Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 15

Problem

Suppose $a \in \mathbb{R}^1$, $f$ is a twice-differentiable real function on $(a,\infty)$, and $M_0,M_1,M_2$ are the least upper bounds of $|f(x)|,|f'(x)|,|f''(x)|$, respectively, on $(a,\infty)$. Prove that

$$ M_1^2 \leq 4M_0M_2.$$

Does $M_1^2 \leq 4M_0M_2$ hold for vector-valued functions too?

Answer

For $x \in (a,\infty)$ and $h>0$, $f(x+2h)=f(x)+2f'(x)h + \frac{1}{2}f''(\xi)(2h)^2$ for some $\xi$ between $x$ and $x+2h$ by Theorem 5.15. Then $|f'(x)| \leq \frac{|f(x+2h) - f(x)|}{2h} + |f''(\xi)h| \leq \frac{M_0}{h} + M_2h$. This holds for all $x$ and $h$ and

$$ \inf_h(\frac{M_0}{h} + M_2h) = 2\sqrt{M_0M_2}$$

Hence, we have $M_1 \leq 2\sqrt{M_0M_2}$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2.$

This inequality also holds for vector-valued functions $f=(f_1,\cdots,f_k)$. We may assume that $M_1 \neq 0$. For a fixed point $y \in (a,\infty)$, define a function

$$ g(x):=f_1'(y)f_1(x) + \cdots + f_k'(y)f_k(x). $$

Then we have

$$|g'(x)|^2 \leq (\sup_x |g'(x)|)^2 \leq 4(\sup_x|g(x)|)(\sup_x|g''(x)|) \\ \leq 4(\sum_{i=1}^k|f_i'(y)|^2)^{1/2}\sup_x(\sum_{i=1}^k|f_i(x)|^2)^{1/2}(\sum_{i=1}^k|f_i'(y)|^2)^{1/2}\sup_x(\sum_{i=1}^k|f_i''(x)|^2)^{1/2} \leq 4 M_1^2M_0M_2.$$

For $y=x$, we have $(\sum_{i=1}^k|f_i'(x)|^2)^2 \leq 4 M_1^2M_0M_2$. This holds for all $x$, and hence $M_1^4 = (\sup_x\sum_{i=1}^k|f_i'(x)|^2)^2 \leq 4 M_1^2M_0M_2$  $\Rightarrow$  $M_1^2 \leq 4M_0M_2$.

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