Prove inequalty Let $f\colon \mathbb{R}\rightarrow \mathbb{R}$ ,$f\in C^3$; $f(x)$ and $f^{'''}(x)$ are bounded.Let $$M_0=\underset{x\in\mathbb{R}}{\sup}\left | f(x) \right |, M_3=\underset{x\in\mathbb{R}}{\sup}\left | f^{'''}(x) \right |$$
(a)Prove that $f^{'}(x)$ is bounded và $$\underset{x\in\mathbb{R}}{\sup}\left | f^{'}(x) \right |\leq \frac{1}{2}(9M_0^2M_3)^{\frac{1}{3}}$$
(b) Is $f^{''}(x)$ bounded?
 A: By, the mean value theorem, 
$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{6}f'''(x+\theta h)$
and, $f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)-\frac{h^3}{6}f'''(x-\theta' h)$
Therefore, $f'(x)=\frac{1}{2h}[f(x+h)-f(x-h)]-\frac{h^2}{12}[f'''(x+\theta h)-f'''(x-\theta' h)]$
i.e., $|f'(x)|\le \frac{M_0}{h}+\frac{h^2}{6}M_3$, for $h>0$
$\implies M_1 \le 3\sqrt[3]{\frac{M_0^2M_3}{24}}=\frac{1}{2}\sqrt[3]{9M_0^2M_3}$
Lets establish the following general boundedness for this problem, if $f:\mathbb{R}\to \mathbb{R}$, is a $p$-times diferentiable function, 
then for each $0\leq k\leq p$, lets denote $M_k:=\underset{x\in\mathbb{R}}{\sup}\left | f^{(k)}(x) \right |$, for $k\ge 0$. Then, $M_k\le 2^{\frac{k(p-k)}{2}}M_0^{1-\frac{k}{p}}M_p^{\frac{k}{p}}$.
The base case is easily verifiable since for $h>0$ and $x\in \mathbb{R}$, the MVT's , $f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x+\theta h)$
$f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x-\theta' h)$ for $\theta,\theta'\in (x-h,x+h)$. 
Consequently, $f'(x)=\frac{1}{2h}(f(x+h)-f(x-h))+\frac{h}{4}(f''(x+\theta h)-f''(x-\theta' h))$
which implies, $|f'(x)|\le \frac{M_0}{h}+\frac {h}{2}M_2$, for $h>0$. Take $h=\sqrt{2\frac{M_0}{M_2}}$, to get $M_1\le \sqrt{2M_0M_2}$.
Now, use induction on $p$, 
$f^{(p-1)}(x+h)=f^{(p-1)}(x)+hf^{(p)}(x)+\frac{h^2}{2}f^{(p+1)}(x+\theta h)$
$f^{(p-1)}(x-h)=f^{(p-1)}(x)-hf^{(p)}(x)+\frac{h^2}{2}f^{(p+1)}(x-\theta' h)$.
Consequently, $|f^{(p)}(x)|=\frac{M_{p-1}}{h}+ \frac{h}{2} M_{p+1}$
Choose, $h=\sqrt{2\frac{M_{p-1}}{M_{p+1}}}$ to get $M_p\le \sqrt{2M_{p-1}M_{p+1}}$
Now by induction hypothesis for $k=p-1$, we get $M_p\le 2^{\frac{p}{2}}M_0^{\frac{1}{p+1}}M_{p+1}^{\frac{p}{p+1}}$
By induction hypothesis, for $0\le k\le p-1$, we have $M_k\le 2^{\frac{k(p-k)}{2}}M_0^{1-\frac{k}{p}}M_p^{\frac{k}{p}}$
Combined with the previous result we have, $M_k\le 2^{\frac{k(p+1-k)}{2}}M_0^{1-\frac{k}{p+1}}M_{p+1}^{\frac{k}{p+1}}$, completing our induction claim.
So, $M_0,M_1,M_2$ are bounded.
