2

I have trouble understanding the bidual space of the normed space $(E, \|\cdot \|)$. Could someone help me understand it?

Boar
  • 125
Giovanni
  • 184
  • 3
    Exactly what is confusing you? – Sujit Bhattacharyya Apr 29 '19 at 05:11
  • Are you comfortable with the dual of a normed space, considered as a normed space itself? – Theo Bendit Apr 29 '19 at 05:19
  • @SujitBhattacharyya it's about its elements, what kind of elements has the bidual space? I am comfortable with the dual space, that is the space of the continuos and linear functional, but I am not able to figure out what are the elements of the bidual space and to recognize them – Giovanni Apr 29 '19 at 06:30

1 Answers1

8
  1. Every normed space $(E,\|\cdot\|)$ has a dual space, denoted by $E^*$, defined as the vector space of all continuous linear functionals on $E$, equipped with the norm $$\|x^*\|:=\sup\{|x^*(x)|: x\in E,\ \ \|x\|\leq 1\}\quad (x^*\in E^*)$$
  2. As $(E^*,\|\cdot\|)$ is also a normed space, with the norm defined in 1., it also has a dual space, denoted $E^{**}$, equipped with the norm $$\|x^{**}(x)\|:=\sup\{|x^{**}(x^*)|: x^*\in E^*,\ \ \|x^*\|\leq 1\}\quad (x^{**}\in E^{**})$$ The normed space $E^{**}$ (equipped with its norm) is called the bidual of $E$.

Some basic properties duality are the following:

A. A dual space is a Banach space. That is, even if $E$ is not a complete normed space, its dual $E^*$ is always complete, hence a Banach space. In particular, the bidual of a normed space is also a Banach space, whether $E$ is complete or not.

B. There is a natural embedding of $E$ into its bidual $E^{**}$. In fact, every $x\in E$ defines a continuous linear functional $i(x)$ on $E^*$ by the formula $$(i(x))(x^*):=x^*(x)\quad x^*\in E^{*}$$ It is not hard to check that the map $x\to i(x)$ is a linear isometry from $E$ into $E^{**}$. It is called the cannonical embedding of $E$ in its bidual.

C. If the cannonical embedding of part B. is also onto, then $E$ is said to be reflexive. A reflexive space is therefore a space $E$ which coincides with its bidual. Note that an incomplete normed space (i.e. not Banach) cannot be reflexive, because the bidual is complete, and isometries preserve completeness. Thus, an incomplete normed space is always "strictly smaller" than its bidual.

uniquesolution
  • 18,541
  • Ok, maybe i got it. Please tell me if I am wrong. For istance if I consider $(E, | \cdot |)=( L^{2}([-1,1]), | \cdot |_{L^{2}([-1,1])} )$, we know that $E'=L^{2}([-1,1])$ from the Fisher-Riesz theorem, and taking $-x^{2}+1\in L^{2}([-1,1])$ and $x\in L^{2}([-1,1])$ I have, using the canonical embedding and the $L^{2}([-1,1])$ norm, that $-x^3+x$ is an element of $(L^{2}([-1,1]))''$. Am I right? – Giovanni Apr 29 '19 at 07:18
  • No, that's not quite right. The product of two functions in $L^2[-1,1]$ is only guaranteed to be in $L^1[-1,1]$ (by the Cauchy-Schwartz inequality). We cannot guarantee that the product will be in $L^2[-1,1]$. – uniquesolution Apr 29 '19 at 07:22
  • Yes, but $-x^{3}+x$ belongs to $L^{2}([-1,1])$. I know that is not guaranteed that the product of two functions is in $L^{2}([-1,1])$, but for that kind of functions I think is straightforward. Actually I am seeking for some example for better understand the definition of the bidual space, if you can provide it would be marvelous! – Giovanni Apr 29 '19 at 07:29
  • I strongly suggest to you to make up your own examples, or else you will never learn anything. – uniquesolution Apr 29 '19 at 07:32
  • I agree. So, what is wrong with the specific example that I showed above? If there's any how can I correct it? – Giovanni Apr 29 '19 at 08:20
  • @uniquesolution Excuse me. Sometimes, it is said in the literature that every normed space is isometrically isomorphic to a dense subspace of a Banach space. Is this Banach space the bidual space of the normed space? Thank you. – Boar Sep 26 '22 at 07:59
  • 1
    @Wombat - Not exactly. The aforementioned Banach space is the closure of the natural embedding of the normed space into its bidual space. See here: https://math.stackexchange.com/questions/2197916/each-normed-space-is-isometrically-isomorphic-to-a-dense-subset-of-a-banach-spac – uniquesolution Sep 26 '22 at 21:17
  • @uniquesolution Thank you. Maybe I was missing out on one important thing: completions of normed spaces are unique up to isometric isomorphisms. So a completion $\hat{E}$ of $E$ is not necessarily $E^{}$ but $\hat{E}$ is always isometrically isomorphic to $E^{}$. – Boar Sep 27 '22 at 01:14
  • @Wombat -- $\hat{E}$ is not always isometrically isomorphic to $E^{}$, it is isometrically isomorphic to a subspace of $E^{}$. A classical example is $L_1$, whose bidual is strictly larger than itself. – uniquesolution Sep 27 '22 at 08:20
  • @uniquesolution What? Isn't $E^{}$ one of the completions of $E$? $E$ is isometrically isomorphic to a dense subspace of the Banach space $E^{}$; according to some books, this entitles $E^{**}$ a completion of $E$. – Boar Sep 27 '22 at 08:43
  • @ 1.In the category of Banach spaces all spaces are complete, i.e., equal to their completions. 2. $L_1$ is a Banach space, but it is not equal to its bidual. 3. From 1,2 it follows that you are wrong. – uniquesolution Sep 27 '22 at 22:36
  • @uniquesolution After reading some proofs, I think I know what you mean. Sorry about the mess. Thank you for being patient all along. Thank you. – Boar Sep 28 '22 at 00:16