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Math Help - 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 || f ||_p = \left[ \int |f|^p d\mu \right]^{1/p},

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

    Is this proof correct or is there a simpler one?


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

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

    Since

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

    and

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

    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 g_{\epsilon}. This would not necessarily hold for f.)

    Therefore

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

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

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

    and letting \epsilon \to 0,

    \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

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

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

    \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

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

    My argument using 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 g_{\epsilon} is monotonic for \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 g_{\epsilon} is monotonic for \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

    \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 \alpha > 0, there exists P such that p \ge P implies

    \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 \epsilon > 0

    \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

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

    To see this, write using the triangle inequality that

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

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

    \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

    \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 p that

    \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 \epsilon that

    \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; August 29th 2007 at 02:48 PM.
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