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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

    (x|y) = \frac{1}{2}(\|x+y\|^2 - \|x\|^2 - \|y\|^2)

    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


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    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


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    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. :)
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