TITLE: If $|f(x)|\leq1$ and $|f''(x)|\leq1$, then $|f'(x)|\leq2$.
QUESTION [9 upvotes]: Let $f:\mathbb{R}\to\mathbb{R}$ be twice differentiable, and supose that for all $x\in\mathbb{R},\,|f(x)|\leq1$ and $|f''(x)|\leq1$. Prove that $|f'(x)|\leq2$ for all $x\in\mathbb{R}.$

I tried working with Taylor polynomials but had no successful idea. 
This question is from the book Berkley Problems in Mathematics.

REPLY [7 votes]: Using Taylor polynomials, you can obtain the estimate:
$$\left|f'(x)\right| \le \frac{2}{h}\|f\|_\infty + \frac{h}{2}\left\|f''\right\|_\infty.$$
This is true for any $h>0$, so in particular:
$$\left|f'(x)\right| \le \inf_{h>0}\left(\frac{2}{h}\|f\|_\infty + \frac{h}{2}\left\|f''\right\|_\infty\right) \le \inf_{h}\left(\frac{2}{h} + \frac{h}{2}\right)$$
Note that the RHS is a function on $h$ with a minimium value of $2$.

EDIT:
Answering to @J.Doe's question, note that due to Taylor's theorem the expansion of $f$ around $x=x_0$ is
$$f(x) = f(x_0) + (x-x_0)f'(x_0) + \frac{1}{2}(x-x_0)^2f''(\xi),$$
where $\xi\in\mathbb{R}$ is some point between $x_0$ and $x$. Now, take $h>0$, denote $x=x+h$ and $x_0=x$ in that expresion and reorder it:
$$f(x+h) = f(x) + hf'(x) + \frac{1}{2}h^2f''(\xi) \implies f'(x) = \frac{f(x+h) - f(x)}{h} - \frac{h}{2}f''(\xi).$$
Finally, just consider absolute values and triangular inequality.