proof the existence of a local minima Let $f:(0,1)\longrightarrow \mathbb R$ be a continuous function. $f$ does not "have" (Don't know if it's the proper term, but couldn't find another way to translate it) global extrema and also $f({1\over 2})=f({3\over 4})$. Show that  $f$ "has" a local minima in the interval $(0,1)$.
I thought of using the extreme value theorem in the close interval $[{1\over 2},{3\over 4}]$, but the question seems to be too easy to prove by using that theorem, and also the choice of such interval seems to be useless since using the theorem would make the information about the equality $f({1\over 2})=f({3\over 4})$ become useless.
However, I can't find an explanation why using the theorem isn't legal.
My question is whether the use of this theorem in the given question is legal, and if not, why?
 A: Hint: the condition $f(\frac{1}{2}) = f(\frac{3}{4})$ ensures that there is a local minimum on $[\frac{1}{2}, \frac{3}{4}]$ that is also a local minimum on $(0, 1)$. Apply the argument that you had in mind to the identity function on $(0, 1)$ to see where it goes wrong without that condition.
A: Hint:
If the (local) minimum value is attained on $(\frac{1}{2}, \frac{3}{4})$, it's done.
If only the (local) maximum value is attained on that interval, then since the function has no global extrema, this value must be attained somewhere else.
A: $f$ has a maximum and a minimum on $[{1\over 2},{3\over 4}]$. If the minimum occurs anywhere in the interior, you are finished as it is a local minimum. Otherwise the minimum occurs at the end points, and $f$ is strictly greater than the minimum in the interior. Let $M$ be the maximum value attained on the interval, and let $x_M$ be a point at which $M$ is attained.
Since $f$ has no global extrema, $f$ must take some value $\alpha>M$ on $[0,1]$, at, say, $x_\alpha$. Suppose $x_\alpha < x_M$, then $f$ attains a minimum value on $[x_\alpha, x_M]$. Since ${1 \over 4} \in (x_\alpha, x_M)$ and $f({1 \over 4}) < M$, we see that the minimum value must occur in $(x_\alpha, x_M)$, and hence this is a local minimum. If $x_\alpha> x_M$, the same reasoning applies mutatis mutandis.
