0
$\begingroup$

Hi, I've read this sentence but I can not understand what it means

[...] $\Phi'$ is the topological dual of some dense space $\Phi$ of $H_{aux}$ [...] Notice that the choice of $\Phi$ is subject to the two conditions: [...] ,On the other hand it must be small enough so that its topological dual $\Phi$ is "sufficiently large" [...]

What the author means by "sufficiently large"? Is it the dimension? Knowing that the spaces under consideration are infinite dimensional

Edit: why does the dual becomes larger when the original space becomes smaller?

Improve this question
$\endgroup$
7
  • 6
    $\begingroup$ Where did you find this sentence? It would probably help if you gave some more context here. $\endgroup$
    – Charles Rezk
    May 1, 2010 at 21:24
  • $\begingroup$ Agreed. As a possible guess, in many contexts "sufficiently large" means "has enough objects in it that we care about." $\endgroup$
    – Qiaochu Yuan
    May 1, 2010 at 21:51
  • $\begingroup$ See arxiv.org/abs/gr-qc/9508015, page 3, starting with Step 5a. There are also references given there that might help. I agree with Charles and Qiaochu. $\endgroup$
    – Jonas Meyer
    May 1, 2010 at 22:20
  • $\begingroup$ My best guess is that this is a functional analysis question (given the wording and the suggestive notation) so I've retagged as such. $\endgroup$
    – Andrew Stacey
    May 2, 2010 at 18:06
  • $\begingroup$ Ok, here is the paper arxiv.org/abs/gr-qc/9504018 it treats the quantization on the diff constraint in LQG and have to tacle between the space $\Phi$ and its dual. $\endgroup$
    – Pedro
    May 3, 2010 at 13:11

1 Answer 1

Reset to default
5
$\begingroup$

To answer your final question: Let $\Phi \supset \Psi$. Consider $\Phi' \subset \Phi^*$, the former is the continuous linear functionals on $\Phi$, and the latter is the set of all linear functionals on $\Phi$. Then the restriction of $\Phi'$ on $\Psi$ is obviously continuous, so $\Phi' \subset \Psi'\subset \Psi^*$.

Therefore if you make a space smaller, you makes its dual bigger.

Intuitively speaking, elements of $\Psi'$ need to be continuous on fewer objects, and hence has fewer constraints; thus $\Psi'$ contains more objects.

For your original question: your interpretation is sort-of okay. The point is that infinite dimensional Hilbert spaces admit dense proper subspaces (hope I am getting the notation correct). And in particular you can have two dense subspaces of a Hilbert space with one strictly contained in the other. You may want to review volume 2 of Reed and Simon.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Okay, here is how I understood it: If a functional A is in $\Phi'$ (i.e continuous on the whole $\Phi$) it will automatically be continuous on $\Psi$ (i.e it is in $\Psi'$) as $\Psi$ inherits the topology of $\Phi$ thus $A\in\Phi'\implies A\in\Psi'$, on the other hand, if A is in $\Psi'$ (i.e continuous on $\Psi$) it can be non-continuous on the whole $\Phi$ (i.e not in $\Phi'$) thus $A\in\Psi'\not\implies A\in\Phi'$. So we have $\Psi\subset\Phi\implies\Phi'\subset\Psi'$ $\endgroup$
    – Pedro
    May 3, 2010 at 15:14
  • $\begingroup$ So "sufficiently large" then means "functions on it are constrained by fewer (topological) constraintes so that they are huge" $\endgroup$
    – Pedro
    May 3, 2010 at 15:15
  • $\begingroup$ So this property applies to topological (continuous) duals and not to algebraic ones? $\endgroup$
    – Pedro
    May 3, 2010 at 15:20
  • 1
    $\begingroup$ depends on what are the bare minimum criterion you are considering. Look at the finite dimensional inner-product space case, and let $W$ be a proper subspace of $V$. Then if you consider the set of objects to be the space of continuous functions on $V$, then there are elements in the algebraic dual of $W$ that is strictly not in the dual of $V$, by virtue of it being non-linear on $V$. For example, the function $f(\|P_W^\perp\|)$ where $P_W^\perp$ is the projection to the orthogonal complement of $W$ in $V$ and $f$ is some arbitrary continuous function will be linear on $W$ (in fact, trivial) $\endgroup$
    – Willie Wong
    May 4, 2010 at 10:07
  • 1
    $\begingroup$ ...but not linear on $V$, so is not an element of the "algebraic dual" $V^*$ while its restriction to $W$ is in $W'$. Going the other way, you should remember Hahn-Banach theorem, which shows that there always is some continuous extension of a continuous linear functional on the subspace. The question is really one of epistemology: is it useful to consider examples as I just described? (I don't know.) Is it useful to consider linear functionals on a Hilbert space that is bounded on a dense subspace but not elsewhere? Evidently yes. $\endgroup$
    – Willie Wong
    May 4, 2010 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.