Proof: Let any element
of be given; by definition, each element of may be approximated by such elements. Let . Then by definition of an orthonormal basis, we find for and for and then resp. such that
- and .
Then note that by the triangle inequality,
- .
Now fix . Then by the triangle inequality,
In total, we obtain that
(assuming that the given sum approximates well enough) which is arbitrarily small, so that the span of tensors of the form is dense in .
Now we claim that the basis is orthonormal. Indeed, suppose that . Then
- .
Similarly, the above expression evaluates to when and . Hence, does constitute an orthonormal basis of .