Use the Intermediate Value Theorem to prove $f:[0,1]\to [0,1]$ continuous and $C\in[0,1]$, there is some $c \in [0,1]$ such that $f(c) = C$.

Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an $f(x) = C$, but I am having trouble proving that a $f(c) = C$.

This is what I have:

Since $f$ is continuous on $[0,1]$ there exists $f(a) = 0$ and $f(b) = 1$.

Let $g(x) = F(x) - C$. We assume that $f(a) < C < f(b)$ $\to$ $0 < C < 1$

when $x = a$, $g(a)$ is negative when $x = b$, $g(b)$ is positive

Therefore, $g(a) < 0 < g(b)$, and since $f$ is continuous on $[0,1]$, so is $g$.

Therefore there exists a $g(x) = 0$, and $f(x) = C$.

How can I prove there is a $f(c) = C$ ? Is this a rule for IVT?


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    $\begingroup$ Ummm....wait, what? What is $c$ (as opposed to $C$)? $\endgroup$ – user5137 Apr 9 '12 at 10:15
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    $\begingroup$ What you say that you can "easily prove", is wrong. Continuous does not imply that 0 and 1 are in the image. $\endgroup$ – Phira Apr 9 '12 at 10:18
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    $\begingroup$ In fact, the theorem as stated is not true. Consider $f(x) = x/2$ and $C=3/4$. You need tnat $f$ is onto $[0,1]$. $\endgroup$ – Patrick Apr 9 '12 at 11:56

The problem in your approach is that your already fix $C$ when you define $g$, whereas we don't know a priori what it will be (and there is also a confusion between $c$ and $C$).

Consider $g(x)=f(x)-x$; $g$ is continuous on $[0,1]$, and $g(0)=f(0)\geq 0$, and $g(1)=f(1)-1\leq 0$ so $g$ take the value $0$ at some $c\in [0,1]$.

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  • $\begingroup$ I'm not sure he's trying to prove the Fixed Point Theorem. Perhaps he is, but what he's stated looks, to me at least, more like the Intermediate Value Theorem applied to $f$ on $[0,1]$, which would work if $f$ was onto $[0,1]$. $\endgroup$ – Patrick Apr 9 '12 at 19:13
  • $\begingroup$ My answer was based on the initial post, and I though there where a confusion between $c$ and $C$. $\endgroup$ – Davide Giraudo Apr 9 '12 at 19:19
  • $\begingroup$ Hi Davide, I took your approach and I think that is the right way to go. I actually left out 'if': the question should actually read: Use the IVT to prove that if f:[0,1] -> [0,1] is continuous, then there exists a C E [0,1], such that f(c) = C. $\endgroup$ – JackReacher Apr 11 '12 at 0:12

You can prove using the intermediate value theorem that if $f:[0,1] \to [0,1]$ is a continuous function then for every $$\lambda \in \left[ \min_{[0,1]}f(x),\max_{[0,1]}f(x)\right] $$ and only for these values of $\lambda$, there exists $c \in [0,1]$ with $f(c)=\lambda$.

The problem is that you pick $C$ arbitrarily, and you can easily define continuous functions on $[0,1]$ which do not take a certain value: constant functions, continuous functions $f: [0,1] \to [0,C/2]$, etc.

The flaw in your argument is that you assume that if $f$ is continuous then there exist $a,b$ such that $f(a)=0$ and $f(b)=1$. You cannot assume such thing unless $f$ is onto

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  • $\begingroup$ yes, f:[0,1] is onto the range [0,1]. In which case, i think Davide is right, the function must intersect with the function g(x) = x, at some point. At this intersection is where f(c) = C. $\endgroup$ – JackReacher Apr 11 '12 at 0:14

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