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Uniform Convergence and integration

  1. Aug 13, 2009 #1
    What can I conclude using the following theorem?

    Let the functions [tex] u_n (x) [/tex] be continuous on the closed interval [tex] a \le x \le b[/tex] and let them converge uniformly on this interval to the limit function [tex] u(x) [/tex]. Then

    [tex]
    \int_a^b u (x) \, dx=\lim_{n \to \infty} \int_a^b u_n (x) \, dx
    [/tex]

    Can I conclude that if [itex] \int_a^b u (x) \, dx \neq \lim_{n \to \infty} \int_a^b u_n (x) \, dx [/itex], then the sequence of functions [itex] u_n (x) [/itex] is NOT uniformly convergent on the interval?

    -Also what is the contrapositive of the above theorem? (I am confused on how to negate [itex] P [/itex]=" Let the functions [itex] u_n (x) [/itex] be continuous on the closed interval [itex] a \le x \le b[/itex] and let them converge uniformly on this interval to the limit function [itex] u(x) [/itex]." )
     
  2. jcsd
  3. Aug 13, 2009 #2
    If you get confused by logic, then try to assign symbols for different statements, and reduce the problem to a simpler form. If you know

    [tex]
    A\quad\textrm{and}\quad B\implies C
    [/tex]

    then you also know

    [tex]
    \textrm{not}\; C\implies (\textrm{not}\;A)\quad\textrm{or}\quad(\textrm{not}\;B)
    [/tex]

    Your example is slightly more confusing than it would need to be, because continuity of [itex]u_n[/itex] is not highly essential. If [itex]u_n[/itex] and [itex]u[/itex] are integrable, and [itex]u_n\to u[/itex] uniformly, then

    [tex]
    \lim_{n\to\infty} \int\limits_a^b u_n(x)dx = \int\limits_a^b u(x) dx.
    [/tex]

    Substituting [itex]A[/itex] to be the knowledge that [itex]u_n[/itex] are all continuous, [itex]B[/itex] to be the knowledge that [itex]u_n\to u[/itex] uniformly, and [itex]C[/itex] to be the knowledge that the limit and integral commute, might bring some clarity.

    I would also set [itex]D[/itex] to be the knowledge that [itex]u_n[/itex] and [itex]u[/itex] are integrable. Then

    [tex]
    B\quad\textrm{and}\quad D\implies C
    [/tex]

    and

    [tex]
    A\quad\textrm{and}\quad B\implies D
    [/tex]

    (additional comment: I just realized that if [itex]u_n[/itex] are integrable and if [itex]u_n\to u[/itex] uniformly, then probably [itex]u[/itex] is integrable too, but I'm not 100% sure of this right now. It could be that some parts of my response are not the most sense making, but I think there is nothing incorrect there anyway.)
     
    Last edited: Aug 13, 2009
  4. Aug 13, 2009 #3
    How would you get [tex] \neg A [/tex]? Do you negate closed and the interval as [itex]a>x>b[/itex]?

    Or in other words,

    " Assume the functions [tex] u_n (x) [/tex] are not continuous on the not closed interval [tex] a > x > b[/tex]"?
     
    Last edited: Aug 13, 2009
  5. Aug 13, 2009 #4
    I would keep the assumption that the functions are of form [itex]u_n:[a,b]\to\mathbb{C}[/itex] untouched, and only let [itex]A[/itex] concern the continuity.

    It should be recognized that there is true ambiguity in questions like this. If I declare that [itex]u_n[/itex] will always have the domain [itex][a,b][/itex], and then state that [itex]A[/itex] means that [itex]u_n[/itex] are all continuous, than [itex]\neg A[/itex] will simply mean that [itex]u_n[/itex] are not all continuous. This is what I meant originally.

    Alternatively one could state that [itex]A[/itex] means that [itex]u_n[/itex] all have the domain [itex][a,b][/itex] and that [itex]u_n[/itex] are all continuous. Then [itex]\neg A[/itex] would mean that [itex]u_n[/itex] don't all have the domain [itex][a,b][/itex] or that [itex]u_n[/itex] are not all continuous.

    You must know yourself what you mean by [itex]A[/itex], before asking what [itex]\neg A[/itex] is.
     
    Last edited: Aug 13, 2009
  6. Aug 13, 2009 #5

    How does one know when to do this?
     
  7. Aug 13, 2009 #6
    My fear is what if I "choose" the wrong statement A like I did in post #3? How would I know what the "right" statement is?
     
  8. Aug 14, 2009 #7
    There are no right or wrong choices for [itex]A[/itex]. What matters is that you know what you have chosen.

    Once [itex]A[/itex] has been fixed, there can be right or wrong deductions, however.

    (There was a movie where somebody said something like "there's no right or wrong decisions. What matters is that you dare to make a decision". It could be this is slightly off topic, though.)
     
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