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Math Help - Pointwise convergence

  1. #1
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    Pointwise convergence

    (1) Let f_n(x)= \frac{nx}{1+nx^2}.

    (a) Find the pointwise limit of (f_n) for all x \in (0,\infty)
    (b) Is the convergence uniform on (0,\infty)?
    (c) Is the convergence uniform on (0,1)?
    (d) Is the convergence uniform on (1,\infty)?

    (2) Let g_n(x)= \frac{nx+sin(nx)}{2n}.

    Find the pointwise limit of (g_n) on \mathbb{R}. Is this convergence uniform on [-10,10]? Is the convergence uniform on all of \mathbb{R}?

    (3) For each n \in \mathbb{N}, find the points on \mathbb{R} where the function f_n(x) = x/(1+nx^2) attains its maximum and minimum values. Use this to prove (f_n) converges uniformly on \mathbb{R}. What is the limit function?

    (4) For each n \in \mathbb{N}, define f_n on \mathbb{R} by
    f_n(x)=\left\{\begin{array}{cc}1,&\mbox{ if }<br />
|x|\geq 1/n\\n|x|, & \mbox{ if } |x|<1/n\end{array}\right.

    (a) Find the pointwise limit of (f_n) on \mathbb{R} and decide whether or not the convergence is uniform.

    (b)Construct an example of a pointwise limit of continuous functions that converges everywhere on the compact set [-5,5] to a limit function that is unbounded on this set.

    If anyone can show me how to do these, I would appreciate it, thanks!
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  2. #2
    Moo
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    Hello !

    Quote Originally Posted by shadow_2145 View Post
    (1) Let f_n(x)= \frac{nx}{1+nx^2}.

    (a) Find the pointwise limit of (f_n) for all x \in (0,\infty)
    That is to say, what is f(x)=\lim_{n \to \infty} f_n(x) ?

    Multiply by \frac{\frac 1n}{\frac 1n}:

    f(x)=\lim_{n \to \infty} \frac{x}{\frac 1n+x^2}=\frac{x}{x^2}=\frac 1x

    -----------------
    Definition of uniform convergence :

    \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \quad \implies \quad \forall x \in \dots ~,~ sup \ |f_n(x)-f(x)|< \varepsilon
    -----------------

    \left|f_n(x)-f(x)\right|=\left|\frac{nx}{1+nx^2}-\frac 1x\right|=\left|\frac{nx^2-(1+nx^2)}{x(1+nx^2)}\right| \boxed{=\frac{1}{x(1+nx^2)}}




    Try to do with that in a first time... Find if the limit is 0 (this is the definition of the < \varepsilon thing) in the given intervals..
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  3. #3
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    Thanks. Could someone show me how to do 1b 1c or 1d as an example please? I am still unsure about the rest too.
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  4. #4
    Moo
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    Quote Originally Posted by shadow_2145 View Post
    (b) Is the convergence uniform on (0,\infty)?
    (c) Is the convergence uniform on (0,1)?
    (d) Is the convergence uniform on (1,\infty)?
    So we were left to :

    |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)}

    For x \in (0, \infty), there exists \alpha >0 such that :
    |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)} \le \frac{1}{\alpha(1+n\alpha^2)}

    And when n \to \infty, \frac{1}{\alpha(1+n\alpha^2)} \to 0, that is to say that \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \implies \forall x \in (0, \infty) ~,~ \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon

    \implies |f_n(x)-f(x)| \le \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon.

    Which is what we're looking for...
    I hope you understand & that it looks correct to you :/
    Same thing for the following questions.
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  5. #5
    Moo
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    Quote Originally Posted by shadow_2145 View Post
    (2) Let g_n(x)= \frac{nx+sin(nx)}{2n}.

    Find the pointwise limit of (g_n) on \mathbb{R}.
    g_n(x)=\frac{x+\tfrac{\sin(nx)}{n}}{2}

    g(x)=\lim_{n \to \infty} g_n(x)=\frac{x+\lim_{n \to \infty} \tfrac{\sin(nx)}{n}}{2}


    -1 \le \sin(nx) \le 1 \implies -\tfrac 1n \le \tfrac{\sin(nx)}{n} \le \tfrac 1n

    \lim_{n \to \infty} -\tfrac 1n = \lim_{n \to \infty} \tfrac 1n=0

    Using the squeeze theorem, we get that \lim_{n \to \infty} \tfrac{\sin(nx)}{n}=0


    Therefore, g(x)=\frac x2.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Moo View Post
    So we were left to :

    |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)}

    For x \in (0, \infty), there exists \alpha >0 such that :
    |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)} \le \frac{1}{\alpha(1+n\alpha^2)}

    And when n \to \infty, \frac{1}{\alpha(1+n\alpha^2)} \to 0, that is to say that \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \implies \forall x \in (0, \infty) ~,~ \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon

    \implies |f_n(x)-f(x)| \le \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon.

    Which is what we're looking for...
    I hope you understand & that it looks correct to you :/
    Same thing for the following questions.
    What you want to show is that \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0 but \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty hence \sup_{x\in (0,\infty)} |f_n(x)-f(x)| does not exist and the sequence can't converge uniformly on (0,\infty).
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  7. #7
    Moo
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    Quote Originally Posted by flyingsquirrel View Post
    Hi


    What you want to show is that \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0 but \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty hence \sup_{x\in (0,\infty)} |f_n(x)-f(x)| does not exist and the sequence can't converge uniformly on (0,\infty).
    That's something I don't understand (someone tried to explain it to me...), when do you know if you fix x or not ? If we can find a sup to the function, which is convergent while n tends to infinity, shouldn't it work ?
    The conditions of a uniform convergence seem to apply to n, not to x ?
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    That's something I don't understand (someone tried to explain it to me...), when do you know if you fix x or not ?
    I'm not sure to understand what you mean. Here we can't fix x since we work with \sup_{x\in(0,\infty)}. What one sometimes does is showing the uniform convergence on every (\alpha,\infty),\,\alpha>0 is that what you mean by "fixing" x ?
    If we can find a sup to the function, which is convergent while n tends to infinity, shouldn't it work ?
    Yes it should.
    The conditions of a uniform convergence seem to apply to n, not to x ?
    Then you'd rather study \sup_{n\in\mathbb{N}}|f_n(x)-f(x)| ?
    Last edited by flyingsquirrel; June 19th 2008 at 02:10 PM. Reason: +s
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  9. #9
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    Quote Originally Posted by flyingsquirrel View Post
    is that what you mean by "fixing" x ?
    Yes, this is what I meant ^^

    Yes it should.
    So how come we don't agree on it ?

    Then you'd rather study \sup_{n\in\mathbb{N}}|f_n(x)-f(x)| ?
    No, it's still \sup_{x \in \dots}, but because x \ge \alpha >0, |f_n(x)-f(x)|<|f_n(\alpha)-f(\alpha)|, because the function is decreasing..
    Isn't it a sup of |f_n(x)-f(x)|?
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  10. #10
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    Moo, The sequence of functions in #1 appears in many texts. There is an extensive discussion of this very problem in Advanced Calculus by Taylor & Mann. Because that is such a classical text, any good mathematics library should have a copy.

    That sequence converges uniformly on any closed interval that does not contain 0 and does not converge uniformly on any interval if 0 is in its closure.
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  11. #11
    Moo
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    Hi Plato,

    Quote Originally Posted by Plato View Post
    Moo, The sequence of functions in #1 appears in many texts. There is an extensive discussion of this very problem in Advanced Calculus by Taylor & Mann. Because that is such a classical text, any good mathematics library should have a copy.

    That sequence converges uniformly on any closed interval that does not contain 0 and does not converge uniformly on any interval if 0 is in its closure.
    I doubt I can find that easily such a book here
    That's what I thought for the convergence, but I didn't see the use of asking for the three intervals... Can you enlighten my blindness here ?
    Basically, is there a difference between asking for convergence in (0, \infty) and [\alpha, \infty[, \alpha>0 ?
    I don't really remember if one can be considered as a closed interval :/ That's a part I don't like in maths lol.


    Thanks:
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  12. #12
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    Yes, this is what I meant ^^
    As uniform convergence on every (\alpha,\infty), \alpha>0 does not imply uniform convergence on (0,\infty), fixing x is useless here.

    Fixing x is useful to show the continuity of the limit : if f_n(x)\underset{n\to\infty}\to f(x) for all x\in(0,\infty), if each f_n is continuous on (0,\infty) and if (f_n) converges uniformly on every (\alpha,\infty) (but not necessarily on (0,\infty)) we get that f is continuous on every (\alpha,\infty). This does imply its continuity on (0,\infty).

    \begin{cases}<br />
\text{continuity on every }(\alpha,\infty),\, \alpha>0 \Rightarrow \text{continuity on }(0,\infty)\\<br />
\text{uniform convergence on every }(\alpha,\infty),\, \alpha>0 \nRightarrow \text{uniform convergence on }(0,\infty)<br />
\end{cases}
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  13. #13
    Moo
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    Yep, now I've understood the whole
    Though I guess it's [alpha, infinity[, since (a,b) = ]a,b[
    Sorry for shadow_2145 to have "spoilt" his thread.
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  14. #14
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    Though I guess it's [alpha, infinity[, since (a,b) = ]a,b[
    It doesn't change anything since uniform convergence on every (\alpha,\infty),\,\alpha>0 \Leftrightarrow uniform convergence on every [\alpha,\infty).
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