A quite tricky problem about mean value theorem.

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be twice differentiable. Suppose $f'(0)=f'(1)=2$ and $\forall x \in [0,1]$,$|f''(x)|\leq 4$. Prove that $|f(1)-f(0)|\leq3$.

I know how to prove that when the bound is 4. It simply needs taylor expansion that
$f(1)=f(0)+f'(0)+\frac{f''(c)}{2}$
$f(1)-f(0)\leq2+\frac{f''(c)}{2}$

And the proof can be obtained immediately. But when the bound is 3 it really confuses me. Thank you for any help or advice.

Answer 1

You are not using the fact that $f'(1) = 2$. Better split the interval $[0, 1]$ in two halves and apply Taylor's theorem twice.

First use Taylor's theorem on $[0, 1/2]$ $$ f(\frac 12) = f(0) + \frac 12 f'(0) + \frac 18 f''(c) $$ to estimate $$ |f(0) - f(\frac 12)| \le 1 + \frac 12 \, . $$ Then estimate $|f(1/2) - f(1)|$ in the same way.

Answer 2

Thanks for Martin R's help! With this I can work out the answer.

$f(\frac{1}{2}) = f(1) - \frac{f'(1)}{2} + \frac{f''(b)}{8}$
$f(\frac{1}{2}) -f(1) = - \frac{f'(1)}{2} + \frac{f''(b)}{8} = -1 + \frac{f''(b)}{8}$
$|f(\frac{1}{2}) -f(1)| = |-1+\frac{f''(b)}{8}| \leq |-1|+|\frac{f''(b)}{8}|\leq1 + \frac{4}{8} = 1+\frac{1}{2}$

$f(\frac{1}{2}) = f(0) + \frac{f'(0)}{2} + \frac{f''(c)}{8}$
$f(\frac{1}{2}) - f(0)= \frac{f'(0)}{2} + \frac{f''(c)}{8}$
$|f(\frac{1}{2}) - f(0)|=|1 + \frac{f''(c)}{8}|\leq|1| + |\frac{f''(c)}{8}|\leq1+\frac{4}{8}\leq 1+\frac{1}{2}$

$|f(1)-f(0)| = |f(1)-f(\frac{1}{2}) -(f(0)-f(\frac{1}{2}))|\leq|f(1)-f(\frac{1}{2})|+|f(0)-f(\frac{1}{2})| $
=$ |f(\frac{1}{2})-f(1)| + |f(\frac{1}{2})-f(0)|\leq 1+\frac{1}{2}+1+\frac{1}{2}=3$

This completes the proof.