Let $f: [-1, 1] \longrightarrow [-1, 1]$ such that $f$ is a class $C^{1}$ function. Prove that there's exist $x_{0} \in [-1, 1]$ such that $|f'(x_{0})| \leq 1$.
I know that $f'([- 1,1])$ is compact, since $f'$ is continuous. Therefore, it is closed and limited. To prove the result, I tried to use the continuity of $f'$ in some sequence and tried to use the Weierstrass theorem, but I could not conclude anything. I would like some suggestion.
On
Assume your claim is false. Your claim combined with the continuity of $f'$ implies that either $f'(x)>1$ is always true or that $f'(x)<-1$ is always true. Also, since $f'$ is continuous, we can take its integral: $$f(1)-f(-1) = \int_{-1}^1 f'(x)\,dx$$ If $f'(x)>1$ then $f(1)-f(-1)>2$. Similarly if $f'(x)<1$ then $f(1)-f(-1)<-2$. Using the triangle inequality, in both cases we have a contradiction.
On
This follows directly from the Mean Value Theorem. In fact, this means that the result does not depend on the continuity of $f'$. By the MVT:
$\frac{|f(1)-f(-1)|}{|2|}=|f(\alpha)|$ for some $\alpha \in (-1,1)$.
But $f(1)$ and $f(-1)$ are contained in $[-1,1]$. Then, $|f(1)-f(-1)|\leq 2$, and the result follows.
Anyway, I believe the desired solution comes from the Fundamental Theorem of Calculus (and then, relies on the continuity of $f'$ to assure it's Riemann-integrability):
Assume by contradiction that $|f'(x)|>1$ for all $x \in [-1,1]$. Let's first assume that $f'$ is positive. By the FTC:
$f(1)-f(-1)=\int_{-1}^1 f'(t)dt$. But, since $f'(x)>1$,it follows that: $f(1)-f(-1)> 2$, which is a contradiction since $f([-1,1])=[-1,1]$. The case where $f'$ is negative follows similarly.
Note that $$\frac{f(1)-f(-1)}{2} = f^\prime(\xi)$$ for some $\xi \in (-1,1)$