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My lecture notes state the following.

definition We say that $x$ in $X$ is a fixed point of a function $f$ from $X$ taking values in $X$ if $f(x)=x$.

theorem If $f:[a,b]\rightarrow :[a,b]$ is continuously differentiable, then $f$ has at least one fixed point.

I am confused because it seems easy to find a counter-example.

counter-example Let $a=3$, $b=4$ and $f(x)=1$. Then $f$ is continuously differentiable and there is no $x$ in $[3,4]$ such that $x=1$.

Am I overseeing something?

Meneer-Beer
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    In your counter-example $f$ is not a function from the interval $[3,4]$ to the interval $[3,4]$ so does not fit the necessary criteria. – Doug M Mar 28 '17 at 20:19
  • How can $f(x) = 1$, given the codomain is $[3,4]$? – quasi Mar 28 '17 at 20:21

1 Answers1

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Yes. You are overseeing the fact that $f$ must have codomain $[a,b]$. $1$ is not in $[3,4]$.

Aloizio Macedo
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