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Consider the intermediate value theorem. It says that a continuous function $f(x)$ on a closed interval $[a,b]$ takes on every value between $f(a)$ and $f(b)$ at least once. Excluding trivialities like $f(x)=\mbox{constant}$, my question is how often can $f(x)$ achieve an intermediate value and still be continuous on $[a,b]$? For any finite number, a continuous function can always be constructed, like sine with a high enough frequency. But is it possible that an intermediate value is achieved by $f(x)$ an infinite number of times? Countable, uncountable number of times?

So looking for a continuous function $f(x)$ on a closed interval $[a,b]$ with $f(a)\neq f(b)$ such that for some $z$ between $f(a)$ and $f(b)$, there exist infinitely many values $c$ in $[a,b]$ for which $f(c)=z$. If such a function is not possible, then perhaps an intuitive argument of its impossibility will help.

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Exactly which functions "like f(x)=constant" do you want to exclude? – Chris Eagle Oct 29 '12 at 21:54
Countably infinitely many $c$ with $f(c)=z$ is certainly possible. What do you think happens with uncountably many? – Andrés E. Caicedo Oct 29 '12 at 22:04
Haha, I should have taken the extra ten second to think this through Andres. So uncountably many isn't possible because either the function will be a constant or discontinuous, right? – Fixed Point Oct 29 '12 at 22:28
Not an answer to the question - but for a radically discontinuous function satisfying the Intermediate Value Theorem you should investigate the Conway Base 13 Function. – Mark Bennet Oct 29 '12 at 22:48
up vote 1 down vote accepted

You can use a non-analitic function... for instance $f(x)=x-1$ $\forall x \in [0,1]$, $f(x)=0$ $\forall x \in [1,2]$ and $f(x)=x-2$ $\forall x \in [2,3]$.

If you think only about analitic functions, well...I can answer using complex analysis. Look here page 32. As you see if there are infinite zeroes then you have the null function so you can't have a function with infinite zeroes because $[a,b]$ is a compact and every infinite subset has an accumulation point.

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(Wrote this before your edit.) Given $a<b$, $c<e<d$, and any $n$, you can find an analytic function such that $f(a)=c$, $f(b)=d$ and $f(z)=e$ has at least $n$ solutions. You cannot replace $n$ with "infinitely many", as the zeros of analytic functions are isolated. – Andrés E. Caicedo Oct 29 '12 at 22:24
So is it possible with non-analytic function? Only continuity is required, not analyticity. – Fixed Point Oct 29 '12 at 22:32
If a function is constant on a sub-interval of $[a,b]$ then it takes that value for uncountably many arguments. – copper.hat Oct 29 '12 at 23:26

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