Bound on first-order derivative I got stuck on the following problem and I hope someone can give me a hint:
Let $f$ be twice-differentiable on $(0, \infty)$. Define
$$M_0\equiv \sup_{x>0}|f(x)|, \quad M_1\equiv\sup_{x>0}|f'(x)|, \quad M_2\equiv\sup_{x>0}|f''(x)|.$$
If both $M_0$ and $M_2$ are finite, prove that $M_1\le\sqrt{2M_0M_2}$.
My situation so far:
The finest bound on $M_1$ I can find is $2\sqrt{M_0M_2}$, which is looser than required and is obtained through the following argument:
By differentiability, for all $x, >0$ we have, by Taylor's theorem,
$$f(y)=f(x)+f'(y)(y-x)+\frac{f''(\xi)}{2}(y-x)^2$$
for some $\xi$ between $x$ and $y$. Thus,
\begin{align}
2M_0 &\ge|f(x)-f(y)| \\
&=\bigg|f'(y)(y-x)+\frac{f''(\xi)}{2}(y-x)^2\bigg| \\
&\ge|f'(y)|\cdot|y-x|-\frac{1}{2}|f''(\xi)|\cdot |y-x|^2 \\
&\ge |f'(y)|\cdot|y-x|-\frac {1}{2} M_2 |y-x|^2
\end{align}
from which we conclude (for $x \ne y$)
$$|f'(y)|\le \frac{2M_0}{|y-x|}+\frac 12\,M_2\,|y-x|$$
Since for any given $y$, $x$ can be chosen freely, $|y-x|$ can be any nonnegative real number, which implies that
$$|f'(y)|\le \inf_{a>0}\bigg\{\frac{2M_0}{a}+\frac 12\,M_2\,a\bigg\}=2\sqrt{\frac{2M_0}{a}\cdot \frac 12\,M_2\,a}=2\sqrt{M_0M_2}$$
It seems that I have made full use of all conditions given, and I suspect whether the proposition I am asked to prove is true (but I failed to find a counterexample). Can anyone give me a hint? Many thanks!
 A: Your problem is similar to Problem 5.15 from Rudin's Principles of Mathematical Analysis, where the problem is to show that $M_1^2\le 4M_0M_2$ for $f:(a,\infty)\rightarrow\mathbb{R}$ which are bounded and have two bounded derivatives, where $a\in\mathbb{R}$. In the problem statement, Rudin mentions that equality can be achieved, e.g. for
$$ f(x) = \left\{\begin{matrix} 2x^2-1 &\text{ for }-1<x<0 \\ \frac{x^2-1}{x^2+1} &\text{ for }x\ge 0\end{matrix}\right.. $$
In this case, the domain of $f$ is $(-1,\infty)$. So the bound that you got is actually tight, and in particular the proposition you are asked to show is false.
A: To add to @Joey Zou's answer, the hint from the book is

If $h>0$, Taylor's theorem shows that $$ f'(x)=\frac1{2h}(f(x+2h)-f(x)]-hf''(\xi)$$ for some $\xi\in(x,x+2h)$. Hence $$ |f'(x)|\leqslant hM_2 + \frac{M_0}h.$$

From there it follows that $$h^2M_2 - h|f'(x)| + M_0 \geqslant 0, $$
and since this is a quadratic in $h$ which is strictly nonnegative, the discriminant must be nonpositive, i.e.
$$|f'(x)|^2 - 4M_2M_0\leqslant 0. $$
Since $x$ is arbitrary we conclude that
$$M_1 \leqslant 2\sqrt{M_0M_2}. $$
