# A few questions about the Hilbert triple/Gelfand triple

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..

The features you mention are an abstraction of important concrete examples, such as Levi-Sobolev spaces $H^1(S^1) \subset L^2(S^1) \subset H^{-1}(S^1)$ on the circle $S^1$. So, actually, part of the reason for the "properties" are that they are what we have in this (and related) examples.
Further, the abstraction does capture features which _turn_out_ to be relevant to doing things. But, first, continuity of linear maps is a thing one should relinquish only with great regret and caution. Second, $V\subset H$ having dense image is not only what we have in examples, but has the positive feature that the adjoint map $H^*\rightarrow V^*$ is again an injection.
The last paragraph you quote from Brezis is exactly looking at the case of Fourier series of functions in the Levi-Sobolev spaces, using Plancherel. Thus, first, the continuity and such are directly verified. Then, there is the further crazy point that these "triples" appear to be in conflict with a thing many people have over-interpreted, namely, the possibility of identifying the dual of a Hilbert space with itself (nevermind complex conjugation, that's not the issue) by Riesz-Fischer. In fact, it is an eminently-do-able exercise to show that isomorphisms $i:V\rightarrow V^*$ and $j:W\rightarrow W^*$ (whether given by Riesz-Fischer or anything else) are compatible with $T:V\rightarrow W$ and its (natural!) adjoint $T^*:W^*\rightarrow V^*$ only when $T$ is injective and is a homeomorphism to its image, which must be a closed subspace. That is, the conditions under which the square $$\matrix{ V & {T}\atop{\rightarrow} & W \cr i\downarrow & & \downarrow j \cr V^* & {T^*}\atop{\leftarrow} & W^* }$$ commutes are very restrictive. The pitfall is in "identifying" a Hilbert space and its dual merely because there is an isomorphism.