Let $f:\mathbb R\to \mathbb R$ be a three-times differentiable function. Suppose $|f(x)|\le 1$ and $|f'''(x)|\le 3$ for any $x\in \mathbb R$. Show that $|f'(x)|\le 1$ for any $x\in \mathbb R$.
The hint to this question is to apply Taylor's theorem on $[x-h,x]$ and $[x,x+h]$ for any $x\in \mathbb R$ and $h>0$. I have $$ \begin{align} f(x-h)&=f(x)-f'(x)h+{f''(x)\over 2}h^2-{f'''(w_1)\over 6}h^3, \\ f(x+h)&=f(x)+f'(x)h+{f''(x)\over 2}h^2+{f'''(w_2)\over 6}h^3 \end{align} $$ Hence $$f(x+h)-f(x-h)=2hf'(x)+{f'''(w_1)+f'''(w_2)\over 6}h^3.$$ However, I don't know how to proceed from here. I only know that $|f(x+h)-f(x-h)|\le 2$. Can anyone help?
$|f'(x) |\le 1$ does not necessarily hold under the given conditions, as the function $f(x) = \sin(ax)$ with $a^3 = 1$ shows.
We can show $|f'(x) |\le 3/2$: $$ f'(x) = \frac{f(x+h)-f(x-h)}{2h} - \frac{f'''(w_1)+f'''(w_2)}{12} h^2 $$ so that $$ |f'(x)| \le \frac 1h + \frac{h^2}{2} $$ for all $h > 0$. Setting $h=1$ gives $|f'(x) |\le 3/2$.
This is a special case of the Landau–Kolmogorov inequality: The sharp bound is $$ \Vert f' \Vert_\infty \le C(3, 1, \Bbb R) \cdot \Vert f \Vert_\infty^{2/3} \cdot \Vert f''' \Vert_\infty^{1/3} $$ where $$ C(3, 1, \Bbb R) = K_2 K_3^{-2/3} $$ and the $K_j$ are the Favard constants. This gives $$ \Vert f' \Vert_\infty \le \frac 12 3^{2/3} \cdot \Vert f \Vert_\infty^{2/3} \cdot \Vert f''' \Vert_\infty^{1/3} \\ \le \frac 12 3^{2/3} \cdot 1^{2/3} \cdot 3^{1/3} = \frac 32 \, . $$
So the upper found $|f'(x) |\le 3/2$ that we determined is indeed best possible.