A pointwise Hadamard-Landau-Kolmogorov inequality for $C^2$ functions

Question

A well-known interpolation inequality proved by Hadamard (1914) and was generalized by Landau (1913) and Kolmogorov (1939) asserts that $$ |f'(x)|^2 \leq 2 \sup_{x \in \mathbf R} |f(x)|\sup_{x \in \mathbf R} |f''(x)| $$ for any $x$ and for any function $f \in C^2(\mathbf R)$. Let us assume additionally $$|f''(x)| \leq 1 \quad\text{for any } x.$$ Then the inequality becomes $$ |f'(x)|^2 \leq 2 \sup_{x \in \mathbf R} |f(x)| \quad\text{for any } x. $$ My question is: can we have $$ |f'(x)|^2 \leq 2 |f(x)| \quad\text{for any } x. $$ This could be too strong, so a weaker version is also meaningful, namely there is some $C>0$ such that $$ |f'(x)|^2 \leq C |f(x)| $$ for all $x \in \mathbf R$.

Answer 1

No, such a pointwise inequality may not hold, even with an arbitrary constant $C$. A simple counterexample is $f(x) = \sin(x)$ with $$ f'(k \pi) = \pm1 \ne 0 = f(k \pi) $$ for all integers $k$.

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If $f$ is nonnegative then $|f''(x)| \le 1$ for all $x \in \Bbb R$ does imply a pointwise inequality $$ f'(x)^2 \leq 2 f(x) \text{ for all } x \in \Bbb R \, . $$ Proof: We start with Taylor's formula: $$ 0 \le f(x-h) = f(x) - f'(x) h + \frac 12 f''(\xi) h^2 \le f(x) - f'(x)h + \frac 12 h^2 \, . $$ It follows that $$ f'(x) \le \frac{f(x)}{h} + \frac h2 $$ and setting $h = \sqrt{2f(x)}$ gives $$ f'(x) \le \sqrt{2f(x)} \, . $$ Similarly one shows $f'(x) \ge -\sqrt{2f(x)}$ by applying Taylor's formula to $f(x+h)$. This proves $$ -\sqrt{2f(x)} \le f'(x) \le \sqrt{2f(x)} $$ for all $x$, i.e. $f'(x)^2 \le 2 f(x)$.

Answer 2

Thanks, Martin R. The final remark (too long for a comment) is as follows:

• the factor $2$ is optimal in the sense that it cannot be replaced by any smaller positive constant and
• the factor $2$ is never achieved.

For the former conclusion, the optimality can be proved by constructing suitable functions, namely given any $C \in (0, 2)$, there is some positive $C^2$ function $f$ such that $$f'(x)^2 > C f(x).$$ Indeed, let $$f(x)=c+\left\{\begin{aligned} & 0 & & \text{if } x \leq -2,\\ & \frac {(x+2)^2}2 & & \text{if } -2 < x \leq -1,\\ & x+\frac 32 & & \text{if } -1<x \leq 1,\\ & \frac {x^2}2 + 2 & & \text{if } 1<x \leq 2,\\ & 2x & & \text{if } x > 2,\\ \end{aligned}\right.$$ for some $c>0$ to be determined later. (This function is not of class $C^2(\mathbf R)$, but twice differentiable except at $\pm 1, \pm 2$, so it serves our proof.) Then, we get $$f'(x)=\left\{\begin{aligned} & 0 & & \text{if } x \leq -2,\\ & x+2 & & \text{if } -2 < x \leq -1,\\ & 1 & & \text{if } -1<x \leq 1,\\ & x & & \text{if } 1<x \leq 2,\\ & 2 & & \text{if } x > 2,\\ \end{aligned}\right.$$ Hence at $-1$ we should have $$f'(-1)^2 - C f(-1) = 1 - C(c+\frac 12) > 0$$ provided $c \in (0, 1/C-1/2)$.

For the later conclusion, I believe that there is no positive function $f$ such that $f'(x) = 2 f(x)$ occurs at some point $x$.