• View
• Print
• Backlinks

First Squeeze Claim:

For any traction {$F$} and any finite composition of tractions, {$G$}, {$$\mathcal S(FG)\le\mathcal S(G)$$}

Proof

The easiest is {$C(a,b,c)=\left(\frac{b+c}2,\frac{a+c}2,\frac{a+b}2\right)$}. In this case, {$\mathcal S(CG)\le \frac12\mathcal S(G)$}. We see, for example, for {$\langle a,b,c\rangle$} and {$\langle a',b',c'\rangle$} in the range of {$G$}, {$d=\mathcal S(G)$}, that {$$\left|\frac{a+c}2 - \frac{a'+c'}2\right|$$} {$$=\left|\frac{1-b}2-\frac{1-b'}2\right|$$} {$$=\frac12|b'-b|\le\frac12d.$$}

The {$L,R,M$} are similar to each other. Here is {$R$} as an example: {$$\left|\frac 1{1+a+b} - \frac 1{1+a'+b'}\right|$$} {$$= \left|\frac 1{2-c} - \frac 1{2-c'}\right|$$} {$$=\left|\frac{c'-c}{(2-c)(2-c')}\right|\le d.$$} and {$$\left|\frac a{1+a+b} - \frac {a'}{1+a'+b'}\right|$$} {$$= \left|\frac {a-a'+ab'-a'b}{(1+a+b)(1+a'+b')}\right|$$} {$$=\left|\frac {a-a'+a(b'-b)+b(a-a')}{(1+a+b)(1+a'+b')}\right|$$} {$$\le\left|\frac {d(1+a+b)}{(1+a+b)(1+a'+b')}\right|$$} {$$\le \frac{d}{1+a'+b'}\le d.$$}