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Uniform Convergence of a Sequence of Functions

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Define

    [itex]f_n : \mathbb{R} \rightarrow \mathbb{R} [/itex] by

    [itex]f_n(x) = \left( x^2 + \dfrac{1}{n} \right)^{\frac{1}{2}}[/itex]


    Show that [itex]f_n(x) \rightarrow |x|[/itex] converges uniformly on compact subsets of [itex]\mathbb{R}[/itex]

    Show that the convergence is uniform in all of [itex]\mathbb{R}[/itex]

    3. The attempt at a solution

    Not quite good at these epsilon proofs, not sure if it needs to go that far but by the root law we have that

    [itex]\lim_{ n\rightarrow \infty} \sqrt{x^2 + \dfrac{1}{n}} = \sqrt{x^2 + \lim_{n \rightarrow \infty} \dfrac{1}{n}} = \sqrt{x^2} = |x|[/itex]. Now this shows that [itex]f_n [/itex] converges uniformly in some subset of [itex]\mathbb{R} [/itex] correct?
     
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  3. Mar 25, 2014 #2

    micromass

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    Well, this only shows that the ##f_n## converge to ##|x|## pointswise. You need some extra argument to show or disprove uniform convergence.
     
  4. Mar 25, 2014 #3
    Well I suppose we should then go to the definition right.

    Let [itex]\epsilon > 0 [/itex], I need to find the right [itex]N(\epsilon)[/itex] such that whenever [itex]n\geq N[/itex], we have that

    [itex] \left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right| < \epsilon [/itex].

    So I have a hard time messing with that

    [itex]\left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right| [/itex]

    any hints?
     
  5. Mar 25, 2014 #4

    micromass

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    Well, you need to prove

    [tex]\textrm{sup}_x |f_n(x) - f(x)| \rightarrow 0[/tex]

    So I would start by calculating this supremum. Maybe try the usual techniques of calculus?
     
  6. Mar 25, 2014 #5
    Just to be clear, I need to find the

    [itex] \lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} | f_n(x) - f(x) | [/itex] and show that as [itex]n \rightarrow \infty [/itex] then [itex]|f_n - f(x) | \rightarrow 0 [/itex]
     
  7. Mar 25, 2014 #6

    micromass

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    You need to show this limit is ##0## by definition of uniform convergence. Do you see why.

    No, that would just be pointswise convergence.
     
  8. Mar 25, 2014 #7
    Thanks, I think I'm gonna have to think about this a little while
     
  9. Mar 26, 2014 #8
    Let me give you a hint: |x| = sqrt(x^2).
     
  10. Mar 26, 2014 #9

    pasmith

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    Use the hint, and show that convergence is uniform on an arbitrary compact subset [itex]S[/itex] of [itex]\mathbb{R}[/itex].

    Note that every [itex]f_n[/itex] is continuous, and [itex]f[/itex] is continuous, so that [itex]g_n = f_n - f[/itex] is continuous, and that [itex]0 < g_{n+1}(x) < g_n(x)[/itex] for every [itex]x \in \mathbb{R}[/itex] and every [itex]n \in \mathbb{N}[/itex].
     
  11. Apr 1, 2014 #10
    Sorry to bring up old news but I haven't really thought about this problem since, but anyway, pasmith all I'm getting from you is that [itex]g_n[/itex] is one-to-one.

    [itex]g_n(x) = f_n(x) - f(x) = \left( x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - |x| = \left(x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - \sqrt{x}[/itex]

    So

    [itex] \lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} \left|f_n(x) - f(x) \right| = \lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} \left| \left(x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - \sqrt{x} \right| [/itex]

    I feel like I need to use the squeeze theorem or something. Am I supposed to use the following argument along the lines of

    [itex] 0 < |g_{n+1}| < |g_n| [/itex] and as [itex]n \rightarrow \infty, |g_n| \rightarrow 0 [/itex]
     
  12. Apr 1, 2014 #11

    jbunniii

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    I assume you intended that last ##\sqrt{x}## to be ##\sqrt{x^2}##, and also I assume you meant
    $$\left(x^2 + \frac{1}{n}\right)^{1/2}$$
    as in the problem statement.

    Try using the fact that
    $$\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$$
    assuming ##a## and ##b## are positive.
     
  13. Apr 1, 2014 #12
    mmmmmmm OK I'll do that! but yes that's what I meant
     
  14. Feb 2, 2017 #13
    Currently working on the same problem for my Intro Real Analysis II course at Rutgers University.

    We know ## x^2 > 0 ## and ## \frac{1}{n} > 0 ## for all ##x \in \mathbb{R}## and for all ##n \in \mathbb{N}##, hence using the hint above we obtain $$ \left | \sqrt{x^2 + \frac{1}{n}} - \sqrt{x^2} \right | = \left | \frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \right | .$$

    I think we need to split the reasoning for "##f_n## converges uniformly to ##f## on ##\mathbb{R}##" into two parts. Consider ##|x| \geq 1##, which implies ##x^2 \geq 1## and certainly ##x^2 + \frac{1}{n} \geq x^2## for natural number ##n##, hence ##\sqrt{x^2 + \frac{1}{n}} \geq \sqrt{1} = 1##. Moreover, ##\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2} \geq \sqrt{x^2 + \frac{1}{n}} \geq 1##. Certainly ##n \geq N## gives ## \frac{1}{n} \leq \frac {1}{N}##, so combining these facts we have that $$ \frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \leq \frac{1}{N}$$ for ##|x| > 1##. So setting ##\frac{1}{N} < \epsilon## gives $$\forall \epsilon > 0, \rm{choose}\ N \in \mathbb{N}\ \rm{such\ that}\ N > \frac{1}{\epsilon} \implies \\
    \forall |x| > 1 \ \forall n \geq N\ \rm{we\ have}\ \left | \frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \right | \leq \frac{1}{n} \leq \frac{1}{N} < \epsilon. $$

    To argue for ##|x|<1##, I am at a loss. Thoughts?

    Attempt:
    We know ##x^2 < 1##; however ##x^2 + \frac{1}{n} > x^2##.
     
    Last edited: Feb 2, 2017
  15. Feb 2, 2017 #14

    Ray Vickson

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    Look at the graph of ##d_n(x) = \sqrt{x^2 + \frac{1}{n}} - |x| > 0##. For any interval ##I## that contains ##0##, what is the value of ##\sup_{x \in I} d_n(x)?##
     
  16. Feb 2, 2017 #15
    I don't remember, but in looking at your argument, why don't you include [\itex]|x|\geq 0[\itex]. Secondly, if you're going to do it for x<0, use the definition of |\cdot|, do the same thing, multiply both top and bottom by the conjugate and I think you have what you got. I hope this helps.
     
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