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Hilbert Spaces

  1. Jun 13, 2005 #1
    A Hilbert Space is a complete inner product space.

    My first question: From the definition above, is it safe to say that every sequence in a Hilbert Space converges? And so can we say that Hilbert Spaces only contain Cauchy sequences?

    Second question: These 'sequences' that we talk about in Hilbert spaces, can they be sequences of anything? Like sequences of functions, sequences of elements (like complex numbers), sequences of inner products?

    Third question: Why is [itex]l_2[/itex], the set of all square-summable sequences, the only [itex]l_p[/itex] space that is a Hilbert space?

    Fourth question: The Cauchy-Schwartz Inequality

    [tex]|\langle x,y \rangle \leq \|x\|\|y\|[/tex]

    This always holds in a Hilbert space right? In which instances does it not hold?

    Fifth question: What are the differences between Hilbert Spaces and Banach Spaces? The only thing I know is that Hilbert spaces are 'nicer' than Banach spaces because of something to do with the parallelogram law - which in turn makes Fourier analysis work better or something...I dont really know.

    That will do for now.
     
    Last edited: Jun 13, 2005
  2. jcsd
  3. Jun 13, 2005 #2
    Consider two spaces:

    1. [tex] X_1 = C[a,b] [/tex]

    2. [tex] X_2 = L_2 [a,b] [/tex]

    Now consider the inner product defined as

    [tex] \langle f,g \rangle = \int_a^b f(t)\overline{g(t)}dt[/tex]

    If we induce this inner product into [itex]X_1[/itex] then we will have an inner product space but NOT a Hilbert space because the axiom fails:

    [tex]\langle f,g \rangle = 0[/tex]

    the inner product can equal zero and f and g dont have to necessarily be 0 themselves.

    But now applying the inner product to [itex]X_2[/itex], this does produce a Hilbert space. But I cant see why this one should work and not the other.
     
  4. Jun 13, 2005 #3

    dextercioby

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    You really lack a textbook,right?

    It's a matter of understanding definitions first and theorems later.

    What's the defintion of complete...?Why would this definition imply your first statement/question...?

    The elements of a Hilbert space are called VECTORS,because,essentially the H space is a vector space.So all the sequences are sequences of vectors.

    For the third question,i'd reccomend Debnath & Mikusinski "Introduction to Hilbert Spaces with Applications".Proving completeness for a space (Banach or Hilbert) is not an easy matter.

    For #4,what are the premises of that proposition ??

    I'll say:"All Hilbert spaces are Banach spaces.Not all Banach spaces are Hilbert spaces".I'll let you dig the literature for the explanation of my assertion.

    Daniel.
     
  5. Jun 13, 2005 #4
    EDIT: Daniel beat me to it!

    No; 'complete' just means that all limit points of convergent sequences in a Hilbert space are contained in the H space. You can still have nonconvergent sequences - they don't define limit points.

    By definition, sequences are composed of elements. Of course your H space can be made up of functions (defined between two other spaces) - in which case you get a sequence of functions; but the functions would be elements of the space.

    I don't know the answer to your 3rd question.

    Yes, it holds on all Hilbert spaces.

    Hilbert spaces have inner products, which induce norms. Banach spaces are a more general (less strict) class of spaces, which have norms but not necessarily inner products.
    further ref: http://mathworld.wolfram.com/BanachSpace.html.
     
  6. Jun 13, 2005 #5
    Thankyou both for replying...

    Dextercioby - you are right I have no textbook and it is making understanding my notes upon revision slightly harder than anticipated.

    I see. My mistake! Ok, so in other words, sequences in Hilbert spaces are closed under taking limits? And Hilbert spaces can still have nonconvergent sequences - but if they ARE convergent, their limit must be in the Hilbert space.

    I see, I see. I have a question about terminology here. You say that a Hilbert space has an inner product which INDUCES an norm. Are ALL norms in a Hilbert space induced solely by inner products?

    So when talking about Hilbert spaces, the norm is defined by the inner product

    [tex]\|x\| = \sqrt{\langle x,x\rangle}[/tex]

    Is this the inner product which induces the norm?

    EDIT: Typo now fixed
     
    Last edited: Jun 13, 2005
  7. Jun 13, 2005 #6

    dextercioby

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    Except for the typo,yes.

    Daniel.
     
  8. Jun 13, 2005 #7
    Why are Hilbert spaces so 'easy' to work with? I mean, I now know that Hilbert spaces are a special case of Banach spaces where the norm is defined by a particular inner product.

    Now when are Banach spaces Hilbert spaces? Banach spaces have norms right? So when this norm satisfies

    [tex]\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2[/tex]

    then the Banach space becomes a Hilbert space with inner product [itex]\langle .,. \rangle[/itex] which induces the norm

    [tex]\|x\| = \sqrt{\langle x,x \rangle}[/tex]

    Is this a suitable assumption? Or am I simply pulling random memories from the semester? :confused:
     
  9. Jun 13, 2005 #8
    EDIT: I'm several posts behind...

    Have you already looked at measure theory and Lebesgue integrals? They're necessary prerequisites to define the L2[a,b] space.
    If [tex]V= \{ \phi (x) \| \phi:(a,b) \rightarrow C[/tex] and [tex]| \phi | ^2[/tex] is mu-integrable over (a,b)}, then you are CORRECT in thinking that under [tex] \langle f,g \rangle = \int_a^b f(t)\overline{g(t)}dt[/tex] it would NOT be a Hilbert space. There is a way around it - a bit of algebra shows that if instead of using regular equalities "=", you define a different equivalence class (say, '~') where
    f~g iff |f - g| = 0 [tex]\mu[/tex]-ae on (a,b) (this stuff is from Cohen, D. p. 15),
    then you do get a Hilbert space, specificaly L2[a,b].
    In a nutshell, if all the functions in your space are square-mu-integrable, then the only way <f,g> = 0 would be if f=g mu-everywhere (everywhere except at some 'points'); so you use your mu-measure to define an equivalence class where f and g, being equal mu-everywhere, are in the same equivalent class, and so <f,g> =0 if and only if f~g; so you get a Hilbert space.

    Could I ask, what textbook are you starting from (and what class will it be for?) I'm using David Cohen's "An introduction to Hilbert space and Quantum Logic), which I'd read for an unnamed elective math class.
     
  10. Jun 13, 2005 #9
    All correct.
     
  11. Jun 13, 2005 #10
    rachmaninoff, I have no textbook - I am working from my notes that I wrote in class (in fact the class had no recommended textbook - which was kind of strange!?).

    The class was called - "Introduction to Hilbert Spaces".

    Well, thats why I cant figure it out!?! I havent been taught either of those topics. But what you wrote is re-assuring that I was on the right track - at least the right axiom!
     
  12. Jun 13, 2005 #11
    I don't believe so, does it necessarily induce a unique inner product satisfying all the conditions for a Hilbert space?
     
  13. Jun 13, 2005 #12
    What if I said it was simply a scalar product?
     
  14. Jun 13, 2005 #13
    What do scalar products have to do with it?
     
  15. Jun 13, 2005 #14
    Well, obviously

    [tex]\|x\| = \sqrt{\langle x,x \rangle}[/tex]

    is the norm in a Hilbert Space. What I am saying is that a Hilbert space can be defined as a special case of a Banach space.

    If you consider the special case when the norm of a complex Banach space satisfies

    [tex]\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2[/tex]

    then, if we only consider this case, the complex Banach space IS a Hilbert space.

    Can I say that a Hilbert space is indistinguishable from a Banach space when this is the case?
     
  16. Jun 13, 2005 #15
    I have a feeling that maybe I should have a good look into Lebesgue integration and measure theory before I can understand everything about Hilbert spaces, especially how Fourier analysis is involved.
     
  17. Jun 13, 2005 #16

    HallsofIvy

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    Yes, a Hilbert Space is a special case of a Banach space, just as a Banach space is a special case of a Frenet space (a complete metric space).
     
  18. Jun 13, 2005 #17
    A Frenet space!? god damn, how many spaces are there!?
     
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