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..


1 Answer 1


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.

  • $\begingroup$ Actually it is Riesz-Fréchet and not Riesz Fischer $\endgroup$
    – Tom Mike
    Jan 15, 2014 at 2:15

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