Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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
add comment

1 Answer

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 $|b|<1+|a|<1+(1+|b|)=2+|b|$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.