1. ## Derivative

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

2. Originally Posted by Markeur
Let $\displaystyle f$ be defined on an interval $\displaystyle I$ of length at least 2 and suppose that $\displaystyle f^{''}$ exists there. If $\displaystyle \vert f(x) \vert \leq 1$ and $\displaystyle \vert f^{''}(x) \vert \leq 1$ for all $\displaystyle x \in I$ show that $\displaystyle \vert f^{'}(x) \vert \leq 2$ 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 $\displaystyle \xi\in I$ at which $\displaystyle f'(\xi)>2.$ Start by using the mean value theorem, and the fact that $\displaystyle |f''|\leqslant1$, to show that if $\displaystyle \eta\in I$ then $\displaystyle |f'(\eta) - f'(\xi)|\leqslant|\eta-\xi|$, and therefore $\displaystyle f'(\eta)\geqslant f'(\xi) - |\eta-\xi|$.

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

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

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

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