If $f\in C^2(\mathbb R)$ then $M_1^2 \le 2M_0 M_2$, where $M_k = \text {sup}_x |(d/dx)^k f(x)|$ for $k=0,1,2.$

Question

I wanna prove this problem. I tried it with Mean Value Theorem but cannot proceed to any plausible result. So could I have some hints?

Answer 1

Mean value theorem generally doesn't work on problems involving $C^2$ functions, but the generalization does. From Taylor's theorem, for any $x,h \in \mathbb R$, \begin{align*} f(x + h) &= f(x) + h\cdot f'(x) + h^2 \cdot \frac{f''(\zeta_1)}{2} \\ f(x - h) &= f(x) - h\cdot f'(x) + h^2 \cdot \frac{f''(\zeta_2)}{2} \end{align*} for some $\zeta_1$ between $x$ and $x + h$ and $\zeta_2$ between $x$ and $x - h$. Subtracting, \begin{align*} f(x+h) - f(x-h) &= 2h \cdot f'(x) - \frac{h^2}{2} \left(\,f''(\zeta_1) - f''(\zeta_2) \right) \end{align*} Solving for $f'(x)$ and bounding, \begin{align*} |f'(x)| &= \left| \frac{1}{2h} \left( f(x+h) - f(x-h) \right) + \frac{h}{4}\left(\,f''(\zeta_1) - f''(\zeta_2) \right) \right| \\ &\leq \frac{M_0}{h} + \frac{hM_2}{2} \end{align*} Taking the sup over all $x$ gives $$M_1 \leq \frac{M_0}{h} + \frac{hM_2}{2}.$$ Rearranging and simplifying, $$M_2h^2 - 2M_1h + 2M_0 \geq 0.$$ I'll let you take it from here.