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Pointwise/Uniform Convergence proof

  • Thread starter Visceral
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Homework Statement


Wondering if you guys could check my proof. This is my first problem with sequences of functions

Let [itex]a > 0 [/itex] and [itex]f_{n}(x) = \frac{nx}{1+nx}. [/itex]

a) Show that the sequence of functions [itex] (f_{n})[/itex] converges pointwise on [itex] [0,∞)[/itex]

b) Show that [itex](f_{n})[/itex] converges uniformly on [itex][a,∞) [/itex]

c) Show that [itex](f_{n})[/itex] does not converges uniformly on [itex][0,a) [/itex]

Homework Equations


ummm, theorems for pointwise convergence and uniform convergence.

edit: definitions given here

http://www.math.psu.edu/wade/M401-notes1.pdf


The Attempt at a Solution



OK so for part a) it seems that the function will converge uniformly on [itex] [0,∞)[/itex] if we let [itex] f(x) = 0 [/itex] for [itex] x=0[/itex] and [itex] f(x) = 1[/itex] for [itex] x>0[/itex]....I believe this is correct

For b), since [itex]a>0[/itex] this implies [itex]f(a) = 1[/itex]. So by the definition of uniform convergence, [itex]|\frac{na}{1+na}-1|< ε[/itex]. This can be simplified down to [itex]|\frac{1}{na}|< ε[/itex].

So now we can let [itex]ε>0[/itex] so that [itex]\frac{1}{aε}>0[/itex]. Then by Archimedes principle there exists an [itex]N(ε)[/itex] in the Reals such that [itex]N(ε)>\frac{1}{aε}[/itex]. Now if [itex]n>N(ε)[/itex] this implies that [itex]|f_{n}(x)-f(x)| = |\frac{1}{na}| = \frac{1}{na} < \frac{1}{N(ε)a}<ε[/itex]. And hence [itex]f_{n}
[/itex] converges uniformly on [itex][a,∞)[/itex].

I wasn't sure about leaving the "a" in there, but I think that might work?

For c)

I let [itex]ε=\frac{1}{2}[/itex], and I let [itex]x = \frac{1}{n}[/itex]. If you go through with this, you end up getting [itex]|f_{n}(x)-f(x)| = \frac{1}{2} = ε[/itex], which can't happen. Therefore it is not uniformly convergent on [itex][0,a)[/itex].

Did I go wrong anywhere? This is from a final exam practice for real analysis, which I have mine on monday next week and there are no answers given.
 

Answers and Replies

  • #2
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Seems alright!
 
  • #3
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Thx micromass. I do have a question, though. It seems as though the functions will converge also for x less than -1. Could we then say it is pointwise convergent on [0,infinity) AND (-infinity,-1), and just let f(x) = -1 for x < -1?
 
  • #4
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f(x)=+1 for x<-1.
 
  • #5
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oops yes you are right. So could this work? The only "bad areas" for the function seem to be when x is in [-1,0)
 
  • #6
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I'm sorry, but there are several problems with your solution.

For part a), the sequence [itex]f_{n}(x) = \frac{nx}{1+nx} [/itex] does not converge uniformly on [itex][0,∞)[/itex]. To show that it does converge pointwise to 1 on [itex](0,\infty)[/itex], fix [itex]c\in(0,\infty)[/itex] and demonstrate that [itex]\lim_{n\rightarrow\infty}f_n(c)=1[/itex]. Since [itex]f_n(0)=0[/itex] for all [itex]n[/itex], the sequence of functions converges to [itex]0[/itex] at [itex]x=0[/itex].

In part b), you make the mistake of assuming the uniform converge to derive a result. You never want to assume the result before you've proved it. You are supposed to use the data ([itex]a>0[/itex], [itex]f_{n}(x) = \frac{nx}{1+nx}[/itex]) to demonstrate the uniform convergence. Here is one way to do so:

Fix [itex]a\in(0,\infty)[/itex]. For each [itex]n[/itex],use elementary methods from calculus to find the absolute max/min (maybe inf/sup) of [itex]f_n[/itex] on [itex][a,\infty)[/itex]. Show that the max/min (or inf/sup) of [itex]f_n[/itex] goes to [itex]1[/itex] as [itex]n\rightarrow\infty[/itex].

You have the right idea in part c), but the part where you say "this can't happen" needs cleaning up. It can happen (and in fact does here!). The point you want to make is that for each [itex]n[/itex] large enough, there is always a point in [itex][0,a)[/itex], namely [itex]\frac{1}{n}[/itex], where [itex]f_n[/itex] is "far away" from [itex]f[/itex].
 
  • #7
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I'm sorry, but there are several problems with your solution.

For part a), the sequence [itex]f_{n}(x) = \frac{nx}{1+nx} [/itex] does not converge uniformly on [itex][0,∞)[/itex]. To show that it does converge pointwise to 1 on [itex](0,\infty)[/itex], fix [itex]c\in(0,\infty)[/itex] and demonstrate that [itex]\lim_{n\rightarrow\infty}f_n(c)=1[/itex]. Since [itex]f_n(0)=0[/itex] for all [itex]n[/itex], the sequence of functions converges to [itex]0[/itex] at [itex]x=0[/itex].

In part b), you make the mistake of assuming the uniform converge to derive a result. You never want to assume the result before you've proved it. You are supposed to use the data ([itex]a>0[/itex], [itex]f_{n}(x) = \frac{nx}{1+nx}[/itex]) to demonstrate the uniform convergence. Here is one way to do so:

Fix [itex]a\in(0,\infty)[/itex]. For each [itex]n[/itex],use elementary methods from calculus to find the absolute max/min (maybe inf/sup) of [itex]f_n[/itex] on [itex][a,\infty)[/itex]. Show that the max/min (or inf/sup) of [itex]f_n[/itex] goes to [itex]1[/itex] as [itex]n\rightarrow\infty[/itex].

You have the right idea in part c), but the part where you say "this can't happen" needs cleaning up. It can happen (and in fact does here!). The point you want to make is that for each [itex]n[/itex] large enough, there is always a point in [itex][0,a)[/itex], namely [itex]\frac{1}{n}[/itex], where [itex]f_n[/itex] is "far away" from [itex]f[/itex].


Thanks for the reply. I have some questions though. For part a) I do not see what I did wrong. I showed that it converged pointwise on [0,∞) to f(x) which I defined as f(x)=1 if x > 0 and f(x) = 0 if x=0 for all x in [0,∞). I know it is not uniformly convergent there.

In b) I did not mean to imply that I assumed what I was trying to prove. I meant to say that I needed to show [itex]| \frac{nx}{1+nx}-1| < ε[/itex] to prove it is uniformly convergent on [a,∞). I was not assuming that and sorry for misleading anyone.

for c)...yeah I guess "this cant happen" was a bad choice of words. I was thinking in my head that that result contradicts uniform convergence, therefore its not uniformly convergent. But you are right, once you pass n, [itex]f_{n}[/itex] is not within ε of f(x), hence its not uniformly convergent.
 
  • #8
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For part a) I do not see what I did wrong. I showed that it converged pointwise on [0,∞) to f(x) which I defined as f(x)=1 if x > 0 and f(x) = 0 if x=0 for all x in [0,∞).
You state (correctly) what the sequence converges to, but you need to also justify/prove this statement.

In b) I did not mean to imply that I assumed what I was trying to prove. I meant to say that I needed to show [itex]| \frac{nx}{1+nx}-1| < ε[/itex] to prove it is uniformly convergent on [a,∞). I was not assuming that and sorry for misleading anyone.
Yes. I see now what you were intending. I'm assuming that [itex]\frac{1}{na}[/itex] is the max of [itex]| \frac{nx}{1+nx}-1|[/itex] on [itex][a,\infty)[/itex]? Also, I'm guessing this is true for [itex]n[/itex] sufficiently large? If this is so, then you're on the right track. But, again, this is something you need to state and prove being careful to include the "[itex]n[/itex] sufficiently large" part.

for c)...yeah I guess "this cant happen" was a bad choice of words. I was thinking in my head that that result contradicts uniform convergence, therefore its not uniformly convergent. But you are right, once you pass n, [itex]f_{n}[/itex] is not within ε of f(x), hence its not uniformly convergent.
Exactly.
 
  • #9
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You state (correctly) what the sequence converges to, but you need to also justify/prove this statement.



Yes. I see now what you were intending. I'm assuming that [itex]\frac{1}{na}[/itex] is the max of [itex]| \frac{nx}{1+nx}-1|[/itex] on [itex][a,\infty)[/itex]? Also, I'm guessing this is true for [itex]n[/itex] sufficiently large? If this is so, then you're on the right track. But, again, this is something you need to state and prove being careful to include the "[itex]n[/itex] sufficiently large" part.



Exactly.
Cool. Thanks for the advice/clarifications
 

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