Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Aamal

  • 2 years ago

can any1 define Schauder basis and Dual space?also an example of sequence space with proof

  • This Question is Closed
  1. Abbie23
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That a Math question?

  2. Aamal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  3. Abbie23
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay :)

  4. Aamal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do u have an idea

  5. nbouscal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Banach spaces... this is a little more advanced than most people here on OS. Some people here may be able to help you, but it might take a bit.

  6. nbouscal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Unless you are using Schauder bases in regular vector spaces, but even then, not sure how many people know these things. I'd have to read up on them myself to help. Looks like elias is replying though, and I think he might know :)

  7. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Example first. Let \[ \Large\cal l_1=\left \{(x_n),\, | \quad\sum_{n=1}^\infty |x_n| < \infty \right \} \] The space defined above is called the Banach space of absolutely summable sequences. It is a sequence space. The set of \[\Large e_1=(1,0,0,0, \cdots,0,0,\cdots )\\ \Large e_2 =(0,1,0,0 \dots,0,0, \cdots\\ \Large e_n = (0,0,0,\cdots,1,\cdots,0, \cdots ),\ n=1,2,3, \cdots \] is a Schauder basis of\[ \Large \cal l_1 \]

  8. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The dual of \[ \Large \cal l_1 \] is \[ \Large \cal l_\infty=\left \{ (x_n),\, | \quad \sup_n |x_n| < \infty \right \} \] the sequence space of bounded sequences.

  9. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    A Schauder basis in a Banach space E is a sequence \( (x_n) \) such that for every element \( x \in E\) there exists a unique sequence \( a_n\) of scalars such that \( x=\sum_{n=1}^\infty a_n x_n\). Every Banach space with a Schauder basis is separable and not every Banach space has a Schauder Basis. For example the sequence space \( \cal l_\infty \) is not separable, so it does not have a Schauder basis. There are also separable Banach spaces with no Schauder Basis. Per Enflo was the first one to discover this space.

  10. nbouscal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    To go with those examples, here are some attempts at understandable definitions: A sequence space is exactly what you would expect it to be. It is just a space of sequences. This is just like how a vector space is a space of vectors. Pretty intuitive. A Schauder basis is just another type of basis, but fills the same role as the normal concept of basis that you are familiar with. It is different from the typical notion of basis because where a regular basis (Hamel basis) consists of finite linear combinations, Schauder bases consist of countably infinite linear combinations. This makes it more appropriate for Banach spaces and other topological spaces. A dual space is the space of all linear functionals of a vector space. A linear functional is a map from a vector space to the field that its scalars come from. Hopefully that all made sense and was accurate, most of it I just gathered from Wikipedia and tried to parse.

  11. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    A dual of a Banach space E is the set \( E^*\) of all continuous linear operator on E equipped with the norm \[ \| x^*\| = \sup_{\| x\| \le 1} \|x^*(x)|| \]

  12. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Replace operator by functional above. A linear functional on E is a linear map from E to the scalar Field ( usually the reals or the complex) Field.

  13. Aamal
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx all of u bt i have to prove that The dual of l1 is l∞

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.