If $|f(x)| \leq 1$ and $|f''(x)| \leq 1$, show $|f'(x)|\leq 2$

Question

Given $f : \mathbb{R} \to \mathbb{R}$, such that $f'(x)$ and $f''(x)$ exist for all $x \in \mathbb{R}$ and for $x \in [0,2]$, the inequalities $|f''(x)| \leq 1$ and $|f(x)| \leq 1$ hold, I am asked to show that for all $x \in [0,2]$, $|f'(x)| \leq 2$.

I am given a hint that says to write down the Taylor expansions of $f(0)$ and $f(2)$ about a point $a \in [0,2]$. With the remainder in Lagrange form, I get:

$$ \begin{align} \exists \xi_0 \in [0,2]\;\;f(0) &= f(a) -af'(a) + \frac{1}{2}a^2f''(\xi_0) \\ \exists \xi_1 \in [0,2]\;\;f(2) &= f(a) +(2-a)f'(a) + \frac{1}{2}(2-a)^2f''(\xi_1) \end{align}$$

I'm not 100% sure on where to go from here. I have tried manipulating the Taylor expansions, using the fact $a \in [0,2]$ to apply the bounds I was given:

$$ \begin{align} f(0) &\leq |f(a) -af'(a) + \frac{1}{2}a^2f''(\xi_0)| \\ &\leq |f(a)| + |af'(a)| + \frac{1}{2}a^2|f''(\xi_0)| \\ &\leq 1 + |af'(a)| + \frac{1}{2}a^2 \end{align} $$

Similarly for $f(2)$: $$ f(2) \leq 1 + |(a-2)f'(a)| + \frac{1}{2}(a-2)^2 $$

But how exactly do I use this to bound $|f'|$? It's probably right there and I've just missed it somehow.

Answer 1

The trick is to subtract the two equations you got to get

$$f'(a) = \frac{f(2) - f(0)}{2} + \frac{1}{4}[a^2f’’(\zeta_0) - (2-a)^2f’’(\zeta_1)]$$

Now using the conditions $|f| \leq 1$, $|f''| \leq 1$ and the triangle inequality we get the desired result

$$|f'(a)| \leq \frac{1+1}{2} + \frac{1}{4}[a^2\cdot 1 + (2-a)^2\cdot 1] \leq 2$$

since $\max_{a\in[0,2]} [a^2 + (2-a)^2] = 4$.

An example of a function satisfying $\max_{a\in[0,2]} |f(a)| = \max_{a\in[0,2]} |f''(a)| = 1$ and where $\max_{a\in[0,2]} |f'(a)| = 2$ is

$$f(x) = \frac{x^2}{2} - 1$$

Answer 2

Hint: $$2\ge f(2)-f(0)=2f'(a)+\frac12((2-a)^2f''(\xi_1)-a^2f''(\xi_0))\ge 2f'(a)-2$$ And similarly for the other bound.