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Referring to "Aronszajn's Criterion for Euclidean Space " by R.D. Arthan: Can any one help me to understand why in lemma one $\|(a+1)p+q\| = \|(a+1)p-q\|$ and again how $\|ap+(b+1)q\|=\|ap-(b+1)q\|$?

Another question is: In lemma 2, sorry but I really don't understand what the following sentence mean "$x$ traverses an arc of the unit circle from $e_1$ to $-e_1$", and why $\|e_1+x\|-\|e_1-x\|$?

Thanks to any one can who can simplify this to me.

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He’s just applying the Aronszajn criterion with

$$\begin{align*} \mathrm{v}_1&=a\mathrm{p}+\mathrm{q}\\ \mathrm{v}_2&=a\mathrm{p}-\mathrm{q}\\ \mathrm{w}_1&=\mathrm{p},\text{ and}\\ \mathrm{w}_2&=\mathrm{p}\;: \end{align*}$$

then $\mathrm{v}_1-\mathrm{w}_1=a\mathrm{p}+\mathrm{q}-\mathrm{p}=(a-1)\mathrm{p}+\mathrm{q}$ and similarly $\mathrm{v}_2-\mathrm{w}_2=(a-1)\mathrm{p}-\mathrm{q}$, so the hypotheses of the Aronszajn criterion are satisfied, and we conclude that $\|\mathrm{v}_1+\mathrm{w}_1\|=\|\mathrm{v}_2+\mathrm{w}_2\|$, i.e., that $\|(a+1)\mathrm{p}+\mathrm{q}\|=\|(a+1)\mathrm{p}-\mathrm{q}\|$. The calculation at the bottom of page $2$ is similar.

In Lemma $2$ the unit circle is simply $S=\{\mathrm{v}\in V:\|\mathrm{v}\|=1\}$. Define $$f(\mathrm{x})=\|\mathrm{e}_1+\mathrm{x}\|-\|\mathrm{e}_1-\mathrm{x}\|\;.$$

When $\mathrm{x}=\mathrm{e}_1$,


and when $\mathrm{x}=-\mathrm{e}_1$,


since $\|\mathrm{e}_1\|=1$. $S$ is connected, and $f$ is a continuous function, so $f[S]$ is a connected subset of $\Bbb R$ containing $-2$ and $2$; as such it must also contain $0$.

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I am the mysterious "R." of the question. Oday e-mailed me and I suggested he post his question so the community could share the answer. And when I got the office Brian had answered it for me! – Rob Arthan Feb 27 '13 at 10:51
Firstly I want to thank you Brian too much, but I am going to ask more questions so I ask you to be patient and if you continuou to simplify to me. – LoveMath Feb 27 '13 at 10:57
now, could you please continuo explain the rest of lemma 2 and also lemma 3 – LoveMath Feb 27 '13 at 10:59
@user50382: It would probably be better to post this as a separate question. And if Rob is willing, it would definitely be better for him to answer: I’m not at all familiar with this material, and it’ll take me a fair bit of work to sort through and explain it. – Brian M. Scott Feb 27 '13 at 11:02
@Rob: There seem to be more questions coming, and since this is material with which I’m not very comfortable, I’d be happy to turn them over to you! – Brian M. Scott Feb 27 '13 at 11:03

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