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Hilbert spaces: Explicit example of an unbounded operator?

  1. Jul 22, 2007 #1
    I'm wondering if someone can give me an example of an everywhere defined unbounded operator on a (separable for simplicity) Hilbert space in a "constructive" manner. Since it's unbounded, simply a dense definition (i.e. on an orthonormal basis) wouldn't work since you can't extend it by continuity. So a Hamel basis would have to be necessary, but in a Hilbert space a Hamel basis would have to be uncountable so that's not even that easy to "get a hold of". (For that matter, can anyone give me an explicit example of a Hamel basis for the continuous functions on the real numbers?)

    I mean if I assumed that the Hamel basis had cardinality at least the continuum (assuming there's nothing between the naturals and the continuum), I could just identify it (or a subset of it) with the reals and then define a multiplication function that multiplied each basis element by its corresponding real number. That would clearly be unbounded and everywhere defined. Although in that case, unless I knew the Hamel basis explicitly, it wouldn't really be that satisfying.

    So basically I'm just wondering how necessary the axiom of choice is for the construction of such an operator. Anyone with any wisdom to share?
     
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  3. Jul 22, 2007 #2
    Though I haven't thought about it much on general Banach spaces, I would be interested to see an explicit everywhere defined operator on a Banach space if not a Hilbert space.
     
  4. Jul 22, 2007 #3

    Hurkyl

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    The derivative operator on the space of smooth real-valued functions on [0, 1] is unbounded.
     
    Last edited: Jul 22, 2007
  5. Jul 23, 2007 #4
    Yes, but under what norm? I.e. under what norm would that space be complete and the derivative operator still unbounded? My analysis is a little rusty, but are the smooth real-valued functions on [0,1] not dense in the continuous function on [0,1] in the infinity norm? In that case, it would not be a Banach space.
     
  6. Jul 30, 2007 #5

    mathwonk

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    how about smooth square summable functions on R, in the L2 norm. the evaluation map at 0, defines an unbounded operator.

    then just take the hilbert completion and extend the operator by zero.
     
  7. Jul 30, 2007 #6

    mathwonk

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    or use hurkyls example aND EXTEND IT BY ZERO to the continuous functions say on [0,1] in the sup norm.
     
  8. Aug 1, 2007 #7
    I've thought about "extension by 0" in the sense of how you guys are describing, but it still doesn't seem to work to my liking. I think the easiest way to do that process would be to just talk about the continuous functions on [0,1] with the infinity norm and start with polynomials as your (not orthonormal) basis. I.e. let e_i = x^i. Then just define an operator that sends e_i to i*e_i. That would give you an operator T defined on the polynomials that is unbounded.

    Now the problem is that "extending by 0" isn't as easy as it seems. Look at the function e^x. It is linearly independent of all the e_i, so you could adjoin it and still have a linearly independent set. In that case you could define Tj=0. But now then what if you decided to adjoin the element e^x-1=j as your next basis element instead of e^x? Then e^x = j + 1 (the unique decomposition), and thus T(e^x) would in fact equal 1.

    Therefore to actually know what the operator does on elements outside the original domain of definition, you'd still have to basically know what the Hamel basis is. So I don't yet really see how you can get around having an explicit Hamel basis, which I don't know how to get without the axiom of choice....you guys following me?
     
  9. Aug 1, 2007 #8

    mathwonk

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    good point. to extend explicitly by zero you need an explicit linear complement of your given subspace. this had escaped me.
     
  10. Aug 1, 2007 #9

    mathwonk

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    it seems to me the word "explicit" is n ot so well defined. if you want something really explicit, yopu need a hilbert space whose e;ements are explicitly given. even the simplest one, square summable sequences of reals or complexes is not that explicit, since one cannot specify all such sequences explicitly.

    on the other hand one can hope for a operation which seems explicit assuming one has the sequence in hand.

    also a space of functions can even less be given fully explicitly, since sequences are merely functions on a countable domain.

    but again one might hope to at least use an explicit word for the operator, like "distribution
    derivative". that is my next guess. but there is still nothing explicit about an existence proof for distribution derivatives.

    or just take the squaring function from the reals to the reals. how explicit is that? a real number is explicitly an arbitrary infinite decimal, again not given explicitly, and its square is an infiniute decimal which serves as a least upper bound for the explicit squares of all finite truncations fo that decomal, proved to exist abstractly.

    there is nothing explicit about uncountable mathematics in any form. that is why constructivists I know work with finite systems.

    it is a complete hoax to tell innocent freshmen that they know how to square real numbers, much less how to define e^x or sinx.
     
    Last edited: Aug 1, 2007
  11. Aug 1, 2007 #10
    Yeah that's basically the conclusion I've come to. I think half the reason that bounded operators work so well with Hilbert/Banach spaces is that you only need to define them on a set of vectors whose span is dense. Therefore if the Hilbert space has a countable orthonormal basis, then "bounded" operators can sort of be thought of as the subset of general operators where "most" things can be done "mostly" constructively (I'll let a logician clarify that statement). You can give recursive definitions of operators that will pretty clearly characterize them in a countable fashion.

    So really it just seems to me that the real reason you add topology to the separable Hilbert space and to restrict to topological (i.e. continuous) operators is to gain back the constructibility that you lose by going to infinite dimensions.
     
  12. Aug 1, 2007 #11

    Hurkyl

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    Well, two of the big ticket items that you get from the Hilbert space structure that you don't have for generic infinite-dimensional vector spaces are:
    (1) The separable Hilbert space is self-dual.
    (2) Functional analysis particularly nice. (See C*-algebra)
     
    Last edited: Aug 1, 2007
  13. Aug 2, 2007 #12
    Yeah I've been spending the last couple years studying functional analysis in its different iterations...this thread was brought about from one of those self reflections about the "basics". Anyway I've enjoyed the thread and the replies. Thanks guys.
     
  14. Aug 9, 2007 #13
    How about l2 to l2?

    Suppose {ai} is dense in [0,1] and ai is not zero for any i, and define an operator T which maps {xi} to {yi} where {xi} and {yi} belongs l2 space by yi=ai*xi. Then it seems the inverse operator of T is infinate since {ai} is dense in [0,1] which means there has a subsequence of {ai} which goes to 0. With this fact, you can easily check the infinate property. I am not good at mathematics,hope it's helpful to you.
     
  15. Aug 11, 2007 #14
    If I'm understanding you correctly, then yeah that operator would be injective and definitely have an unbounded inverse, but in that case the inverse wouldn't be everywhere defined which is kind of what I was looking for.
     
  16. Dec 12, 2008 #15
    A few points:

    1. You cannot extend an unbounded operator on an infinite separable Hilbert space. Unbounded is the same as non continuous at zero, and by linearity non continuous at 0 is the same as non continuous everywhere.

    2. Hellinger Toeplitz theorem: everywhere defined symmetric ==> unbounded (a consequence of the open mapping theorem). In other words there cannot be any symmetric unbounded operator on the Hilbert space which is everywhere defined.

    G.
     
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