Uniform convergence ( understanding how to apply a theorem)

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The discussion revolves around demonstrating that the sequence of functions f_n = e^{-nx} does not converge uniformly on the domain [0, ∞). It is established that the pointwise limit of f_n converges to a discontinuous function F, which takes the value 1 at x=0 and 0 for x in (0, ∞). According to the theorem discussed, if a sequence of continuous functions converges to a discontinuous function, then it cannot converge uniformly. Participants clarify the distinction between pointwise and uniform convergence, emphasizing that uniform convergence requires all sequences f_n(x) to converge simultaneously for every x in the domain. The conclusion is that since the limit function F is not continuous, the convergence of f_n is only pointwise, not uniform.
bennyska
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Homework Statement


show a function f_n is not uniformly convergent using a theorem:


Homework Equations


if f_n converges uniformly to F on D and if each f_n is cont. on D, then F is cont. on D


The Attempt at a Solution


not really sure what to do. use the contrapositive? would that be, then, if F is not continuous on D, then either f_n is not uniformly convergent on D or each f_n is not cont. on D? specifically, the function is f_n=e-nx. i think if i understand how to apply the theorem, it shouldn't be too hard.
 
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bennyska said:

Homework Statement


show a function f_n is not uniformly convergent using a theorem:


Homework Equations


if f_n converges uniformly to F on D and if each f_n is cont. on D, then F is cont. on D


The Attempt at a Solution


not really sure what to do. use the contrapositive? would that be, then, if F is not continuous on D, then either f_n is not uniformly convergent on D or each f_n is not cont. on D? specifically, the function is f_n=e-nx. i think if i understand how to apply the theorem, it shouldn't be too hard.

Well, for f_n(x)=e^{-nx}, you should first find the pointswize limit (this will be the uniform limit if the convergence were to be uniform). Is the pointswize limit continuous??
 
i'm still having a little trouble understanding the difference between pointwise and uniform convergence (although this problem may have helped me understand it).
my domain is [0, infinity). if x=0, then the function converges pointwise to F(x)=1. if i fix x in (0,infinity), then it converges to F(x)=0. so F(x) is not continuous?
when i asked someone else this (original) question, the reply i got was, "to restate the theorem, if a sequence of continuous functions converges to a discontinuous function, then it does not converge uniformly."
so, if i understand this concept, for uniform, no matter what x i choose, it converges to the same function, where as for pointwise, the function it converges to depends on how i fix my x.
 
bennyska said:
i'm still having a little trouble understanding the difference between pointwise and uniform convergence (although this problem may have helped me understand it).
my domain is [0, infinity). if x=0, then the function converges pointwise to F(x)=1. if i fix x in (0,infinity), then it converges to F(x)=0. so F(x) is not continuous?
when i asked someone else this (original) question, the reply i got was, "to restate the theorem, if a sequence of continuous functions converges to a discontinuous function, then it does not converge uniformly."

Yes, that's basically it. If the convergence were uniform, then the limit would be continuous. But it isn't, so the convergence is only pointswize.

so, if i understand this concept, for uniform, no matter what x i choose, it converges to the same function, where as for pointwise, the function it converges to depends on how i fix my x.

Hmm, I don't really understand what you said, but I don't think that's what uniform convergence means.
Pointswize convergence is very easy, just calculate the limit for any x. The limit of other x-values don't matter for pointswize convergence.
Uniform convergence is stronger than pointswize convergence. Meaning: the other x-values will matter!

In pointswize convergence, you only want fn(x) to be very close to f(x). For other x-values, it may happen that fn(x) is very far from f(x).
But uniform convergence demands that fn(x) is very close to f(x) AND it will demand that for other x-values it occurs that fn(x) is as close to f(x).

HINT: download the free program on www.padowan.dk/ You can use this to simulate convergence of functions:
- First type Ctrl+F and add n and 1 to the list.
- Then type Ins and add a function to your choice, for example e^(-n*x)
- Then go to Calculate -> Animate (or something like that), and you can view an animation of the convergence of functions.

Try to view the convergence of several sequences of functions and try to get a feeling for uniform/pointswize convergence. Try the following functions:

e^{-nx},~x^n,~\sin^n(x),~\frac{1}{n}\sin(nx)

this should give you a feeling of uniform convergence...
 
i'm having a hard time doing latex on this board, so i'll just sum up what i did like this.
my domain is [0, infinity). i showed if x=0, then fn(x) converges to 1, and if x is in (0, infinity), then fn(x) converges to 0. so my F, defined one way when x=0 and another when x>0, is not continuous. then the theorem says that fn does not converge uniformly. (a latex pdf is attached)
 

Attachments

Yes, that is correct. However, in your pdf, you should write [0,\infty) and not [0,\infty]...


To use LaTeX here, just put your statements inside [ tex ] ... [ /tex ] (without spaces), instead of $ ... $
 
bennyska said:
so, if i understand this concept, for uniform, no matter what x i choose, it converges to the same function, where as for pointwise, the function it converges to depends on how i fix my x.
( t being epsilon-close to c means that | t - c | < epsilon , after fixing some arbitrary epsilon [ this is an example on R ] )
( t being in a delta -neighbourhood of c means that after fixing some delta, 0 < | t - c | < delta [on R again ] )

Uniform convergence is about this: given functions f_n on some domain A, you have a different sequence for each input x; this is clear because evaluating each function at x just gives you a number and thus you get a sequence of all outputs of f_n at some x. So to be clear, to have a sequence of functions f_n, you will have many more sequences f_n( c ) after evaluating each function at c ( where c is something in your domain A, i.e. a different sequence f_n( c_k ) for different c_k's ). When your functions f_n converge uniformly on A, it means that every one of these many sequences f_n( c ) [ for some arbitrary c ] converge to their limit "at the same time" ( i.e. for the same N, each n > N are epsilon-close to their respective limits ). To think about this, imagine the sequences were not going to infinity, but rather, they were going to some point a. Then, a "uniform convergence" of the sequences f_n( x_i ) means that every sequence f_n ( x_i ) for each x_i , get epsilon-close to their respective limits, as long as n is in some delta-neighbourhood around n ( any n in this delta-neighbourhood of a should allow f_n( x_k ) to be epsilon-close to its respective limit L ).

To be clear: It might not be the case that you will be able to find some nice non-piecewisely defined function such that a sequence of functions f_n ( x ) converges to it for all x; but that is okay. The "f(x) " represents an individual limit of a sequence of numbers f_n( x ) after fixing x. A function f may be defined encompassing every limit of the sequences f_n ( x ) ; since each f_n is a function, we have well-definedness of such a function f, since the same x for every f_n will not produce different values.


The continuity of a limiting function f given that each f_n is continuous and converge uniformly to f is a result of this theorem:

suppose "f_n converges uniformly to f" on some domain A. Furthermore, for each f_n,
lim x -> t f_n ( x) exists. Then, lim x -> t f(x ) exists, and

lim x -> t ( lim n-> infinity f_n ( x ) ) = lim n -> infinity ( lim x -> t f_n ( x ) )

In the case of each f_n being continuous, we get that f is continuous:
lim n -> infinity ( lim x -> t f_n ( x ) ) = lim n -> infinity f_n ( t ) = f( t ) = lim x -> t ( lim n-> infinity f_n ( x ) ) = lim x -> t f ( x ) which shows that f is continuous.
 
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thanks for your help. it will take me a few reads to understand that last post.
 
you're welcome, I edited it though to clear something up: you have to make sure that uniform convergence is ultimately dependent on the particular domain for which the property holds; so each sequence f_n ( x ) converges "at the same time", given that each x is in a domain A.
 
  • #10
Also, since you were asking, I would like to add something. The difference between point-wise convergence and uniform convergence is this: point-wise convergence is a property of a single sequence f_n(c ) after being given a c. Uniform convergence is a "collective property" of an entire interval [a,b] or domain A. That is, a point-wise convergence just tells you that if you have functions f_n, and each f_n is defined at some c, so evaluating each f_n at c gives you a sequence of numbers f_n( c ). If f_n is pointwise convergent at c, then the sequence of numbers f_n(c) converges. Uniform convergence means that an entire domain A does whatever I said it would do in my previous post. ( So "uniform convergence" of a single point is pretty much meaningless, it is reduced to mean the same thing as pointwise convergence of some point )
 
  • #11
wisvuze said:
Also, since you were asking, I would like to add something. The difference between point-wise convergence and uniform convergence is this: point-wise convergence is a property of a single sequence f_n(c ) after being given a c. Uniform convergence is a "collective property" of an entire interval [a,b] or domain A. That is, a point-wise convergence just tells you that if you have functions f_n, and each f_n is defined at some c, so evaluating each f_n at c gives you a sequence of numbers f_n( c ). If f_n is pointwise convergent at c, then the sequence of numbers f_n(c) converges. Uniform convergence means that an entire domain A does whatever I said it would do in my previous post. ( So "uniform convergence" of a single point is pretty much meaningless, it is reduced to mean the same thing as pointwise convergence of some point )

Pointswise convergence is also a "collective property". I can easily say that f_n(x)=x/n converges pointswise to 0 on the interval [0,1].
 
  • #12
you are right, but I think that you can still look at that as an "individual" property, since to say
f_n ( x ) = x/n converges pointwise to 0 on [a,b] requires each x in [a,b ] to individually pointwise convergence , a criterion that doesn't rely on the pointwise convergence of anything else in the interval. So pointwise convergence can be a "collective property" as a collection of things that satisfy an individual property. At any rate, the uniform convergence criterion requires knowledge of the convergence of other sequences f_n( c ) for c in the same domain, whereas the criterion for pointwise convergence requires no such knowledge
 
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