point convergence and uniform convergence definition question

**transgalactic** · April 12th 2012, 06:37 AM

i was told that if there is a point x0 from [a,b]
and $lim_{n->infty} f_n{x0}=0$
then it converges to a point

first part of the question:
here i got $f_n(x)=\sqrt{x^2+\frac{1}{n}}$
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??

**emakarov** · April 12th 2012, 07:28 AM

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

Originally Posted by transgalactic

i was told that if there is a point x0 from [a,b]
and $lim_{n->infty} f_n{x0}=0$
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 $f_n$ converges at point x: this means that $\lim_{n\to\infty}f_n(x)$ exists (it is not necessarily zero). Also, the sequence $f_n$ converges pointwise if it converges at every point.

Originally Posted by 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.

**JakeBarnes** · April 12th 2012, 07:35 AM

I don't entirely understand your question. By definition, $f_n \to f$ pointwise if for each fixed x in the domain, the pointwise limit $\lim_{n \to \infty} f_n(x) = f(x)$ . For a pointwise limit, you're considering a sequence of points, $x_1, x_2, x_3, ...$ where the jth point in your sequence is just $f_j(x)$ . So for each x and $f_n = \sqrt{x^2 + \frac{1}{n}}$ , we have that $f_n(x) \to |x|$ since $\lim_{n \to \infty} \sqrt{x^2 + \frac{1}{n}} = \sqrt{x^2}} = |x|$ (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 $f_n \to f$ uniformly if $\sup{|f_n(x) - f(x)|} \to 0$ as $n \to \infty$ . 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.

**transgalactic** · April 12th 2012, 07:36 AM

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
$lim_{n->\infty} sup|fn-f| ->0$
this is the definition for uniforn convergence

if x=9
$lim_{n->\infty} sup|fn-0|>=|3-0| not 0$

**emakarov** · April 12th 2012, 07:43 AM

Originally Posted by transgalactic

ok so convegence to a point is if the limit exists

Originally Posted by 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 $f_n$ converges at point x

Originally Posted by transgalactic

if x=9
$lim_{n->\infty} sup|fn-0|>=|3-0| not 0$

Originally Posted by emakarov

I also don't understand what fn-0 is

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

**transgalactic** · April 12th 2012, 07:48 AM

Originally Posted by JakeBarnes

I don't entirely understand your question. By definition, $f_n \to f$ pointwise if for each fixed x in the domain, the pointwise limit $\lim_{n \to \infty} f_n(x) = f(x)$ . For a pointwise limit, you're considering a sequence of points, $x_1, x_2, x_3, ...$ where the jth point in your sequence is just $f_j(x)$ . So for each x and $f_n = \sqrt{x^2 + \frac{1}{n}}$ , we have that $f_n(x) \to |x|$ since $\lim_{n \to \infty} \sqrt{x^2 + \frac{1}{n}} = \sqrt{x^2}} = |x|$ (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 $f_n \to f$ uniformly if $\sup{|f_n(x) - f(x)|} \to 0$ as $n \to \infty$ . 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
?

**emakarov** · April 12th 2012, 08:19 AM

Originally Posted by 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.

**transgalactic** · April 12th 2012, 08:52 AM

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

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

**emakarov** · April 12th 2012, 10:25 AM

Originally Posted by 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?

Originally Posted by transgalactic

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

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

Originally Posted by 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 $f_n(x)$ converge to f(x) (here f(x) = 0) for all points x.

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

**transgalactic** · April 12th 2012, 11:28 AM

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?

**JakeBarnes** · April 12th 2012, 11:34 AM

Originally Posted by 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 $f_n(x) \to y$ as $n \to \infty$ 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.

**transgalactic** · April 12th 2012, 11:46 AM

Originally Posted by 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 $f_n(x)$ converge to f(x) (here f(x) = 0) for all points x.

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

**emakarov** · April 12th 2012, 12:03 PM

Originally Posted by transgalactic

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

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

Originally Posted by 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 $f_n$ . Rather, given both a sequence $f_n$ and a function f we say whether $f_n$ and f are in the relationship " $f_n$ converges to f."

Originally Posted by transgalactic

why you looked only on the maximum of fn?

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

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?

**transgalactic** · April 12th 2012, 12:28 PM

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

how did you saw that the maximum is 0.5?

**emakarov** · April 12th 2012, 12:34 PM

Originally Posted by 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.

Originally Posted by transgalactic

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

how did you saw that the maximum is 0.5?

Using WolframAlpha

But this is not hard to see by taking a derivative.

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