# An inequality involving $L^1$ and $L^\infty$ norms of a function of a real variable and its derivative

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:

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

• 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 $N$. – Aleksei Kulikov Dec 31 '18 at 8:09
• @AlekseiKulikov You're right! A hypothesis was missing. I need F to be nondecreasing. I'll correct the question. – 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.