Rudin Principles of Mathematical Analysis Chapter 5 Exercise 15 I tried to solve it using generalized mean value theorem (Theorem 5.9 in the book), but ran into a wrong conclusion.
Let $f$ be twice-differentiable real function on $(a, \infty)$. Let $M_{0}, M_{1}, M_{2}$ be the least upper bounds of $|f(x)|, |f'(x)|, |f''(x)|$, respectively, on $(a, \infty)$. Prove that $$M_{1}^{2} \leq 4M_{2}M_{0}$$
Fix $y\in (a,\infty)$. Then for any $z\in(a,\infty)$ where $y<z$, for some $t\in(y,z)$,
$$[f(z)-f(y)]f'(t)=[f'(z)-f'(y)]f(t)$$
Since $y\neq z$,
$$\frac{[f(z)-f(y)]f'(t)}{z-y}=\frac{[f'(z)-f'(y)]f(t)}{z-y}$$
If I take the limit as $z$ goes to $y$, then it becomes:
$$[f'(y)]^{2}=f''(y)f(y)$$
which implies:
$$[f'(y)]^{2} \leq M_{2}M_{0}$$ and therefore
$$M_{1}^{2} \leq M_{2}M_{0}$$
But clearly, this is wrong from the hint in the book since $M_{1}^{2}= 4M_{2}M_{0}$ can actually happen. I'm not sure what I did wrong. Could you point to where the error occurred?

Your error is in this line
$$[f(z)-f(y)]f'(t)=[f'(z)-f'(y)]f(t)$$
which is not true.
