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Thread: Pointwise convergence

  1. #1
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    Pointwise convergence

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

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

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

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

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

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

    (a) Find the pointwise limit of $\displaystyle (f_n)$ on $\displaystyle \mathbb{R}$ 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 $\displaystyle [-5,5]$ 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 $\displaystyle f_n(x)= \frac{nx}{1+nx^2}$.

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

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

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

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

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

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




    Try to do with that in a first time... Find if the limit is 0 (this is the definition of the $\displaystyle < \varepsilon$ 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
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    Quote Originally Posted by shadow_2145 View Post
    (b) Is the convergence uniform on $\displaystyle (0,\infty)$?
    (c) Is the convergence uniform on $\displaystyle (0,1)$?
    (d) Is the convergence uniform on $\displaystyle (1,\infty)$?
    So we were left to :

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

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

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

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

    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
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    Quote Originally Posted by shadow_2145 View Post
    (2) Let $\displaystyle g_n(x)= \frac{nx+sin(nx)}{2n}$.

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

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


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

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

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


    Therefore, $\displaystyle g(x)=\frac x2$.
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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 :

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

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

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

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

    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 $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0$ but $\displaystyle \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty$ hence $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)|$ does not exist and the sequence can't converge uniformly on $\displaystyle (0,\infty)$.
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  7. #7
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    Quote Originally Posted by flyingsquirrel View Post
    Hi


    What you want to show is that $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0$ but $\displaystyle \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty$ hence $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)|$ does not exist and the sequence can't converge uniformly on $\displaystyle (0,\infty)$.
    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
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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 $\displaystyle x$ since we work with $\displaystyle \sup_{x\in(0,\infty)}$. What one sometimes does is showing the uniform convergence on every $\displaystyle (\alpha,\infty),\,\alpha>0$ is that what you mean by "fixing" $\displaystyle x$ ?
    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 $\displaystyle \sup_{n\in\mathbb{N}}|f_n(x)-f(x)|$ ?
    Last edited by flyingsquirrel; Jun 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" $\displaystyle x$ ?
    Yes, this is what I meant ^^

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

    Then you'd rather study $\displaystyle \sup_{n\in\mathbb{N}}|f_n(x)-f(x)|$ ?
    No, it's still $\displaystyle \sup_{x \in \dots}$, but because $\displaystyle x \ge \alpha >0$, $\displaystyle |f_n(x)-f(x)|<|f_n(\alpha)-f(\alpha)|$, because the function is decreasing..
    Isn't it a sup of $\displaystyle |f_n(x)-f(x)|$?
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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
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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 $\displaystyle (0, \infty)$ and $\displaystyle [\alpha, \infty[$, $\displaystyle \alpha>0$ ?
    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 $\displaystyle (\alpha,\infty)$, $\displaystyle \alpha>0$ does not imply uniform convergence on $\displaystyle (0,\infty)$, fixing $\displaystyle x$ is useless here.

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

    $\displaystyle \begin{cases}
    \text{continuity on every }(\alpha,\infty),\, \alpha>0 \Rightarrow \text{continuity on }(0,\infty)\\
    \text{uniform convergence on every }(\alpha,\infty),\, \alpha>0 \nRightarrow \text{uniform convergence on }(0,\infty)
    \end{cases}$
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  13. #13
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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 $\displaystyle (\alpha,\infty),\,\alpha>0 \Leftrightarrow $ uniform convergence on every $\displaystyle [\alpha,\infty)$.
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