# every Hilbert space has an orthonormal basis

Theorem - Every Hilbert space $H\neq\{0\}$ has an orthonormal basis.

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Proof : As could be expected, the proof makes use of Zorn’s Lemma. Let $\mathcal{O}$ be the set of all orthonormal sets of $H$. It is clear that $\mathcal{O}$ is non-empty since the set $\{x\}$ is in $\mathcal{O}$, where $x$ is an element of $H$ such that $\|x\|=1$.

The elements of $\mathcal{O}$ can be ordered by inclusion, and each chain $\mathcal{C}$ in $\mathcal{O}$ has an upper bound, given by the union of all elements of $\mathcal{C}$. Thus, Zorn’s Lemma assures the existence of a maximal element $B$ in $\mathcal{O}$. We claim that $B$ is an orthonormal basis of $H$.

It is clear that $B$ is an orthonormal set, as it belongs to $\mathcal{O}$. It remains to see that the linear span of $B$ is dense in $H$.

Let $\overline{\mathrm{span}\,B}$ denote the closure of the span of $B$. Suppose $\overline{\mathrm{span}\,B}\neq H$. By the orthogonal decomposition theorem we know that

 $H=\overline{\mathrm{span}\,B}\oplus(\overline{\mathrm{span}\,B})^{\perp}$

Thus, we conclude that $(\overline{\mathrm{span}\,B})^{\perp}\neq\{0\}$, i.e. there are elements which are orthogonal (http://planetmath.org/OrthogonalVectors) to $\overline{\mathrm{span}\,B}$. This contradicts the maximality of $B$ since, by picking an element $y\in(\overline{\mathrm{span}\,B})^{\perp}$ with $\|y\|=1$, $B\cup\{y\}$ would belong belong to $\mathcal{O}$ and would be greater than $B$.

Hence, $\overline{\mathrm{span}\,B}=H$, and this finishes the proof. $\square$

Title every Hilbert space has an orthonormal basis EveryHilbertSpaceHasAnOrthonormalBasis 2013-03-22 17:56:05 2013-03-22 17:56:05 asteroid (17536) asteroid (17536) 4 asteroid (17536) Theorem msc 46C05