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Math Help - Uniform Convergence

  1. #1
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    Uniform Convergence

    Consider f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)} .

    (a) For what values of x does the series converge?

    By comparison test,  f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)} converges  \forall x \in \mathbb{R} \{  0 }.

    (b) On what intervals of x in the form  (a,b) does the series converge uniformly?

    Now since  \left |\frac{1}{n(1+nx^2)} \right|=\frac{1}{n+n^2x^2}\leq \frac{1}{n^2x^2}  \forall x \in \mathbb{R} \{  0 }, by M-Weierstrass Test the series converges uniformly on  (- \infty, 0) and  (0, \infty) .

    (c) On what intervals of x in the form  (a,b) does the series fail to converge uniformly?

    0?

    (d) Is f continuous at all points where the series converges?

    No idea how to do this.

    Could someone please check the answers that I have and help me for (d)?
    Last edited by nmatthies1; November 15th 2009 at 01:54 PM.
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  2. #2
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    Quote Originally Posted by nmatthies1 View Post
    Consider f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)} .


    (b) On what intervals of x in the form  (a,b) does the series converge uniformly?

    Now since  \left |\frac{1}{n(1+nx^2)} \right|=\frac{1}{n+n^2x^2}\leq \frac{1}{n^2x^2}  \forall x \in \mathbb{R} \{  0 }, by M-Weierstrass Test the series converges uniformly on  (- \infty, 0) and  (0, \infty) .

    Be careful, you need a uniform bound for 1/x^2 for applying M-test. Your argument is correct for the compact subsets of (-\infty,0)\cup (0,\infty), when such a bound always exists.

    Just observe that this gives you an answer for (d), for each
    x\neq 0 you have that there exists x\in[a,b]\subseteq \mathbb{R}\setminus\{0\} such that the series converges uniformly on [a,b].
    Now, recall that the uniform convergence of continuous functions preserves continuity. Actually M-test gives continuity whenever the terms of the series are.
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  3. #3
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    Quote Originally Posted by Enrique2 View Post
    Be careful, you need a uniform bound for 1/x^2 for applying M-test. Your argument is correct for the compact subsets of (-\infty,0)\cup (0,\infty), when such a bound always exists.

    Just observe that this gives you an answer for (d), for each
    x\neq 0 you have that there exists x\in[a,b]\subseteq \mathbb{R}\setminus\{0\} such that the series converges uniformly on [a,b].
    Now, recall that the uniform convergence of continuous functions preserves continuity. Actually M-test gives continuity whenever the terms of the series are.
    What do you mean by uniform bound?
    Also, I can't quite seem to get my head around (d), why is it that uniform convergence guaratees preservation of continuity?
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  4. #4
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    An uniform bound on [a,b] (or (a,b) is a bound for every x\in [a,b]. It is a necessary condition in M test that f_n(x)\leq M_n for each x\in [a,b] and \sum_nM_n being convergent. Now look that in an interval (0,b) you have
    that the bound \frac{1}{xn^2} depends on x and goes to infinity as x goes to 0. This is not allowed in the M-test, you need a bound independent of x, i.e. valid for all the interval.

    The fact of that uniform convergence of a sequence of continuous functions is again continuous is a well known theorem, I don't remember the name of the author. That is because C(K) endowed with the uniform norm is a Banach space if K is compact. Sure that you can check it in a book of Analysis, I suppose in Apostol for example.
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