Thread: Uniform convergence vs pointwise convergence

Hi guys,

Can someone please explain to me the advantages of uniform convergence over pointwise convergence? I was told that pointwise convergence does not preserve continuity while uniform convergence does. Also, pointwise convergence depends on x while uniform convergence doesn't. But whats the big deal about it? I am not getting the whole picture of it. We are learning Fourier Series by the way.

Thanks.

For example, if f_n : [ a , b ] -> IR is a sequence of continuous functions such that f_n -> f uniformly on [a , b ] then

Int_a^b ( f_n ) -> Int_a^b ( f )

as n -> + infty

perhaps I didn't understand the question but...

the difference between uniformly convergence and convergence by the point's you can best describe if u look at some sequence of functions (f_n (x) )_{n\in \mathbb{N}} defined in R where f_n (x) = x/n

as you see functions are lines that passes thru coordinate (0,0) and with x-axes have less and less angle as x grows (as shown on image down) (uploading attachments doesn't work )



(hmmm perhaps image is to small and I can't fix it right now.... )


but, as you see you have n functions there so how does you decide does that sequence converge ? well it's very important you notice that if you chose some fix value for "x" you get common numerical sequence that you can test for convergence with tools what did you use anyway by that point (values x=1 and x=2 are shown on images to the right )

notice also that if you draw strait line up at x= 1 let's say, you will get at the intersections of that line and line of the functions actually values of that now numerical sequence

notice that you now have

lim_{n \to \infty} \frac {1}{n} ==== converges to zero (for x = 1)

lim_{n \to \infty} \frac {2}{n} ==== converges to zero (for x = 2)

but also notice that if u chose x to be any number M (irrelevant how big it is, just to be fixed ) it will be

lim_{n \to \infty} \frac {M}{n} ==== converges to zero (for x = M )

and this type of convergence you call convergence by points or you say

(\forall x \in A)(\forall \varepsilon >0 )(\exists n_0 \in N) (\forall n \in N) (n>=n_0 \Rightarrow |f_n (x) -f(x) |<\varepsilon )

as you see from this n_0 depends from x and \varepsilon (with big point at depending on X)

but if you have sequence where you can find some function that all functions from the sequence converge then you have uniformly convergence ....

must be ::

(\forall \varepsilon >0 )(\exists n_0 \in N)(\forall x \in A) (\forall n \in N) (n>=n_0 \Rightarrow |f_n (x) -f(x) |<\varepsilon )


and from that you see that if sequence is uniformly converge than it is also and by point but if you have "convergence by point" you don't have (in most cases) uniformly convergent sequence

when you say some sequence is uniformly convergent than you know that in any point of region on where sequence is defined (let's say set A subset R) you will have that sequence converge to the same function

\lim_{n\to \infty} f_n(x) = f(x)

can somebody help me with this problem please??

Show that the sequence {f_n} where f_n=nxe^(-nx^2 ) for n=1,2,3,… converges pointwise in [0,1].Is the convergence uniform?Justify

Originally Posted by chath

Show that the sequence {f_n} where f_n=nxe^(-nx^2 ) for n=1,2,3,… converges pointwise in [0,1].Is the convergence uniform?Justify

Prove that for all and .


P.S. It is better a new thread for every new problem.