1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this function in Hilbert space?

  1. May 21, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) For what range of ##\nu## is the function ##f(x) = x^{\nu}## in Hilbert space, on the interval ##(0,1)##. Assume ##\nu## is real, but not necessarily positive.

    (b) For the specific case ##\nu = \frac{1}{2}##, is ##f(x)## in Hilbert space? What about ##xf(x)##? What about ##\frac{d}{dx}f(x)##?

    2. Relevant equations

    3. The attempt at a solution

    If the function ##f(x) = x^{\nu}## belongs in Hilbert space, then ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

    (assuming that, by 'Hilbert space', the question means 'the vector space of square-integrable functions,' as that is the only Hilbert space of interest to physicists).

    Therefore,

    ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

    ## = \int_{0}^{1} x^{2 \nu} dx < \infty## since ##\nu## is real

    ## = \int_{0}^{1} e^{{\rm ln} x^{2 \nu}} dx < \infty##

    ## = \int_{0}^{1} e^{2 \nu {\rm ln} x} dx < \infty##

    ## = [ \frac{2 \nu}{x} e^{2 \nu {\rm ln} x} ]_{0}^{1} < \infty##

    ## = [ 2 \nu e^{2 \nu {\rm ln} 1} - \frac{2 \nu}{0} e^{2 \nu {\rm ln} 0} ] < \infty##

    The only way for this the last sentence to be valid is for ##\nu## to be 0.

    Am I on the right track?
     
  2. jcsd
  3. May 21, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi,
    No, you derail when you do the integration. You can easily check by taking ##\nu = 1## .
    There are easier ways to integrate ##x^{2\nu}## (if you differentiate it, you'll probably see it)
     
  4. May 21, 2015 #3
    Thanks for pointing it out! Don't know why I made my life harder when it was already an easy problem.

    So, here's my new attempt:

    If the function ##f(x) = x^{\nu}## belongs in Hilbert space, then ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

    (assuming that, by 'Hilbert space', the question means 'the vector space of square-integrable functions,' as that is the only Hilbert space of interest to physicists).

    Therefore,

    ##\int_{0}^{1} |x^{\nu}|^{2} dx < \infty##

    ## = \int_{0}^{1} x^{2 \nu} dx < \infty## since ##\nu## is real

    ## = [ \frac{x^{2 \nu + 1}}{2 \nu + 1} ]_{0}^{1} < \infty##

    ## = \frac{1}{2 \nu + 1} < \infty##

    Therefore, ##2 \nu + 1 \neq 0##, so that ##\nu \neq - \frac{1}{2}##.

    Therefore, the function ##f(x) = x^{\nu}## is in Hilbert space, on the interval ##(0,1)##, for all real values of ##\nu## except ##-\frac{1}{2}##.

    P.S.: I'd assume that for the function ##f(x) = x^{\nu}## to ##\textit{really}## be in Hilbert space (on the interval ##(0,1)##), we must also show that there is an inner product (that converges to a complex number) defined on the Hilbert space, and that the Hilbert space is complete. I have no idea, though, as to how I can form an inner product with only the one function ##f(x) = x^{\nu}##.

    (b) ##\int_{0}^{1} |x^{\frac{1}{2}}|^{2} dx##

    ## = \int_{0}^{1} x dx##

    ## = [ \frac{x^{2}}{2} ]_{0}^{1}##

    ## = \frac{1}{2}##

    ##< \infty##

    So, for the specific case ##\nu = \frac{1}{2}##, ##f(x)## in Hilbert space.


    Am I on the right track?
     
  5. May 21, 2015 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    On the right track allright, but overshooting the length of the rails a little. Look again at your primitive. Can you carry out the integration for e.g. ##\nu = -2## ?

    And (b) is right, so far, but there are two more things asked there...
     
  6. May 21, 2015 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re inner product: you can try looking at your integral as the inner product ##f(x) \cdot f(x)##...

    And in a fancy notation it would be ##<\nu|\nu>## ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Is this function in Hilbert space?
  1. Hilbert spaces (Replies: 19)

  2. Hilbert spaces (Replies: 5)

  3. Hilbert space (Replies: 2)

  4. Hilbert Spaces (Replies: 2)

  5. Hilbert Space (Replies: 3)

Loading...