Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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..

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
Thanks for the answer. – soup Dec 12 '12 at 16:02
Actually it is Riesz-Fréchet and not Riesz Fischer – user118351 Jan 15 '14 at 2:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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