I got to the following inequality by a (hopefully correct) tortuous argument:

If $F:[a,b] \to \mathbb{R}$ is a absolutely continuous monotone function then: $$ \|F'\|_1^2 \leq 4 \|F\|_1 \, \|F'\|_\infty $$

Question 1: Does this inequality have a name?

Question 2: Is there a short proof?

Remark: Cases of equality should be with $\pm F$ like that:

enter image description here

(The plateau in the middle may vanish; in this case $F$ is affine with zero mean.)

  • 1
    $\begingroup$ Are you sure your inequality is correct? Take $[a, b] = [0, N]$ for big $N$ and let $F(x) = \sin(x)$. Then l.h.s. is of order $N^2$, while r.h.s is of order $N$. $\endgroup$ – Aleksei Kulikov Dec 31 '18 at 8:09
  • $\begingroup$ @AlekseiKulikov You're right! A hypothesis was missing. I need F to be nondecreasing. I'll correct the question. $\endgroup$ – Jairo Bochi Dec 31 '18 at 9:32

For question 2: assume w.l.o.g. $F$ nondecreasing. If $F(a)\ge 0$ $$\|F'\|_1^2=(F(b)-F(a))^2\le F(b)^2-F(a)^2=\int_a^b 2FF'\le 2\|F\|_1\|F'\|_\infty.$$ Similarly if $F(b)\le 0$. If $F$ changes sign, take $c\in[a,b]$ such that $F(c)=0$ and apply the above to $F$ restricted to $[a,c]$ and $[c,b]$ (using the fact that $(A+B)^2\le 2A^2+2B^2$ and the additivity of the $L^1$-norm).

For equality to occur, $F$ must change sign and we must have $|F'|=\|F'\|_\infty$ a.e. on the two intervals where F is nonzero. Also, the $L^1$-norm of $F$ must be the same on the two intervals, so the equality case is precisely the one mentioned in the question.


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