Let $f\colon [0,1]\to\mathbb{R}, 0<f(x)<1$ for all $x \in[0,1]$.

$f(x)$ continuous.

Prove there exists $c$ such that $f(c)=c$

My attempts:

As $x \in[0,1]$ then $0\leq x \leq 1$ so $0 < x < 1$ and $0 < f(x) < 1$

So $0< f(x)-x < 0$ but that doesn't make much sense.

Next I tried setting $g(x)=f(x)-x$ and try to use Bolzano's theorem, but then I have to show $g(x)$ is positive somewhere and negative somewhere else.




Would suffice?

  • $\begingroup$ I think you need continuity. You might define $g(x)=f(x)-x$ and work with that. $\endgroup$ – Michael Apr 3 '15 at 2:02
  • $\begingroup$ @Michael True, I forgot to mention that. I thought of using Bolzano's theorem, but didn't know where to go from there. $\endgroup$ – YoTengoUnLCD Apr 3 '15 at 2:03
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    $\begingroup$ I think the downvote came because you didn't show us any of your own ideas... However, it is difficult to show ideas if one does not know how to get started at all. Hint: Consider Michael's comment and your comment mentioning Bolzano's Theorem. $\endgroup$ – sranthrop Apr 3 '15 at 2:11
  • $\begingroup$ The downvotes are almost certainly for the reason @sranthrop mentioned. You'd be surprised just how much you can improve a question (both in terms of its reception and the kind of answers it gets) by showing even the smallest amount of work. Sometimes even stating an idea as to how to start will save you from closure or downvotes (maybe even garner you upvotes). $\endgroup$ – Daniel W. Farlow Apr 3 '15 at 2:18
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    $\begingroup$ In your attempts: The fact $0 \leq x \leq 1$ certainly does not imply $0 < x < 1$, so I do not understand your first use of the word "so." Next, I do not understand how you conclude $0 < f(x) - x < 0$, that is incorrect. If $1 < 2$ and $2 < 3$, you cannot "subtract inequalities" to get $-1<-1$. You can add inequalities, but not subtract them. $\endgroup$ – Michael Apr 3 '15 at 2:22

Let $g(x)=f(x)-x$. Clearly $g(x)$ is continuous.

Since $g(0)>0,g(1)<0$, by the Intermediate Value Theorem of continuous function, there is $c\in[0,1]$, such that $g(c)=0$. So $f(c)=c.$


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