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!

Question about L2 which is not L-infinity

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Construct a function u of the space H'(B), where B is a unit sphere in R^3, which does not belong to L∞(B).


    2. Relevant equations

    (relevant facts)
    all L2 are hilbert, therefore the problem reduces for us finding an L2 function which is not L-infinity.


    3. The attempt at a solution
    u=((1/(x+1))^(1/c))-1where c>2.

    please note that this is adjusted because our domain is limited to a unit sphere. the unadjusted equation looks like this:

    u=(1/x)^(1/c) where c>2.

    we don't even know if this is correct. spent 2 days thinking about the problem. any help in finding this function would be much appreciated. thank you!
     
  2. jcsd
  3. Feb 16, 2009 #2
  4. Feb 16, 2009 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Don't just guess! You want to find some function that, over a given domain, does not satisfy the definition of membership in [itex]L_{\infty}[/itex] space but does satisfy the definition of membership in [itex]L_2[/itex] space. As a hint, [itex]L_2[/itex] space comprises the set of square-integrable functions over that domain. What, in a nutshell, characterizes [itex]L_{\infty}[/itex] space?
     
  5. Feb 17, 2009 #4
    [itex]L_{\infty}[/itex] = inf{C>=0: |f(x)| <=C for almost every x}
    f is an element of [itex]L_{\infty}[/itex]=||f||p as p approaches infinity is finite

    from here, we derived the fact that the function u that we stated is of a p-series with p<1 for L2, therefore, the integration is finite. therefore is an element of L2.

    on the other hand, the function u we stated above is of p-series with p>1 for ||f||p as p approaches infinity. therefore its integration is infinite, therefore is not an element of [itex]L_{\infty}[/itex].

    i just want second opinions. i just think our analysis of our function u has a lot of loopholes, given the domain constraints (unit sphere).

    can you suggest a function that satisfies the condition?
     
  6. Feb 17, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Forget the details. You can think of L_infinity as just 'bounded'. So you want a square integrable function on the sphere that is unbounded? Should be easy enough. I don't get your examples though. What's x? If you are working on a ball, it might be easier to use spherical coordinates.
     
  7. Feb 17, 2009 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    In short, a function is in L-infinity if it is bounded throughout the domain except possibly for a set of measure zero. So, you are looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere.

    Some questions:
    • Why are you so bothered by the domain constraints?
      This is good old R3, so you are simply looking at a volume integral over the unit sphere.

    • What do you mean by x in "u=((1/(x+1))^(1/c))-1 where c>2"?
      • If you mean [itex]x\in {\mathcal R}^3, ||x||<=1[/itex], then what does 1/(x+1) mean?
      • If you mean x, as in x,y,z, you should say so.

    • Why do you need to subtract 1?
      For a finite domain, if f(x) is in Lp, then so is g(x)=f(x)+c.

    Hint: That the domain is the unit sphere in R3 suggests a spherically symmetric function, thereby reducing the problem to R1.
     
  8. Feb 17, 2009 #7
    @dick

    i was just thinking in terms of x,y to make things easier to imagine. i restricted my domain as a unit circle in the x,y plane.

    an unbounded f that is square integrable on a unit sphere...

    ****

    thinking thinking thinking......

    f(x,y)=exp(1/xy)???

    ?? it is unbounded as x,y approaches 0.

    still thinking......

    ****
     
  9. Feb 17, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Think polar coordinates if you are working on the disk.
     
  10. Feb 17, 2009 #9
    i'm just so concerned because some integrals of functions become finite because of the domain restrictions....

    ok. i'll try thinking of a spherically symmetric func, to make things simpler.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about L2 which is not L-infinity
  1. The space l infinity (Replies: 15)

Loading...