Prove inequalty

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.