Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Banach Space that is NOT Hilbert

  1. Oct 2, 2008 #1
    I know that all Hilbert spaces are Banach spaces, and that the converse is not true, but I've been unable to come up with a (hopefully simple!) example of a Banach space that is not also a Hilbert space. Any help would be appreciated!
     
  2. jcsd
  3. Oct 2, 2008 #2
    Hint: a necessary and sufficient for a Banach space [tex](\mathcal{B},\|\cdot\|)[/tex] to be a Hilbert space is for the norm to satisfy the parallelogram identity:

    [tex] \| a+b \|^2 + \|a-b \|^2 = 2 \|a\|^2 + 2\| b\|^2 [/tex]

    for each [tex] a,b \in \mathcal{B}[/tex]. Now think of some simple Banach spaces and check the above.
     
  4. Oct 2, 2008 #3
    Thank you, Anthony; I was aware of this condition but could not come up with anything. I believe I'm looking for a complete normed vector space (Banach) that is NOT complete under the inner product norm (as required for Hilbert), but I can't quite get my brain around how this would look.
     
  5. Oct 2, 2008 #4
    Speaking about spaces that are not Hilbert spaces is not quite clear always. Strictly speaking, if you don't define any inner product onto a given Banach space, then it is not a Hilbert space! But if you are asked to give an example of a Banach space that is not an Hilbert space, most surely it means that you want a kind of space that cannot even be made into a Hilbert space. That means that an inner product that would give the original norm, doesn't exist.
     
  6. Oct 2, 2008 #5
    I got stuck there. How do you prove the bilinearity of

    [tex]
    (x|y) = \frac{1}{2}(\|x+y\|^2 - \|x\|^2 - \|y\|^2)
    [/tex]

    by using the parallelogram only? :confused: (I'm dealing with real vector spaces first.)
     
  7. Oct 2, 2008 #6
    This completeness business is little bit misdirection, since dealing with finite dimensional spaces, where completeness is trivial, is enough. One can device a two dimensional norm space which is not an inner product space, and then you have a Banach space which is not a Hilbert space.
     
  8. Oct 2, 2008 #7
    This is exactly what I'm seeking; can someone please provide an example? Thank you.
     
  9. Oct 2, 2008 #8
    Think of a space of functions: Say you fix some interval, look at the continuous functions...
     
  10. Oct 2, 2008 #9

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Or just consider the p-norms on R^2, and use the parallelogram law. It's very easy to prove that the only p-norm that comes from an inner product is the 2-norm.
     
  11. Oct 3, 2008 #10
    If [tex]\| \cdot \|[/tex] satisfies the parallelogram identity, then the induced inner product is given by the polarization identity:

    [tex] (a,b) = \frac{1}{4} \left( \| a+b\|^2 - \|a-b\|^2\right)[/tex]

    The only tough thing to prove is linearity. Here's a hint that should lead you on the right direction:

    [tex]\begin{align*} \| (a+b)+c\|^2 + \| (a+b)-c\|^2 &= 2\left( \| a+b\|^2 + \|c\|^2 \right) \\
    \| (a-b)+c\|^2 + \| (a-b)-c\|^2 &= 2\left( \| a-b\|^2 + \|c\|^2 \right)\end{align*}[/tex]

    for [tex]a,b,c\in\mathcal{B}[/tex], which both follow from the parallelogram identity. Now subtract them and use the definition of [tex](\cdot, \cdot)[/tex] given by the polarization identity.

    Old Guy: I would follow morphism's advice and try out some norms you know of in [tex]\mathbf{R}^n[/tex].
     
  12. Oct 3, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why do you want to prove it? Your original problem was to find a Banach space that is not a Hilbert space. It was pointed out that a norm that does not satisfy the "parallelogram" inequality is not an innerproduct space. It was also pointed out that any Lp[/sup] for p other than 2 is a Banach space that is not a Hilbert space. There is no need to prove that "if the parallelgram inequality is satisified, a Banach space is a Hilbert space.
     
  13. Oct 3, 2008 #12
    HallsofIvy, it was Old Guy, not jostpuur, who posed the original problem. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Banach Space that is NOT Hilbert
  1. Banach space C(X) (Replies: 6)

  2. Matrix Banach Space (Replies: 3)

  3. Rigged Hilbert space (Replies: 5)

Loading...