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Let $H$ be a Hilbert space and $\{e_j\}_{j \in \Gamma} \subset H$ be a orthonormal system. Then $\{e_j\}_{j \in \Gamma}$ is called an orthonormal basis of $H$ if

$\overline{\operatorname{span}\{e_j\: | \: j \in \Gamma \}} = H$.

Furthermore, we have $\{e_j \: | \: j \in \Gamma \}^\perp = \overline{\operatorname{span}\{e_j\: | \: j \in \Gamma \}}^\perp$

I have a lot of trouble to understand what would go wrong if we would not consider the closure in these both statements. Is there a counterexample for this? I appreciate any help.

Diglett
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  • Your first and second sentence don't quite make sense. Perhaps the last 3 words of the first sentence need to be replaced by "a set of elements in $H$". – John Hughes Nov 12 '17 at 23:29
  • Thanks, I fixed it now! – Diglett Nov 13 '17 at 04:30
  • The formula involving the orthogonal complement also works without the closure. The reason is that for all sets $A \subset H$, you have $A^\perp = \overline{A}^\perp$. – gerw Nov 13 '17 at 07:03

1 Answers1

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That is a very good question to ask yourself. First of all by definition you are only allowing finite linear combinations, that`s the main reason why you need that closure. For more on this please look at this question.

Now we can look at your question. Can you find a vector for which a finite linear combination of basisvectors would never give you the vector you`re searching for?

Hint: Look at the space $l^2$, the space of sequences $(z_n)_n$ for which $\sum_n |z_n|^2 < \infty$, notice that the "standard basis" from linear algebra, generalises to this space, and notice that the sequence $(\frac{1}{n})_n$ lives inhere. Can you write it as a finite linear combination?

Aweygan
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Verbe
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  • Sorry for my late reply. I always thought that the span is the set of finite linear combinations even when we don't have the closure? That's why I can't see why the closure is so important... – Diglett Nov 17 '17 at 15:59
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    The span is the set of finite linear combinations. Look at the example in my answer, how can you write that sequence as a finite linear combination? That is impossible. But you can get arbitrary close. That is why the closure is so important :-) – Verbe Nov 17 '17 at 21:43
  • Ah, because of the closure we can consider the limits of convergent sequences in $H$, right? So $(\frac{1}{n})n$ lies in the closure because it is the limit of $(x_n)_n$ where $x_n = \sum{k=1}^n \frac{1}{k} e_k$ and $e_k$ denotes the $k$-th unit vector. And this limit would not lie in the span without the closure, is that correct? – Diglett Nov 19 '17 at 17:23
  • That is correct! :-) – Verbe Nov 19 '17 at 21:41