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Thread: P norm and essential supremum

  1. #1
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    P norm and essential supremum


    If ||f ||p= {x |f|p dm}1/p is p norm of measurable complex function
    ' f ' , m denoting
    the measure on set X ,what is the relationship between lim|| f ||p (p) (if exists)
    and || f || the essential supremum of f ?
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  2. #2
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    I've never seen this stated, but I think when $\displaystyle || f ||_p = \left[ \int |f|^p d\mu \right]^{1/p},$

    $\displaystyle \lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = || f ||_{\infty}.$

    Is this proof correct or is there a simpler one?


    Let $\displaystyle M = || f ||_{\infty} = \text{ess }\sup |f(t)| = \inf\ \{ N:\ \mu\{ t:|f(t)| > N \} = 0\ \}.$

    Define for $\displaystyle 0 < \epsilon < M,$ $\displaystyle g_{\epsilon}(t) = \min \{ |f(t)|,M-\epsilon \}.$

    Since

    $\displaystyle
    \lim_{p \to \infty} \left| \frac{g_{\epsilon}(t)}{M-\epsilon}\right|^p = \begin{cases}
    1 & \text{if } g_{\epsilon}(t) = M-\epsilon \\
    0 & \text{otherwise} \\
    \end{cases}
    $

    and

    $\displaystyle \mu\{ t : g_{\epsilon}(t) = M -\epsilon \} > 0$ by definition of $\displaystyle M,$

    $\displaystyle 0 < \lim_{p \to \infty} \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \le 1.$

    (Note: getting a positive expression here so the next limit equals 1 is the whole reason for introducing the function $\displaystyle g_{\epsilon}.$ This would not necessarily hold for $\displaystyle f.$)

    Therefore

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1,$

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int |g_{\epsilon}|^p \ d\mu \right]^{1/p} = {M-\epsilon},$

    $\displaystyle \lim_{p \to \infty} \frac{|| g_{\epsilon} ||_p}{\mu(X)^{1/p}} = {M-\epsilon},$

    and letting $\displaystyle \epsilon \to 0,$

    $\displaystyle \lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = M = || f ||_{\infty}.$
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  3. #3
    Super Member Rebesques's Avatar
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    is there a simpler one?

    Not that I know of.

    I only have the same question every time: Why are the limits for epsilon and for p interchangeable?
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  4. #4
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    Quote Originally Posted by Rebesques View Post
    Not that I know of.

    I only have the same question every time: Why are the limits for epsilon and for p interchangeable?
    Good question. I chewing on it.
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    Quote Originally Posted by Rebesques View Post
    I only have the same question every time: Why are the limits for epsilon and for p interchangeable?
    Thanks for questioning the proof. I think the limits are not interchangeable and the proof is flawed.

    The goal of the proof is to show

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$

    If $\displaystyle \mu\{ t : |f(t)| = M \} = 0$, then $\displaystyle |f(t)/M| < 1$ almost everywhere and

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right] = 0$.

    Then the desired limit has the indeterminate form

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}
    = 0^0.$

    My argument using $\displaystyle g_{\epsilon}$ to try to get around this is invalid.

    I don't see the solution and it seems to me to be a non-trivial complication. I find this odd because the proof (or a simpler version of it) is left as an exercise for the reader in a textbook I have.
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  6. #6
    Super Member Rebesques's Avatar
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    I also think the solution is simple (maybe we have the same textbook? )


    I e-mailed a friend (who happens to be a better mathematician ) and he says you are justified to interchange the limits, as $\displaystyle g_{\epsilon}$ is monotonic for $\displaystyle \epsilon$.

    Other than this, the expert says the limits can switch, as long as the quantity is nonnegative; If one of the double limits is infinity, the other will be the same. This last one, I must admit, has got me thinking
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  7. #7
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    Quote Originally Posted by Rebesques View Post
    I also think the solution is simple (maybe we have the same textbook? )

    I e-mailed a friend (who happens to be a better mathematician ) and he says you are justified to interchange the limits, as $\displaystyle g_{\epsilon}$ is monotonic for $\displaystyle \epsilon$.

    Other than this, the expert says the limits can switch, as long as the quantity is nonnegative; If one of the double limits is infinity, the other will be the same. This last one, I must admit, has got me thinking
    I think your friend is not looking at the problem in enough detail.

    The basic goal is to prove

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$

    This means given $\displaystyle \alpha > 0$, there exists $\displaystyle P$ such that $\displaystyle p \ge P$ implies

    $\displaystyle \left| \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha.$

    We can try using as my "proof" does that for fixed $\displaystyle \epsilon > 0$

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1$

    and

    $\displaystyle \lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ but it does not help.

    To see this, write using the triangle inequality that

    $\displaystyle
    \left|\left[\frac{1}{\mu(X)}\int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}-1\right| \le$
    $\displaystyle \left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} \right|$ $\displaystyle +
    \left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} - 1 \right| .$

    Now $\displaystyle \lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ guarantees that for each $\displaystyle p$ we can find $\displaystyle \epsilon(p)$ such that

    $\displaystyle \left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} \right| < \alpha/2.$

    But there is no corresponding guarantee that

    $\displaystyle \left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha/2.$

    For that we would need for given $\displaystyle p$ that

    $\displaystyle \lim_{\epsilon \to 0} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$

    Instead we have for given $\displaystyle \epsilon$ that

    $\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$

    Oops.
    Last edited by JakeD; Aug 29th 2007 at 02:48 PM.
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