Can someone please help me understand how, in the solution below, we can get to $g(1)<0$ and $g(0)>0$ from the fact that $g(1) \neq 0 $ and $ g(0) \neq 0 $?
Thanks.
Rudin Chapter 4, question 14 (Exercise 4.14)
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Hint: what are the ranges of values that $f(0)$ and $f(1)$ can be? – Stromael Nov 24 '13 at 11:12
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1where did you get solutions to rudin ? – ILoveMath Nov 24 '13 at 11:17
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1I think there is typo. The '$0 \neq f(x) \neq 1$' should be '$0 \leq f(x) \leq 1$'. – user10676 Nov 24 '13 at 11:24
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2Continuous function from the unit interval into the unit interval has a fixed point and Show that a continuous function has a fixed point – Martin Sleziak Nov 24 '13 at 12:24
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@DonAnselmo When I googled for rudin "suppose f is continuous mapping of I into I" I found this file. It seems it might be from there. – Martin Sleziak Nov 24 '13 at 13:51
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I think that is because f is a map to [0,1]
and therefore f can get a max value of 1 and min value of 0
because we assume that $f(1)\ne1$ it clear that $f(1)<1$
and $f(0)\ne0$ so it clear that $f(0)>0$
and we get
$g(1)= f(1)-1 <1-1 =0$
$g(0) =f(0)-0 =f(0) >0$
have a nice day
Martin Sleziak
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david
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For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Nov 24 '13 at 13:47
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One of the hypothesis is that $f$ maps $I$ onto $I$, this means that for all $x\in I$, it holds that $0\leq f(x)\leq 1$. If $f(1)-1\neq 0$ and $f(0)-0\neq 0$, then $f(1)<1$ and $f(0)>0$, thus $g(1)<0$ and $g(0)>0$.
Git Gud
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