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8. Completeness of \(\ell^2\)

≪ 7. Continuity and Compactness | Table of Contents | 9. Homework 3, Problems 5 and 6 ≫

Some people in my Tuesday office hours have asked me to provide a full proof of the completeness of \(\ell^2\). I thought this proof was pretty standard and available in most textbooks, but upon further inspection it seems that this is instead a standard exercise in most textbooks. How tragic.

Anyways, the critical problem with proving the completeness of \(\ell^2\) is the inability to switch limits, as we’ll see later. We must go through great lengths to justify this.

To show that \(\ell^2\) is complete, we must show that every Cauchy sequence converges. So, we start by letting \(\left\lbrace S_N \right\rbrace _{N=1}^{\infty}\) be a Cauchy sequence. Since each \(S_N\) is a sequence, we may write \(S_N=\left\lbrace a _{n,N} \right\rbrace _{n=1}^{\infty}\). Here, \(a _{n, N}\) represents the \(n\) term of the \(N\)-th sequence. We can put these guys in a table: \[\begin{array}{c|cccccc} & n=1 & n=2 & n=3 & n=4 & n=5 & \cdots \\ \hline S_1 & a _{1,1} & a _{2,1} & a _{3,1} & a _{4,1} & a _{5, 1} & \cdots \\ S_2 & a _{1,2} & a _{2,2} & a _{3,2} & a _{4,2} & a _{5, 2} & \cdots \\ S_3 & a _{1,3} & a _{2,3} & a _{3,3} & a _{4,3} & a _{5, 3} & \cdots \\ S_4 & a _{1,4} & a _{2,4} & a _{3,4} & a _{4,4} & a _{5, 4} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \\ S_N & a _{1,N} & a _{2,N} & a _{3,N} & a _{4,N} & a _{5, N} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{array} \] This is reminiscent of the table from homework 2, problem 3! Unlike that problem, though, we claim that each of the columns already converge to a real number.

Fix an \(n\in \mathbb{N}\) (i.e., focus our view on a single column of the table). For any \(N, M\in \mathbb{N}\), we have that \[\left\lvert a _{n,N} - a _{n, M}\right\rvert = \sqrt{\left( a _{n,N} - a _{n,M} \right)^2} \leq \sqrt{ \sum _{m=1}^{\infty} \left( a _{m,N} - a _{m, M} \right)^2}= d\left( S_N, S_M \right). \] Let \(\epsilon>0\). Since the \(S_N\)’s form a Cauchy sequence in \(\ell^2\), there exists some \(N _{\epsilon}\) such that for all \(N, M> N _{\epsilon}\), \(d\left( S_N, S_M \right)< \epsilon\). From the above inequality, it follows that for all \(N,M> N _{\epsilon}\), \(\left\lvert a _{n,N}-a _{n,M} \right\rvert< \epsilon\). This, in particular, means that the sequence \(\left\lbrace a _{n,N} \right\rbrace _{N=1}^{\infty}\) (i.e. the sequence formed by going down the \(n\)-th column of the table) is a Cauchy sequence. Hence, it converges to some limit.

For each \(n\), define \[a _{n,\infty}= \lim _{N\to\infty} a _{n,N}.\] That is, \(a _{n,\infty}\) is the limit “going down the \(n\)-th column”. This forms a sequence \(S_\infty=\left\lbrace a _{n,\infty} \right\rbrace _{n=1}^{\infty}\).

We have two things to do: first, we need to verify \(S_\infty\in\ell^2\). Then, we need to show that \(S_N\to S_\infty\) with respect to the \(\ell^2\) metric.

We will start with \(S_\infty\in\ell^2\); this amounts to showing that \(\lim _{K\to\infty} \sum _{n=1}^{K} a _{n,\infty}^2<\infty\). We first need a technical lemma:

Proof

If we abuse notation and denote \(0\) as the sequence of all zeroes, we have \[d\left( S_N, 0 \right) \leq d\left( S_N, S_M \right)+ d\left( S_M,0 \right)\] by the triangle inequality. Then, we get after some rearrangements that \[\left\lVert S_N \right\rVert-\left\lVert S_M \right\rVert = d\left( S_N, 0 \right)- d\left( S_M,0 \right) \leq d\left( S_N,S_M \right).\] Rehashing this argument with \(N\) and \(M\) swapped yields \[\left\lVert S_M \right\rVert-\left\lVert S_N \right\rVert\leq d\left( S_N,S_M \right).\] Thus, \(\left\lvert \left\lVert S_N \right\rVert-\left\lVert S_M \right\rVert \right\rvert\leq d\left( S_N,S_M \right)\). For all \(\epsilon>0\), there exists some \(N_\epsilon\) such that if \(N,M>N_\epsilon\), then \[\left\lvert \left\lVert S_N \right\rVert-\left\lVert S_M \right\rVert \right\rvert\leq d\left( S_N,S_M \right)<\epsilon,\] with the last inequality from Cauchiness. The claim follows. \(\square\)

(I’m very sorry about the super ugly triple bars in this lemma…)

In particular, \(\lim _{N\to\infty} \left\lVert S_N \right\rVert\) exists. The naïve thing to do here is to say \[\sum _{n=1}^{\infty}a _{n,\infty}^2=\lim _{N\to\infty} \sum _{n=1}^{\infty} a _{n,N}^2 = \lim _{N\to\infty} \left\lVert S_N \right\rVert^2\] and call it a day. However, the summation is secretly a limit, and you cannot do this without justifying this in some way.

Rather, we shall fix some \(K\in \mathbb{N}\). Then, consider that \[\sum _{n=1}^{K}a _{n,\infty}^2 = \sum _{n=1}^{K}\lim _{N\to\infty} a _{n, N}^2 =\lim _{N\to\infty}\sum _{n=1}^{K} a _{n,N}^2 \leq \lim _{N\to\infty}\sum _{n=1}^{\infty}a _{n, N}^2 = \lim _{N\to\infty}\left\lVert S_N \right\rVert^2. \] We can interchange the limit with the sum in the second equality since it’s just a finite sum. But this bound works for every single \(K\)! We have thus shown that the sequence \(\left\lbrace \sum _{n=1}^{K} a _{n,\infty}^2 \right\rbrace _{K=1}^{\infty}\) is bounded above, and it’s clearly monotone (nondecreasing). Therefore, it must converge to something finite, i.e. \[\sum _{n=1}^{\infty} a _{n,\infty}^{2} < \infty\implies S_\infty\in\ell^2.\]

The last thing we need to do is verify that \(S_N\to S_\infty\) in \(\ell^2\). As with the above, a difficulty arises from having multiple limits.

Fix any \(\epsilon>0\). Since the \(S_N\)’s are Cauchy, there exists some \(N_\epsilon\) such that if \(N,M>N_\epsilon\), \(d\left( S_N,S_M \right)< \epsilon\). We claim that for any \(N>N_\epsilon\), \(d\left( S_N,S_\infty \right)\leq\epsilon\), which implies that \(S_N\to S_\infty\)!

Fix any such \(N>N_\epsilon\). We have that \[d\left( S_N,S_\infty \right)^2=\lim _{K\to\infty}\sum _{n=1}^K \left( a _{n,N}-a _{n,\infty} \right)^2.\] We could rewrite these \(a _{n,\infty}\)’s as limits, but we will still run into the same problem as before — we can’t justify swapping the two limits around. Rather, we claim that every term in this sequence is bounded above by \(\epsilon\). If \(K\in \mathbb{N}\), we have \[\sum _{n=1}^{K} \left( a _{n,N}-a _{n,\infty} \right)^2 = \sum _{n=1}^{K}\lim _{M\to\infty} \left( a _{n,N}-a _{n,M} \right)^2 = \lim _{M\to\infty} \sum _{n=1}^{K} \left( a _{n,N}-a _{n,M} \right)^2.\] Of course these are just partial sums of \(d\left( S_N,S_M \right)\); thus we have \[\sum _{n=1}^{K} \left( a _{n,N}-a _{n,\infty} \right)^2 \leq \lim _{M\to\infty}d\left( S_N,S_M \right)^2\leq \epsilon^2.\] This last inequality is because \(d\left( S_N,S_M \right)<\epsilon\) for all \(M>N_\epsilon\)! You should verify that this last limit does exist; it’s very similar to the lemma given prior.

But again we get that \(\left\lbrace \sum _{n=1}^{K} \left( a _{n,N}- a _{n,\infty} \right)^2 \right\rbrace _{K=1}^{\infty}\) is bounded above by \(\epsilon^2\) and is monotone nondecreasing, so it converges to something at most \(\epsilon^2\)! We conclude that \(d\left( S_N,S_\infty \right)\leq \epsilon\) for all \(N>N_\epsilon\), and it follows that \(S_N\to S_\infty\).

We conclude that every Cauchy sequence in \(\ell^2\) converges to some limit in \(\ell^2\), thus \(\ell^2\) is complete. \(\square\)