Mixing
Understanding the proof of $l_2$ being complete.
$begingroup$
Let $l_2$ be the collection of bounded real sequences $x = (x_n)$ for which $sum_{n=1}^{infty}|x_n|^2<infty$. I have to prove that $l_2$ is complete (that every Cauchy sequence in $l_2$ converges to a point in $l_2$).
Proof:
Let $(f_n)$ be a sequence in $l_2$, where now we write $f_n = (f_n(k))_{k=1}^{infty}$, and suppose that $(f_n)$ is Cauchy in $l_2$. That is, suppose that for each $epsilon > 0$ there is a $n_0$ such that $||f_n - f_m||_2 < epsilon$ whenever $m,n geq n_0$. We now want to show that $(f_n)$ converges, in the metric of $l_2$, to some $fin l_2$.
1) First show that $f(k) = lim_{nto infty},f_n(k)$ exists in $mathbb{R}$ for each k:
To see why, note that $|,f_n(k) - f_m(k)|leq ||,f_n - f_m||_2$ for any k, and hence $(f_n(k))_{k=1}^{infty}$ is Cauchy in $mathbb{R}$ for each k. Thus, $f$ is the obvious candidate for the limit of $(f_n)$, but we still have to show that the convergence takes place in the metric space $l_2$; that is, we need to show that $fin l_2$ and that $||,f_n - f||_2 to 0$ (as $n to infty$).
2) Now show that $fin l_2$; that is, $||,f||_2 < infty$. We know that $(f_n)$ is bounded in $l_2$; say, $||f_n||leq B$ for all n. Thus, for any fixed $N < infty$, we have:
$sumlimits_{k=1}^{N}|,f(k)|^2 = lim_{ntoinfty}sumlimits_{k=1}^{N}|,f_n(k)|^2 leq B^2$.
Since this holds for any N, we get that $||,f||_2leq B$.
3) Now we repeat step 2 (more or less) to show that $f_n to f$ in $l_2$.
Given any $epsilon > 0$ choose $n_0$ such that $||,f_n - f_m||_2 < epsilon$ whenever $m,n > n_0$. Then, for any N and any $ngeq n_0$,
$sumlimits_{k=1}^{N}|,f(k) - f_n(k)|^2 = lim_{n to infty}sumlimits_{k = 1}^{N}|,f_m(k) - f_n(k)|^2 leq epsilon^2$.
Since this holds for any N, we have $||,f - f_n||_2 leq epsilon$ for all $ngeq n_0$. That is, $f_n to f$ in $l_2$.
Questions:
- I think I understand the general idea of the proof. You take a sequence and suppose it's Cauchy. You then have to show that it has a limit and thus converges, and show that the limit lies in $l_2$, which would mean that any Cauchy sequence in $l_2$ is convergent to a point in $l_2$. So why is step 3 necessary? I must be mistaken, but it seems to me that in step 1 and 2 it has already been proven that the chosen sequence has a limit and that it lies in $l_2$.
- I don't understand the notation that is used in this proof: "Let $(f_n)$ be a sequence in $l_2$, where we now write $f_n = (f_n(k))_{k=1}^{infty}$". Why is the letter $k$ added to the sequence and what does it mean? If you have a sequence $x_n$, the subscript is the argument right? Can't you write $x_n$ like $x,(n)$?
- Why does $|,f_n(k) - f_m(k)|leq||,f_n - f_m||_2$ imply that $f_n(k)$ has a limit? Perhaps this will be clear once I understand the use of the letter $k$.
Thanks in advance!
real-analysis sequences-and-series limits convergence continuity
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
$endgroup$
add a comment |
$begingroup$
Let $l_2$ be the collection of bounded real sequences $x = (x_n)$ for which $sum_{n=1}^{infty}|x_n|^2<infty$. I have to prove that $l_2$ is complete (that every Cauchy sequence in $l_2$ converges to a point in $l_2$).
Proof:
Let $(f_n)$ be a sequence in $l_2$, where now we write $f_n = (f_n(k))_{k=1}^{infty}$, and suppose that $(f_n)$ is Cauchy in $l_2$. That is, suppose that for each $epsilon > 0$ there is a $n_0$ such that $||f_n - f_m||_2 < epsilon$ whenever $m,n geq n_0$. We now want to show that $(f_n)$ converges, in the metric of $l_2$, to some $fin l_2$.
1) First show that $f(k) = lim_{nto infty},f_n(k)$ exists in $mathbb{R}$ for each k:
To see why, note that $|,f_n(k) - f_m(k)|leq ||,f_n - f_m||_2$ for any k, and hence $(f_n(k))_{k=1}^{infty}$ is Cauchy in $mathbb{R}$ for each k. Thus, $f$ is the obvious candidate for the limit of $(f_n)$, but we still have to show that the convergence takes place in the metric space $l_2$; that is, we need to show that $fin l_2$ and that $||,f_n - f||_2 to 0$ (as $n to infty$).
2) Now show that $fin l_2$; that is, $||,f||_2 < infty$. We know that $(f_n)$ is bounded in $l_2$; say, $||f_n||leq B$ for all n. Thus, for any fixed $N < infty$, we have:
$sumlimits_{k=1}^{N}|,f(k)|^2 = lim_{ntoinfty}sumlimits_{k=1}^{N}|,f_n(k)|^2 leq B^2$.
Since this holds for any N, we get that $||,f||_2leq B$.
3) Now we repeat step 2 (more or less) to show that $f_n to f$ in $l_2$.
Given any $epsilon > 0$ choose $n_0$ such that $||,f_n - f_m||_2 < epsilon$ whenever $m,n > n_0$. Then, for any N and any $ngeq n_0$,
$sumlimits_{k=1}^{N}|,f(k) - f_n(k)|^2 = lim_{n to infty}sumlimits_{k = 1}^{N}|,f_m(k) - f_n(k)|^2 leq epsilon^2$.
Since this holds for any N, we have $||,f - f_n||_2 leq epsilon$ for all $ngeq n_0$. That is, $f_n to f$ in $l_2$.
Questions:
- I think I understand the general idea of the proof. You take a sequence and suppose it's Cauchy. You then have to show that it has a limit and thus converges, and show that the limit lies in $l_2$, which would mean that any Cauchy sequence in $l_2$ is convergent to a point in $l_2$. So why is step 3 necessary? I must be mistaken, but it seems to me that in step 1 and 2 it has already been proven that the chosen sequence has a limit and that it lies in $l_2$.
- I don't understand the notation that is used in this proof: "Let $(f_n)$ be a sequence in $l_2$, where we now write $f_n = (f_n(k))_{k=1}^{infty}$". Why is the letter $k$ added to the sequence and what does it mean? If you have a sequence $x_n$, the subscript is the argument right? Can't you write $x_n$ like $x,(n)$?
- Why does $|,f_n(k) - f_m(k)|leq||,f_n - f_m||_2$ imply that $f_n(k)$ has a limit? Perhaps this will be clear once I understand the use of the letter $k$.
Thanks in advance!
real-analysis sequences-and-series limits convergence continuity
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
$endgroup$
add a comment |
$begingroup$
Let $l_2$ be the collection of bounded real sequences $x = (x_n)$ for which $sum_{n=1}^{infty}|x_n|^2<infty$. I have to prove that $l_2$ is complete (that every Cauchy sequence in $l_2$ converges to a point in $l_2$).
Proof:
Let $(f_n)$ be a sequence in $l_2$, where now we write $f_n = (f_n(k))_{k=1}^{infty}$, and suppose that $(f_n)$ is Cauchy in $l_2$. That is, suppose that for each $epsilon > 0$ there is a $n_0$ such that $||f_n - f_m||_2 < epsilon$ whenever $m,n geq n_0$. We now want to show that $(f_n)$ converges, in the metric of $l_2$, to some $fin l_2$.
1) First show that $f(k) = lim_{nto infty},f_n(k)$ exists in $mathbb{R}$ for each k:
To see why, note that $|,f_n(k) - f_m(k)|leq ||,f_n - f_m||_2$ for any k, and hence $(f_n(k))_{k=1}^{infty}$ is Cauchy in $mathbb{R}$ for each k. Thus, $f$ is the obvious candidate for the limit of $(f_n)$, but we still have to show that the convergence takes place in the metric space $l_2$; that is, we need to show that $fin l_2$ and that $||,f_n - f||_2 to 0$ (as $n to infty$).
2) Now show that $fin l_2$; that is, $||,f||_2 < infty$. We know that $(f_n)$ is bounded in $l_2$; say, $||f_n||leq B$ for all n. Thus, for any fixed $N < infty$, we have:
$sumlimits_{k=1}^{N}|,f(k)|^2 = lim_{ntoinfty}sumlimits_{k=1}^{N}|,f_n(k)|^2 leq B^2$.
Since this holds for any N, we get that $||,f||_2leq B$.
3) Now we repeat step 2 (more or less) to show that $f_n to f$ in $l_2$.
Given any $epsilon > 0$ choose $n_0$ such that $||,f_n - f_m||_2 < epsilon$ whenever $m,n > n_0$. Then, for any N and any $ngeq n_0$,
$sumlimits_{k=1}^{N}|,f(k) - f_n(k)|^2 = lim_{n to infty}sumlimits_{k = 1}^{N}|,f_m(k) - f_n(k)|^2 leq epsilon^2$.
Since this holds for any N, we have $||,f - f_n||_2 leq epsilon$ for all $ngeq n_0$. That is, $f_n to f$ in $l_2$.
Questions:
- I think I understand the general idea of the proof. You take a sequence and suppose it's Cauchy. You then have to show that it has a limit and thus converges, and show that the limit lies in $l_2$, which would mean that any Cauchy sequence in $l_2$ is convergent to a point in $l_2$. So why is step 3 necessary? I must be mistaken, but it seems to me that in step 1 and 2 it has already been proven that the chosen sequence has a limit and that it lies in $l_2$.
- I don't understand the notation that is used in this proof: "Let $(f_n)$ be a sequence in $l_2$, where we now write $f_n = (f_n(k))_{k=1}^{infty}$". Why is the letter $k$ added to the sequence and what does it mean? If you have a sequence $x_n$, the subscript is the argument right? Can't you write $x_n$ like $x,(n)$?
- Why does $|,f_n(k) - f_m(k)|leq||,f_n - f_m||_2$ imply that $f_n(k)$ has a limit? Perhaps this will be clear once I understand the use of the letter $k$.
Thanks in advance!
real-analysis sequences-and-series limits convergence continuity
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
$endgroup$
Let $l_2$ be the collection of bounded real sequences $x = (x_n)$ for which $sum_{n=1}^{infty}|x_n|^2<infty$. I have to prove that $l_2$ is complete (that every Cauchy sequence in $l_2$ converges to a point in $l_2$).
Proof:
Let $(f_n)$ be a sequence in $l_2$, where now we write $f_n = (f_n(k))_{k=1}^{infty}$, and suppose that $(f_n)$ is Cauchy in $l_2$. That is, suppose that for each $epsilon > 0$ there is a $n_0$ such that $||f_n - f_m||_2 < epsilon$ whenever $m,n geq n_0$. We now want to show that $(f_n)$ converges, in the metric of $l_2$, to some $fin l_2$.
1) First show that $f(k) = lim_{nto infty},f_n(k)$ exists in $mathbb{R}$ for each k:
To see why, note that $|,f_n(k) - f_m(k)|leq ||,f_n - f_m||_2$ for any k, and hence $(f_n(k))_{k=1}^{infty}$ is Cauchy in $mathbb{R}$ for each k. Thus, $f$ is the obvious candidate for the limit of $(f_n)$, but we still have to show that the convergence takes place in the metric space $l_2$; that is, we need to show that $fin l_2$ and that $||,f_n - f||_2 to 0$ (as $n to infty$).
2) Now show that $fin l_2$; that is, $||,f||_2 < infty$. We know that $(f_n)$ is bounded in $l_2$; say, $||f_n||leq B$ for all n. Thus, for any fixed $N < infty$, we have:
$sumlimits_{k=1}^{N}|,f(k)|^2 = lim_{ntoinfty}sumlimits_{k=1}^{N}|,f_n(k)|^2 leq B^2$.
Since this holds for any N, we get that $||,f||_2leq B$.
3) Now we repeat step 2 (more or less) to show that $f_n to f$ in $l_2$.
Given any $epsilon > 0$ choose $n_0$ such that $||,f_n - f_m||_2 < epsilon$ whenever $m,n > n_0$. Then, for any N and any $ngeq n_0$,
$sumlimits_{k=1}^{N}|,f(k) - f_n(k)|^2 = lim_{n to infty}sumlimits_{k = 1}^{N}|,f_m(k) - f_n(k)|^2 leq epsilon^2$.
Since this holds for any N, we have $||,f - f_n||_2 leq epsilon$ for all $ngeq n_0$. That is, $f_n to f$ in $l_2$.
Questions:
- I think I understand the general idea of the proof. You take a sequence and suppose it's Cauchy. You then have to show that it has a limit and thus converges, and show that the limit lies in $l_2$, which would mean that any Cauchy sequence in $l_2$ is convergent to a point in $l_2$. So why is step 3 necessary? I must be mistaken, but it seems to me that in step 1 and 2 it has already been proven that the chosen sequence has a limit and that it lies in $l_2$.
- I don't understand the notation that is used in this proof: "Let $(f_n)$ be a sequence in $l_2$, where we now write $f_n = (f_n(k))_{k=1}^{infty}$". Why is the letter $k$ added to the sequence and what does it mean? If you have a sequence $x_n$, the subscript is the argument right? Can't you write $x_n$ like $x,(n)$?
- Why does $|,f_n(k) - f_m(k)|leq||,f_n - f_m||_2$ imply that $f_n(k)$ has a limit? Perhaps this will be clear once I understand the use of the letter $k$.
Thanks in advance!
real-analysis sequences-and-series limits convergence continuity
real-analysis sequences-and-series limits convergence continuity
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
asked Oct 27 '17 at 12:31
titusAdamtitusAdam
1,100313
1,100313
add a comment |
add a comment |
2 Answers
2
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oldest
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Question $1)$ you have a sequence of functions $f_nin l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).
Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $sum_{n=1}^infty |a_n|^2<infty$. The second is that elements in $l_2$ are functions $f:mathbb{N}rightarrowmathbb{C}$ satisfying that $sum_{n=1}^infty |f(n)|^2<infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.
Question $3)$ this is the tricky part because we have two different norms in here.
The first is the $l_2$ norm that is $|f|_2 =sum_{n=1}^infty |f(n)|^2$ and the second is the norm on $mathbb{C}$, for every given $k$, $f(k)$ is an element in $mathbb{C}$. The inequality $|f_n(k)-f_m(k)|leq |f_n-f_m|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $mathbb{C}$), therefore you can use the fact that $mathbb{C}$ is complete.
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
$endgroup$
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
add a comment |
$begingroup$
Let's start with this $k$ in $f_n(k)$: The sequence $f_nin l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example
$$(f_1)_{kinmathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,ldots).$$
To simplify the expression and have a connection to maps $mathbb{N}rightarrowmathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.
Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence:
$$forall varepsilon>0 forall k, exists n_0=n_0(varepsilon,k)inmathbb{N} mbox{ such that } forall ngeq n_0: |f_n(k)-f(k)|<varepsilon.$$
Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm
$$|(f)_{kinmathbb{N}}|^2=sum_{k=1}^infty (f(k))^2.$$
So $f_nrightarrow f$ in $l_2$ is by definition
$$forall varepsilon>0 exists n_0=n_0(varepsilon)mbox{ such that }forall ngeq n_0 |f_n-f|_{l_2}^2:=sum^infty_{k=1}(f_n(k)-f(k))^2leq varepsilon.$$
Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.
Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
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2 Answers
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2 Answers
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$begingroup$
Question $1)$ you have a sequence of functions $f_nin l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).
Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $sum_{n=1}^infty |a_n|^2<infty$. The second is that elements in $l_2$ are functions $f:mathbb{N}rightarrowmathbb{C}$ satisfying that $sum_{n=1}^infty |f(n)|^2<infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.
Question $3)$ this is the tricky part because we have two different norms in here.
The first is the $l_2$ norm that is $|f|_2 =sum_{n=1}^infty |f(n)|^2$ and the second is the norm on $mathbb{C}$, for every given $k$, $f(k)$ is an element in $mathbb{C}$. The inequality $|f_n(k)-f_m(k)|leq |f_n-f_m|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $mathbb{C}$), therefore you can use the fact that $mathbb{C}$ is complete.
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
$endgroup$
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
add a comment |
$begingroup$
Question $1)$ you have a sequence of functions $f_nin l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).
Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $sum_{n=1}^infty |a_n|^2<infty$. The second is that elements in $l_2$ are functions $f:mathbb{N}rightarrowmathbb{C}$ satisfying that $sum_{n=1}^infty |f(n)|^2<infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.
Question $3)$ this is the tricky part because we have two different norms in here.
The first is the $l_2$ norm that is $|f|_2 =sum_{n=1}^infty |f(n)|^2$ and the second is the norm on $mathbb{C}$, for every given $k$, $f(k)$ is an element in $mathbb{C}$. The inequality $|f_n(k)-f_m(k)|leq |f_n-f_m|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $mathbb{C}$), therefore you can use the fact that $mathbb{C}$ is complete.
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
$endgroup$
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
add a comment |
$begingroup$
Question $1)$ you have a sequence of functions $f_nin l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).
Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $sum_{n=1}^infty |a_n|^2<infty$. The second is that elements in $l_2$ are functions $f:mathbb{N}rightarrowmathbb{C}$ satisfying that $sum_{n=1}^infty |f(n)|^2<infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.
Question $3)$ this is the tricky part because we have two different norms in here.
The first is the $l_2$ norm that is $|f|_2 =sum_{n=1}^infty |f(n)|^2$ and the second is the norm on $mathbb{C}$, for every given $k$, $f(k)$ is an element in $mathbb{C}$. The inequality $|f_n(k)-f_m(k)|leq |f_n-f_m|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $mathbb{C}$), therefore you can use the fact that $mathbb{C}$ is complete.
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
$endgroup$
Question $1)$ you have a sequence of functions $f_nin l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).
Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $sum_{n=1}^infty |a_n|^2<infty$. The second is that elements in $l_2$ are functions $f:mathbb{N}rightarrowmathbb{C}$ satisfying that $sum_{n=1}^infty |f(n)|^2<infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.
Question $3)$ this is the tricky part because we have two different norms in here.
The first is the $l_2$ norm that is $|f|_2 =sum_{n=1}^infty |f(n)|^2$ and the second is the norm on $mathbb{C}$, for every given $k$, $f(k)$ is an element in $mathbb{C}$. The inequality $|f_n(k)-f_m(k)|leq |f_n-f_m|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $mathbb{C}$), therefore you can use the fact that $mathbb{C}$ is complete.
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
edited Jan 23 at 12:47
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
answered Oct 27 '17 at 12:54
YankoYanko
7,7751830
7,7751830
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
add a comment |
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Thank you for your reply! I don't understand your analogy in your answer to question 2 though. You say that it's almost like $1$ is a limit of $1-frac{1}{n}$ but that $(0,1)$ is not complete because $1$ is not in it. However in step 2 we already show that $fin l_2$. So a correct analogy would be: it's like $1$ is a limit of $1-frac{1}{n}$ so $[0,1]$ is complete because $1$ is in it. Right?
$endgroup$
– titusAdam
Oct 28 '17 at 9:52
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
Another point:)! I assume that with $mathbb{C}$ you mean the set of complex numbers? Why is there any norm on $mathbb{C}$, isn't $l_2$ the set of all bounded real sequences?
$endgroup$
– titusAdam
Oct 28 '17 at 10:02
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
@titusAdam, about your first question, yes you are right I just wanted to explain why it is important to show that $fin l_2$. About your second question.. it depends.. some prefer to use the complex numbers and some prefer to use the reals.. it doesn't really change anything. So if you want you can replace $mathbb{C}$ with $mathbb{R}$.
$endgroup$
– Yanko
Oct 28 '17 at 11:12
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
I understand that $fin l_2$ is necessary, but this is already shown in step 2 right? So the analogy you use isn't correct?
$endgroup$
– titusAdam
Oct 28 '17 at 11:29
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
$begingroup$
yes, being in $l_2$ just means that $|f|_2 <infty$.
$endgroup$
– Yanko
Oct 28 '17 at 11:30
add a comment |
$begingroup$
Let's start with this $k$ in $f_n(k)$: The sequence $f_nin l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example
$$(f_1)_{kinmathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,ldots).$$
To simplify the expression and have a connection to maps $mathbb{N}rightarrowmathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.
Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence:
$$forall varepsilon>0 forall k, exists n_0=n_0(varepsilon,k)inmathbb{N} mbox{ such that } forall ngeq n_0: |f_n(k)-f(k)|<varepsilon.$$
Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm
$$|(f)_{kinmathbb{N}}|^2=sum_{k=1}^infty (f(k))^2.$$
So $f_nrightarrow f$ in $l_2$ is by definition
$$forall varepsilon>0 exists n_0=n_0(varepsilon)mbox{ such that }forall ngeq n_0 |f_n-f|_{l_2}^2:=sum^infty_{k=1}(f_n(k)-f(k))^2leq varepsilon.$$
Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.
Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
$endgroup$
add a comment |
$begingroup$
Let's start with this $k$ in $f_n(k)$: The sequence $f_nin l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example
$$(f_1)_{kinmathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,ldots).$$
To simplify the expression and have a connection to maps $mathbb{N}rightarrowmathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.
Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence:
$$forall varepsilon>0 forall k, exists n_0=n_0(varepsilon,k)inmathbb{N} mbox{ such that } forall ngeq n_0: |f_n(k)-f(k)|<varepsilon.$$
Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm
$$|(f)_{kinmathbb{N}}|^2=sum_{k=1}^infty (f(k))^2.$$
So $f_nrightarrow f$ in $l_2$ is by definition
$$forall varepsilon>0 exists n_0=n_0(varepsilon)mbox{ such that }forall ngeq n_0 |f_n-f|_{l_2}^2:=sum^infty_{k=1}(f_n(k)-f(k))^2leq varepsilon.$$
Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.
Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
$endgroup$
add a comment |
$begingroup$
Let's start with this $k$ in $f_n(k)$: The sequence $f_nin l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example
$$(f_1)_{kinmathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,ldots).$$
To simplify the expression and have a connection to maps $mathbb{N}rightarrowmathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.
Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence:
$$forall varepsilon>0 forall k, exists n_0=n_0(varepsilon,k)inmathbb{N} mbox{ such that } forall ngeq n_0: |f_n(k)-f(k)|<varepsilon.$$
Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm
$$|(f)_{kinmathbb{N}}|^2=sum_{k=1}^infty (f(k))^2.$$
So $f_nrightarrow f$ in $l_2$ is by definition
$$forall varepsilon>0 exists n_0=n_0(varepsilon)mbox{ such that }forall ngeq n_0 |f_n-f|_{l_2}^2:=sum^infty_{k=1}(f_n(k)-f(k))^2leq varepsilon.$$
Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.
Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
$endgroup$
Let's start with this $k$ in $f_n(k)$: The sequence $f_nin l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example
$$(f_1)_{kinmathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,ldots).$$
To simplify the expression and have a connection to maps $mathbb{N}rightarrowmathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.
Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence:
$$forall varepsilon>0 forall k, exists n_0=n_0(varepsilon,k)inmathbb{N} mbox{ such that } forall ngeq n_0: |f_n(k)-f(k)|<varepsilon.$$
Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm
$$|(f)_{kinmathbb{N}}|^2=sum_{k=1}^infty (f(k))^2.$$
So $f_nrightarrow f$ in $l_2$ is by definition
$$forall varepsilon>0 exists n_0=n_0(varepsilon)mbox{ such that }forall ngeq n_0 |f_n-f|_{l_2}^2:=sum^infty_{k=1}(f_n(k)-f(k))^2leq varepsilon.$$
Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.
Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
answered Oct 27 '17 at 12:57
humanStampedisthumanStampedist
2,228214
2,228214
add a comment |
add a comment |
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