# Hilbert space: Proof the formula similar to Parallelogram Law

Let $H$ be a Hilbert space, and $x_1,x_2,x_3\in H$. I have to show that $$\|x_1-x_2\|^2+\|x_1-x_3\|^2+\|x_2-x_3\|^2=3(\|x_1\|^2+\|x_2\|^2+\|x_3\|^2)-\|x_1+x_2+x_3\|^2.$$ I think it similar to Parallelogram Law, we can expand everything in terms of $⟨\cdot,\cdot⟩$ to verify.

And more general, let $x_1,\ldots,x_n\in H$, we have $$\sum_{i<j}\|x_i-x_j\|^2=n\sum_{i=1}^n\|x_i\|^2-\left\|\sum_{i=1}^nx_i\right\|^2.$$

• You can always expand everything in terms of $\langle \cdot, \cdot \rangle$'s and verify. – user384138 Dec 26 '16 at 2:09

\begin{align} &\hspace{6mm}\|x_1-x_2\|^2+\|x_1-x_3\|^2+\|x_2-x_3\|^2+\|x_1+x_2+x_3\|^2\\ &=\langle x_1-x_2,x_1-x_2\rangle + \langle x_1-x_3,x_1-x_3\rangle+\langle x_2-x_3,x_2-x_3\rangle\\ &\hspace{4mm}+\langle x_1+x_2+x_3,x_1+x_2+x_3\rangle\\ &= \|x_1\|^2-\langle x_2,x_1\rangle - \langle x_1,x_2\rangle+\|x_2\|^2\\ &\hspace{4mm}+\|x_1\|^2-\langle x_3,x_1\rangle - \langle x_1,x_3\rangle+\|x_3\|^2\\ &\hspace{4mm}+\|x_2\|^2-\langle x_2,x_3\rangle - \langle x_3,x_2\rangle+\|x_3\|^2\\ &\hspace{4mm}+\|x_1\|^2+\|x_2\|^2+\|x_3\|^2\\ &\hspace{4mm}+\langle x_1,x_2\rangle + \langle x_1,x_3\rangle+\langle x_2,x_1\rangle\\ &\hspace{4mm}+ \langle x_2,x_3\rangle+\langle x_3,x_1\rangle + \langle x_3,x_2\rangle\\ &=3\bigl(\|x_1\|^2+\|x_2\|^2+\|x_3\|^2\bigr)\,. \end{align}

In general:

$$\Biggl(\sum_{\substack{i,j\\1\leq i<j\leq n}}\|x_i-x_j\|^2\Biggr)+\Biggl\|\sum_{i=1}^n x_i\Biggr\|^2=n \sum_{i=1}^n \|x_i\|^2\,.$$

• I realized the question now include the general case that I posted and also the comment from @OpenBall. Are you able to solve it? – Siong Thye Goh Dec 26 '16 at 10:21