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Function sequence convergence

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that a sequence ##f_n \to f \in C[0,1]## with the sup norm ##|| ||_\infty##, then ##f_n \to f \in C[0,1]## with the integral norm.

    3. The attempt at a solution

    given ##\epsilon > 0 \exists n_0 \in N## s.t

    ##||(fn-f) (x)|| < \epsilon \forall n > n_0## with ## x \in [a,b]##

    ie ##\forall \epsilon > 0 \forall n_0 \in N## s.t

    ##sup |(f_n-f)(x)|=sup|f_n(x)-f(x)|## with ##x \in C[0,1]## and ##n \in \mathbb{N} \implies f_n \to f \in C[0,1]##

    if this is correct, do I attempt the same for the integral norm?
     
  2. jcsd
  3. Mar 12, 2012 #2

    Mark44

    Staff: Mentor

    It looks to me like there are two words missing in your problem statement - "if" and "then".
    I believe what the problem is saying is this:
    IF fn → f using the || || norm, THEN fn → f using the integral norm.

    If my surmise is correct, what you have done is prove the hypothesis, which is done only in if and only if proofs.
     
  4. Mar 12, 2012 #3
    Yes, your surmise is correct. So what you recommend I do? I need to show some link between between the sup and integral norm, right? Not sure how to do that...thanks
     
  5. Mar 12, 2012 #4

    Mark44

    Staff: Mentor

    Given that for any ε > 0, there is an N0 such that, if n > N0, then ||fn - f|| < 0,

    show that for any ε > 0, there is an M0 such that, if n > M0,
    [tex]\int_0^1 f~f_n~dt < \epsilon[/tex]

    You'll need to use the definition of the infinity norm to establish the second inequality.
     
  6. Mar 13, 2012 #5
    The definition of an infinity norm in a function space is

    ##||f||_\infty= sup | f(x)|## where ##x \in C[a,b]##

    Not sure how to proceed further especially when I see a product of ##f*f_n##...?
     
  7. Mar 13, 2012 #6

    Mark44

    Staff: Mentor

    How is the integral norm defined? I just took a guess at how it would look.
     
  8. Mar 13, 2012 #7

    LCKurtz

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    He means ##< \epsilon##

    Hopefully you could figure out he meant$$
    \int_0^1|f_n(x)-f(x)|\, dx <\epsilon$$
     
  9. Mar 13, 2012 #8

    Mark44

    Staff: Mentor

    Yes, I definitely meant ϵ, not 0.

    And I misremembered the integral norm. I think I confused it with one of the inner products.
     
  10. Mar 13, 2012 #9
    I cant see the link. All thats going through my head is for the integral norm

    ##\displaystyle \int_0^1 |f_n(x)-f(x)| dx = | \frac{|f_n(x)-f(x)|^{n+1}}{n+1}|_0^1## which would evaluate to some value < ε if the assumption ##f_n(x) \to f(x)## is true

    and for the infinity norm

    ## sup|f_n(x)-f(x)| = max{|f_n(x)-f(x)|}## for ##x \in C[0,1]## which would evaluate to some value < ε if the assumption ##f_n(x) \to f(x)## is true also...?
     
  11. Mar 13, 2012 #10

    LCKurtz

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    Oh my. After seeing that I think the best suggestion I can make is for you to schedule a meeting with your instructor. I don't see how to continue helping you with this problem without working it for you. Do you have the appropriate prerequisites for the course these problems are coming from?
     
  12. Mar 13, 2012 #11

    Mark44

    Staff: Mentor

    Exactly my thought.

    IIRC, Bugatti79 is studying this stuff on his own.
     
  13. Mar 14, 2012 #12

    I know well this in not correct mathematically, I shouldn't have wrote that. I was just trying to indicate that if we integrate this function between the limits 0 and 1 we should get some value less than some (ε>0) that we have chosen. Of course, that doesn't make me progress any further.
     
  14. Mar 14, 2012 #13
    Mark is this was you aim at:

    [tex] \int_0^1 |f_n(x)-f(x)| dx \leq \int_0^1 ||f_n(x)-f(x)||_{\infty} dx =\epsilon\ \forall n \geq N_0 [/tex]
     
  15. Mar 14, 2012 #14
    Should that be "##< \epsilon##" not "##= \epsilon##"?
     
  16. Mar 14, 2012 #15
  17. Mar 31, 2012 #16
    I have just put the information together for completeness, I am still wondering how to finish it.?

    Show that a sequence fn→f∈C[0,1] with the sup norm ||||∞, then fn→f∈C[0,1] with the integral norm.

    Given that for any ε > 0, there is an N0 such that, if n > N0, then ||fn - f||∞ < 0,

    show that for any ε > 0, there is an M0 such that, if n > M0 then

    $$\int_0^1|f_n(x)-f(x)|\, dx <\epsilon$$..?
     
  18. Mar 31, 2012 #17

    jgens

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    The relevant inequality for this problem is [itex]|f_n(x)-f(x)| \leq ||f_n-f||_{\infty}[/itex] for all [itex]x \in [0,1][/itex]. But honestly, based on your threads in the homework help forums, you really lack the mathematical maturity to be dealing with convergence in function spaces. It would be worthwhile for you to go back and review basic convergence in [itex]\mathbb{R}[/itex].
     
  19. Apr 3, 2012 #18
    I have looked through my notes and through Erwin Kreyszig's 'Introductory Functional Analysis'. Can you be more specific what you mean by basic convergence in R, so I can check the book.

    Thanks
     
  20. Apr 3, 2012 #19

    jgens

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    Unless you have some secret talent you are not letting on to here, then that book is much too difficult for you.

    I should have been more precise. I meant convergence in the p-norms on [itex]\mathbb{R}^n[/itex].
     
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