The proof starts like this:
I can't prove this inequality. What I've tried is split into two cases: If $|s-t|\ge 2$ I can prove it. If $|s-t|<2$ I have $\varphi (s)=|s+2m|$ and $\varphi (t)=|t+2n|$ for some integers $m,n$ such that $|m-n|=0$ or $1$. Thus
$$|\varphi (s)-\varphi (t)|=||s+2m|-|t+2n||\le |s+2m-(t+2n)|\le |s-t|+2|m-n|$$
If $|m-n|=1$ I don't know what to do.
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Manuel Ocaña
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Since $\varphi(x)\in[0,1]$, we always have $|\varphi(s)-\varphi(t)|\leq1$. So the only non-trivial case occurs when $|s -t|<1$. In this case, by subtracting an adequate multiple of $2$ from both, we may assume without loss of generality that either $s,t\in[-1,1]$ or $s,t\in[0,2]$:
$s,t\in[-1,1]$. If both $s,t$ have the same sign we have $|\varphi(s)-\varphi(t)|=|s -t|$. If signs are opposite, say $s<0$ and $t>0$, then $$|\varphi(s)-\varphi(t)|=|-s-t|\leq|s -t| . $$
$s,t\in[0,2]$. Now $\varphi(x)=1-|x-1|$. Then $$ |\varphi(s)-\varphi(t)|=\big|\,|t-1|-|s-1|\,\big| $$ and so we may replace $s,t$ with $s-1,t-1$, which takes us to case 1 again.
Martin Argerami
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can you expand on how you assumed that without loss of generality? – Manuel Ocaña Mar 17 '24 at 23:18
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There are $x,y \in (-1,1)$ such that $\varphi(x)=\varphi(t)$ and $\varphi(y) = \varphi(s)$ – ExcitedMathematician Mar 17 '24 at 23:20
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But how do you know that $|x-y|\le |t-s|$ – Manuel Ocaña Mar 17 '24 at 23:26
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You mean $||x|-|y||\leq |s-t|$? – ExcitedMathematician Mar 17 '24 at 23:32
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No, but do you have an answer to that? what is? – Manuel Ocaña Mar 17 '24 at 23:36
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Sure, note that if $|s-t|<2$ we get equality. Otherwise $|s-t|>2$ but $|x-y|\leq 2$ – ExcitedMathematician Mar 17 '24 at 23:54
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@ManuelOcaña: I have changed the answer a bit, to account for the fact that subtracting a multiple of $2$ might not put you in $[-1,1]$. – Martin Argerami Mar 18 '24 at 00:04
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@ExcitedMathematician You get equality where? And why? – Manuel Ocaña Mar 18 '24 at 00:10
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Let $G:\mathbb{R}\rightarrow \mathbb{R}$ be $G(x)=|x-i+1|$ on $[i-2,i]$ for $i\geq 1$ and $G(x)=|x+i+1|$ on $[-i-2,-i]$ for $i\geq 1$
Now $G(x+2)=\varphi(x)$ for all $x$ and $0\leq G \leq 1$
Now a case by case analysis should work.
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The equality $G(x+2)=\varphi(x)$ only works for $x\in[-1,1]$, and so you don't have $|G(s)-G(t)|=|\varphi(s)=\varphi(t)|$ for all $s,t$. The problem, that I had to correct in my answer, is that by subtracting multiples of $2$ you are not guaranteed to get both numbers in $[-1,1]$. Say for example that $s=2.9$, $t=3.8$. Then $s-2=0.9$, $t-2=1.8$, and $s-4=-1.1$. You have $$|\varphi(s)-\varphi(t)|=|0.9-0.2|=0.7,$$ while $$|G(s)-G(t)|=|0.9-1.8|=0.9.$$ – Martin Argerami Mar 18 '24 at 03:45
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@MartinArgerami If $x\in \mathbb{R}$, then $G(x+2)=|(x+2)-2|=|x|=\varphi(x)$. Clearly this holds for all $x\in \mathbb{R}$. – monoidaltransform Mar 18 '24 at 05:30
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not sure what you mean. Take $x=6$, then $G(x+2)=G(8)=|6|=6$, while $\varphi(6)=0$. – Martin Argerami Mar 18 '24 at 05:42
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@MartinArgerami $\varphi(6)=6$. Recall, $\varphi(x)=|x|$. I'm just using $G$ to denote the 'extension' of $\varphi$ to $\mathbb{R}$. – monoidaltransform Mar 18 '24 at 06:20
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no, $\varphi(6)=0$. The function $\varphi$ only agrees with the absolute value on $[-1,1]$. The extension is periodic. – Martin Argerami Mar 18 '24 at 13:14
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