Let $ g \in C^2 [x,y] $ and $ P $ be its interpolation linear polynomial at $ a_0 $ and $ a_1 $ in $ [x,y] $. Prove that $ \lVert g-P\rVert _\infty < 1/8(a_1-a_0)^2B $ where $ \lvert g''(x)\rvert<B $ for all $ x \in [a,b] $.
I can't think of how to prove this. I am not sure which method to use.
Let $g(x)=(x-a_0)(x-a_1)$, $g(x)$ is not bounded when $x \mapsto \infty$. So there is a real number $u$ such that $g(u)>1/4 (a_0-a_1)^2$, and such that $a_0,a_1 \in [-u,u].$
For all $x \in [-u,u], g''(x)=2$, so $B=2$.
$g(a_0)=g(a_1)=0$ implies $P=0$.
So $\|g-P\|_{\infty}\geq g(u)>1/4 (a_0-a_1)^2=1/8(a_0-a_1)^2B$
So, it seems to be false.