Convergent sequence of functions?

pivoxa15

1. The problem statement, all variables and given/known data
How do you show a sequence of functions in terms of n is convergent (in general)?




3. The attempt at a solution
Do you presuppose the value of the function say v then show d(fn,v)->0 for n large?

v is determined by looking at the sequence of functions and guessing what they may be approaching.

Is there an easier way?

quasar987

I assume you're asking this purely out of curiosity so I'll just give the fact and not much of the details. If you want a more serious answer, just pick any real analysis book.

There are two types or "notions" of convergence that we can be interested in here: pointwise and uniform.

In pointwise, we fix x to say x0 and we look at the resulting numerical sequence {f_n(x0)}. This is a sequence like you know them. You can either guess the value to which it converge and prove that you're right with epsilon & N. Or, if you're only interested in knowing whether or not the sequence converge and not to its actual limit value is, you can try to find out whether or not the sequence is Cauchy.

In uniform, we wonder if there is a function f(x) to which the sequence {f_n(x)} converges pointwise for all values of x in some set, and does so "essentially as fast" at all points. A little more technically speaking, given an epsilon, we ask that there is an N such that FOR ALL x in the set, |f_n(x)-f(x)|<epsilon as soon as n>N. There is also a Cauchy criterion for uniform convergence and other criteria as well.

pivoxa15

So all uniformly convergent functions are pointwise convergent?

There is a function with a max value in the limit that is non zero but the function as a whole for all x in the domain converges to f(x)=0. What do you make of that?

Does it mean its pointwise convergent since the max value occurs at the end of the domain.

matt grime

yes

that sentence doesn't make sense. A function cannot converge. A sequence converges. So let's say we have a sequence of functions f_n. Now you hav just said it converges without specifying which kind of convergence which means that

doesn't make sense.

So, you're claiming to have a sequence f_n(x) that pointwise converges to f(x) which is the zero function.

Now, what does 'a function with a max value in the limit that is non-zero' mean?

Do you meant that that if we set S_n to be the sup(f_n(x)), then S_n does not converge to zero?

Certainly there are functions like that.

pivoxa15

Yes that is what I mean. The thing is the non zero max value is in the domain of the function. i.e S_n(x) is non zero with x in the domain.

That is very unintutive. The whole sequence for every x goes to 0. i.e f_n converges to 0, however.

matt grime

First, the 'non-zero max value' is not in the domain of the function. Second, the maximal value need not exist: sup is not max. It is the Sup that you want to think about. Whether the function attains the sup is immaterial, and possibly causing you confusion.

The standard examples look something like this:

let f_n=x^n on the interval (0,1).

The Sup of f_n is 1 for all n (though it doesn't attain that value anywhere on the domain).

The pointwise limit is the zero function.

This is not the uniform limit. It does not converge in the uniform metric.

pivoxa15

Why isn't the 'non-zero max value' not in the domain of the function? It could be? i.e function is defined on [0,1] and max x occurs when x = n/(n+1) so for large n the non zero max value is certainly in this interval.

Data

It can't be in the domain of the limit function; otherwise the limit function would have to share the discrepancy. Other examples abound.

For example, for each n>0 take [itex]f_n(x) = ne^{-(x-n)^2}[/itex] on the reals. Then for every [itex]x \in \mathbb{R}[/itex] the sequence [itex](f_n(x))_{n=1}^\infty[/itex] converges pointwise to 0, but the sequence of functions doesn't converge uniformly on, say, [itex](0, \infty)[/itex].

Here [itex]f_n(n)[/itex] is always n; Thus the "nonzero max value" is actually going to infinity with n. Simultaneously, increasing n moves the "bump" farther out, resulting in the pointwise convergence. It's always there, so the sequence of functions can't converge uniformly.

matt grime

*If* a function attains its sup, then the value of the sup is in *range*, for pete's sake, not the bloody domain.

matt grime

No, the point at which it attains its maximum is in the domain (and is for all positive n in this fictitious example, I don't know why you think it is for 'large n'). The value of the maximum is whatever f(n/(1+1)) is, and is not in the domain.

pivoxa15

"The point at which it attains its maximum is in the domain." is what I am trying to say. The fact that the maximum which is in the range is non zero in the limit suggest f dosen't converge to 0 for all x in the domain. But some f do and with a point in the domian where it attains a non zero maximum in the limit.

matt grime

I'm very confused by what you're talking about. You keep talking about *the* maximum as if there is only one maximal value at play here. Surely you mean on for each f_n. f is a function. f does not converge. I repeat: a sequence converges (or diverges). A function does not.

Stop writing f when you mean a sequence f_n, and be clearer with what has a maximum (or sup for preference). And if you're going to say converge, at least indicate in what sense you mean converge - L_1, L_2, ..., L_infinity etc. (Pointwise or uniform would be a start.)

Actually, is there a question here any more?

pivoxa15

My question is

f_n converges pointwise to 0 for all x in [0,1] but there exists a non zero sup in the limit in the range of f_n. How can that be?

Is this possible because this condition would imply that f_n always gets lowered but not neccessary reaching 0 and the sup always gets shifted towards the right or left so that as n increase f_n is lower than f_n for smaller n, for all x in the domain. So it doesn't mean f_n will reach the 0 value for all x hence sup f_n can still be nonzero in the lmit.

matt grime

It can't happen. If f, the pointwise limit is 0, then sup(f(x))=0. f is identically zero.

I am assuming that sup lim (f_n) is what you meant by the phrase:


It is certainly true that if f_n converges pointwise to f, that the two quantities

lim sup(f_n)

and

sup lim (f_n)

need not be equal. Though not in the case that the domain of f is compact.

pivoxa15

Take f_n(x)=nx^n(1-x) for x in [0,1]

sup lim (f_n)=1/e

but f_n converges pointwise for all x in the domain to 0

matt grime

Hold on. You've got things the wrong way round.

sup lim f_n

is zero, since lim f_n is the zero function (taking the limit pointwise).


You mean

lim sup f_n =1/e,

but why would expect sups to commute with pointwise limits?

pivoxa15

So in this case lim sup(f_n) does not equal sup lim (f_n) but the domain is compact?

It is not very intuitive but maybe one can think of it as the sup(f_n) as n increase keep shifting to the right or left so at each x, f_n(x) converges pointwise to 0 but some points will never ever reach it as lim sup(f_n) is non zero.