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Show that if $f$ is continuous on $[0,1]$ and $0\leq f(x) \leq 1$ for all $ x \in [0,1]$, then there exists one point $c \in [0,1]$ at which $f(c)=c$.

(Hint: Apply the Intermediate Value Theorem to the function $g(x)=x−f(x)$.)

I have some idea how to do this, but the area where I am getting stumped is...the value of $f(x)$?

How can I find the value of $f(x)$? Do I even need to find $f(x)$?

Edit:

okay I got something...

I only need f(0)>0 and f(1)<1 to show g(0)<0

So then what happens for f(0)=0 or f(1)=1 ?

Willie Wong
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Martin
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1 Answers1

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To begin, some clarification: under the stated conditions, $f(x)$ may have more than one fixed point. To see this, take $f(x) = \cos^2 (n \pi x)$ for sufficiently large $n$.

This being the case:

You don't need to know the value of $c$ such that $f(c) = c$, only that such a $c$ exists. Without more detailed knowledge of $f(x)$, $c$ cannot be determined in any event; but its existence may be affirmed. Let $I$ denote the closed unit interval $[0, 1]$: $I = [0, 1]$, and let $f:I \to I$ be continuous. (Note that we need $f:I \to I$ here; to see this, consider $f:I \to \Bbb R$ given by, e.g. $f(x) = -x - 2$.) First look at the values of $f(x)$ at $0$ and $1$. If $f(0) = 0$ and/or $f(1) = 1$, we are done. Assuming then that neither of these possibilities hold, since $f:I \to I$, we must have $f(0) > 0$ and $f(1) < 1$. Next, look at $g(x) = f(x) - x$. $g(x)$ is manifestly continuous, with $g(0) > 0$ and $g(1) < 0$. By the intermediate value theorem, there must be a $c \in (0, 1)$ with $g(c) = 0$. For such a $c$, we have $f(c) = c$, i.e., $c$ is the sought-for fixed point of $f(x)$. QED.

Hope this helps. Cheerio,

and of course,

Fiat Lux!!!

Robert Lewis
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