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!

Quick question about convergence

  1. Mar 20, 2013 #1

    Zondrina

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    Let ##s_n(x) = \frac{1}{n} e^{-(nx)^2}##. Show there is a function ##s(x)## such that ##s_n(x) → s(x)## uniformly on ##ℝ## and that ##s_n'(x) → s'(x)## for every x, but that the convergence of the derivatives is not uniform in any interval which contains the origin.

    2. Relevant equations

    ##s_n(x) → s(x)## as ##n→∞##

    3. The attempt at a solution

    For any real x, ##s_n(x) → 0 = s(x)## as ##n→∞## so we have pointwise convergence.

    ##\forall ε>0, \exists N(ε) \space | \space n>N \Rightarrow |s_n(x)-s(x)| < ε, \forall x \in ℝ##

    ##|s_n(x) - s(x)| ≤ 1/n##

    So choosing ##n > 1/ε## means we have uniform convergence.

    ##s_n'(x) = -2xne^{-(nx)^2}##

    Now, for all real x ##s_n'(x) → 0 = s'(x)## as ##n→∞##.

    ##|s_n'(x) - s'(x)| = 2|x|ne^{-(nx)^2}##

    I'm stuck on showing the convergence of the derivatives is not uniform in any interval containing the origin. Don't really know how to argue this one.
     
  2. jcsd
  3. Mar 20, 2013 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Let ##(-\varepsilon,\varepsilon)## be an interval around the origin.

    1) Can you calculate

    [tex]A_n=\sup_{x\in (-\varepsilon,\varepsilon)} |s_n^\prime(x) - s^\prime(x)|[/tex]

    2) Does ##A_n\rightarrow 0##?

    3) What does (2) say about uniform convergence?
     
  4. Mar 20, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

    [itex]A_n=\sup_{x \in (-ε, ε)} |s_n'(x) - s'(x)| = \sup_{x \in (-ε, ε)} 2|x|ne^{-(nx)^2}[/itex]

    That's some new notation on me. I've never seen sup used in such a way.

    (2) I would assume not.

    (3) Not uniform, but pointwise.
     
  5. Mar 20, 2013 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Let's take a compact interval ##[-\varepsilon,\varepsilon]## instead then. Then the ##\sup## is actually a ##\max## then.

    So, when I'm asking for ##\sup_{x \in [-\varepsilon,\varepsilon]} 2|x|ne^{-(nx)^2}##, I'm looking for the maximum value of this function.

    So use calculus to find where the function ##g_n(x)= 2|x|e^{-(nx)^2}## takes on its maximum and calculate the maximum.
     
  6. Mar 20, 2013 #5

    Zondrina

    User Avatar
    Homework Helper

    Alright, that makes sense. I'll even pick the real open interval (-ε,ε) containing 0 since i see it doesn't matter now.

    With how the interval and ##|s_n(x) - s(x)|## are defined I have 2 possible functions to examine.

    Finding the maximums of these functions with regular calculus seems to be far beyond my understanding.

    ##|s_n(x) - s(x)| = 2xne^{-(nx)^2}## for ##0 ≤ x < ε##
    ##|s_n(x) - s(x)| = -2xne^{-(nx)^2}## for ##-ε < x ≤ 0##

    How do I find the maximum? It's not like I can take the derivative and find criticals of this monster.
     
  7. Mar 20, 2013 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Why not?
     
  8. Mar 20, 2013 #7

    Zondrina

    User Avatar
    Homework Helper

    It looks terrible, but for the case where x is positive I took the derivative and found ##x = ± \frac{1}{ n \sqrt{2}}## to be the roots of the derivative.

    For the case where x is negative, I took the derivative and found the same roots.
     
  9. Mar 20, 2013 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Ok, so the maximum is reached in ##1/(n\sqrt{2})##. The value in that point is

    [tex]f(1/(n\sqrt{2})) = \frac{2}{\sqrt{2}}e^{-1/2}[/tex]

    So, we see that

    [tex]\sup_{x\in (-\varepsilon,+\varepsilon)} |s_n^\prime(x)-s^\prime(x)| = \frac{2}{\sqrt{2}} e^{-1/2}[/tex]

    Does this converge to ##0## if n goes to infinity?? (obviously not)

    Can you deduce from this that convergence is not uniform?
     
  10. Mar 20, 2013 #9

    Zondrina

    User Avatar
    Homework Helper

    Ohhhhhh I see where you were going with this now. So since the supremum over (-ε,ε) of ##|s_n(x) - s(x)|## doesn't go to zero we know that the convergence of the derivatives can't possibly be uniform.

    If it were to be uniform, the difference would go to zero if we subtract the sequence from the limiting function ( or vice versa ).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Quick question about convergence
Loading...