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 f
n(x) to be very close to f(x). For other x-values, it may happen that f
n(x) is very far from f(x).
But uniform convergence demands that f
n(x) is very close to f(x) AND it will demand that for other x-values it occurs that f
n(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...