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Lp space examples

  1. Apr 15, 2007 #1
    Can anyone give me examples of functions in one space but not another?

    For instance, some f(x) with f(x) in L2 but not in L1?

    The best I can come up with so far is:
    f(x) = 1/sqrt(x). Thus is unbounded at 0, but would be in L1 since its integral on 0 to 1 is bounded.

    But I can't seem to come up with any examples of a function that would be in L2 but not L1. I've tried polynomial, trig, and some complex functions.

    Also, it appears that Ln for some finite n is "larger" than Linfinity. This seems to follow from my above example, since that f(x) wouldn't be in Linfinity but would be in L1, right?

    But if Linfinity is the limit of Ln as n goes to infinity, then there must be some inflection point where some Lm is "smaller" than the Ln for m > n.

    Is this accurate?

    Or is it that L1 is "larger" than L2? This makes more sense, for 1/sqrt(x) is in L1. But (1/sqrt(x))^2 = 1/x is not in L2. So f(x) is not in L2.

    Last edited: Apr 15, 2007
  2. jcsd
  3. Apr 15, 2007 #2


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    There are roughly two reasons why an integral might not converge:

    1) it has some finite values of x where it blows up to infinity, and if it blows up too fast, then the integral diverges

    2) it doesn't decay fast enough as x goes to infinity (analogously, the sum of 1/n diverges but the sum of 1/n2 converges, because 1/n2 decays "fast enough")

    Raising a function to a higher power helps it to decay faster at infinity, but it makes the areas of finite x where the function blows up worse.
  4. Apr 16, 2007 #3
    Lp spaces come with an interval attached, usually. This interval is important.

    For any bounded interval [a,b], if f is in L2[a,b], then f is also in L1[a,b]. This is not true for unbounded intervals.

    Take the function

    [tex]f(x)=\left\{\begin{array}{cc}\frac{1}{x} & x \geq 1 \\ 0 & x < 1\end{array}[/tex]

    This is in L2(R) but not in L1(R).
  5. Apr 17, 2007 #4
    Thank you both. That's exactly the clarification I was looking for.
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