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Pointwise and uniform convergence of sequence of functions

  • Thread starter Jncik
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  • #1
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Homework Statement



show if
gif.gif
has a uniform convergence of pointwise

also we know that x gets values from 0 to 1


The Attempt at a Solution



for the pointwise I think its easy to show that limfn(x) as n->infinity is 0

but Im really stuck in uniform convergence

I know that fn converges uniformly to f if and only if for all ε>0 there is a natural number N so that for all n>=N |fn - f|<ε

if f(x) = 0

we get

gif.gif


now from this what can we do??

thanks in advance
 

Answers and Replies

  • #2
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So we have to decide whether

[tex]\sup\left|\frac{x^2}{(1-nx)^2+x^2}\right|\rightarrow 0[/tex].

To decide this, you'll need to calculate the maximum value of [tex]\left|\frac{x^2}{(1-nx)^2+x^2}\right|[/tex]. And then you'll need to check whether this converges to 0 or not...

To calculate the maximum value, you'll have to do the usual: calculate the first derivative, see when it is 0, etc....
 
  • #3
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I guess it's really difficult to compute this by hand then....

it's really odd, my professor solves this in 3 lines but I'm 100% he's either wrong or doesn't explain the solution at all

he says that

for all n, there is a n0>n so that x0 = 1/n0

then he says

|fn0(x0) - f(x0)| = ... = 1

because of this we dont have uniform convergence...

i dont really understand it..
 
  • #4
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3,281
No, it's not difficult to compute by hand at all!! You just need to find the maximum of the function...

And what your professor proposed is indeed correct, but he doesn't really give much details. Your professor claims that for every n, there is an x such that fn(x)=1. This implies that the maximum of the function fn is greater then 1! But then the maximum cannot converge to 0. This implies that there cannot be uniform convergence...
 
  • #5
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No, it's not difficult to compute by hand at all!! You just need to find the maximum of the function...

And what your professor proposed is indeed correct, but he doesn't really give much details. Your professor claims that for every n, there is an x such that fn(x)=1. This implies that the maximum of the function fn is greater then 1! But then the maximum cannot converge to 0. This implies that there cannot be uniform convergence...
thanks I think I understand what he proposed now

but for the maximum of the function

i find that derivative is the following

[PLAIN]http://img24.imageshack.us/img24/9923/21098361.gif [Broken]

for x = 0 it is 0

how can we figure out if the max is over 0?
 
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  • #6
22,097
3,281
thanks I think I understand what he proposed now

but for the maximum of the function

i find that derivative is the following

[PLAIN]http://img24.imageshack.us/img24/9923/21098361.gif [Broken]

for x = 0 it is 0

how can we figure out if the max is over 0?
Well, you'll need to figure out when the expression above is 0. So, for what x, does

[tex]\frac{2x(nx-1)}{(n^2x^2-2nx+x^2+1)^2}=0[/tex].

This is equivalent to the question: for what x, does

[tex]2x(nx-1)=0[/tex]...
 
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  • #7
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well it's 0 for x = 0 and also it can be 0 for x = 1 and n = 1

but now I should figure out where's the max of the function

should I get values for x = -1, 0.5, 1.5 to find out what's the sign of f' or is there any other easier way?

for x = -1 I get -2*(n+1)/(n^2+2n+2)^2, because n is a natural this number is negative

for x = 0.5 I get (1-0.5n)/(0.25n^2-n+1.25)^2 and it depends on the n.. so i cant figure out if its positive or negative..

i ve never computed maximum of a function with n and x, our professor never showed us an example on this case and its really difficult for me to understand whats going on
 
  • #8
22,097
3,281
well it's 0 for x = 0 and also it can be 0 for x = 1 and n = 1
No, that's not correct. Remember that n is a fixed value, you can't just set it to 1!!
If 2x(nx-1)=0, then x=0 or nx-1=0. In this last case, we have x=1/n.
So for fn, we have the following possibilities for a maximum: 0 and 1/n. Now you will have to calculate both |fn(0)| and |fn(1/n)| to see which value is higher. This is the maximum of the function!
 
  • #9
103
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No, that's not correct. Remember that n is a fixed value, you can't just set it to 1!!
If 2x(nx-1)=0, then x=0 or nx-1=0. In this last case, we have x=1/n.
So for fn, we have the following possibilities for a maximum: 0 and 1/n. Now you will have to calculate both |fn(0)| and |fn(1/n)| to see which value is higher. This is the maximum of the function!
so because we get 0 and 1 and 1>0 this is the max, it would only converge if it was 0

thank you very much for your help, all I needed was a simple explanation like yours.... thanks again :)
 

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