Understanding the proof of $l_2$ being complete.

Question

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 $f\in l_2$.

1) First show that $f(k) = lim_{n\to \infty}\,f_n(k)$ 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 $f\in l_2$ and that $||\,f_n - f||_2 \to 0$ (as $n \to \infty$).

2) Now show that $f\in l_2$; 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:

$\sum\limits_{k=1}^{N}|\,f(k)|^2 = lim_{n\to\infty}\sum\limits_{k=1}^{N}|\,f_n(k)|^2 \leq B^2$.

Since this holds for any N, we get that $||\,f||_2\leq B$.

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 $n\geq n_0$,

$\sum\limits_{k=1}^{N}|\,f(k) - f_n(k)|^2 = lim_{n \to \infty}\sum\limits_{k = 1}^{N}|\,f_m(k) - f_n(k)|^2 \leq \epsilon^2$.

Since this holds for any N, we have $||\,f - f_n||_2 \leq \epsilon$ for all $n\geq n_0$. 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!

Answer 1

Question $1)$ you have a sequence of functions $f_n\in l_2$ 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 $f:\mathbb{N}\rightarrow\mathbb{C}$ 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.

Answer 2

Let's start with this $k$ in $f_n(k)$: The sequence $f_n\in l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example $$(f_1)_{k\in\mathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,\ldots).$$ To simplify the expression and have a connection to maps $\mathbb{N}\rightarrow\mathbb{R}$ 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: $$\forall \varepsilon>0\ \forall k,\ \exists n_0=n_0(\varepsilon,k)\in\mathbb{N} \mbox{ such that } \forall n\geq n_0:\ |f_n(k)-f(k)|<\varepsilon.$$ 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 $$\|(f)_{k\in\mathbb{N}}\|^2=\sum_{k=1}^\infty (f(k))^2.$$ So $f_n\rightarrow f$ in $l_2$ is by definition $$\forall \varepsilon>0\ \exists n_0=n_0(\varepsilon)\mbox{ such that }\forall n\geq n_0\ \|f_n-f\|_{l_2}^2:=\sum^\infty_{k=1}(f_n(k)-f(k))^2\leq \varepsilon.$$ 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.