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Sep 27, 2022 at 11:20 vote accept math110
Feb 7, 2021 at 16:14 comment added Yaakov Baruch @MattF. If I understood your question, the result is trivial since $|x+y|\ge|x|-|y|$ it even follows that $3|2a+b|+3|a+2b|\ge 3|a|+3|b|$.
Feb 5, 2021 at 22:58 comment added LSpice I found the reference to "on $[0, 1]$" in the title confusing since there is no free variable. I did a bit of proofreading in the text, and changed the title so that it did not appear to refer to a free variable but, I think, without otherwise changing its meaning. I hope that this was all right.
Feb 5, 2021 at 22:57 history edited LSpice CC BY-SA 4.0
Miscellaneous TeX; changed title to be hopefully more descriptive
Feb 5, 2021 at 22:42 answer added fedja timeline score: 20
Feb 5, 2021 at 15:16 answer added Iosif Pinelis timeline score: 14
Feb 5, 2021 at 14:44 comment added user44143 Suppose $f(x)=a$ on $[0,\frac12]$ and $f(x)=b$ on $(\frac12,1]$. Then the inequality for $n=3$ is $$\frac{|3a|+3|2a+b|+3|a+2b|+|3b|}{8}\ge \frac{|a|+|b|}{2}$$ which reduces to $$3|2a+b|+3|a+2b|\ge|a|+|b|$$That inequality is true, but is there a way to make it obvious?
Feb 5, 2021 at 12:56 review Suggested edits
Feb 5, 2021 at 13:39
Feb 5, 2021 at 10:07 comment added Neil Strickland Here are some preliminary reductions. Step functions are $L^1$ dense, so we can assume that $f$ is a step function. Then we can apply a measure-automorphism of $[0,1]$ to ensure that $f$ is weakly increasing. Then if we want we can do another $L^1$ approximation to reduce to the case where $f$ is strictly increasing and piecewise-linear. This means that $f=F'$ for some strictly convex $F$, which can be normalised to have $\min(F)=0$. I don't know if any of that helps.
Feb 5, 2021 at 9:52 history edited user44143 CC BY-SA 4.0
put the question at the top
Feb 5, 2021 at 6:28 history asked math110 CC BY-SA 4.0