Local maxima or minima of a continuous functions I'm trying to disprove/prove the following claim:
Let $f:[0,1] \to R$ be a continious function, such that:
$$f(0.5) < f(0) < f(1) $$
Prove that there exists a local minima or maxima to $f$.
It seems so simple to prove. Since its continuous then $f(0)$ is a maxima or otherwise there must be a point near it which is, since its point which is lower then others in the given region.
Unfortunately, I failed to figure out a formal proof.
 A: Since your function is continuous in $[0,1]$, it has a minimum and a maximum necessarily.
Lemma 1:
Since $f$ is continuous, there is no $\xi$ such that $\lim _{x\rightarrow \xi} f$ that does not exist or is equal to $\pm \infty$, since for all $\xi \in \mathbb{R}$, the limit exists and is equal to $f(\xi)$. Hence, $f$ is upper and lower bounded.
Take a line parallel to the $x$ axis, that is $y=a\in\mathbb{R}$, and find the $a$ which is the upper bound of $f$. There is no way that $\lim_{x\rightarrow x_0} f=a$ and $f(x_0)\neq a$, since $f$ is continuous. This function is also continuous at 0 and at 1, so you don't have to worry about what happens at 0 and at 1.
In your problem, specifically, you can find a $\xi \in (0, 1)$ such that $f$ has a global minimum at $\xi$. The function cannot have a global minimum at 0, or 1, since you have already found a value that is less than both. But $f$ has a global min, we have shown this before. $\therefore,\ \exists \xi \in (0, 1):\ f $ has a global minimum at $\xi$.
A: Since $f$ is continuous on $[0,1]$, it has at least one global minimum in this interval. Since $f(0.5)\lt f(0)$ and $f(0.5)\lt f(1)$, any such global minimum is not at an endpoint, so it must be a local minimum.
