Bounding $|f'|$ given a bound for $|f|$ and $|f''|$.

Question

I came across this problem on a Berkeley preliminary exam, and have yet to come up with a solution.

Suppose that $f$ is a twice-differentiable real-valued function on the real line such that $|f(x)| \leq 1$ and $|f''(x)| \leq 1$ for all $x$. Find, with proof, a constant $b$ s.t. $|f'(x)| < b$ for all $x$.

So, mean value theorem along with the bound on $f''$ tells us that

$$ |f'(x) - f'(y)| \leq |x - y| $$

but I don't see how to proceed from here. I also tried working with Taylor's theorem, but I didn't get anywhere with that.

Someone told me that given $|f(x)| \leq M$ and $|f''(x)| \leq M'$ then $|f'(x)| \leq \sqrt{M\cdot M'}$, but I've been unable to prove such a result, and my Google abilities have failed me. Any help is appreciated.

Also related: http://math.stackexchange.com/questions/1095378/showing-that-a-function-is-in-l1 — Jack D'Aurizio, Feb 26 '15 at 15:47

score 2 · Accepted Answer · edited Feb 26 '15 at 15:57

For $h > 0$,

$$f(x + h) = f(x) + f'(x)h + \frac{f''(\xi)}{2}h^2$$

for some $\xi \in (x, x + h)$. Thus

$$f'(x) = \frac{f(x + h) - f(x)}{h} - \frac{f''(\xi)}{2}h,$$

which implies

$$|f'(x)| \le \frac{|f(x + h)| + |f(x)|}{h} + \frac{|f''(\xi)|}{2}h \le \frac{2}{h} + \frac{h}{2}.$$

Since $h$ was arbitrary,

$$|f'(x)| \le \inf_{h > 0} \left(\frac{2}{h} + \frac{h}{2}\right) = 2\sqrt{\frac{2}{h}\cdot\frac{h}{2}} = 2.$$

Hence, any $b > 2$ will do.

Bounding $|f'|$ given a bound for $|f|$ and $|f''|$.

1 Answers1