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


    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

    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......


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

    still thinking......

  9. Feb 17, 2009 #8


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook