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We know that

$$|l(x+u)-l(x)|<1 \text{ for } x\geq y>0 \text{ and } u\in[0,1]$$

Why does:

$$|l(y+u)|<1+|l(y)|,x \in (y+1,y+2)$$

imply that

$$|l(x)| \leq 1 + |l(y+1)| \leq 2+|l(y)|$$

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What part of this question is not covered by @Joseph's answer below? – Did May 3 '12 at 12:54
The part that Antonio Vargas explained below. – Chris May 3 '12 at 13:48
up vote 1 down vote accepted

This is true for any inequality of the form $|a-b|<1$. Since $|a-b|=|b-a|$ we can say both that $|a|<1+|b|$ and $|b|<1+|a|$. Then combining the two statements give that $|b|<1+|a|<1+(1+|b|)=2+|b|$

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Ok, that explains the second part of the last double inequality, but why does $|l(x)| \leq 1 + |l(y+1)|$ follow? – Chris May 3 '12 at 3:17
what is $l(x)$? an arbitrary function? or a linear function? – Joseph Skelton May 3 '12 at 3:26
@Chris, if $x \in (y+1,y+2)$ then $x = y+1+u$ for some $u \in [0,1]$. Your assumption then gives $$ |l(x)| = |l(y+1+u)| < 1 + |l(y+1)|. $$ – Antonio Vargas May 3 '12 at 5:18
Ahhh, got it now! – Chris May 3 '12 at 13:43

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