Thread: Prove that f(x)=f(x^2)

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We have function f:[0,1]->R, which is continuous and f(0)=f(1). Prove that exsists x (0,1) such that f(x) = f(x^2).

I even don't know how to start solving this...

Hi happygirl!

I'd start with trying to find an example where it works out.
After that we'll see.

So, suppose f(x) = C for some constant C.
Does it have an x in (0,1) such that f(x) = f(x^2)?

Suppose f(x) = |x - (1/2)|.
Does it have an x in (0,1) such that f(x) = f(x^2)?

First note that since f(x) is continuous, it has a maximum and a minimum on the closed interval [0,1]. (One or both of them might be equal to f(0) = f(1)). However, we can assume the function is non constant, otherwise it is trivial and the result would follow immediately. Therefore, f(x) has either an absolute minimum or an absolute maximum on the open interval (0,1), or both.

Let g(x) = f(x) - f(x^2)

Then g(0) = g(1) = 0. You now need to show that there is some other x such that g(x) = 0.

Il show this by contradiction. Lets assume that there is no such x. That means that either f(x) > f(x^2) for all x in (0,1) OR f(x) < f(x^2) for all x in (0,1), due to the intermediate value theorem.

Assume f(x) > f(x^2) for all x in (0,1). If f(x) has an absolute maximum, occuring at x = k, then this inequality will by definition fail to hold for x = sqrt(k). Similarly, if f(x) has an absolute minimum, occuring at x=k, then the inequality will fail to hold for x = k. Therefore, it is not possible that f(x) > f(x^2) for all x in (0,1).

Equivalently, assume f(x) < f(x^2) for all x in (0,1). If f(x) has an absolute maximum, occuring at x = k, then this inequality will by definition fail to hold for x = k. Similarly, if f(x) has an absolute minimum, occuring at x = k, then the inequality will fail to hold for x = sqrt(k). Therefore, it is not possible that f(x) < f(x^2) for all x in (0,1).

Since neither of those is possible, it has to be the case that f(x) = f(x^2) somewhere in (0,1).