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L∞(E) Norm

  1. Nov 17, 2013 #1
    Real Analysis, L∞(E) Norm as the limit of a sequence.

    [itex]|| f ||_{\infty}[/itex] is the lesser real number [itex] M [/itex] such that [itex]| \{ x \in E / |f(x)| > M \} | = 0 [/itex] ([itex] | \cdot | [/itex] used with sets is the Lebesgue measure).

    For every [itex]1 \leq p < \infty[/itex] and for every [itex]E[/itex] such that [itex] 0 < | E | < \infty [/itex] we define:
    [itex]N_{p} [f] = (\frac{1}{ | E | } \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} [/itex]

    a) [itex]p_{1} < p_{2} \Rightarrow N_{p_{1}} [f] \leq N_{p_{2}} [f][/itex]
    b) [itex]N_{p} [f+g] \leq N_{p} [f] + N_{p} [g][/itex]
    c) If [itex]\frac{1}{p} + \frac{1}{q} = 1[/itex] [itex]\frac{1}{ | E | } \displaystyle \int_{E} | f g | \leq N_{p} [f] N_{q} [g][/itex] [itex][/itex]
    d) [itex]\lim_{p \to \infty} N_{p} [f] = | | f | |_{\infty}[/itex]

    I need help with the a), I couldn't solve it. I didn't got anything good in my attemps, so I'll show my work in the other three points (Also I'm taking the opportunity to someone find any mistake I could have overlooked).

    b) By triangular inequality of the norm [itex]|| \cdot ||_{p}[/itex] and observing that:
    [itex]| E |^{\frac{1}{p}} N_{p} = || \cdot ||_{p}[/itex]
    We have:
    [itex]| E |^{\frac{1}{p}} N_{p} [f+g] \leq | E |^{\frac{1}{p}} N_{p} [f] + | E |^{\frac{1}{p}} N_{p} [g][/itex]
    Divide both sides by [itex]| E |^{\frac{1}{p}}[/itex] the work is done.

    c) By Hölder Inequality we have: [itex]\displaystyle \int_{E} | f g | \leq (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}}[/itex]
    We divide both sides by [itex]| E |[/itex] and knowing that as [itex]\frac{1}{p}+\frac{1}{q}=1[/itex] we have:
    [itex]\frac{1}{| E |} \displaystyle \int_{E} | f g | \leq \frac{1}{| E |} (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}} = (\frac{1}{| E |} \int_{E} | f |^{p} )^{\frac{1}{p}}(\frac{1}{| E |} \int_{E} | g |^{q})^{\frac{1}{q}} [/itex]

    d) Let be the sequence of nonnegative simple functions [itex]\{ \varphi_{n} \}_{n \in \mathbb{N}}[/itex] with [itex]\varphi_{n} \leq \varphi_{n+1} \forall n \in \mathbb{N}[/itex] and:
    [itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex].
    It's clear that [itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex] if and only if [itex]\lim_{n \to \infty} \varphi_{n}^{p} = | f |^{p}[/itex] as:
    [itex]\phi_{n} = \displaystyle \sum_{i=1}^{m_{n}} \alpha_{i,n} X_{E_{i,n}} \Longrightarrow \phi_{n}^{p} = \displaystyle \sum_{i=1}^{m_{n}} \alpha_{i,n}^{p} X_{E_{i,n}}[/itex]
    Let be [itex]\varphi[/itex] a non negative simple function such that:
    [itex]\varphi \leq | f |[/itex]

    [itex]\varphi = \displaystyle \sum_{i=1}^{m} \alpha_{i} X_{E_{i}}[/itex] with [itex] | E_{i} | > 0 \forall i: 1 \leq i \leq m[/itex] and:
    [itex]\alpha_{1} < \alpha_{2} < ... < \alpha_{m} [/itex]
    [itex]\displaystyle \int_{E} \varphi^{p} = \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex]
    As [itex]p \to \infty[/itex] [itex]\alpha_{m}^{p} | E_{m} |[/itex] grows faster than [itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex], to every [itex] \varepsilon > 0 \exists p \in \mathbb{R}[/itex] such that:
    [itex]| \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 | < \varepsilon[/itex]
    As all coefficients of the sumatory are positive, we can drop the absolute value bars.
    [itex] \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 < \varepsilon[/itex]
    [itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} | < (1+\varepsilon) \alpha_{m}^{p} | E_{m} | [/itex]
    Then we have:
    [itex]\alpha_{m}^{p} | E_{m} | \leq \int_{E} \varphi^{p} < (1 + \varepsilon) \alpha_{m}^{p} | E_{m} |[/itex]
    We divide by [itex]| E |[/itex] sides and take power to [itex]\frac{1}{p}[/itex] and we have:
    [itex]((1 + \varepsilon) \frac{E_{m}}{| E |})^{1/p} \alpha_{m}[/itex]
    As [itex]p \to \infty[/itex] this converges to [itex]\alpha_{m}[/itex]
    Now by Beppo-Levi we have:
    [itex]\lim_{n \to \infty} \int_{E} \varphi_{n}^p = \int_{E} |f|^{p}[/itex]
    (We are using the simple functions [itex]\varphi_{n}^p[/itex] to aproximate [itex]| f |^{p}[/itex]
    We power both sides to [itex]\frac{1}{p}[/itex] and we are almost done.
    The only thing left it to prove that [itex]|| f ||_{\infty} = \lim_{n \to \infty} (\lim_{p \to \infty} N_{p} [ \varphi_{n} ])[/itex] which is easy, because to every set with positive measure, taking [itex]n[/itex] big enough we'll have a [itex]\varphi_{n}[/itex] with that set, then the norm infinity is the supremum of that sequence, and as this is an increasing sequence then it's the limit.
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 18, 2013 #2
    part a. See the following (the norm he defines here is the ##L_p## norm):

    Moreover, if we're on a bounded domain, we also have the relatively standard result that if f∈Lp1 for some p1∈[1,∞), then it is in Lp for every p≤p1 (which can be shown using Hölder's inequality). http://math.stackexchange.com/quest...ble-for-a-function-to-be-in-lp-for-only-one-p

    This is more general.
    Let M be the set of all measures of subsets of your space. Then, if p<q, then Lp < Lq if M has a positive lower bound, and Lq < Lp if M has a finite upper bound. http://www.johndcook.com/blog/2012/11/11/how-lp-spaces-nest/
  4. Nov 18, 2013 #3
    Thanks, tomorrow I'll ask my teacher about this problem.
  5. Nov 18, 2013 #4
    Isn't the [itex]L^{p}(E)[/itex] norm:
    [itex]| | f | | = (\displaystyle \int_{E} | f |^{p} )^{\frac{1}{p}}[/itex]
    This one is different, it has the measure of [itex]E[/itex] dividing before the root.
  6. Nov 18, 2013 #5
    You are right, the E is inside the root, which I didn't quite see. Well, I found you a proof. The finite measure of E is necessary, and shown by the counterexample at the bottom. The division by |E| falls out of his final inequality.


    Theorem Let X be a finite measure space. Then, for any 1≤p<q≤+∞
    The proof follows from Hölder inequality. Note that 1/p=1/q+1/r, with r>0. Hence
    ∥f∥Lp≤meas (X)1/r∥f∥Lq.

    The case reported on the wikipedia link of commenter answer follows from this, since of course, if X does not contain sets of arbitrary large measure, X itself can't have an arbitrary large measure.

    For the counterexample: f(x)=1/x belongs to L2([1,+∞)), but clearly it does not belong to L1([1,+∞)).

    Has your prof discussed measure in detail?
  7. Nov 18, 2013 #6
    What do you mean by "discussed measure in detail"? (I wouldn't know, I have not any reference course to compare).
    I think we did, our first set of practices were enterely of Lebesgue measure and we spend roughly 1/3 of the course (5-6 weeks) doing it.
    Later we had a brief views about "abstract measures"

    Elemental sets -> [itex]\sigma[/itex] - elemental sets -> Borelians and a couple of ways of generate them.
    External lebesgue measure.
    Lebesgue Measure
    Internal lebesgue measure, non-measurable sets.

    That was more or less what we covered on measure before we moved to measurable functions.

    We saw that for sets of finite measure [itex]L^{p}(E) \subset L^{q}(E)[/itex] if [itex]q \leq p[/itex]
    For infinite measure I know it's false because I already had the oportunity to solve the following exercise:
    Let be [itex]E = [0,\infty)[/itex].
    Prove that [itex]f(x) = x^{\frac{1}{2}} (1+|ln(x)|)^{-1} \in L^{2} (E)[/itex] but [itex]f \notin L^{p} (E) \forall p \in [0,\infty) - \{ 2 \}[/itex]
    Last edited: Nov 18, 2013
  8. Nov 18, 2013 #7
    I'll transcribe here what I understood from the links.
    [itex]| E | < \infty[/itex]
    Let be [itex]p,q / 1 \leq p \leq q \Longrightarrow 1 < \frac{q}{p}[/itex]
    Then there exists [itex]s \in \mathbb{R}[/itex] such that:
    [itex]1 = \frac{p}{q} + \frac{1}{s}[/itex]
    There exists a [itex]r \in \mathbb{R}[/itex] such that [itex]s= \frac{r}{p}[/itex]
    [itex]1 = \frac{p}{q} + \frac{p}{r}[/itex]
    Now we apply the Hölder inequality:
    [itex]\displaystyle \int_{E} | f |^{p} = \displaystyle \int_{E} | f |^{p} \cdot X_{E} \leq | E |^{\frac{p}{r}} ( \displaystyle \int_{E} (| f |^{p})^{\frac{q}{p}} )^{\frac{p}{q}} = | E |^{\frac{p}{r}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{p}{q}}[/itex]
    We power to [itex]\frac{1}{p}[/itex]:
    [itex]( \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq | E |^{\frac{1}{r}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}[/itex]
    [itex]\frac{1}{r} = \frac{1}{p} - \frac{1}{q} \Longrightarrow | E |^{\frac{1}{r}} = \frac{| E |^{\frac{1}{p}}}{| E |^{\frac{1}{q}}}[/itex]
    [itex]\frac{1}{| E |^{\frac{1}{p}}} ( \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq \frac{1}{| E |^{\frac{1}{q}}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}[/itex]
    [itex]\therefore ( \frac{1}{| E |} \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq ( \frac{1}{| E |}\displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}[/itex]
  9. Nov 19, 2013 #8
    It sounds like you did indeed cover a lot of measure theory -- certainly more than I ever did. I never took a course in it specifically; covered Lp spaces in a course either on real or functional analysis. So maybe you should be helping me???

    It looks to me like your interpretation of the sketchy proof I sent you is correct.

    I don't think you need the line about s. If p/q < 1 then there has to be a real number that fills in the gap, and p/r is as good as any (since r can be whatever works). This doesn't invalidate your proof in any way. It's just a little roundabout and where you can simplify that kind of thing it's a good idea.
  10. Nov 19, 2013 #9
    The course is of real analysis, both we spend a lot of time in the first chapter.

    And I found that you can directly apply the following generalization of Hölder inequality:
    [itex]\displaystyle \sum_{i=1}^{m} p_{i} = \frac{1}{r} \wedge f_{i} \in L^{p_{i}} \forall i : 1 \leq i \leq m \Longrightarrow || \displaystyle \prod_{i=1}^{m} f_{i} ||_{r} \leq \displaystyle \prod_{i=1}^{m} | | f_{i} | |_{p_{i}}[/itex]

    Thank you very much for your help again, I think this is the hardest course I ever toke and I really need help (As I don't speak English very well I really feel I need to say it explicity because I may sound a little rude to people helping me but it's only because my English is quite basic).
    Last edited: Nov 19, 2013
  11. Nov 19, 2013 #10
    I wish my real analysis course had spent more time on measures. You have not sounded rude to me, but it is nice to get thanked. I really enjoy helping with problems, and I learn a lot too.
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