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Possible Duplicate:
Implication of an inequality

For $k>0,x\geq y>0$ the following holds $|l(x)| \leq k+|l(y)|,$

for $x \in (y+k-1,y+k)$. Now, why does the same inequality hold also for $x \in (y,y+k)$?

Edit: The different posts where part of the same proof. The objective was to show that a function $l(x)$ is bounded if $l(x+u)-l(x)|<1 \text{ for } x\geq y>0 \text{ and } u\in[0,1]$.

The proof (answers to the previous posts are included) goes as follows:

Let $x=y,\ y+u=z$ for every $z \in (y,y+1)$ and we have $|l(x+u)-l(x)|=|l(z)-l(y)|<1.$

Using the triangle inequality we have:

$|l(z)|=|l(z)-l(y)+l(y)|\leq |l(z)-l(y)|+|l(y)|<1+|l(y)|.$

Using the same argument we have for $x \in (y+1,y+2)$:

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

For positive $k$ we have:

$|l(x)| \leq k+|l(y)|,$

for $x \in (y+k-1,y+k)$.

Now the thing is, that I need this to be true for $x \in (y,y+k)$ as well and I do not understand why it is so.

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    I've edited the question.2012-05-03

0 Answers 0