1. ## Derivative

Let $f$ be defined on an interval $I$ of length at least 2 and suppose that $f^{''}$ exists there. If $\vert f(x) \vert \leq 1$ and $\vert f^{''}(x) \vert \leq 1$ for all $x \in I$ show that $\vert f^{'}(x) \vert \leq 2$ on the interval.

2. Originally Posted by Markeur
Let $f$ be defined on an interval $I$ of length at least 2 and suppose that $f^{''}$ exists there. If $\vert f(x) \vert \leq 1$ and $\vert f^{''}(x) \vert \leq 1$ for all $x \in I$ show that $\vert f^{'}(x) \vert \leq 2$ on the interval.
What do you think? It's pretty clear which theorem we should use, no?

3. Mean Value Theorem?How do I start from there?

4. The only way I can see to do this is by contradiction. Suppose that there is a point $\xi\in I$ at which $f'(\xi)>2.$ Start by using the mean value theorem, and the fact that $|f''|\leqslant1$, to show that if $\eta\in I$ then $|f'(\eta) - f'(\xi)|\leqslant|\eta-\xi|$, and therefore $f'(\eta)\geqslant f'(\xi) - |\eta-\xi|$.

Now let $s\in I$ with $s\geqslant\xi$. Then

. . . . . . . \begin{aligned}f(s) &= f(\xi) + \int_\xi^sf'(\eta)\,d\eta \\ &\geqslant f(\xi) + \int_\xi^s\bigl(f'(\xi) - (\eta-\xi)\bigr)\,d\eta \\ &= f(\xi) + (s-\xi)f'(\xi) -\tfrac12(s-\xi)^2.\end{aligned}

A similar calculation shows that if $r\leqslant\xi$ then $f(r)\leqslant f(\xi) - (\xi-r)f'(\xi) + \tfrac12(\xi-r)^2.$ Subtract that inequality from the previous one to get $f(s) - f(r) \geqslant (s-r)f'(\xi) - \tfrac12\bigl((s-\xi)^2 + (\xi-r)^2\bigr).$

Now the interval $I$ has length at least 2, so we can choose r and s so that $s-r=2$. Notice that $(s-\xi)^2 \leqslant (s-\xi)(s-r) = 2(s-\xi)$, and similarly $(\xi-r)^2\leqslant 2(\xi-r).$ Therefore $(s-\xi)^2 + (\xi-r)^2 \leqslant 2(s-r) = 4.$ It follows from the inequality in the previous paragraph that $f(s) - f(r)\geqslant 2f'(\xi) - 2 > 2$ (because $f'(\xi)>2$). But that is impossible if $|f(r)|\leqslant1$ and $|f(s)|\leqslant1.$

That contradiction shows that $f'(x)\leqslant2$ for all $x\in I$. A similar argument shows that $f'(x)\geqslant-2$ (or you can save work by applying the above argument to the function $-f$).