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?



Topological dual and the notions of "smaller" and "larger" than...
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. 


