# Norm of sum of $n$ vectors with norm $\leq 1$ is equal to $n$

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?

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 $$\|\sum^{n}_{k=1}\xi_{k}\|\leq n-2+\sqrt{2(1+\langle\xi_{i},\xi_{j}\rangle)^{2}} As this is a contradiction we find that all $$\xi_{i}$$ are the same.