3

I've seen, during some proofs, in many texts an argument as the following:

Consider $x\in H,$ $x\neq 0,$ $H$ a complex Hilbert space. The orthonormal set $\{\frac{x}{||x||}\}$ can be extended to an orthonormal basis of $H.$

Another kind of is: if $\{e_{1},\ldots,e_{n}\}$ is an orthonormal set, then such set can be extended to an orthonormal basis.

I know that every Hilbert space has orthonormal basis. Even more, if $H$ is separable, then every orthonormal basis has to be numerable.

I was thinking, in the last case, if we have an independent set, we can use Gram-Schmidt process to get the desire basis, but what about in the above cases? I begin to believe in the use of Zorn's lemma, but I'm not sure.

Any kind of help is thanked in advanced.

Squird37
  • 1,357
  • 7
  • 15

2 Answers2

7

Zorn's lemma is the way to go.

But if you know every Hilbert space has an orthonormal basis, one can deduce that every orthonormal set extends to an orthonormal basis.

Let $(e_i)_{i\in I}$ be an orthonormal set in a Hilbert space $H$. Then the closed linear span $H_1$ of the $e_i$ is a closed subspace of $H$. Then $H$ is the orthogonal direct sum of $H_1$ and $H_2$, where $H_2$ is the orthogonal complement of $H_1$. Then $H_2$ has an orthonormal basis, and the union of this with $(e_i)_{i\in I}$ is an orthonormal basis of $H$,

Angina Seng
  • 161,540
4

If $H$ is a Hilbert space, and $A\subset H$ is an orthonormal set, then we define $$ \mathcal S=\{B\subset H: A\subset B\,\,\text{and}\,\, B\,\,\text{orthonormal set}\} $$ Then, Zorn's Lemma provides that $\mathcal S$ possesses a maximal element (i.e. a maximal orthonormal set, containing $A$), with respect to "$\subset$", which turns out to be an orthonormal basis of $H$.

Yiorgos S. Smyrlis
  • 86,581