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Suppose $A$ is a normed vector space and $a_i \in A$ and we have the following sequence $$ \Vert a_1\Vert \leq \Vert a_2 \Vert \leq \cdots \leq \Vert a_n \Vert $$

for any $1\leq i<n-1$. How can I show the following is not necessarily true $$\Vert a_{i+1} -a_i \Vert \leq \Vert a_{i+2} - a_i \Vert ?$$ It's easy to see this is not true for $\mathbb{R}$ any other ways of seeing this.

Thanks.

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3 Answers

up vote 2 down vote accepted

It mostly relies on the triangle inequality.

If you have $$||a_1|| \leq \cdots \leq ||a_n||$$ then by substracting by $||a_i||$ for a given $i$ gives : $$0 \leq ||a_{i+1}|| - ||a_i|| \leq ||a_{i+2}|| - ||a_i||$$

But the triangle inequality only tells you $||a_{i+1}|| - ||a_i|| \leq ||a_{i+1} - a_i||$ and $||a_{i+2}|| - ||a_i|| \leq ||a_{i+2} - a_i||$ respectively. I don't see any way to go any further.

You can visualize this a bit: imagine $a_i$ and $a_{i+1}$ have close norms but "point to" very different directions, and that $a_i$ and $a_{i+2}$ have very different norms but sensibly the same direction. Then in that case you can see it wouldn't work.

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Let $A$ be $\mathbb R^2$ with the Euclidean norm and let $n=3$.

Let $a_1=(0,1)$, $a_2=(1,0)$, and $a_3=(0,2)$.

Then $a_1$ and $a_2$ have norm $1$ and $a_3$ has norm $2$. That is, your condition on the norms is fulfilled.

However, $\Vert a_2-a_1\Vert=\sqrt{2}>1=\Vert a_3-a_1\Vert$.

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That's exactly an illustration of my comment, thanks! –  Alp Mestanogullari Jan 29 '11 at 0:39
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it easy to see this is not true for R any other ways of seeing this

If you already know it is not true for $\mathbb{R}$, you can take any counterexample in $\mathbb{R}$ and multiply by a fixed nonzero vector in $A$ (assuming $A$ has a nonzero vector, of course). For example, multiplying $1,-1$, and $2$ by $x\neq 0$ gives $\|x\|=\|-x\|\lt\|2x\|$ but $\|-x-x\|\gt\|2x-x\|$.

But perhaps by "other ways" you meant that you had already considered this method and wanted a different type of example. In that case, Rasmus's nice example is definitely a better answer.

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+1 Oh, right... I misread the question and thought there were no counterexamples in $\mathbb R$ (and didn't check carefully). Given that it fails already for $\mathbb R$ it's actually not really interesting to look for more complicated counterexamples... Also your example is better in the sense that the desired inequality fails in a more dramatic way (the left side is twice the right side -- that's probably the worst that can happen). –  Rasmus Jan 29 '11 at 2:11
    
@Rasmus: Thanks. Your example is nice, though, as it shows that you don't have use vectors going in opposite directions to get examples. I don't know, maybe $1,-1$, and $1$ is more dramatic. –  Jonas Meyer Jan 29 '11 at 2:20
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