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Thread: convergence in Lp, convergence for p smaller infinity

  1. #1
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    convergence in Lp, convergence for p smaller infinity

    Let X=[0,1] , A=$\displaystyle [0,1]\cap B$. (B is the Borel-$\displaystyle \sigma$ algebra), $\displaystyle \lambda$ the lebesgue measure. In $\displaystyle (X,A,\lambda)$ explain why these statements are true or false:

    1) if $\displaystyle \exists 1 \le p < \infty$ with $\displaystyle f_n \in L_p$ and $\displaystyle ||f_n||_p \rightarrow 0$ then $\displaystyle ||f_n||_\infty \rightarrow 0$

    2) if $\displaystyle f_n \in L_p$ and $\displaystyle ||f_n||_p \rightarrow 0 \forall 1 \le p < \infty$ then $\displaystyle ||f_n||_\infty \rightarrow 0$


    Well I honestly do not have much ideas how to solve this. It's clear 1) isn't true because 2) wouldn't make sense if 1) was true. But a formal explanation? A prove? I don't know.

    All I tried was finding functions that are for example in L_1 but not in L_2 . But I couldn't find a contradiction. Also there exist Lebesgue-integrable functions that are not bounded...??

    Could please someone give me a hint?
    Thanks from girdav
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  2. #2
    Super Member girdav's Avatar
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    Re: convergence in Lp, convergence for p smaller infinity

    Try $\displaystyle f_n(x):=\frac 1n\chi_{(0,1/n)}(x)$.
    Thanks from huberscher
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    Re: convergence in Lp, convergence for p smaller infinity

    $\displaystyle \int \frac{1}{n^p}* \chi_{(0,\frac{1}{n})} d\mu=\frac{1}{n^{p+1}} < \infty \forall n \in \mathbb{N} \Rightarrow f_n \in L_p \forall 1 \le p < \infty $
    and $\displaystyle lim_{n \to \infty} (\frac{1}{n^{p+1}})^{\frac{1}{p}} =0 $
    but why doesn't converge


    $\displaystyle ||f_n||_{\infty} $ to 0 ?

    $\displaystyle \forall y \in [\frac{1}{3n},\frac{1}{2n}] : |f_n (y)| = |\frac{1}{n}|$and
    $\displaystyle \mu([\frac{1}{3n},\frac{1}{2n}])=\frac{1}{6n} >0 \forall n \in \mathbb{N}$


    I'm not sure because for me $\displaystyle (0,\frac{1}{n})$ converges to the empty set so why doesn't $\displaystyle ||f_n||_{\infty} $ go to 0 ?
    Last edited by huberscher; Oct 21st 2012 at 07:14 AM.
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    Super Member girdav's Avatar
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    Re: convergence in Lp, convergence for p smaller infinity

    Sorry, I meant $\displaystyle n\chi_{(0,n^{-1})}$ for the first example. For the second one, take $\displaystyle f_n(x)=\frac{\log x}n$.
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    Re: convergence in Lp, convergence for p smaller infinity

    Hi

    Well, $\displaystyle \int n \chi_{(0,\frac{1}{n})}d\mu=1 \forall n$ but $\displaystyle \lim_{n\rightarrow \infty} (n^{p-1})^{\frac{1}{p}}=\infty \forall p > 1 $ ???

    And how can I calculate $\displaystyle \int (\frac{log(x)}{n})^p d\mu $ ?? It's not a Riemann integral...?

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    Super Member girdav's Avatar
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    Re: convergence in Lp, convergence for p smaller infinity

    Maybe $\displaystyle f_n=\sqrt n\chi_{(0,n^{-1})}$ is better for the first question.

    For the computation of the integral, use the substitution $\displaystyle t=\log x$.
    Last edited by girdav; Oct 21st 2012 at 12:06 PM.
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    Re: convergence in Lp, convergence for p smaller infinity

    Hi

    What does $\displaystyle \sqrt{(0,\frac{1}{n})}$ mean? I've never seen this before. The square root of an interval???

    The substitution didn't help here either. I don't know what to do because this is not a Riemann integral ????

    $\displaystyle \frac{dt}{d\mu}=0$ so in the end this integral is simply 0 ???

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    Super Member girdav's Avatar
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    Re: convergence in Lp, convergence for p smaller infinity

    I've edited for the first; for the second is just an improper integral (I "cheated" a little as the function is not in $\displaystyle L^{\infty}$).
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    Re: convergence in Lp, convergence for p smaller infinity

    Hi

    Thank you.
    1st:
    So $\displaystyle (0,\frac{1}{n})$ isn't a zero set for all n so it isn't one when n goes to infinity as well?

    then for all $\displaystyle x\in (0,\frac{1}{n}) |f_n(x)|=n^{\frac{1}{2}}$ and this is not bounded.

    This is the correct reason for that yes?


    2nd:
    $\displaystyle \int |f_n|^p d\mu(x)=\int \frac{1}{n^p}*\log(x)^p d\mu(x)= \frac{1}{n^p}*\int t^p*e^t d\mu(t)$
    $\displaystyle = \frac{1}{n^p} (e-p(e-(p-1)(e-(p-2)(e-...)))) $ (*)

    But what is $\displaystyle (e-p(e-(p-1)(e-(p-2)(e-...))))$ ? Is it smaller than infinity?

    But here when n goes to infinity (*) goes to zero and so does the pth root of (*) or what did I do wrong here?

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    Re: convergence in Lp, convergence for p smaller infinity

    Yes asb $\displaystyle \int_0^{+\infty}e^{-t}t^pdt$ is convergent. But I cheated a little as I took an unbounded function.
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    Re: convergence in Lp, convergence for p smaller infinity

    Hi

    1st: You just didn't answer my question. What about my arguments concerning the first part?

    2nd: Why isn't $\displaystyle ||f_n||_{\infty}\to 0$ for n to infinity?

    I don't understand ??
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    Re: convergence in Lp, convergence for p smaller infinity

    You are right for the first question (I forget answering you before, sorry...).

    For the second one, $\displaystyle f_n$ is not essentially bounded (otherwise $\displaystyle t\mapsto \log t$ on $\displaystyle (0,1)$ would be so).
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    Re: convergence in Lp, convergence for p smaller infinity

    Hi

    But your "cheating" isn't a problem at all. The function has to be in L_p for all p only... it can be unbounded?
    Last edited by huberscher; Oct 22nd 2012 at 12:20 PM.
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  14. #14
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    Re: convergence in Lp, convergence for p smaller infinity

    Yes, as $\displaystyle \log t\geq n$ if $\displaystyle 0<t\leq e^{-n}$, we should have $\displaystyle \lVert \log \cdot\rVert_{\infty}\geq n$ for each integer $\displaystyle n$, which is not possible.
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    Re: convergence in Lp, convergence for p smaller infinity

    So the reason here is formally said for the 2nd part:

    $\displaystyle f_n\in L_p \forall 1 \le p < \infty$ but

    $\displaystyle \mu((0,\frac{1}{n}))=\frac{1}{n}>0 \forall n$ and
    $\displaystyle \forall x \in (0,\frac{1}{n}) \Rightarrow f_n (x) \ge \frac{\log{\frac{1}{n}}}{n} = 0$

    ---doesn't work
    Last edited by huberscher; Oct 22nd 2012 at 12:38 PM.
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