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Let $\cal{H}$ be a (separable) Hilbert space and $\xi_1$, ..., $\xi _n \in \cal H$ be vectors with $\left\Vert \xi_i \right\Vert \leq 1$ for $i= 1, ..., n$. Assume that $\left\Vert \sum_{i=1}^n \xi _i \right\Vert=n$. Does this already imply that the $\xi_i $ are all equal and $\left \Vert \xi_1 \right\Vert=1$?

It seems obvious if one visualizes the situation. But how do I see this rigorously?

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First note that by the triangle inequality we have that $$\|\sum^{n}_{i=1}\xi_{i}\|\leq \sum^{n}_{i=1}\|\xi_{i}\|\leq \sum^{n}_{i=1}1=n$$ Hence $\|\sum^{n}_{i=1}\xi_{i}\|=n$ if and only if $\|\xi_{i}\|=1$. Now suppose $\xi_{i}\neq \xi_{j}$. Note that if there is a $\lambda\in\mathbb{R}$ such that $\lambda\xi_{i}=\xi_{j}$, then $\lambda=\pm 1$ as $$1=\|x_{j}\|=|\lambda|\|\xi_{i}\|=|\lambda|$$ and since $\xi_{i}\neq \xi_{j}$ we have $\lambda=-1$. Then $$\|\sum^{n}_{k=1}\xi_{k}\|=\|\sum_{k\in\{1,...,n\}\setminus\{i,j\}}\xi_{k}\|\leq n-2.$$ So $\xi_{i}$ and $\xi_{j}$ are linearly independent. It follows that \begin{align*} \|\sum^{n}_{k=1}\xi_{k}\|&\leq n-2+\|\xi_{i}+\xi_{j}\|=n-2+\|(1+\langle\xi_{i},\xi_{j}\rangle\xi_{i})\xi_{i}+(\xi_{j}-\langle\xi_{i},\xi_{j}\rangle\xi_{i})\|\\ &=n-2+\sqrt{(1+\langle\xi_{i},\xi_{j}\rangle)^{2}+\|\xi_{j}-\langle\xi_{i},\xi_{j}\rangle\xi_{i}\|^{2}}\\ &\leq n-2+\sqrt{(1+\langle\xi_{i},\xi_{j}\rangle)^{2}+(\|\xi_{j}\|+\|\langle\xi_{i},\xi_{j}\rangle\xi_{i}\|)^{2}}\\ &\leq n-2+\sqrt{2(1+\langle\xi_{i},\xi_{j}\rangle)^{2}} \end{align*} Since by Cauchy-Schwarz we have that $\langle\xi_{i},\xi_{j}\rangle=\|\xi_{i}\|\|\xi_{j}\|\leq1$ with equality if and only if $\xi_{i}$ and $\xi_{j}$ are linearly dependent we have $$\|\sum^{n}_{k=1}\xi_{k}\|\leq n-2+\sqrt{2(1+\langle\xi_{i},\xi_{j}\rangle)^{2}}<n.$$ As this is a contradiction we find that all $\xi_{i}$ are the same.

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  • $\begingroup$ Great, thank you! $\endgroup$ – worldreporter14 Sep 26 at 11:04

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