2
$\begingroup$

Let $Y$ be an inner-product space, and let $A$ be an orthonormal system. We're trying to find a case to demonstrate the fact that even if for any given $x$ in $Y$ there's some $u$ in $A$ such that $\langle u,x \rangle \neq 0$, that doesn't imply that the system is complete.

We've asserted that $A$ should not be a subset of a complete orthonormal system, and thus could not be completed to one. That led us to the realization that it's essential that Gram-Schmidt's operation could not be applied, which probably means we need to find an IPS whose dimension is at least the continuum cardinality.

However, we were not able to actually construct such an example, any tips will be welcomed...

Rephrase: My wording seems to have caused some confusion, so I'll try again: Find an IPS Y and an orthogonal system $A\subset Y$ such that $\forall x\in Y \exists u\in A: \langle x,u\rangle \ne0$ yet A is not a complete system (e.g. not an orthonormal base, e.g. it's span is not dense in Y, e.g. It doesn't always conform to Parseval's equality. These are all equivalent for any IPS).

  • 0
    If the span of $A$ is not dense then $X = \overline{\mathrm{span}}\,{A} \subsetneqq Y$. Thus there exists $0 \neq z \perp X$, but this means in particular that $\langle z,u\rangle =0$ for all $u \in A$, so you can't find a system as you want.2011-07-27
  • 0
    That's not true. The space not being dense doesn't mean that it's complement is contained in it's perpendicular space. That's exactly the eccentricity of non-countable spaces I'm trying to demonstrate. To be more precise, you can't imply that $z\notin Cl(Span(A)) \implies z\in \perp A$, that can only be proven for vector spaces of countable dimensions...2011-07-27
  • 0
    I don't understand what you're saying and I didn't say anything of the sort you're writing. There is $z' \in Y \smallsetminus X$, right? Now take the orthogonal projection $z''$ of $z'$ onto $X$ (this exists because $X$ is closed and convex, for example). The element $z = z' - z'' \neq 0$ and by definition $z \perp X$. Countability of the (Hilbert-)dimension has absolutely nothing to do with this argument. [Also, you should be a bit careful whom you're talking to when making assertions such as "that can only be proven for vector spaces of countable dimensions..."]2011-07-27
  • 0
    Wait a minute! What you've done here is to prove that given $z'\in Y\setminus Y$ we get that **it's orthogonal projection to $Y$** (which you denoted $z$) is perpendicular to $Y$. This proves nothing. In fact, if you can prove that for any $z'$ (for given $A$ and $Y$) the projection is essential (that is, $z''!=0$) this proves my point exactly... I'll repeat this again to be clear - just because the complement of the closure of the span is not empty doesn't mean that the perpendicular space is not trivial.2011-07-27
  • 0
    You're simply wrong here and you misunderstood the argument. Once again: If $p: Y \to X$ is the orthogonal projection then we have for *every* $z' \in Y \smallsetminus X$ that $z'' = p(z') \neq z'$ (since $X$ is closed we have $p(x) = x$ if and only if $x \in X$). Therefore $z = z' - z'' = z' - p(z') \neq 0$ and $p(z) = p(z') - p(p(z')) = 0$ (since a projection satisfies $p^2 = p$). In other words, $0 \neq z \perp X$. Once again in words: $z'$ is an element of $Y \smallsetminus X$ and $z''$ is its projection to $X$. The difference $z = z'-z''$ is the component of $z'$ *orthogonal* to $X$.2011-07-27
  • 0
    By the way: you may want to read Theorem 3.1.7 on [page 82](http://books.google.com/books?id=a1R0livwR9AC&pg=PA82) of Pedersen's *[Analysis Now](http://books.google.com/books?id=a1R0livwR9AC)*, where you find the details of what I'm saying here. Note also that in my notation $X = (A^{\perp})^{\perp}$ (by Cor. 3.1.8 there), so if $A^{\perp} = 0$ then $Y = X$.2011-07-27
  • 0
    As noted above, you've asserted completeness which was never assumed, which caused all this misunderstanding.2011-07-28
  • 0
    Noted. I apologize if it seems that I'm being disrespectful or unappreciative of the time you spend helping me with the problem. I have the utmost respect and gratitude for anyone who is willing to help, difference of opinions aside...2011-07-28
  • 0
    Thank you. Once again, my apologies for all the fuss. I removed my previous comment in view of your last one, and I hope I can be of more help next time than this time. Two remarks concerning your question: As I noted in a comment to paul's answer: his example has countable dimension (as a vector space over $\mathbb{R}$ or $\mathbb{C}$). A second remark, assuming completeness, there is a difference in Hilbert dimension and ordinary linear algebra dimension. It follows from the Baire category theorem that every Hilbert space has either finite or uncountable dimension (as a vector space).2011-07-28

2 Answers 2

4

If the inner-product space $Y$ is complete (a.k.a. "Hilbert"), then having trivial orthogonal complement (the question's hypothesis) implies that the orthogonal or orthonormal set is complete (as Mariano and Theo observe, above). That is, as M. and T. note, there is no example of the sort requested.

Is it conceivable that having the inner-product space $Y$ itself not be complete is seriously relevant to the questioner's issues? Of course, then there are complications in the usual harmony of Hilbert spaces (separable or not). But the resulting "counter-examples" are more pranks than serious. E.g., let $Y$ be the subspace of $\ell^2$ ("algebraically") spanned by the usual $e_1,e_2,\ldots$, but with one $e_{i_o}$ removed, replaced by a vector $v$ having all non-zero entries. Take $A=\{e_i\}$ with $e_{i_o}$ missing. If I'm not mistaken, this ruse (or similar) contrives a situation formally meeting the question's requirement.

However, surely this is not of operational interest, due to its artificiality (despite formal, literal correctness)?

  • 0
    Ah, so *that's* what this question is after? Still, I don't understand the objections made to my comments. Of course, I was assuming completeness all along (but who would think about *complete* orthonormal systems in a context of non-complete spaces?) By definition completeness of an ONS means that your space is isometrically isomorphic to $\ell^2$ of the index set.2011-07-27
  • 0
    @Theo B., I don't know, but was trying to make something constructive out of it. It might be that some source's definition is "broken", in the sense that it posits Plancherel-Parseval as the criterion of "completeness", without requiring the space be Hilbert? Anyway, there seems to be no _enforcement mechanism_ for definitions, eh? :)2011-07-27
  • 0
    Yes, there definitely isn't such a thing :) To add one more "constructive point": the *algebraic* dimension of your space is is countable, so this doesn't have much to do with cardinality.2011-07-27
  • 0
    So let me get this straight... You go ahead and say that I'm wrong and such a space does not exist. Then you go and give an example of such a space, but say it doesn't count because it's not interesting enough? Well excuse me, but as "artificial" as this example may be, it still stands to the posed requirements, which means that I was right all along and such an example does exist, in spite of what Theo tried to assert...2011-07-28
  • 0
    I apologize if it seems that I'm being disrespectful or unappreciative of the time you spend helping me with the problem. I have the utmost respect and gratitude for anyone who is willing to help, difference of opinions aside...2011-07-28
2

I am not quite sure I understand what you want... This answers what I guess you are asking, though:

In an infinite dimensional, separable Hilbert space, like $\ell^2$, an orthonormal set is at most countable (because it is discrete). Yet $\ell^2$ is of uncountable dimension.

The set of 'canonical' vectors in $\ell^2$ is an orthonormal set which does not span $\ell^2$ yet its orthogonal subspace is trivial.

  • 0
    An orthonormal system $\{e_n\}$ is called *complete* by physicists if $\sum |\langle f, e_n \rangle|^2 = \|f\|^2$ for all $f$ from the Hilbert space. Equivalently, if it is an orthonormal basis. Under this definition there is no non-complete orthonormal system with trivial complement, of course. :)2011-07-27
  • 0
    Some things in the question were eaten by the parser due to the use of `<` and `>`. However, I still don't really understand what is being asked even after fixing this.2011-07-27
  • 0
    What about $\{\delta_x | x \in [0,1]\}$ for $l^2[0,1]$, isn't it an orthonormal system which isn't countable?2011-07-27
  • 0
    @user: $l^2[0,1]$ isn't separable. Yes, your functions are an uncountable orthonormal system, but that proves non-separability.2011-07-27
  • 0
    So when you said $l^2$ you've meant over a discrete set?2011-07-27
  • 0
    So when you said $l^2$ you've meant over a discrete set? Considering $l^2\{0,1\}$ then $\delta_x$ becomes the set of sequences $\{\{\delta_{n,m}\}_{m=1}^{\infty}\}_{n=1}^{\infty}$. This set is countable, yet its span is dense in (the non countable) $l^2\{0,1\}$, which proves that it is separable. I think the confusion arises from the fact that for an orthonormal system to be complete it suffices that it's span is dense in Y.2011-07-27
  • 0
    @user: There seem to be many more confusions here. First, I have no idea what you mean by "$\delta_x$ becomes the set of sequences ...". Second do you mean the *square-summable* sequences indexed by $[0,1]$ (usually denoted by $l^2[0,1]$) or the *square-integrable functions* (usually denoted by $L^2[0,1]$). The former isn't separable, the latter is. However, the latter doesn't contain the Dirac $\delta$'s as you seem to imply.2011-07-27
  • 0
    I was talking about the former. I'm not familiar with $L^2$-spaces (with a capital L) yet. Maybe my wording was misleading, I'll edit the question to be more accurate.2011-07-27