Results 1 to 2 of 2

Math Help - [SOLVED] Integrability of an infinite series of functions

  1. #1
    Member
    Joined
    Mar 2008
    From
    Acolman, Mexico
    Posts
    118

    [SOLVED] Integrability of an infinite series of functions

    I need to prove that if the sequence of functions \{ f_n \}
    converges uniformly to f on [0,1], and each f_n is integrable, then the limit function f is also integrable on [0,1].

    Does this help at all??
    Since \{ f_n \} converges uniformly to f, then
    \exists N such that n>N \Rightarrow |f(x)-f_n(x)|<\varepsilon

    \Rightarrow -\varepsilon<f(x)-f_n(x)<\varepsilon<br />
\Rightarrow f_n(x)-\varepsilon<f(x)<f_n(x)+\varepsilon
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by akolman View Post
    I need to prove that if the sequence of functions \{ f_n \}
    converges uniformly to f on [0,1], and each f_n is integrable, then the limit function f is also integrable on [0,1].

    Does this help at all??
    Since \{ f_n \} converges uniformly to f, then
    \exists N such that n>N \Rightarrow |f(x)-f_n(x)|<\varepsilon

    \Rightarrow -\varepsilon<f(x)-f_n(x)<\varepsilon<br />
\Rightarrow f_n(x)-\varepsilon<f(x)<f_n(x)+\varepsilon
    First let us remark that \lim_{N\to\infty}S_N=\sum_{n=1}^{\infty}a_n where S_N represents the Nth partial sum. So now \sum{a_n}\text{ converges}\leftrightarrow~S_N\text{ converges}

    1. So now we know that for a sequence \left\{S_N\right\} to be convergent it must be montonic and bounded. Therefore \exists{M}~\backepsilon\lim_{N\to\infty}S_N\leqsla  nt{M}\stackrel{\text{in turn}}{\implies}\sum{a_n}\leqslant{M}. So for every point where \sum{f_n(x)} converges we have by neccessity \exists{M}\backepsilon~\sum{f_n(x)}\leqslant{M}. And if \forall{x}\in[0,1]~\sum{f_n(x)}\stackrel{\text{uniformly}}\longright  arrow{f} then \forall{x}\in[0,1]\exists{M}\backepsilon~f\leqslant{M}. Thus on [0,1] f is bounded.

    2. Before we start the next step we are given that f_n(x) is Riemann integrable, thus bounded and continous.

    3.Next we just use the theorem that if \forall{x}\in[a,b]~~\sum{f_n(x)}\stackrel{\text{uniformly}}\longrigh  tarrow{f} and f_n(x) is continuous on [a,b] then f is continous on [a,b]. So the fact that \sum{f_n(x)}\stackrel{\text{uniformly}}\longrighta  rrow{f}\wedge{f_n(x)}\in\mathcal{C}~~\forall{x}\in[0,1] tells us that f is continuous on [0,1]

    4. So we have shown that on [0,1] f is both bounded and continous. So it is Riemann integrable.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative of an infinite series of functions
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 6th 2011, 10:12 AM
  2. sum of infinite series of functions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 29th 2010, 10:49 PM
  3. [SOLVED] Infinite series #3
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 23rd 2009, 07:29 PM
  4. [SOLVED] 2 infinite series
    Posted in the Calculus Forum
    Replies: 8
    Last Post: July 12th 2008, 11:03 AM
  5. [SOLVED] Sum of infinite series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 22nd 2008, 05:57 AM

Search Tags


/mathhelpforum @mathhelpforum