point convergence and uniform convergence definition question

i was told that if there is a point x0 from [a,b]

and $\displaystyle lim_{n->infty} f_n{x0}=0$

then it converges to a point

__first part of the question:__

here i got $\displaystyle f_n(x)=\sqrt{x^2+\frac{1}{n}}$

and the solution tell that it conveges to |x|

so the limit is |x|

but x could be x=7 then it wont converge to a point because the limit will not be zero.

__second half of the question:__

further more i could say

the it does converge to a point for x0=0.

if we want to find out if it converges uniformly

for x=9

lim_{n->\infty} sup|fn-0|>=lim_{n->\infty}|3+1/n|=3 not zero

but the book says that it does converge uniformly

why??

Re: point convergence and uniform convergence definition question

You can write [tex]\lim_{n\to\infty} f_n(x_0)=0[/tex] for $\displaystyle \lim_{n\to\infty} f_n(x_0)=0$.

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**transgalactic** i was told that if there is a point x0 from [a,b]

and $\displaystyle lim_{n->infty} f_n{x0}=0$

then it converges to a point

You need to review the definitions. As far as I know, there is no definition for when a sequence of functions converges to a point. There is a definition when a sequence of functions $\displaystyle f_n$ converges **at** point x: this means that $\displaystyle \lim_{n\to\infty}f_n(x)$ exists (it is not necessarily zero). Also, the sequence $\displaystyle f_n$ converges pointwise if it converges at every point.

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**transgalactic** if we want to find out if it converges uniformly

for x=9

lim_{n->\infty} sup|fn-0|>=lim_{n->\infty}|3+1/n|=3 not zero

Uniform convergence can't be checked at one point (such as x = 9); it's a property of functions viewed on the whole set. I also don't understand what fn-0 is and what 3+1/n has to do with anything.

Re: point convergence and uniform convergence definition question

I don't entirely understand your question. By definition, $\displaystyle f_n \to f $ pointwise if for each fixed x in the domain, the pointwise limit $\displaystyle \lim_{n \to \infty} f_n(x) = f(x) $. For a pointwise limit, you're considering a sequence of points, $\displaystyle x_1, x_2, x_3, ...$ where the jth point in your sequence is just $\displaystyle f_j(x)$. So for each x and $\displaystyle f_n = \sqrt{x^2 + \frac{1}{n}}$, we have that $\displaystyle f_n(x) \to |x|$ since $\displaystyle \lim_{n \to \infty} \sqrt{x^2 + \frac{1}{n}} = \sqrt{x^2}} = |x| $ (this follows from the square root function being continuous, or a simple epsilon estimate if necessary).

Notice in the definition that the pointwise limit is entirely a local property of a sequence of functions. It does not depend on any other point in the domain. In fact, it's entirely possible for the f_n's to only converge for one point in their domain. By definition, we say that $\displaystyle f_n \to f$ uniformly if $\displaystyle \sup{|f_n(x) - f(x)|} \to 0 $ as $\displaystyle n \to \infty$. From the definition, it's clear that a uniform limit implies a pointwise limit. However, the uniform condition is strictly stronger because it's a global condition on the function. All points in the domain must converge to the limiting function at some uniform rate. It does not make sense to as if a sequence of functions converges uniformly at a point.

Re: point convergence and uniform convergence definition question

ok so convegence to a point is if the limit exists

i showed that for x=0 fn(x) converges to sero

so if it convegers uniformyle then for every x

$\displaystyle lim_{n->\infty} sup|fn-f| ->0$

this is the definition for uniforn convergence

if x=9

$\displaystyle lim_{n->\infty} sup|fn-0|>=|3-0| not 0$

Re: point convergence and uniform convergence definition question

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**transgalactic** ok so convegence to a point is if the limit exists

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**emakarov** As far as I know, there is no definition for when a sequence of functions converges to a point. There is a definition when a sequence of functions $\displaystyle f_n$ converges **at** point x

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**transgalactic** if x=9

$\displaystyle lim_{n->\infty} sup|fn-0|>=|3-0| not 0$

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**emakarov** I also don't understand what fn-0 is

Who said $\displaystyle f_n(9)$ must converge to 0?

Re: point convergence and uniform convergence definition question

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**JakeBarnes** I don't entirely understand your question. By definition, $\displaystyle f_n \to f $ pointwise if for each fixed x in the domain, the pointwise limit $\displaystyle \lim_{n \to \infty} f_n(x) = f(x) $. For a pointwise limit, you're considering a sequence of points, $\displaystyle x_1, x_2, x_3, ...$ where the jth point in your sequence is just $\displaystyle f_j(x)$. So for each x and $\displaystyle f_n = \sqrt{x^2 + \frac{1}{n}}$, we have that $\displaystyle f_n(x) \to |x|$ since $\displaystyle \lim_{n \to \infty} \sqrt{x^2 + \frac{1}{n}} = \sqrt{x^2}} = |x| $ (this follows from the square root function being continuous, or a simple epsilon estimate if necessary).

Notice in the definition that the pointwise limit is entirely a local property of a sequence of functions. It does not depend on any other point in the domain. In fact, it's entirely possible for the f_n's to only converge for one point in their domain. By definition, we say that $\displaystyle f_n \to f$ uniformly if $\displaystyle \sup{|f_n(x) - f(x)|} \to 0 $ as $\displaystyle n \to \infty$. From the definition, it's clear that a uniform limit implies a pointwise limit. However, the uniform condition is strictly stronger because it's a global condition on the function. All points in the domain must converge to the limiting function at some uniform rate. It does not make sense to as if a sequence of functions converges uniformly at a point.

entirely a local property of a sequence of functions"

It does not depend on any other point in the domain. In fact, it's entirely possible for the f_n's to only converge for one point in their domain

"

if it converge on only one point then its not pointwise convergent,because you said that

"pointwise if for each fixed x in the domain"

so for fn to be point wise convergent does it has to convege for all the point in the domain or not

?

Re: point convergence and uniform convergence definition question

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**transgalactic** so for fn to be point wise convergent does it has to convege for all the point in the domain or not?

Yes, it has to converge at every point of the domain, but the rates of conversion at different points don't have to be coordinated at all.

Re: point convergence and uniform convergence definition question

ok so in uniform convergence for every point in the domain we converge to some function

but in uniform convergence for every point we converge to the same function.

what does it has to do with the supremum of subtraction??

for example:

in $\displaystyle f_n(x)=\frac{nx}{1+n^2x^2}$

i know that there is a pointwise convergence

but why its not uniform covergent

how to link between the supremum of this subtraction and the fact that not all points converge to the same function

??

$\displaystyle lim_{n->\infty} sup|fn-f|$

Re: point convergence and uniform convergence definition question

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**transgalactic** ok so in uniform convergence for every point in the domain we converge to some function

but in uniform convergence for every point we converge to the same function.

Um, what?

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**transgalactic** what does it has to do with the supremum of subtraction??

The definition of uniform convergence uses the supremum of the difference.

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**transgalactic** how to link between the supremum of this subtraction and the fact that not all points converge to the same function

??

I am not sure what "not all points converge to the same function" means. Here there is a pointwise convergence, so $\displaystyle f_n(x)$ converge to f(x) (here f(x) = 0) for all points x.

The sequence $\displaystyle f_n$ converges to f uniformly if for every $\displaystyle \varepsilon$, the graph of $\displaystyle f_n$ is within the band of width $\displaystyle 2\varepsilon$ around the graph of f for all n starting from some N. In the example above, the maximum of all $\displaystyle f_n$ is 1/2. So, one cannot find N such that for all n > N the graphs of $\displaystyle f_n$ lie within the band {(x, y) : |y| <= 1/2}.

Re: point convergence and uniform convergence definition question

sorry i made a typing mistake

"pointwise convergence ,is when for every point in the domain we converge to some function f"

ok thats make sense

for all n the graph of fn is within 2epsilon from f

in my example we look for the supremum of the difference.

what is f?

how did you find it?

why you looked only on the maximum of fn?

Re: point convergence and uniform convergence definition question

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**transgalactic** entirely a local property of a sequence of functions"

It does not depend on any other point in the domain. In fact, it's entirely possible for the f_n's to only converge for one point in their domain

"

if it converge on only one point then its not pointwise convergent,because you said that

"pointwise if for each fixed x in the domain"

so for fn to be point wise convergent does it has to convege for all the point in the domain or not

?

I did not say that if the sequence $\displaystyle f_n(x) \to y $ as $\displaystyle n \to \infty$ for only one x in its domain, then these functions are pointwise convergent. What I was attempting to get across is that this limit may or may not exist for individual values of x. This is not the case for a uniform limit.

Re: point convergence and uniform convergence definition question

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**emakarov** Um, what?

The definition of uniform convergence uses the supremum of the difference.

I am not sure what "not all points converge to the same function" means. Here there is a pointwise convergence, so $\displaystyle f_n(x)$ converge to f(x) (here f(x) = 0) for all points x.

The sequence $\displaystyle f_n$ converges to f uniformly if for every $\displaystyle \varepsilon$, the graph of $\displaystyle f_n$ is within the band of width $\displaystyle 2\varepsilon$ around the graph of f for all n starting from some N. In the example above, the maximum of all $\displaystyle f_n$ is 1/2. So, one cannot find N such that for all n > N the graphs of $\displaystyle f_n$ lie within the band {(x, y) : |y| <= 1/2}.

ok so in my example

for x=0 and for other axes fn converges to sero.

so why its not uniformly convergent( its the definition)

???

(i know about looking at the supremum of the difference,i just saw that for every x fn converges to sero)

Re: point convergence and uniform convergence definition question

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**transgalactic** in my example we look for the supremum of the difference.

Yes, and $\displaystyle \sup|f_n(x)-f(x)|<\varepsilon$ iff the graph of $\displaystyle f_n$ is within $\displaystyle \varepsilon$ of the graph of f, i.e., the graph of $\displaystyle f_n$ lies within a $\displaystyle 2\varepsilon$-wide band around the graph of f.

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**transgalactic** what is f?

how did you find it?

Exactly the same questions can be asked about the pointwise limit, which you say you understand. The definition of the limit (whether pointwise or uniform) does not require finding f given $\displaystyle f_n$. Rather, given both a sequence $\displaystyle f_n$ and a function f we say whether $\displaystyle f_n$ and f are in the relationship "$\displaystyle f_n$ converges to f."

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**transgalactic** why you looked only on the maximum of fn?

Where in the definition of the limit did you see the maximum of $\displaystyle f_n$?

I am puzzled by your questions. Which of the following best describes your difficulty?

(1) You don't understand the definition of the uniform convergence because it uses vague terms, unfamiliar terms or some reasoning that you can't follow. (Definitions rarely use proofs and I don't think this one does.)

(2) You understand the definition but can't form a mental picture or intuition about what it means.

(3) Something else.

I can't see how (1) can be the case because the definition uses plain language phrases like "for all," "if ..., then ..." and mathematical operations like subtraction and absolute value, which I am sure you know well. Well, there is supremum, but the same can be said about the definition of supremum.

If (2) is the case, what do you think about the description in terms of a band around a graph?

Re: point convergence and uniform convergence definition question

ahhh ok so f is the function which for every x0 in the domain fn(x0) converge to.

now we look at sup|fn-f|

so in my example

$\displaystyle f_n(x)=\frac{nx}{1+n^2x^2}$

how did you saw that the maximum is 0.5?

Re: point convergence and uniform convergence definition question

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**transgalactic** ahhh ok so f is the function which for every x0 in the domain fn(x0) converge to.

Yes, a pointwise limit is the only candidate for a uniform limit.

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**transgalactic** so in my example

$\displaystyle f_n(x)=\frac{nx}{1+n^2x^2}$

how did you saw that the maximum is 0.5?

Using WolframAlpha (Smile) But this is not hard to see by taking a derivative.