 0$ and $g(b)\le 0$, contradiction. Thus for some $x$ we have $g(x)=0$ and $x$ is the desired fixed point.

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  • $\begingroup$ How did you know that f(a)≥a and f(b)≤b? $\endgroup$ – dexter Jun 14 '15 at 10:28
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    $\begingroup$ Because by hypothesis $f$ takes values in $[a,b]$. $\endgroup$ – Mizar Jun 14 '15 at 10:34

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