Uniform Convergence of a Sequence of Functions

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  • #1
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Homework Statement



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]

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?
 

Answers and Replies

  • #2
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[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?
Well, this only shows that the ##f_n## converge to ##|x|## pointswise. You need some extra argument to show or disprove uniform convergence.
 
  • #3
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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?
 
  • #4
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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?
 
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  • #5
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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]
 
  • #6
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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]
You need to show this limit is ##0## by definition of uniform convergence. Do you see why.

and show that as [itex]n \rightarrow \infty then |f_n - f(x) | \rightarrow 0 [/itex]
No, that would just be pointswise convergence.
 
  • #7
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Thanks, I think I'm gonna have to think about this a little while
 
  • #8
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Let me give you a hint: |x| = sqrt(x^2).
 
  • #9
pasmith
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Homework Statement



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]

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?
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].
 
  • #10
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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].
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]
 
  • #11
jbunniii
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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]
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.
 
  • #12
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mmmmmmm OK I'll do that! but yes that's what I meant
 
  • #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:
  • #14
Ray Vickson
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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?
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)?##
 
  • #15
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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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