1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted