4

Let $H$ be a hilbert space. And let$ B$ be a basis of $H$. I think a orthonormal set$ S$ to be a basis iff $|S| =|B|$. (But I'm not sure about this) Am I right? If this is wrong, is the same argument right under the condition $H$ is separable?

The original problem is this.

Let ${e_n}$ is an orthonormal basis for a separable Hilbert space. ${f_n}$ is an orthnormal set such that $\Sigma ||e_n - f_n|| <1 $. Then prove ${f_n}$ is a basis.

Can I ask you a Hint for this? Thank you :)

user277793
  • 237

2 Answers2

3

Your argument is not correct, even when $H$ is a separable Hilbert space. Take, as a counter example, $H=\ell^2(\mathbb{N})$. Then, $(e_i)_{i \in \mathbb{N}}$ is an orthonormal basis, but $(e_{2i})_{i \in \mathbb{N}}$ is not.

For your original problem, see this question: orthonormal system in a Hilbert space

Community
  • 1
User
  • 742
2

HINT:

Let $v$ be a vector perpendicular to all the $f_n$'s.

We have $$||v||^2 = \sum_n |\langle v, e_n\rangle|^2 = \sum_n |\langle v, e_n -f_n\rangle |^2 \le ||v||^2 \cdot \sum_n ||e_n - f_n||^2$$

Now $\sum_n ||e_n - f_n||^2 < 1$ ( this weaker inequality is all we need) and from the above get $||v||^2\le 0$, and so $v=0$.

orangeskid
  • 53,909
  • Then we do not need the condition $f_n$ is orthonormal. Thank you :) – user277793 Dec 05 '15 at 16:19
  • The remaining part is using {f_n} perp is 0 and M perp perp is closure of span( M). Right? – user277793 Dec 05 '15 at 16:21
  • @user277793: No worries...Yes, indeed, we simply prove that the set $(f_n)$ is total. Now $f_n$ orthonormal is used if we want to check that it's an orthonormal basis ( we would also use the result you quoted above). – orangeskid Dec 05 '15 at 16:28