I am attempting to fully understand Hilbert triples by reading Brezis' Function Analysis book.
Consider $V \subset H \subset V^*$, where $V$ is Banach and $H$ is Hilbert. $V$ is dense in $H$.
Why do we need density of $V$?
Assume the injection $V \subset H$ is continuous. There is a canonical map $T:H^* \to V^*$ that just restricts functionals on $H$ to take arguments restricted to $V$.
$T$ has the properties: (1) $|Tf|_{V^*} \leq C|f|_{H^*},$ (2) $T$ is injective, (3) $R(T)$ is dense in $V^*$ if $V$ is reflexive.
Why do we need $V \subset H$ to be continuous? What's the need for these three properties? I'm not asking "why are they true" but what is the significance of these properties for this discussion? The second one is fine, I suppose. I guess the third property is nice as it says we can get close as want to to an element of $V^*$ by elements on $H^*$, but so what?
Identifying $H^*$ with $H$ and using $T$ as a canonical embedding from $H^*$ into $V^*$, we write $V \subset H \equiv H^* \subset V^*$, where all injections are continuous and dense.
Why is continuous and dense worth pointing out?
The situation is more delicate if $V$ turns out to be a Hilbert space with its own inner product. We could identify $V$ and $V^*$ with this inner product, but then the Hilbert triple becomes absurd. We cannot simulataneously identify both $V$ and $H$ with their dual spaces. Here is a very instructive example.
Let $H = \ell^2$, with $(u,v)_H = \sum u_nv_n$ and $V = \{u : \sum n^2u_n^2 < \infty\}$ with $(u,v)_V = \sum n^2u_nv_n.$ Clearly $V \subset H$ is dense and continuous injection. We identify $H$ with $H^*$ while $V^*$ is identified with $V^* = \{f : \sum \frac{1}{n^2}f_n^2 < \infty \}$ which is bigger than $H$. The scalar product $\langle , \rangle_{V^*, V}$ is $\langle f, v \rangle_{V^*, V}= \sum f_nv_n$.
Can somebody explain this "instructive example" to me as I don't understand the point.
Sorry for so many questions but I really do not understand this topic well. Thanks for any help. I already read the other threads on this topic btw..