Pythagorean Theorem (Hilbert Space)

Theorem

Let $H$ be a Hilbert space with inner product norm $\left\|{\cdot}\right\|$.

Let $f_1, \ldots, f_n \in H$ be pairwise orthogonal.

Then:

$\displaystyle \left\|{\sum_{i=1}^n f_i}\right\|^2 = \sum_{i=1}^n \left\|{f_i}\right\|^2$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\Vert{\sum_{i=1}^n f_i}\right\Vert^2\)	\(=\)	\(\displaystyle \left\langle{\sum_{i=1}^n f_i, \sum_{j=1}^n f_j}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of inner product norm
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^n \sum_{j=1}^n \left\langle{f_i, f_j}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Linearity of inner product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^n \sum_{j=1}^n \begin{cases}\left\langle{f_i, f_i}\right\rangle & i = j\\ 0 & i \ne j\end{cases}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	The $f_i$ are pairwise orthogonal
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^n \left\Vert{f_i}\right\Vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of inner product norm

$\blacksquare$

Note

If $H$ is $\R^2$ with the usual inner product, and $n=2$, this theorem reduces to the well-known Pythagoras's Theorem.

Sources

John B. Conway: A Course in Functional Analysis (1990)... (previous)... (next) $I.2.2$