$|f''(x)|\leq M$ implies $|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|\leq \frac{1}{8} M(b-a)^2$

Question

I am stuck with this problem.

$f : [a,b] \to \mathbb R$ is double differentiable on $[a,b]$ and suppose $|f''(x)|\leq M ~ \forall x \in [a,b]$. Let $c \in (a,b)$. Then $|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|\leq \frac{1}{8} M(b-a)^2$

I can prove this when the upper bound is $M(b-a)^2$.

$|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|=|c-a| |\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}|$. This equals $|c-a||\frac{f'(\xi_1)-f'(\xi_2)}{\xi_1-\xi_2}||\xi_1-\xi_2|$ for some $\xi_1\in (a,c)$ and $\xi_2 \in (a,b)$. This equals $|c-a||f''(\psi)||\xi_1-\xi_2|\leq (b-a)^2M$

Although I know this is weak and I haven't exploited the $\xi '$s fully. I cannot find the solution afterer lots of tries, so any help is appreciated.

You may want to try using Taylor's Theorem. (The link to the theorem FYR: https://en.wikipedia.org/wiki/Taylor%27s_theorem) — Ken Hung, Aug 08 '20 at 15:58
Does this answer your question? Proving inequation with maximum and 2nd derivative – you would apply it to $g(x) = f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}$. — Martin R, Aug 08 '20 at 16:24

score 1 · Accepted Answer · answered Aug 08 '20 at 16:21

1

Hint: The polynomial $$ p(x)=f(a)+(x-a)\frac{f(b)-f(a)}{b-a} $$ interpolates $(a,f(a))$ and $(b,f(b))$. What can you say about $$ f(x)-p(x)-(f(\xi)-p(\xi))\cdot\frac{(x-a)(x-b)}{(\xi-a)(\xi-b)}? $$

answered Aug 08 '20 at 16:21

user10354138

33,239

I dont know interpolates, anyway thanks – Aug 08 '20 at 16:32

$|f''(x)|\leq M$ implies $|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|\leq \frac{1}{8} M(b-a)^2$

1 Answers1