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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.

  • 0
    What is $l(x)$?2012-05-03
  • 1
    Suppose $x \in (y,y+1)$. Then $|l(x)| \leq 1 + |l(y)| \leq k + |l(y)|$. Can you fill in the rest?2012-05-03
  • 0
    Why is this not a [duplicate](http://math.stackexchange.com/q/140215/6179)?2012-05-03
  • 0
    @Didier, I suspect it would take at least as much work to explain why it is a duplicate as to just solve it.2012-05-03
  • 1
    @Gerry I see no reason to encourage OPs to ask over and over basically the same question.2012-05-03
  • 0
    Agreed. I'm not convinced OP knows it's the same question. Perhaps the equivalence can be explained to OP.2012-05-03
  • 0
    I've edited the question.2012-05-03

0 Answers 0