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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 $x \in \mathbf R$.

QA Ngô
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2 Answers2

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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 $k$.

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. $f'(x)^2 \le 2 f(x)$.

Martin R
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  • It seems that if $f >0$, then the pointwise inequality is true. Here is an argument. By Taylor's expansion, there are numbers $\xi_+$ and $\xi_-$ such that $$f(x+h)=f(x)+f'(x)h +\frac{f''(\xi_+)}2h^2$$ and $$f(x-2h)=f(x)-2f'(x)h +2f''(\xi_-)h^2$$ for $h>0$. Adding these two identities gives $$0 \leq f(x+h)+f(x-2h) = 2f(x) - f'(x)h + h^2 \Big(\frac{f''(\xi_+)}2+2f''(\xi_-)\Big) \leq 2f(x) - hf'(x) +\frac 52 h^2.$$ Hence $$hf'(x) \leq 2f(x) + \frac 52 h^2.$$ Now we choose $h=\sqrt{f(x)}$ to get $$f'(x) \leq \frac 92 \sqrt{f(x)}.$$ What do you think? – QA Ngô Mar 17 '24 at 17:13
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    @QANgô: That can even be simplified: $0 \le f(x+h)=f(x)-f'(x)h +\frac{f''(\xi)}2h^2$ suffices to conclude that $f'(x) \le 2 \sqrt{f(x)}$. But you still need $\sqrt{f(x)} \le C f(x)$. – Martin R Mar 17 '24 at 17:36
  • Thanks. For the reverse inequality, we argue similarly. Indeed, we have $$f(x+2h)=f(x)+2f′(x)h+2f''(\xi_+)h^2$$ and $$f(x−h)=f(x)-f′(x)h+\frac{f''(\xi_−)}2 h^2$$ for $h>0$ . Adding these two identities gives $$0 \leq f(x+2h)+f(x-h) = 2f(x) + f'(x)h + h^2 \Big(\frac{f''(\xi_-)}2+2f''(\xi_+)\Big) \leq 2f(x) + hf'(x) +\frac 52 h^2.$$ Choosing $h=\sqrt{f(x)}$ gives $$f'(x) \geq -\frac 92 \sqrt{f(x)}.$$ – QA Ngô Mar 17 '24 at 17:42
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    @QANgô That is correct but the factor can be improved, see the updated answer. It is a simplified version of your ideas. – Martin R Mar 17 '24 at 19:15
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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 $x$.

QA Ngô
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    Here https://math.stackexchange.com/a/3623256/42969 I reproduced Landau's example of a $C^2$ function that the factor $2$ is best possible. – Martin R Mar 18 '24 at 05:13