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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!

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.

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.