Pointwise vs. Uniform Convergence.

rumjum

1. The problem statement, all variables and given/known data

I need to understand as to why the following series fn(x) = x/(1+n*x^2) is point wise convergent (as mentioned in the book of Ross) and not [obviously] uniformly convergent.

2. Relevant equations

The relevant equation used is that lim (n-> infinity) sup|(fn(x) - f(x))|= 0 implies uniform convergent.---- (1)

3. The attempt at a solution

It is obvious lim (n-> infinity) fn(x) = f(x) = 0 for x not equal to zero. And when x=0, fn(0) = 0 and hence as n-> infinity, fn(0) = f(0) =0.

As mentioned in Ross and ( I can see) that fn(x) is pointwise convergent. But, it looks like for all x, the function fn(x) converges to f(x)=0. So, I am unclear as to why it is not uniform convergent? Please clarify.

The book further uses the theorem (1) to prove uniform converfence, which I can understand.

Office_Shredder

which is the definition of pointwise convergence. Note that a good way to find

sup|f_n(x)-f(x)| is to take the derivate of the difference and look for local maxima (since in this case the difference tends to zero as x goes to zero and infinity). Although to start, the question should have a domain in which the uniform convergence of the function is to be determined, I'm just assuming from context it's something like [0,infinity).

So [tex] \frac {d}{dx}(f_n(x)-f(x)) = \frac {d}{dx}f_n(x) = \frac {d}{dx} \frac {x}{1+n*x^2} = \frac {1+n*x^2 - x(2nx)}{(1+n*x^2)^2} = 0 \rightarrow 1-nx^2=0 \rightarrow x= +/- \frac{1}{ \sqrt{n}} [/tex]

Obviously this is a maximum, as the function is zero at 0 and infinity and clearly positive on (0,infininty), so we can just plug it in to find sup|f[sub]n[/sub(x)|

1/sqrt(n)/(1+n/n) = 1/(2sqrt(n)) => 0

So in fact it is uniformly convergent

rumjum

Well, I understood the part of finding the sup using derivatives. And yes you are right that the domain is set of Real number (R). But, now I am a bit more confused about the difference between pointwise and uniform convergence. I am under the impression (and correct me if am wrong) that the difference between pointwise and uniform is simply that for pointwise , there is an n >N such that for some x fn(x) -> f(x).
But, for uniform convergence, there is an n > N such that for all x fn(x) -> f(x).
So, if in the above example, for all x, the lim (fn(x) ->0)) then is it not a uniform convergence?

Thanks for your response.

Office_Shredder

Pointwise convergence is, for any x in the domain, for all epsilon>0, there exists N where N depends on x such that n>N implies |f_n(x)-f(x)|<epsilon

Uniform convergence is, for any epsilon>0, there exists N such that for the same N, no matter which x you pick n>N implies |f_n(x)-f(x)|<epsilon.
One example of a sequence that doesn't converge uniformly is xⁿ on [0,1]. You can see this as, if x<1, xⁿ converges pointwise to zero, i.e. for a given point, it converges to zero. If x=1, xⁿ converges to 1 obviously. But for a given epsilon, say 1/4, for any N I can find a point (1/2)^1/N so if x is that, x^N = 1/2

Another example of a sequence that doesn't converge uniformly is arctan(nx) on all of R. Clearly, it converges for a given x to 0 if x=0, or pi/2 if x>0, and -pi/2 if x<0. This is the pointwise convergence, i.e. you pick a point and see where it converges. But arctan(nx) does not converge uniformly to that function, as picking epsilon=1, for any N if x=1/(10000N) then arctan(Nx) = arctan(1/10000) which is less than pi/2-1.

The best way to visualize this is to see that uniform convergence essentially implies that for large n, f_n(x) is very very very very veeeery close to f(x) as a graph everywhere, whereas pointwise convergence simply says that tracking the value of a single point over time will give the right value

rumjum

I re-read your statement on pointwise convergence of the previous post and referred to the text book. I understood what you were saying about pointwise convergence. Thanks for further clarifying the same and for your example. Appreciate it.