Suppose that $|f(x)|\le1$ and $|f'''(x)|\le3$ for any $x\in \mathbb{R}$. Prove that $|f'(x)|\le 3/2$ for any $x\in \mathbb{R}$.

Asked May 17 '23 at 13:27

Active May 17 '23 at 13:44

Viewed 47 times

I encountered this problem: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function which is three times differentiable on $\mathbb{R}$. Suppose that $|f(x)|\le1$ and $|f'''(x)|\le3$ for any $x\in \mathbb{R}$. Prove that $|f'(x)|\le \frac{3}{2}$ for any $x\in \mathbb{R}$. (Hint: Apply Taylor’s Theorem to f on $[x-h,x]$ and on $[x,x+h]$ for any $x\in\mathbb{R}$ and $h>0$ .)

I think the main trick will be in both intervals; degree 2 is the same, so we can derive the inequality condition. However, I cannot figure out the correct way to use $|f(x)|\le1$ and $|f'''(x)|\le3$. Can anyone help me? Thanks in advance!

asked May 17 '23 at 13:27

Lumos

Here's a similar question that should be of great assistance. You need to go one level deeper than the given question because you have $f'''$ instead of $f''$, but the idea is the same. – Sarvesh Ravichandran Iyer May 17 '23 at 13:33
Also see this question. – Sarvesh Ravichandran Iyer May 17 '23 at 13:34
1

All these estimates are particular cases of the Landau–Kolmogorov inequality. – Martin R May 17 '23 at 13:53

Suppose that $|f(x)|\le1$ and $|f'''(x)|\le3$ for any $x\in \mathbb{R}$. Prove that $|f'(x)|\le 3/2$ for any $x\in \mathbb{R}$.

0 Answers0