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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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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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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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