Banach spaces

Principe2
a well-known theorem affirms that

each Hilbert space is isomorphic to an $l^2(A)$.

Is it true that each Banach space is isomorphic to an $l^p(A)$?

Risposte
david_e1
I understand and I agree with you now.

Fioravante Patrone1
"david_e":

For the subscript on the sup I put it as the supremum is taken on the finite subsets $A$ of $[-1,1]$....


I understood it. But in my opinion it should not be used.
The "sup" is defined for a set. Since you describe explicitly the set there is no need to add the subscript. Moreover, the symbol "$A$" used in the set is not a variable, but just a signpost, so for me it is "deprecated" its use outside the span of its proper use. I mean: "outside" of the curly brackets (or braces), formally you are not allowed to refer to "$A$".

I could understand the following use (if I were in LaTeX I would use "displaymath" for a better rendering :-D ):

$ \text{sup}_{A \ \text{finite subset of } [-1,1] \} \ \sum_{a \in A} x_a^2 < \infty $

Bye

david_e1
Yes it was a typo. Obviously a sum over a finite set is finite.

For the subscript on the sup I put it as the supremum is taken on the finite subsets $A$ of $[-1,1]$....

Camillo
New interesting abbreviations, at least for me :

* lub = least upper bound
* glb = greatest lower bound

Fioravante Patrone1
"david_e":
Thanks.

Let me see if I have understood correctly. $l^2([-1,1])$ is the set of all the tuples of real numbers $x$, with index $a \in [-1,1]$ such that:

$ \text{sup}_A \{ \sum_{a \in A} x_a^2 < \infty \ A \ \text{finite subset of } [-1,1] \} < \infty $

Is it right?



Just to be disgustingly formalist, I would write it in this way:

$ \text{sup } \{ \sum_{a \in A} x_a^2 \ : \ \ A \ \text{finite subset of } [-1,1] \} < \infty $,

since I don't understand why one should put the subscript "A" to the "sup" in this case, and the "$< oo$" inside.

For a broader point of view, you should look at the way in which it is defined the sum of an arbitrary set of numbers. This definition comes from there.



B.t.w., I "automatically" used "sup", being a native Italian speaker. In english is often used "lub" (least upper bound). As "glb" (greatest lower bound) is used instead of the Italian "inf". But also "sup" and "inf" are in use.

david_e1
Thanks.

Let me see if I have understood correctly. $l^2([-1,1])$ is the set of all the tuples of real numbers $x$, with index $a \in [-1,1]$ such that:

$ \text{sup}_A \{ \sum_{a \in A} x_a^2 < \infty \ A \ \text{finite subset of } [-1,1] \} < \infty $

Is it right?

Fioravante Patrone1
$l^2(A)$ is the set of all A-tuple of real numbers (I assume we are on $RR$) s.t. their squares are unconditionally summable.

Otherwise said, sup$ { \sum_{a \in A_F} x_a^2 : A_F $ is a finite subset of $A } \in RR$.


Notice that a necessary condition for being summable is that the set of the $a$ in $A$ s.t. $x_a != 0$ is (at most) countable. Clearly, this countable set will depend on the element $(x_a)_{a \in A}$ that is considered.

david_e1
"irenze":
every SEPARABLE Hilbert space is isomorphic to $l^2(ZZ)$, but here he has a generic set $A$...

Ops... I knew only the "4 dummies" version of this therem. :-D

So it's isomorphic to $L^2(A)$?

I think that I don't know what $l^2(A)$ is... (if it's not $L^2$).

irenze
every SEPARABLE Hilbert space is isomorphic to $l^2(ZZ)$, but here he has a generic set $A$...

david_e1
"ubermensch":
each Hilbert space is isomorphic to an $l^2(A)$.

This holds also for non separable Hilbert spaces? It seems to me that an isomorphic space of a separable space is separable...

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